Efficient method for orbital launch trajectories

Abstract

A method is provided for efficient orbital launch trajectories. A payload (e.g., satellite) is launched as high as a first radius with respect to the center of the Earth. The method decreases the payload altitude in response to a gravitational pull of the Earth, and ultimately the payload attains a stable orbit around the Earth at a second radius with respect to the center of the Earth. If the first radius is twice the second radius, the payload acquires a gravitational first potential energy at the first radius, and in the stable orbit the payload has a second (potential and kinetic) energy equal to the gravitational first potential energy, with the second kinetic energy being equal to the energy required to maintain a stable orbital velocity. Advantageously, the stable orbit can potentially be at any orbital inclination angle in the range between 0 and 360 degrees.

Claims

1. A method for efficient orbital launch trajectories, the method comprising: launching a payload as high as a first radius height (r.sub.0), defined with respect to the center of the Earth, associated with a first altitude, defined with respect to sea level; the payload attaining a gravitational first potential energy at the first radius height; decreasing the payload height in response to a gravitational pull of the Earth; and, the payload attaining a stable orbit around the Earth at a second radius height (r.sub.ORBIT), defined with respect to the center of the Earth, associated with a second altitude, defined with respect to sea level.

2. The method of claim 1 wherein launching the payload includes launching the payload to a first radius height equal to 2 times the height of the second radius height.

3. The method of claim 2 wherein the payload attaining the stable orbit includes the payload having a total energy that is the sum of a gravitational second potential energy and kinetic energy, with the payload total energy being equal to the gravitational first potential energy.

4. The method of claim 1 wherein the payload attaining the stable orbit includes attaining a stable orbit at any orbital inclination angle in the range between 0 and 360 degrees.

5. The method of claim 1 wherein launching the payload includes launching the payload from any latitude on a surface of the Earth in the range between 90 and 90 degrees.

6. The method of claim 3 wherein the payload attaining the gravitational first potential energy at the first radius height includes the gravitational first potential energy being expressed as: $-G\frac{\mathrm{mM}}{r_0};$ where G is Newton's gravitational constant; where m is the mass of the payload; and where His the mass of the Earth.

7. The method of claim 6 wherein attaining the stable orbit includes the gravitational second potential energy being expressed as: $-\frac{\mathrm{GmM}}{r_{ORBIT}}.$

8. The method of claim 7 wherein the payload attaining the stable orbit and having the total energy equal to the gravitational first potential energy includes the relationship being expressed as: $-G\frac{mM}{r_0}=\frac{1}{2}m{v}_{\mathrm{ORBIT}}^2-\frac{GmM}{r_{ORBIT}};$ where .sub.ORBIT is the velocity of the payload in the stable orbit.

9. The method of claim 8 further comprising: subsequent to the payload attaining the first radius height, initiating a rocket (payload) assisted descent maneuver; wherein decreasing the payload height in response to the gravitational pull of the Earth includes decreasing the payload height at a first angle of descent tangential to the stable orbit.

10. The method of claim 9 further comprising: subsequent to the payload attaining the first radius height, performing a payload roll maneuver; and, wherein decreasing the payload height at the first angle of descent includes the first angle of descent being in a first angle of orbital inclination responsive to the roll maneuver.

11. The method of claim 9 wherein launching the payload as high as the first radius height includes launching the payload to a third radius height (r.sub.h) defined with respect to the center of the Earth, where
r.sub.h=s.sub.f.Math.r.sub.ORBIT where 1.1<s.sub.f<2; the method further comprising: subsequent to the rocket assisted descent maneuver, applying a velocity adjustment force to the payload; and, wherein decreasing the payload height at the first angle of descent includes modifying a descending velocity of the payload in response to the velocity adjustment force.

12. The method of claim 11 wherein the payload attaining the third radius height includes the payload having a total energy at the third radius height equal to its gravitational potential energy; and, wherein applying the velocity adjustment force to the payload includes the applied velocity adjustment force being responsive to the gravitation potential energy at the third radius height.

13. The method of claim 12 further comprising: subsequent to the rocket assisted descent maneuver, permitting the payload to fall, under only the force of gravity, to a fourth radius height, defined with respect to the center of the Earth, higher than the second radius height; and, wherein applying a velocity adjustment force to the payload includes applying the velocity adjustment force as the payload approaches the fourth radius height.

14. A method for efficient orbital launch trajectories, the method comprising: launching a payload from a surface of the Earth; the payload attaining a first radius height (r.sub.0), defined with respect to the center of the Earth; the payload attaining a gravitational first potential energy at the first radius height; decreasing the payload height in response to the gravitational first potential energy; performing a payload roll maneuver; creating a first angle of descent responsive to the roll maneuver; and, the payload attaining a stable orbit around the Earth at a second radius height (r.sub.ORBIT), defined with respect to the center of the Earth.

15. The method of claim 14 wherein the payload attaining the first radius height includes creating a relationship between the first radius height and second radius height expressed as:
r.sub.0=s.sub.f.Math.r.sub.ORBIT where 1.1<s.sub.f<2.

16. The method of claim 5 wherein the payload gravitational first potential energy at the first radius height is expressed as: $-G\frac{mM}{r_0}=\frac{1}{2}m{v}_{\mathrm{ORBIT}}^2-\frac{GmM}{r_{ORBIT}};$ where s.sub.f=2; where .sub.ORBIT is the velocity of the payload in the stable orbit; where G is Newton's gravitational constant; where m is the mass of the payload; and where M is the mass of the Earth.

17. A method for obtaining efficient orbital launch trajectories, the method comprising: for a payload, identifying a first radius height (r.sub.0) with respect to the center of the Earth; identifying a second radius height (r.sub.ORBIT) with respect to the center of the Earth, less than the first radius height; identifying a first height difference between the first radius height and the second radius height; launching the payload to the first radius height from a surface of the Earth; using an Orbital Descent software program, identifying a direction .sub.F, thrust F.sub.D, and descent burn time T.sub.BD from the first radius height; decreasing the payload height in response to the first height difference, direction .sub.F, thrust F.sub.D, and descent burn time T.sub.BD; and, the payload obtaining a stable orbital velocity at the second radius height.

18. The method of claim 17 wherein the first height difference is expressed with a scaling factor S, where:
r.sub.0=s.sub.f.Math.r.sub.ORBIT where 1.1<s.sub.f<2.

19. The method of claim 18 further comprising: the payload attaining a gravitational first potential energy at the first radius height expressed as: $-G\frac{mM}{r_0}=\frac{1}{2}m{v}_{\mathrm{ORBIT}}^2-\frac{GmM}{r_{ORBIT}},$ where s.sub.f=2; where .sub.ORBIT is the velocity of the payload in the stable orbit; where G is Newton's gravitational constant; where m is the mass of the payload; and where M is the mass of the Earth.

Description

BRIEF DESCRIPTION OF THE DRAWINGS

(1) FIG. 1A depicts the orbital speed in [m/s] (meter per second) versus radius measured with respect to the stationary coordinate system where its origin is in the center of Earth and altitude measured from sea level.

(2) FIG. 1B depicts the orbital speed in Mach number, versus radius measured with respect to the stationary coordinate system where its origin is in the center of Earth and altitude measured from sea level, where 1 Mach is speed of sound at sea level, 1 Mach=340 [m/s] and the numbers 1, 5, and 10 correspond to the row numbers in Table 2.

(3) FIG. 1C depicts the orbital period in hours versus radius measured with respect to the stationary coordinate system where its origin is in the center of the Earth and altitude measured from sea level.

(4) FIG. 1D depicts the escape velocity as a function of radius for the Earth along with markings of the orbital altitudes given in Table 2.

(5) FIG. 1E depicts the density of air as a function of altitude.

(6) FIG. 1F depicts orbital inclination as a function of launch azimuth from 6 locations, representing launch from the Equator, along with the numberings corresponding with the launch latitudes shown in the first column of Table 2.

(7) FIG. 2A depicts the gravitational acceleration g(h) as function of altitude h from sea level and distance from the earth's centered radius g(r), where r=h+r.sub.EARTH.

(8) FIG. 2B depicts the gravitational acceleration normalized to its sea level value as a function of altitude g(h) and the distance from the Earth's centered radius r, g(r), with the radius of the Earth taken at the equatorial radius of r.sub.EARTH=6,378 km.

(9) FIG. 2C depicts the gravitational acceleration normalized to its sea level value as a function of altitude g(h) with the maximum burn altitude h.sub.B(.sub.f=0) for T2W=1.1, 2, 4 and 8 marked on the curve.

(10) FIG. 3A depicts the gravitational potential energy as a function of radius r.sub.ORBIT with respect to Earth's center, along with the kinetic energy when in orbit at the radius r.sub.ORBIT, and its total energy.

(11) FIG. 3B depicts the same plots in FIG. 3A, where x axes plotted in logarithmic scale.

(12) FIG. 3C depicts the orbital velocity as a function of orbital altitude with the velocity gained when dropping from 2.Math.r.sub.ORBIT to r.sub.ORBIT along with markings of the orbital altitudes given in Table 2 which matches the orbital velocity as a function of orbital altitude curve.

(13) FIG. 3D depicts the launch altitude as a function of the orbital altitude along with markings of the orbital altitudes given in Table 2.

(14) FIG. 3E depicts the launch altitude to the orbital altitude ratio along with markings of the orbital altitudes given in Table 2 where the x axis is plotted in logarithmic scale.

(15) FIG. 3F depicts the fall times from (2.Math.r.sub.ORBIT to r.sub.ORBIT), (2.Math.r.sub.ORBIT to r.sub.EARTH) and (r.sub.ORBIT to r.sub.EARTH).

(16) FIG. 3G depicts the velocity that the payload gains, which was at rest at a radius of r.sub.h in terms of the orbital velocity .sub.ORBIT for a circular orbit at the radius of r.sub.ORBIT as a function of s.sub.f when falling from r.sub.h to r.sub.ORBIT,

(17) $sf=\frac{r_h}{r_{\mathrm{ORBIT}}}.$

(18) FIG. 3H depicts the rocket thrust assisted velocity reached .sub.B, having rocket thrust being in the same direction of the constant gravitational acceleration shown as .sub.B.sup.+g (in rocket powered vertical descent) and rocket thrust being in the opposite direction of the constant gravitational acceleration shown as .sub.B.sup.g (in rocket powered vertical ascent) using the solution of modified Tsiolkovsky's rocket equation for T2W=1.1 and 4 versus .sub.f.

(19) FIG. 3I depicts the rocket thrust assisted distance covered having rocket thrust being in the same direction of the constant gravitational acceleration shown as h+(in rocket powered vertical descent) and rocket thrust being in the opposite direction of the constant gravitational acceleration shown as h.sub.B.sup.9 (in rocket powered vertical ascent) using the solution of modified Tsiolkovsky's rocket equation for T2W=1.1 and 4 versus .sub.f.

(20) FIG. 3J is same data depicted in FIG. 3I in logarithmic x axes to show the covered distances clearer for larger .sub.f values.

(21) FIG. 3K depicts the ratios between the velocities reached .sub.VB(.sub.f)=.sub.B.sup.+g/.sub.B.sup.g and ratios of the distances covered .sub.HB(.sub.f)=h.sub.B.sup.+g(.sub.f, T2W)/h.sub.B.sup.9 (.sub.f, T2W) for T2W=1.1.

(22) FIG. 3L depicts the ratios between the velocities reached .sub.VB(.sub.f)=.sub.B.sup.+g/.sub.B.sup.g and ratios of the distances covered .sub.HB(.sub.f)=h.sub.B.sup.+g (.sub.f, T2W)/h.sub.B.sup.g(.sub.f, T2W) for T2W=4.

(23) FIG. 4A depicts .sub.B(.sub.f, T2W) for thrust-to-weight ratio T2W=1.1 and 8 with a constant acceleration g

(24) FIG. 4B depicts .sub.B(.sub.f, T2W) for thrust-to-weight ratio T2W=1.1 and 8 at a constant acceleration g, where the horizontal axis is in logarithmic scale.

(25) FIG. 5A depicts h.sub.B(.sub.f, T2W) for thrust-to-weight ratio T2W=1.1, 2, 4, and 8 with a constant acceleration g.

(26) FIG. 5B depicts h.sub.B(.sub.f, T2W) for thrust-to-weight ratio T2W=1.1 and 8 with a constant acceleration g, where the horizontal axis is in logarithmic scale.

(27) FIG. 6A depicts the climb altitude h(.sub.f, T2W) for the constant gravitational acceleration of g=9.8 [m/s.sup.2] with thrust-to-weight ratio T2W=1.1 and 8 where the vertical axis is in logarithmic scale.

(28) FIG. 6B depicts burn height h.sub.B(.sub.f, T2W) and coast height h.sub.C(.sub.f, T2W) for the constant gravitational acceleration of g=9.8 [m/s.sup.2] with thrust-to-weight ratio T2W=1.1 and 8 where the vertical axis is in logarithmic scale.

(29) FIG. 7A depicts asymptote x intercept .sub.ASYM(T2W) vs. T2W and .sub.MIN(T2W) by solving the non-linear equation (6.7).

(30) FIG. 7B depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field with asymptotes where vertical axis is in logarithmic scale for thrust-to-weight ratio T2W=1.1 and 8.

(31) FIG. 7C depicts burn height h.sub.BINV2(.sub.f, T2W) and coast height h.sub.CINV2(.sub.f, T2W) at T2W for the inverse square gravitational field with asymptotes where the vertical axis is in logarithmic scale.

(32) FIG. 7D depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field with asymptotes and h(.sub.f, T2W) for the constant gravitational acceleration of g=9.8 [m/s.sup.2] where the vertical axis is in logarithmic scale for thrust-to-weight ratio T2W=1.1 and 8.

(33) FIG. 7E depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field with asymptotes and h(.sub.f, T2W) for the constant gravitational acceleration of g=9.8 [m/s.sup.2] where the vertical and horizontal axes are in logarithmic scale for thrust-to-weight ratio T2W=1.1 and 8.

(34) FIG. 7F depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field with asymptotes for <.sub.ASYM(T2W) with thrust-to-weight ratio T2W=1.1, 2, 4, and 8.

(35) FIG. 8 depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field and climb altitude h(.sub.f, T2W) with the constant gravitational acceleration of g=9.8 [m/s.sup.2] vs. T2W, where .sub.f=0.1, 0.2, 0.3, and 0.4.

(36) FIG. 9A depicts climb altitude h.sub.INV2(.sub.f, T2W) and inverse square gravitational field h(.sub.f, T2W) functions for T2W=1.1 and 4 constructed over 400 uniformly spaced discrete sampling points.

(37) FIG. 9B depicts the functions plotted in FIG. 9A constructed with 40 discrete sampling points marking the curves to show a numerical algorithm for implementing the geometrical solution method in a computer program based on interpolation between the discrete sampling points.

(38) FIG. 9C depicts a detailed view of the IR 2 region showing the discrete sampling points in the neighborhood of intersection with h.sub.GIVEN=40 km of the

(39) h.sub.INV2(.sub.f, T2W=1.1) and h(.sub.f, T2W=1.1) functions.

(40) FIG. 10A depicts a detailed view of the IR 1 and IR 2 regions showing discrete sampling points in the neighborhood of intersection between h.sub.GIVEN=40 km and 400 km for the h.sub.INV2(.sub.f, T2W=1.1) and h(.sub.f, T2W=1.1) functions.

(41) FIG. 10B depicts a detailed view showing discrete sampling points in the neighborhood of the intersection region JR 1 between h.sub.GIVEN=20,200 km with the

(42) h.sub.INV2(.sub.f, T2W=1.1) and h.sub.INV2(.sub.f, T2W=4) functions.

(43) FIG. 10C shows the convergence properties of Newton's method employed in solving rocket equations.

(44) FIG. 11 depicts the skin of a propellant tank.

(45) FIG. 12 depicts the mass of the propellant tank m=f (r, t.sub.skin) for the skin thickness of t.sub.skin=3, 6, and 10 cm carrying 20 tons of propellant mass.

(46) FIG. 13A depicts k=f (r, t.sub.skin) for an aluminum skin thickness of t.sub.skin=3, 6, and 10 cm when carrying 20 tons of propellant mass.

(47) FIG. 13B depicts the plot of relation (8.12), which gives the minimum value of k=k.sub.MAX that needs to be satisfied to give a desired value of .sub.f.

(48) FIG. 13C depicts the height h.sub.INV2(k.sub.MAX, T2W) function, an important relationship between the height that the rocket can reach and the k.sub.MAX value that the rocket design must satisfy, graphically permitting the determination of k.sub.MAX for h.sub.GIVEN=400 km, 20,200 km, and 400,000 km.

(49) FIG. 13D depicts the SpaceX Starship Super Heavy (loaded with 3,400 tons of propellant) and the ULA Atlas first stage (loaded with 284 tons of propellant) booster heights with a t.sub.SKIN thickness of 3 [cm], with respective diameters of 9 [m] and 3.81 [m].

(50) FIG. 13E depicts h.sub.INV2(r, T2W), where r is the rocket diameter of the SpaceX Starship Super Heavy loaded with 3,400 Tons of propellant having t.sub.SKIN=3 [Cm].

(51) FIG. 13F depicts altitude as a function of time for rocket launching 1,000 [kg] of payload to 40, 100 and 400 [km] altitudes for T2W=1 and 4, k=0.01, NOT an orbital launch.

(52) FIG. 13G depicts propellant and total mass as a function of time for launching 1,000 [kg] of payload to 40 [km] altitude for T2W=1, and 4, k=0.01.

(53) FIG. 13H depicts altitude as a function of time for rocket launching of 1,000 [kg] of payload to 20,200 and 35,786 [km] (GPS and GEO) altitudes for T2W=1.1 and 4, k=0.01, NOT an orbital launch

(54) FIG. 13I depicts altitude as a function of time for rocket launching of 1,000 [kg] of payload to a circular orbital altitude of 400 [km] (ISS) with s.sub.f=1.5 and 2 for T2W=1.1 and 4, k=0.01.

(55) FIG. 13J depicts propellant and total mass as a function of time in logarithmic horizontal axes for launching 1,000 [kg] of payload to a circular orbit at 400 [km] (ISS) altitude with s.sub.f=2 for T2W=1.1 and 4, k=0.01.

(56) FIG. 13K depicts propellant and total mass as a function of time in logarithmic horizontal axes for launching 1,000 [kg] of payload to a circular orbit at 400 [km] (ISS) altitude with s.sub.f=1.5 and 2 for T2W=1-1 and 4, k=0.01.

(57) FIG. 13L depicts altitude as a function of time for rocket launching of 1,000 [kg] of payload to a circular orbital altitude of 20,200 [km] (GPS) with s.sub.f=2 for T2W=1.1 and 4, k=0.01.

(58) FIG. 13M depicts altitude as a function of time for rocket launching of 1,000 [kg] of payload to a circular orbital altitude of 20,200 [km] (GPS) with s.sub.f=1.5 and 2 for T2W=1.1 and 4, k 0.01,

(59) FIG. 13N depicts propellant and total mass as a function of time in logarithmic horizontal axes for launching 1,000 [kg] of payload to a circular orbital altitude of 20,200 [km](GPS) with s.sub.f=2 for T2W=1.1 and 4, k=0.01.

(60) FIG. 13O depicts propellant and total mass as a function of time in logarithmic horizontal axes for launching 1,000 [kg] of payload to a circular orbital altitude of 20,200 [km](GPS) with s.sub.f=1.5 and 2 for T2W=1.1 and 4, k=0.01,

(61) FIG. 14 depicts a rocket being launched into orbit using a conventional pitchover or gravity turn maneuver (prior art).

(62) FIG. 15 is a flowchart illustrating a method for efficient orbital launch trajectories.

(63) FIG. 16 is a drawing depicting the relationship between the first radius (r.sub.0), second radius (r.sub.ORBIT), the payload, and the Earth.

(64) FIG. 17 is a diagram depicting the relationships shown in FIG. 16, as referenced to Earth's sea level.

(65) FIG. 18 is a plan view of the Earth centered on the polar axis.

(66) FIG. 19 is a diagram the relationship between (r.sub.0=2.Math.r.sub.ORBIT), second radius (r.sub.ORBIT), the payload, the Earth, and (r.sub.h) 1900.

(67) FIGS. 20A through 20C are drawings depicting a velocity adjustment applied to the payload during or after the rocket assisted descent maneuver.

(68) FIG. 21 is a flowchart illustrating a method for minimizing the energy required for an orbital launch.

(69) FIG. 22 is a diagram depicting a variety of exemplary stable orbits that can be obtained using the disclosed method.

DETAILED DESCRIPTION

(70) The Problem to be Solved.

(71) This invention is related to putting satellites into orbit with smaller rockets employing less propellant compared to the standard rockets with less demanding rocket engines for launching the same orbital mass into orbit. Therefore, an introduction to some very basic orbital mechanics is presented in Section 2 for a satellite having a circular orbit around the earth.

(72) From the German rocket work, which goes back to 1930's, the importance of having a low mass rocket to reach a given altitude is a well-known fact and it became a key factor in any rocket design [7]. To reach the speeds and altitudes that are desired in rocket applications, the rocket equation enforces that most of the initial rocket mass must be propellant mass m.sub.prop [1-3]. Propellant is contained in the propellant tank and let m.sub.tank is the mass of the propellant tank and it also includes the mass of the overall volume of the rocket housing the propellant. Let the parameter k is the ratio of the propellant to the propellant tank given as,

(73) $\begin{array}{cc}k=\frac{m_{tank}}{m_{prop}}& \left(1.3\right)\end{array}$

(74) Using the parameter k, mass of the rocket engine m.sub.RE, and the payload mass m.sub.PAY, the initial mass of the rocket m.sub.0, can be given as,
m.sub.0=m.sub.prop(1+k)+m.sub.RE+m.sub.PAY(1.4)

(75) For the rocket to be able to lift-off vertically from the pad its initial thrust-to-weight ratio T2W must be greater than 1, giving the following thrust equation,

(76) $\begin{array}{cc}\frac{F_T}{g.Math.T2W}=m_{prop}\left(1+k\right)+m_{RE}+m_{\mathrm{PAY}}& \left(1.5\right)\end{array}$

(77) Where F.sub.T is the thrust generated by the rocket engine and g is the gravitational acceleration at the launch pad. Thrust, F.sub.T obtained from the solution of the equation (1.5) must be a positive value. It is shown that once m.sub.prop satisfies the rocket equation, which enables it to reach the desired height, positive thrust value is only possible if the k parameter is less than a specific value defined as k.sub.MAX. In its simplest form k.sub.MAX is given as,

(78) $\begin{array}{cc}{k}_{\mathrm{MAX}}<\frac{{}_f}{\left(1-{}_f\right)}& \left(1.6\right)\end{array}$

(79) Where .sub.f is the ratio between the final mass value of the rocket m.sub.f and its initial rocket mass m.sub.0 given as,

(80) 0$\begin{array}{cc}{}_f=\frac{m_f}{m_0}& \left(1.7\right)\end{array}$

(81) Which .sub.f satisfies the non-linear rocket equation formulated including gravitational potential acting upon the rocket to reach a given height above sea level, h.sub.GIVEN. Therefore, it is a solved quantity and determines the maximum value of k, noted as k.sub.MAX given in (1.7) in its simplest form. This mathematically relates the k value giving the maximum altitude that the rocket can reach and must be satisfied. If the physical construction of the rocket gives a k larger than k.sub.MAX, for the given value of .sub.f, the rocket cannot reach the altitude corresponding to .sub.f.

(82) It is also shown that the opposite is also true. A given k value which can be measured or calculated from some simple rocket geometry, propellant physical and chemical properties define the maximum altitude that the rocket can reach for a given payload, thrust-to-weight ratio at the launch, and rocket engine thrust-to-weight ratio. In its simplest form, the minimum value of .sub.f that can be achieved with this given k value becomes,

(83) $\begin{array}{cc}{}_f>\frac{k}{\left(1+k\right)}& \left(1.8\right)\end{array}$

(84) Most of the rocket propellant tank shapes determine the shape of the rocket and typically they are cylindrical. For cylindrical geometry, the k value is also a function of rocket radius, therefore rocket diameter can be related to the maximum altitude that it can reach for a given thrust-to-weight ratio at the launch, and rocket engine thrust-to-weight ratio. The detailed derivations on this subject are given in Sections 7 and 8 below.

(85) Another advantage of the disclosed method is that it allows any launch site on earth to achieve any orbital inclination angle from any launch site [8]. The inclination angle of a satellite orbit is one of the five orbital parameters defining its orbit. It is the angle between the satellite orbital plane and the Earth's equatorial plane. As an example, the International Space Station (ISS) has an inclination of 51.6. Due to certain safety limitations all the launch sites give only a limited range of azimuth angle (angle from the true north) launches. Inclination angle i, is related to the latitude of the launch site and the launch azimuth A with a simple formula given as,
COS(i)=Sin(A).Math.Cos()(1.9)

(86) This relation puts limits on the inclination angle for the satellite orbit launched from any site. As an example, for all launches from Cape Canaveral in central Florida (=28.4 North), the launch azimuth angle limits are 35-120 giving the corresponding inclination angle limits of 57-39. Similarly, for all launches from Vandenberg Air Force Base in California (=34.7 North), the launch azimuth angle limits are 140-201 giving the corresponding inclination angle limits of 56-104. Therefore, the most popular Polar orbital launches cannot be done from Cape Canaveral, but rather, must be done from Vandenberg Air Force. It is obvious that having no inclination angle restriction is a great advantage and has great commercial value by itself.

(87) Table 1 gives the most popular launch site latitudes and dates on which they became operational. FIG. 1E gives the plot of relation (1.9) for all most popular launch sites with the Azimuthal Limitations shown on the plot for Cape Kennedy and Vandenberg.

(88) TABLE-US-00001 TABLE 1 Most popular launch site latitudes and dates when they became operational. FIG. 1E Operational reference Launch Site Latitude Since 1 Kennedy Space Center, FL, USA 28.60N 1962 2 Vandenberg Space Force Base, 34.77N 1958 CA, USA 3 Baikonur Cosmodrome, Russian 45.95N 1955 Federation-Kazakhstan 4 ESA Space Center, Kourou, 5.16N 1964 French Guiana 5 SpaceX Star Base, Boca Chica, 25.99N 2018 TX, USA

(89) In the disclosed method the satellite is launched to a radius of r.sub.h which could be in any permitted azimuth from the launch site. From there a rocket assisted descent can be made to any direction to get to get to any inclination desired. It is obvious to see that by employing this launch strategy, satellites can be launched into most popular Polar orbits from even Cape Canaveral, French Guiana, or anywhere on Earth.

(90) Anything related to space has always been a very popular subject in the public and the orbital launch of satellites is becoming a big and competitive industry where profitability is becoming an important part of it. Many published technical materials related to the subject exist in public domain, but unfortunately publications in this area, like in many other areas, take shortcuts in mathematics and derivations. This hides many of the facts and causes one to miss the key important details. As an example, without defining m.sub.prop in (1.4) the key results of (1.5) to (1.7) cannot be generated.

(91) Another goal of this work is to present some of the orbital launch and rocket design related mathematics and mechanics by taking no short cuts. Therefore, in this work there are some chapters which are dedicated to the derivation of the basic principles.

(92) Simple Orbital Mechanics

(93) English mathematician and the founder of mechanics among many other earth-shaking discoveries, Sir Isaac Newton's (1642-1726) second law of mechanics allows anyone to calculate the tangential speed required to put a satellite into a circular or elliptical orbit around the earth. at an altitude h from sea level. All the information needed to perform these calculations is in Philosophie Naturalis Principia Mathematica, which is the foundation of mechanics, integral and differential calculus. It was first published in 1687 with the financial help of Edmond Halley (1656-1742), another famous English astronomer and mathematician named with a comet named after him.

(94) One cannot mention differential calculus without mentioning German mathematician Gottfried Wilhelm von Leibnitz (1646-1716). Leibnitz invented differential calculus independently from Newton with a more convenient notation than used by Newton. Everything mathematically presented in this work is entirely based on the inventions of these two giants of science 334 years after.

(95) Newton's gravitational attraction force F.sub.G between 2 spherical masses m and M, having a distant d between their centers is given as,

(96) $\begin{array}{cc}{F}_G=G\frac{mM}{d^2}& \left(2.1\right)\end{array}$

(97) Where G is Newton's gravitational constant. This law is also known as the inverse square law of gravitational attraction force or field. This attraction force on the surface of the earth can be defined with the expression,
F.sub.G=Mg.sub.0(2.2)

(98) Where go is the gravitational acceleration constant at the surface of the earth and it can be formulated as,

(99) $\begin{array}{cc}{g}_0=G\frac{M}{r_{\mathrm{EARTH}}^2}& \left(2.3\right)\end{array}$

(100) Where in (2.3) now M and r.sub.EARTH represent the mass and radius of the earth. For a satellite to stay in a circular orbit, the centrifugal force of the satellite at a radius of r.sub.ORBIT measured from the center of the earth, must be equal to the Earth's gravitational attraction force as given by (2.3) as,

(101) $\begin{array}{cc}m\frac{v_{\mathrm{ORBIT}}^2}{r_{\mathrm{ORBIT}}}=G\frac{mM}{r_{\mathrm{ORBIT}}^2}& \left(2.4\right)\end{array}$

(102) Where m and v.sub.ORBIT are the mass of the satellite and the tangential velocity of the satellite in the circular orbit around the earth at a radius of r.sub.ORBIT measured from the center of the earth, respectively.

(103) Simplifying (2.4) gives the necessary tangential velocity to keep the satellite in orbit at a radius of r.sub.ORBIT, measured from the center of the earth as,

(104) $\begin{array}{cc}{v}_{\mathrm{ORBIT}}^2=\frac{GM}{r_{\mathrm{ORBIT}}}& \left(2.5\right)\end{array}$

(105) Getting the square root of both sides gives,

(106) $\begin{array}{cc}{v}_{\mathrm{ORBIT}}=\sqrt{\frac{GM}{r_{\mathrm{ORBIT}}}}& \left(2.6\right)\end{array}$

(107) In space terminology one usually specifies the orbital altitude h referenced from the sea level giving the relation between r.sub.ORBIT and h as,
r.sub.ORBIT=r.sub.EARTH+h(2.7)

(108) FIG. 1A is the orbital velocity as a function of radius and altitude as given in (2.6) in [m/s] along with the escape velocity from the Earth. FIG. 1B is the orbital velocity as a function of radius and altitude as given in (2.6) in Mach numbers, which is a speed defined in terms of speed of sound at sea level which is 340 [m/s]. The markings on the curves are the orbital velocities of satellites in circular orbit at orbital altitudes given in Table 2. As can be seen, the speed of any satellite in any of the orbits given in Table 2 is hypersonic.

(109) The numerical values of the constants in (2.3) are,
Universal Gravitational Constant G=6.67410.sup.11[m.sup.3.Math.kg.sup.1] or [N.Math.m.sup.2.Math.kg.sup.2]
Mass of Earth M=5.97210.sup.24[kg]

(110) The Earth's Polar and Equatorial radiuses are slightly different and are, r.sub.EPolar=6,357 [km] and r.sub.EEqutorial=6,378 [km]

(111) Since the radius of the orbit and the constant speed of the satellite is known, one can easily calculate the time T that it takes for the satellite to complete one revolution at the circular orbit as,

(112) $\begin{array}{cc}T=\frac{2r_{\mathrm{ORBIT}}}{v_{\mathrm{ORBIT}}}=\frac{2r_{\mathrm{ORBIT}}}{\sqrt{\frac{GM}{r_{\mathrm{ORBIT}}}}}=\frac{2{\left(r_{\mathrm{ORBIT}}\right)}^{\frac{3}{2}}}{\sqrt{GM}}& \left(2.8\right)\end{array}$

(113) This is nothing more than Johannes Kepler's (1571-1630) third law applied for a circular orbit. Kepler, a German mathematician, used Tyco Brahe (1546-1601), a Danish astronomer known for the most accurate astronomical data to date collected from his observatory at Benatky nad Jizerou, near Prague of today's Czech Republic, for validating his three laws of planetary motion while he was Tyco's assistant.

(114) The orbital period as a function of radius as given in (2.8) and altitude is plotted in FIG. 1C. The markings on the curves are the orbital periods of the satellites in circular orbit at orbital altitudes given in Table 2.

(115) FIG. 1D depicts the escape velocity as a function of radius for the Earth along with markings of the orbital altitudes given in Table 2.

(116) FIG. 1E depicts the density of air as a function of altitude.

(117) FIG. 1F depicts orbital inclination as a function of launch azimuth from 6 locations, representing launch from the Equator, along with the numberings corresponding with the launch latitudes shown in the first column of Table 1.

(118) Regarding expressions (2.6) and (2.8), Aristarchus of Samos (310 BC-230 BC), an ancient Greek mathematician and astronomer, was the first to propose that the sun is the center of the Solar system, and the Earth is rotating around sun. Before him, Aristotle (384 BC-322 BC) another ancient Greek philosopher postulated that earth was round and was at the center of the universe, and everything rotated around it. Claudius Ptolemy (100-170), another ancient Greek mathematician supported the geocentric earth model. Copernicus (1473-1543), the Polish mathematician and astronomer, brought back the idea of the earth revolving around the sun.

(119) All the debates related to the subject were settled after Halley asked Newton what the trajectory of the Halley Comet was, which changed the flow of history. To this question, Newton simply said, (i)t is an Eclipse. and when Halley asked how he knew that Newton simply replied, I calculated it, which was an unheard-of task then. Halley asked how it was calculated and after Newton showed how he calculated it, a completely surprised and impressed Halley told him that he must publish all this, which Newton said it is too expensive to publish and he couldn't afford it. Totally impressed, Halley decided to finance Newton's monumental work in Philosophie Naturalis Principia Mathematica which undoubtedly became the most important scientific publication in human history, which is the mathematics for explaining the entire mechanics of objects.

(120) Another useful parameter related to the subject is escape velocity, which gives a good idea of the speeds required in the space launch business. Escape velocity .sub.ESCAPE in its simplest form is derived between two spherical objects with masses m and M and radiuses for each are noted as r.sub.m and r.sub.M respectively. In the formulation it is also assumed that Mm and r.sub.Mr.sub.m and the larger mass, M will also assumed to be stationary and both masses also non-rotating. If the initial center-to-center distances between these spherical masses is d, the escape velocity .sub.ESCAPE is the initial velocity that needs to be provided to the smaller mass m in the opposite direction to the gravitational attraction force between these two masses, which will allow it to escape the gravitational attraction force of M and reaching center-to-center distance d.fwdarw.. For this condition escape velocity .sub.ESCAPE is given as,

(121) $\begin{array}{cc}{v}_{\mathrm{ESCAPE}}=\sqrt{\frac{2GM}{d}}& \left(2.9\right)\end{array}$

(122) As an example, for an object like a spaceship launched straight up from the surface of the earth, the escape velocity .sub.ESCAPE relation given in (2.9) will become,

(123) $\begin{array}{cc}{v}_{\mathrm{ESCAPE}}=\sqrt{\frac{2G{m}_{\mathrm{EARTH}}}{r_{\mathrm{EARTH}}}}& \left(2.1\right)\end{array}$

(124) Where r.sub.EARTH and m.sub.EARTH are radius and the mass of the earth. The escape velocity .sub.ESCAPE as given in (2.9) can be calculated by applying conservation of energy principle, and by equating the kinetic energy to the potential energy for the mass m for the two-mass system with properties explained as m and M above. Since the method disclosed herein is derived using the energy conservation law, it is useful to derive the escape velocity using this very important concept in detail.

(125) Potential energy PE(r.sub.A) of the object with the mass m at a radius of r.sub.A, measured from the center of the reference mass M, can be calculated with the work integral W applied to the inverse square law gravitational attraction force field from infinite to r.sub.A as [21,23],

(126) 0$\begin{array}{cc}W=\mathrm{PE}\left(r_A\right)={}_{}^{r_A}G\frac{mM}{r^2}\mathrm{dr}=-G\frac{mM}{r_A}& \left(2.11\right)\end{array}$

(127) In other words, relation (2.11) is the work done for bringing the mass m with zero initial speed from infinite to a distance of r.sub.A away from the stationary mass M. During this free-fall from infinite to r.sub.A, the mass m will gain speed and will have a value of .sub.A. As a result, its kinetic energy KE(r.sub.A), becomes,

(128) $\begin{array}{cc}\mathrm{KE}\left(r_A\right)=\frac{1}{2}m{v}_A^2& \left(2.12\right)\end{array}$

(129) The energy conservation law states that the total energies at initial and final states must be equal, which can be expressed with an equation,
PE(r.sub.A)+KE(r.sub.A)=PE()+KE()(2.13)

(130) Energy conservation equation (2.13) explicitly becomes,

(131) $\begin{array}{cc}\frac{1}{2}m{v}_A^{2}G\frac{mM}{r_A}=0+0& \left(2.14\right)\end{array}$

(132) Solving .sub.A from (2.14) gives,

(133) $\begin{array}{cc}{v}_A=\sqrt{\frac{2GM}{r_A}}& \left(2.15\right)\end{array}$

(134) Due to reciprocity, this is also the magnitude of the escape velocity .sub.ESCAPE, in the opposite direction. Escape velocity from earth then can be formulated by substituting r.sub.EARTH instead of r.sub.A giving the same expression as given in (2.10) as,

(135) $\begin{array}{cc}{v}_{\mathrm{ESCAPE}}=\sqrt{\frac{2GM}{r_{\mathrm{EARTH}}}}& \left(2.16\right)\end{array}$

(136) FIG. 1A shows the escape velocity from earth along with the orbital velocity as a function of radius and altitude. As can be seen, escape velocity is greater than any of the orbital velocities. Since escape velocity has the radius of the Earth in its formulation it should also be a function of altitude as well.

(137) FIG. 1D depicts the escape velocity as a function of radius for Earth along with markings of the orbital altitudes given in Table 2.

(138) At this point it is very interesting to note that if .sub.ESCAPE, as given by relation (2.10), is equated to speed of light C, it becomes the non-relativistic definition of a Black Hole which was first conceived by an English scientist John Mitchell (1724-1793) in 1783. The mass/radius calculations of his Dark Star, derived from (2.15) is not very different than the relativistic calculations based on the solution of Einstein's field equations describing his general relativity work of 1915. Einstein's field equation describing general relativity, which was first solved by Karl Schwarzschild in 1915 gives the Schwarzschild Radius of the event horizon as,

(139) $\begin{array}{cc}{R}_S=\frac{2GM}{c^2}& \left(2.17\right)\end{array}$

(140) Where, C is the speed of light. Although John Mitchell's result is 122 years pre-special relativity, with no restrictions on the maximum velocity of anything, gives identical numerical value to (2.16) which is another remarkable coincidence in science.

(141) FIG. 1A depicts the orbital speed in [m/s] (meter per second) versus radius measured with respect to the stationary coordinate system where its origin is in the center of Earth and altitude measured from sea level.

(142) FIG. 1B depicts the orbital speed in Mach number, versus radius measured with respect to the stationary coordinate system where its origin is in the center of Earth and altitude measured from sea level, where 1 Mach is speed of sound at sea level (1 Mach=340 [m/s]) and the numbers 1, 5, and 10 correspond to the row numbers in Table 2. As is conventional, speed is represented herein as a Mach number. One of the early questions to be explored in rocketry was to determine if rocket propulsion could generate velocities sufficient to put an object into orbit, or sufficient to create an escape velocity .sub.ESCAPE from the Earth. To illustrate, escape velocity .sub.ESCAPE is shown superimposed in FIGS. 1A.

(143) FIG. 1C depicts the orbital period in hours versus radius measured with respect to the stationary coordinate system where its origin is in the center of the Earth and altitude measured from sea level. This is a way of presenting the extreme hypersonic speeds needed for putting a satellite into orbit in terms of orbital periods. For low Earth orbits (LEOs) the orbital periods are on the order of 90 minutes, meaning that satellites in LEOs rotate around the Earth every hour or two. The orbital periods of some well-known satellites are given in Table 2, with related data.

(144) A list of some orbital radiuses of well-known satellites and Van Allen Radiation Belts are presented in Table 2. These orbital radiuses between the Earth and the Moon, along with orbital periods, velocities, their orbital masses, and free fall times to the Earth surface, are calculated by the analytical formula given in [4-5] from the orbits that are presented in table. These orbitals are marked on the curves in FIG. 1B, FIG. 1C, and FIG. 2B.

(145) FIG. 2A depicts the gravitational acceleration g(h) as function of altitude h from sea level and distance from the earth's centered radius g(r), where r=h+r.sub.EARTH.

(146) FIG. 2B depicts the gravitational acceleration normalized to its sea level value as a function of altitude g(h) and the distance from the Earth's centered radius r, g (r), with the radius of the Earth taken at the equatorial radius of r.sub.EARTH=6,378 km.

(147) FIG. 2C depicts the gravitational acceleration normalized to its sea level value as a function of altitude g(h) with the maximum burn altitude h.sub.B(.sub.f=0) for T2W=1.1, 2, 4 and 8 marked on the curve.

(148) TABLE-US-00002 TABLE 2 Some Satellite altitudes and critical distances between the Earth and the Moon along with orbital periods, velocities, typical orbital masses, calculated fall times to Earth, and the ratio of the gravitational acceleration ratio to the g.sub.0 at the Earth's surface. Free Fall time to Satellite or Sea Level with critical Distances no air resistance Orbital Orbital Between Earth [minutes] and Period Velocity and the Moon Altitude[km]/mass[kg] g/g.sub.0 [Hour] [m/sec]/[Mach] 1 255/1,000 37.23/0.9246 1.49 7,751/22.8 2 400/Multiple Launches 38.75/0.8854 1.54 6,668/22.55 3 540/11,110 40.22/0.8499 1.59 7,590/22.32 4 781/689 42.78/0.7937 1.67 7,461/21.94 5 20,200/1,630 352.8/0.05758 11.98 3,872/11.39 6 35,786/4,276 711.6/0.02288 23.94 3,074/9.04 7 1,000 (~0.2 YEARTH)/NA 45.17/0.7472 1.75 7,350/21.62 8 12,000 (~2 YEARTH)/NA 200/0.12 6.89 4,657/13.70 9 13,000 (~3 YEARTH)/NA 217/0.1083 7.46 4,535/13.34 10 60,000 (~10 EARTH)/NA 1,412/0.00923 47.28 2,450/7.21
Included in the table with corresponding numbering in column 1 are: GOCE 1,000 [kg], ESA Gravity Field and Steady-State Ocean Circulation Explorer, ISS International Space Station Hubble 11,110 [kg], Iridium 689 [kg] GPS Block 11F 1,630 [kg] Latest Geo-Synchronous satellite 4,276 [kg]

(149) Inner Van Allen (Min): Lower altitude of the Van Allen inner radiation belt, due to interaction between Earth's magnetic field and incoming high speed charged particles, mainly from the sun. Inner Van Allen (Max): Higher altitude of the Van Allen inner radiation belt Outer Van Allen (Min): Lower altitude of the Van Allen outer radiation belt Outer Van Allen (Max): Higher altitude of the Van Allen outer radiation belt
Applying Energy Conservation Principle to the New Satellite Launch Method

(150) All the space launch systems today are very large systems and have rocket motors with very large thrust values giving them large thrust to weight ratios, always much greater than 1 to achieve the orbital velocities only achievable with very large burn rates as can be seen in Tables 4A and 4B. The potential energy of a satellite with a mass m at a circular orbit with a radius of r.sub.ORBIT, measured from the center of the earth can be calculated with the integral as done in (2.10) is [21-23],

(151) $\begin{array}{cc}\mathrm{PE}\left(r_{\mathrm{ORBIT}}\right)={}_{}^{r_{\mathrm{ORBIT}}}G\frac{mM}{r^2}\mathrm{dr}=-G\frac{mM}{r_{\mathrm{ORBIT}}}& \left(3.1\right)\end{array}$

(152) The kinetic energy of the satellite at a circular orbit with a radius measured from the center of the earth r.sub.ORBIT is,

(153) $\begin{array}{cc}KE\left(r_{\mathrm{ORBIT}}\right)=\frac{1}{2}m{v}_{\mathrm{ORBIT}}^2& \left(3.2\right)\end{array}$

(154) Since satellite is orbiting the earth at a circular orbit with a radius measured from the center of the earth r.sub.ORBIT, it must satisfy the .sub.ORBIT.sup.2 relation derived earlier in (2.5). Substituting (2.5) in (3.2) gives,

(155) $\begin{array}{cc}KE\left(r_{\mathrm{ORBIT}}\right)=\frac{1}{2}m\left(\frac{GM}{r_{ORBIT}}\right)& \left(3.3\right)\end{array}$

(156) The total energy of the satellite TE(r.sub.ORBIT) is the sum of its kinetic (3.3) and potential (3.1) energies giving,

(157) $\begin{array}{cc}\mathrm{TE}\left(r_{\mathrm{ORBIT}}\right)=\frac{1}{2}G\frac{mM}{r_{ORBIT}}-G\frac{mM}{r_{ORBIT}}=-\frac{1}{2}G\frac{mM}{r_{ORBIT}}& \left(3.4\right)\end{array}$

(158) Now, the question is to find what radius r.sub.0, measured from the center of the Earth, will give the same total energy (3.4) to the satellite with the same mass m while it is stationary. Since we assume the satellite is stationary at radius r.sub.0, its kinetic energy is zero, meaning that its total energy is equal to its potential energy,

(159) 0$\begin{array}{cc}TE\left(r_0\right)=PE\left(r_0\right)=-G\frac{mM}{r_0}& \left(3.5\right)\end{array}$

(160) Equating (3.4) to (3.5) r.sub.0 can be calculated solving it from,

(161) $\begin{array}{cc}-G\frac{mM}{r_0}=-\frac{1}{2}G\frac{mM}{r_{ORBIT}}& \left(3.6\right)\end{array}$

(162) Solving r.sub.0 from (3.6) gives surprisingly simple result of,
r.sub.0=2.Math.r.sub.ORBIT(3.7)

(163) Equation (3.7) means that the satellite in a circular orbit at a radius of r.sub.ORBIT, measured from the center of the earth, will have the same total energy while it is stationary at a radius of 2.Math.r.sub.ORBIT.

(164) FIG. 3A shows apparent three curves, but there are four curves, which two of them are on top of each other. The four curves plotted in FIG. 3A are the gravitational potential energies for a mass m equal to 1 [kg] at radiuses at r.sub.ORBIT and 2r.sub.ORBIT marked as PE(r.sub.ORBIT) and PE(2r.sub.ORBIT), kinetic energy KE(r.sub.ORBIT) and total energy TE(r.sub.ORBIT) of the mass m at a circular orbit at a radius of r.sub.ORBIT from the center of the earth. For convenience the mass m is taken as 1 [kg] in energy calculations. The marked points on the total energy curve TE(r.sub.ORBIT) are the total energies of the satellite with a mass of m at a circular orbit at orbital altitudes given in Table 2 following the same numbering on the plots just to show that TE(r.sub.ORBIT) and PE(2r.sub.ORBIT) exactly match, which is the numerical verification of the proof given in Sections 3.0 and 3.1 in detail. To have a clearer view of the tight spacings between the markings for the Table 2 orbital altitudes, the FIG. 3A data is presented by having its x axes presented in the logarithmic scale in FIG. 3B.

(165) FIG. 3C depicts the orbital velocity as a function of the orbital altitude .sub.ORBIT(h.sub.ORBIT) as shown in FIG. 1A with the velocities gained by dropping from r.sub.0=2r.sub.ORBIT to r.sub.ORBIT for the orbital altitudes given in Table 2. As can be seen the markers are exactly on the .sub.ORBIT(h.sub.ORBIT) curve.

(166) The only problem is that the velocity vector of the falling object with a mass m is towards the center of the stationary spherical body with a mass M, not in the tangent direction to the desired orbital trajectory. Therefore, during the descent a rocket thrust may be needed to steer/divert the descending spacecraft or satellite into the desired orbital trajectory.

(167) Substituting (2.7), which gives the relation between the orbital altitude h of the satellite and r.sub.ORBIT into (3.7) gives,
r.sub.0=2.Math.(r.sub.EARTH+h)(3.8)

(168) Typically, satellite orbits are given as their altitude from sea level instead of radius from the center of the earth. Therefore writing (3.8) as a function of the required altitude above sea level h.sub.0 corresponding to r.sub.0 gives,
h.sub.0=2.Math.(r.sub.EARTH+h)r.sub.EARTH=r.sub.EARTH+2h(3.9)

(169) Close examination of (3.9) shows that for small orbital altitudes compared to the radius of earth, like Low Earth Orbit (LEO) launches, the initial launch altitude is much larger than the orbital altitude h. In other words, for LEO launches the rocket must travel far longer distances compared to the conventional method of launching a satellite into orbit (h.sub.0h).

(170) FIG. 3D depicts the launch altitude as a function of the orbital altitude along with markings of the orbital altitudes given in Table 2. FIG. 3D displays the relation (3.9), but not very clearly for LEO launches, as the marked orbital launch altitudes are very tightly spaced and difficult to distinguish from each other. Proposed first launch altitude h.sub.0 to the desired orbital altitude h ratio displays this fact much clearly giving,

(171) $\begin{array}{cc}\left(h\right)=\frac{r_{\mathrm{EARTH}}+2h}{h}& \left(3.1\right)\end{array}$

(172) FIG. 3E depicts the launch altitude to the orbital altitude ratio along with markings of the orbital altitudes given in Table 2 where the x axis is plotted in logarithmic scale. FIG. 3E depicts (h) function where the x axis is drawn logarithmic to show the LEO ratios much clearly. As can be seen for LEO the ratios are above 25, even for large values of desired orbital altitudes h, the ratio approaches 2.

(173) FIG. 3F depicts the fall times from (2.Math.r.sub.ORBIT to r.sub.ORBIT), (2.Math.r.sub.ORBIT to r.sub.EARTH and (r.sub.ORBIT to r.sub.EARTH).

(174) FIG. 3G depicts the velocity that the payload gains, which was at rest at a radius of r.sub.h in terms of the orbital velocity v.sub.ORBIT for a circular orbit at the radius of r.sub.ORBIT as a function of s.sub.f when falling from r.sub.h to r.sub.ORBIT,

(175) $sf=\frac{r_h}{r_{\mathrm{ORBIT}}}$

(176) To demonstrate the derived mechanics for an orbital altitude h=400 [km] with a real numerical example will be useful.

(177) a. Launch the satellite to an altitude h.sub.0 from sea level, same altitude of International Space Station (ISS) as given by relation (3.9),
h.sub.0=6,378+4002=7,178[km](3.11)

(178) The rocket is launched from the surface of the earth such that it climbs to 7,178 [km] of altitude from sea level straight up, in radial direction opposite to the gravitational attraction force of the earth and when it reaches this altitude, it will become stationary, which means it has zero kinetic energy. At this point its potential energy is equal to the total energy, and it is equal to the total energy if it was in an orbit at 400 [km]. It can reach this altitude with any speed profile, and as long as it becomes stationary at the altitude h.sub.0, it will have the same total energy as if it was in a circular orbit around the earth at an altitude 400 [km] above sea level.

(179) ii) In this step the satellite falls back freely towards earth. During the descent, the satellite's speed will only have a radial component, towards to the center of the earth and when it falls to the desired orbital altitude of 400 [km], the magnitude of its speed will be equal to the speed needed to put it in the orbit at 400 [km] altitude from sea level, but its speed will be in a radial direction, not in a direction tangential to the desired orbital trajectory, and it will continue to fall towards the Earth.

(180) iii) Since during the free fall explained above, the descent velocity vector direction is typically never tangential to the orbital trajectory to put the satellite into orbit. Therefore, during the descent some energy must be supplied to the satellite through a proper rocket thrust in the proper direction to provide the change in the descent trajectory.

(181) This concept, even the math seems to be correct, is counter intuitive to conventional launch strategy. As opposed to conventional logic, a satellite can be put to an orbit of 400 [km] above the sea level (for example), without climbing to the same orbital altitude with a roll and a gravity turn as has been done since 1957, with an orbital velocity tangent to the orbit, by first climbing to an altitude of 7,178 [km] above sea level, which is 17.945 times higher than 400 [km], and attaining an energy equal to the total orbiting energy.

(182) FIG. 13I demonstrates the methodology employed for two different values of s.sub.f=1.5 and 2 along with the related mass versus time plots of FIG. 13J and FIG. 13K.

(183) Similar graphical demonstration is given for 20,200 [km] (GPS) orbital launch in FIG. 13M with the related mass versus time plots FIG. 13N and FIG. 13O.

(184) At first it may seem counter intuitive and very unnecessary launch strategy of going up for 7,178 [km] to put a satellite to an orbit at 400 [km], as opposed to reaching 400 [km] of altitude directly with a curved trajectory which becomes tangential to the orbit with the velocity vector tangential to the orbital trajectory at the calculated orbital speed. So, it is a good practice to verify this claim by substituting r.sub.ORBIT in the total energy equation (3.4) for the orbital radius r.sub.ORBIT as,

(185) $\begin{array}{cc}\mathrm{TE}\left(r_{\mathrm{ORBIT}}\right)=\frac{1}{2}m{v}^2-G\frac{mM}{r_{ORBIT}}=0-\frac{1}{2}G\frac{mM}{r_{ORBIT}}& \left(3.12\right)\end{array}$$\begin{array}{cc}\frac{1}{2}m{v}^2=\frac{1}{2}G\frac{mM}{r_{\mathrm{ORBIT}}}& \left(3.13\right)\end{array}$

(186) Giving the same orbital velocity derived and given in (2.6) as,

(187) $\begin{array}{cc}v=v_{ORBIT}=\sqrt{\frac{GM}{r_{ORBIT}}}& \left(3.14\right)\end{array}$

(188) This is not exactly a proof, because the formula was derived using the same energy conservation law. While the energy conservation law is well understood, this exercise shows that there is no arithmetic error in the calculations. The result is counter intuitive, and additional proof can be derived using a different approach, like Newton's equation of motion for a free-falling body in an inverse square law gravitational field to have additional confidence in the derivation.

(189) a. Analytical Solution of Newton's Equation of Motion to Derive Velocity as Function of Radius of a Free-Falling Mass in an Inverse Square Law Gravitational Field

(190) Newton's equation of motion for a free-falling object from a radius of r.sub.0 in an inverse square law gravitational field given as,

(191) $\begin{array}{cc}F=-G\frac{Mm}{r^2}=m\frac{dv}{\mathrm{dt}}& \left(3.15\right)\end{array}$

(192) Where,

(193) $\begin{array}{cc}v=\frac{dr}{\mathrm{dt}}& \left(3.16\right)\end{array}$

(194) Analytically and if,

(195) $\begin{array}{cc}v\left(r_{\mathrm{ORBIT}}\right)=\sqrt{\frac{GM}{r_{\mathrm{ORBIT}}}}\mathrm{where}{r}_0=2.Math.r_{\mathrm{ORBIT}},& \left(3.17\right)\end{array}$ then the proof is certain.

(196) Where F, G, m, M, r, v and t are the gravitational force between mases m and M, the universal gravitational constant, mass of the object in motion being modeled with the inverse square gravitational field of object with a mass of M, which is assumed to be stationary due to its very large mass with respect to m, radius, which is the center-to-center distance between the two spherical objects with masses m and M, velocity, and time respectively. Since a space launch system from the Earth is the interest, M is the mass of the Earth and m is the mass of the satellite or the spacecraft.

(197) Although the ordinary differential equation (3.13) is simple, its analytical solution is not very straightforward. Applying the Leibnitz's chain rule [22-26], which was shown to be an especially useful trick derived in [22] for the analytical solution of the non-linear Poisson's equation in the MOS (Metal-Oxide-Semiconductor), gives,

(198) $\begin{array}{cc}\frac{dv}{\mathrm{dt}}=\frac{dv}{dr}\frac{dr}{\mathrm{dt}}& \left(3.18\right)\end{array}$

(199) Substituting (3.16) in (3.18) gives,

(200) 0$\begin{array}{cc}\frac{dv}{\mathrm{dt}}=\frac{dv}{dr}v& \left(3.19\right)\end{array}$

(201) Substituting (3.19) in (3.15) gives,

(202) $\begin{array}{cc}-G\frac{\mathrm{mM}}{r^2}=\mathrm{mv}\frac{\mathrm{dv}}{\mathrm{dr}}& \left(3.2\right)\end{array}$

(203) Simplifying (3.20) gives,

(204) $\begin{array}{cc}-G\frac{M}{r^2}=v\frac{\mathrm{dv}}{\mathrm{dr}}& \left(3.21\right)\end{array}$

(205) Rearranging (3.21) gives,

(206) $\begin{array}{cc}\mathrm{GM}\frac{\mathrm{dr}}{r^2}=\mathrm{vdv}& \left(3.22\right)\end{array}$

(207) Equation (3.22) becomes a quite simple differential equation that can be solved either with definite integrals or by solving the integration constant with the boundary condition. The rocket equation derivation is done employing many definite integrals, just to add a different flavor to the work, the solving the integration constant gives the following equation,

(208) $\begin{array}{cc}\mathrm{GM}\frac{1}{r}=\frac{1}{2}{v}^2+C& \left(3.23\right)\end{array}$

(209) Where C is an arbitrary integration constant which needs to satisfy the initial conditions of the problem. Since at t=0 the radius is r=r.sub.0 and the initial velocity is .sub.0, the integration constant C must satisfy the equation,

(210) $\begin{array}{cc}\mathrm{GM}\frac{1}{r_0}=\frac{1}{2}{v}_0^2+C& \left(3.24\right)\end{array}$

(211) Giving,

(212) $\begin{array}{cc}C=\mathrm{GM}\frac{1}{r_0}-\frac{1}{2}{v}_0^2& \left(3.25\right)\end{array}$

(213) Substituting (3.25) into (3.23) gives the complete solution as,

(214) $\begin{array}{cc}\mathrm{GM}\frac{1}{r}=\frac{1}{2}{v}^2+\mathrm{GM}\frac{1}{r_0}-\frac{1}{2}{v}_0^2& \left(3.26\right)\end{array}$

(215) Rearranging terms in (3.26) gives,

(216) $\begin{array}{cc}{v}^2=2\left(\mathrm{GM}\frac{1}{r}-\mathrm{GM}\frac{1}{r_0}+\frac{1}{2}{v}_0^2\right)& \left(3.27\right)\end{array}$

(217) Finally, the (r) relation that we are looking for becomes,

(218) $\begin{array}{cc}v\left(r\right)=\sqrt{2\left(\mathrm{GM}\frac{1}{r}-\mathrm{GM}\frac{1}{r_0}+\frac{1}{2}{v}_0^2\right)}& \left(3.28\right)\end{array}$

(219) Writing (3.28) in more compact form gives,

(220) 0$\begin{array}{cc}v\left(r\right)=\sqrt{2\mathrm{GM}\left(\frac{1}{r}-\frac{1}{r_0}\right)+v_0^2}& \left(3.29\right)\end{array}$

(221) For a falling mass to the earth the bounds of r and r.sub.0 can be given as,
r.sub.Err.sub.0 and r.sub.Er.sub.0(3.30)

(222) For .sub.0=0 (3.29) gives,

(223) $\begin{array}{cc}v\left(r\right)=\sqrt{2\mathrm{GM}\left(\frac{1}{r}-\frac{1}{r_0}\right)}& \left(3.31\right)\end{array}$
r=r.sub.0=2.Math.r.sub.ORBIT(3.32)

(224) (3.31) becomes,

(225) $\begin{array}{cc}v\left(r_{\mathrm{ORBIT}}\right)=\sqrt{2\mathrm{GM}\left(\frac{1}{r_{\mathrm{ORBIT}}}-\frac{1}{r_0}\right)}& \left(3.33\right)\end{array}$

(226) Finally, the counter intuitive same answer that is being sought appears as,

(227) $\begin{array}{cc}v\left(r_{\mathrm{ORBIT}}\right)=\sqrt{\frac{\mathrm{GM}}{r_{\mathrm{ORBIT}}}}& \left(3.34\right)\end{array}$

(228) To write the result in text, it can be said that if an object in an inverse square law gravitational field falls from any radius r.sub.0 with respect to a larger mass M, with a zero initial velocity to half of its initial radius r.sub.0 with respect to the larger mass M, the speed that it will gain is equal to the speed obtained when it is in a circular stable orbital rotational speed around the larger mass M at an orbital radius of r.sub.ORBIT. In the process of proving the relation (3.15) beyond any doubt, a very general and handy (r) relation is also given as (3.27) which is also used later in this work. Since more confidence exists in the (3.7) it is also useful to know the fall times between the first and second radiuses. Analytical formulas are available [4, 5] and FIG. 3F depicts the fall times from (2.Math.r.sub.ORBIT to r.sub.ORBIT), (2.Math.r.sub.ORBIT to r.sub.EARTH) and (r.sub.ORBIT to r.sub.EARTH) as a function of altitude up to 60,000 [km]. As can be seen, all of the fall times related to LEO satellites are very tightly spaced and difficult to distinguish from each other, which are shown on the left-hand side of FIG. 3F. Table 3A lists the calculated fall times from 2.Math.r.sub.ORBIT to r.sub.ORBIT for the satellites listed in Table 2, with the corresponding numbering.

(229) TABLE-US-00003 TABLE 3A Calculated Fall times from 2 .Math. r.sub.ORBIT to r.sub.ORBIT for the satellites listed in Table 2, with the corresponding numbering. As can be seen if the time to orbit is an issue, the method presented in this disclosure is not necessarily the best choice. Numbering Orbital Launch Fall Time Fall Time in Altitude Altitude from from TABLE 2 [km] h.sub.0[km] h.sub.0 .fwdarw. h[s] h.sub.0 .fwdarw. h[hour] 1 255 6,888 2,199 0.611 2 400 7,178 2,272 0.631 3 540 7,458 2,343 0.651 4 781 7,940 2,466 0.685 5 20,200 46,778 17,644 4,901 6 35,786 77,950 35,257 9,793
a. Ideal Gas Law, Gas Density

(230) The gas density calculations will be useful in Section 7 where some basic rocket fuel and oxidizer calculations are given. Modifying the well-known ideal gas law,
PV=nRT(3.35)

(231) Or more conveniently written as,
PVM.sub.mol=mRT(3.36)

(232) Where P, V, n, R, T, m and M are pressure, volume, mol, ideal gas constant, temperature, mass of the gas, and molecular mass respectively. Dividing both sides of (3.34) with V gives the ideal gas law as,

(233) $\begin{array}{cc}{\mathrm{PM}}_{\mathrm{mol}}=\frac{m}{V}\mathrm{RT}=\mathrm{RT}& \left(3.37\right)\end{array}$

(234) Where is the density of the gas which leads to the gas density formula,

(235) $\begin{array}{cc}=\frac{{\mathrm{PM}}_{\mathrm{mol}}}{\mathrm{RT}}& \left(3.38\right)\end{array}$

(236) Gases related to this work are H.sub.2, N.sub.2, O.sub.2, CO.sub.2, and He. Molecular masses M for these gases can be calculated based on their atomic masses giving approximately 2, 28, 32, 44, and 4 [gr/mol] respectively. The gas density relation as given in (3.38) and its function of altitude as shown in FIG. 1E gets into the numerical solution of the equation of motion for including the air drag during the rocket's travel in the atmosphere. Air drag force F.sub.D is expressed by the well-known relation,

(237) $\begin{array}{cc}{F}_D=\frac{1}{2}{k}_{D}S{v}^2& \left(3.39\right)\end{array}$

(238) Where , k.sub.D, S, and v are the air density, drag coefficient, surface area perpendicular to the air flow and air speed respectively.

(239) If the satellite in orbit faces air drag, as formulated in (3.39), it will either burn out due to frictional heat generated or it will lose speed and energy and eventually spiral down to Earth and will crash to the Earth's surface. As can be seen in FIG. 1E density of air goes very close to zero after altitudes of 100 [km], which gives practically zero drag force, and the satellite can stay in orbit forever after reaching the orbital velocity with the right direction. This is the only reason of putting satellites into orbit higher than 100 [km], or in other words above the atmosphere. Sixty-seven years after the launch of the first satellite in 1957, still many news and social media news sources very incorrectly use zero g, no gravity, or zero gravity terminology at the satellite or space craft launch news. As can be seen in FIGS. 2A and 2B the gravitational pull of earth in all LEOs is not less than 0.75 of its value at sea level, not even close to zero gravity as often reported. Zero gravity is only possible at an infinite distance from a mass.

(240) Rocket Thrust and Mass.

(241) Rockets create thrust by ejecting parts of its mass with high velocity, which can be mathematically formulated using the conservation of momentum principle, and can be considered a straight-forward calculus exercise [1-3]. There are many scientists who need to be credited for this derivation going back to 1810 [3]. The first record of the derivation of the rocket equation is known to have been done by the British mathematician William Moore in his work Theory on the motion of Rockets and Treatise on the Motion of Rockets and an assay on Naval Gunnery, which was published in 1813. The minister William Leitch, another British scientist also independently derived the fundamentals of rocketry in 1861. Robert Goddard in the USA also independently derived the rocket equation in 1912. Hermann Oberth in Germany derived the same equation studying the feasibility of space travel in 1920's.

(242) The Russian scientist Konstantin Tsiolkovsky's (1857-1935) derivation of the rocket equation in 1897 is accepted as being the first to consider whether rockets could achieve the speeds necessary for space travel and therefore it is named as Tsiolkovsky Rocket Equation, which he called formula of aviation. He is also the inventor of multi-stage rockets based on the very interesting mathematical properties of the rocket equation. He is also the inventor of the space elevator and many other things related to rocket science and is therefore accepted as being the father of rocket science. Hermann Julius Oberth (1894-1989) a German rocket scientist also derived the rocket equation while working in Peenemunde during Second World War for the V-2 ballistic rocket with his student Werner von Braun who ended up running the Apollo program at NASA.

(243) Tsiolkovsky's Rocket Equation.

(244) Several forms of derivation are possible [1-3]. The equation of motion for the rocket in vector notation can be written as,

(245) $\begin{array}{cc}{\stackrel{.fwdarw.}{F}}_{\mathrm{EXT}}=m\left(t\right)\frac{d\stackrel{.fwdarw.}{v}}{\mathrm{dt}}+v_e\frac{d\left[m\left(t\right)\right]}{\mathrm{dt}}\stackrel{.fwdarw.}{u}& \left(5.1\right)\end{array}$ where {right arrow over (F)}.sub.EXT, m(t), , .sub.e, and {right arrow over (u)} are the sum of external forces, the mass of the rocket, which is a function of time, velocity of the rocket, exhaust gas velocity relative to the rocket, and the unit exhaust gas velocity vector with respect to the rocket, respectively. The time dependent mass of the rocket for constant fuel burn rate b.sub.r can be given as,
m(t)==m.sub.0b.sub.rt(5.2) where m.sub.0 and b.sub.r are the initial total mass of the rocket and fuel burn rate in [kg/s]. With no external forces and in scalar notation (5.1) can be written as,

(246) $\begin{array}{cc}m\left(t\right)\frac{dv}{\mathrm{dt}}=-v_e\frac{d\left[m\left(t\right)\right]}{\mathrm{dt}}& \left(5.3\right)\end{array}$ where .sub.e is the effective exhaust velocity in [m/s], which has a range of 2,500-4,500 [m/s] based on the propellant used, the rocket engine design, its convergent-divergent nozzle geometry, and injection-mix efficiency of the propellant into the thrust chamber for liquid propellants. The range of these some values of .sub.e for some popular rockets are presented in Table 4A.

(247) TABLE-US-00004 TABLE 4A List of exhaust velocity and propellant types of some well-known rockets at lift-off. RP-1 is a rocket fuel based on kerosene and LOX is Liquid OXygen. Mass Thrust to Rocket Engine V.sub.E[m/s] Flow weight Country of Launch Fuel Specific Thrust Rate (T2W) Origin Vehicle Oxidizer Impulse[s] [kN] [kg/s] Dry weight V-2 Rocket A-4 (V-2) (75% Alcohol 1,989 m/s 245 133 kg/s Engine 25% water) NAZI Germany LOX 203 s Aerojet TITAN II Hypergolic 2,528 m/s 1,900 LR-87-AJ-1 GLV Propellant 258 s (2 Nozzle) (Gemini (Oxidizer ignites USA Launch at contact) Veichle) Hydrezeine (Aerozine-50) and N.sub.2O.sub.4 Rocketdyne F-1 Saturn V RP-1 2,580 m/s 7,740 2,577 kg/s 94 USA First Stage LOX 263 s 8,400 kg (5 F-1s giving total thrust of 35,000 [kN]) Energomash Energia RP-1 3,030 m/s 7,250 82 RD-170 LOX 309 s 9,750 kg (4 nozzle) Russia Enorgamash Atlas V RP-1 3,050 m/s 3,830 1,250 kg/s 78.44 RD-180 LOX 311 s 5,480 kg (2 nozzle) Russia Rocketdyne Space Liquid Hydrogen 4,460 m/s 5,250 RS-25 Shuttle LOX 455 s USA (3X RS-25) Thiokol/Pratt- Space Solid Fuel 2,370 m/s 2 6,500 250 Ton 45 Tons Empty Withney SRB Shuttle PBNA-APCP 242 s Propellant Diameter 3.71 m USA 2 Solid Tb = 127 s Height 45.46 m Rocket Boosters Blue Horizon ULA Vulcan CH4 3,087 m/s BE-4 USA Centaur LOX 315 s Vulcain 2 Ariene-5 Liquid Hydrogen 1,390 European Space First Stage LOX Agency P241 Ariene-5 Solid Fuel 2 7,080 273 Ton 33 Tons Empty European Space First Stage 2 AP Aluminum Gross Diameter 3.06 m Agency Solid Rocket HTBP Tb = 140 s Height 31.6 m Boosters SpaceX Merlin SpaceX RP-1 2,770 m/s 854 kN 184 USA Falcon LOX 311 s 470 kg SpaceX Raptor SpaceX CH.sub.4 3,400 m/s 2,210 kN 650 kg/s 200 USA Starship LOX 1,500 kg Heavy 33X Raptors 72MN

(248) Multiplying both sides of (5.3) with dt gives,
m(t)d=.sub.ed[M(t)](5.4)

(249) On the other hand, differentiating (5.2) the generated thrust, F.sub.T in scalar form in [Newtons] becomes,
F.sub.T=.sub.eb.sub.r(5.5)

(250) The negative sign indicates that the generated thrust is in the opposite direction of the exhaust gas flow relative to the rocket. Multiplying both sides of (5.4) with dt eliminates the time dependency and gives the very simple differential equation where its analytical solution is trivial as,

(251) $\begin{array}{cc}dv=-v_e\frac{dm}{m}& \left(5.6\right)\end{array}$

(252) Integration of both sides of (5.6) using the proper limits gives,

(253) 0$\begin{array}{cc}{}_{v_0}^{v_f}dv=-v_e{}_{m_0}^{m_f}\frac{dm}{m}& \left(5.7\right)\end{array}$ where m.sub.0, m.sub.f, .sub.0, and .sub.f are the initial and final total mass of the rocket in [kg], and initial and final velocity of the rocket, respectively in [m/s]. Finally, the solution of (5.7) leads rocket equation to its most common form as,

(254) $\begin{array}{cc}v=v_e\ln \left(\frac{m_0}{m_f}\right)=I_{SP}{g}_0\ln \left(\frac{m_0}{m_f}\right)& \left(5.8\right)\end{array}$ where,
=.sub.f.sub.0(5.9) where , I.sub.SP, and g.sub.0 are the difference in velocity, specific impulse in seconds [s], and standard gravity in [m/s.sup.2] respectively. Since in a rocket m.sub.0m.sub.f, is always larger than the mass ejection velocity or exhaust gas exit velocity from the rocket engine nozzle, which is in the range of 2,500-4,500 [m/s] as can be seen in Table 4A. As can be seen relations (5.9) and (5.11) enable the rocket to achieve very large orbital velocities or even escape velocity, which is 11,000 [m/s] on Earth's surface, and theoretically propel the rocket to an infinite distance away from the Earth ignoring other gravitational effects present.

(255) For constant fuel burn rate b.sub.r the fuel burn time T.sub.B relates the final mass m.sub.f to the burn rate b.sub.r as,
m.sub.f=m.sub.0m.sub.prop=m.sub.0b.sub.rT.sub.B(5.10) where m.sub.prop is the mass of the (fuel+oxidizer), which is consumed until the end of fuel burn time T.sub.B. One of the most important and useful applications of the rocket equation is in relating the initial and final mass of the rocket as a function of desired speed difference . Taking the first part of (5.8), it can be written as,

(256) $\begin{array}{cc}v=v_e\ln \left(\frac{m_0}{m_f}\right)& \left(5.11\right)\end{array}$

(257) Dividing both sides with .sub.e gives,

(258) $\begin{array}{cc}\frac{v}{v_e}=\ln \left(\frac{m_0}{m_f}\right)& \left(5.12\right)\end{array}$

(259) (5.12) can also be written as,

(260) $\begin{array}{cc}{e}^{\frac{v}{v_e}}=\frac{m_0}{m_f}& \left(5.13\right)\end{array}$

(261) Solving m.sub.f from (5.13) gives,

(262) $\begin{array}{cc}{m}_f=m_0{e}^{-\frac{v}{v_e}}& \left(5.14\right)\end{array}$

(263) Substituting the first part of (5.10),
m.sub.f=m.sub.0m.sub.prop(5.15)

(264) Into (5.14) gives,

(265) $\begin{array}{cc}{m}_0{e}^{-\frac{v}{v_e}}=m_0-m_{prop}& \left(5.16\right)\end{array}$

(266) Solving m.sub.prop from (5.16) gives,

(267) $\begin{array}{cc}{m}_{prop}=m_0\left(1-e^{-\frac{v}{v_e}}\right)& \left(5.17\right)\end{array}$

(268) The equation (5.17) is very useful because by entering 2 numbers into it, it is possible to calculate m.sub.prop, the fuel needed as a percentage of the initial mass m.sub.0 to gain a given speed and rocket exhaust velocity for the case of no external forces.

(269) 5.1 Derivation of Tsiolkovsky's Rocket Equation for Constant Gravitational Acceleration.

(270) For a rocket moving straight up vertically against Earth's gravitational force the most important force to consider is the Earth's gravitational force. The resulting gravitational acceleration g acting upon is formulated as [4-5, 15, 18-21],

(271) $\begin{array}{cc}g=G\frac{m_{EARTH}}{r^2}\mathrm{where}r=r_{EARTH}+h& \left(5.18\right)\end{array}$ where r, G, m.sub.EARTH, h, r.sub.EARTH are object distance to the center of the Earth, Newton's constant of gravitation, mass of the Earth, and altitude measured from the surface of the Earth. Following are the numerical values in (5.18) as,
G=6.67410.sup.11[m.sup.3.Math.kg.sup.1] or [N.Math.m.sup.2.Math.kg.sup.2]
Mass of Earthm.sub.EARTH=5.97210.sup.24[kg]

(272) The Earth's Polar and Equatorial radiuses are slightly different and are,
r.sub.Epolar=6,357[km] and r.sub.EEqutorial=6,378[km].

(273) As shown in FIGS. 2A, 2B, and 2C, for all practical purposes the gravitational acceleration g for the range of altitudes involved in this analysis can be assumed constant with a numerical value of 9.81 [m/s.sup.2] at sea level. Since rocket propulsion can deliver the very high velocities required to put a satellite into orbit, it became the only practical means of doing so. In this mode the Earth's gravitational acceleration g must be included in the rocket equation [1-3].

(274) For this case the rocket equation of motion becomes,

(275) $\begin{array}{cc}{\stackrel{.fwdarw.}{F}}_{\mathrm{EXT}}=m\frac{d\stackrel{.fwdarw.}{v}}{\mathrm{dt}}-v_e\frac{dm}{\mathrm{dt}}\stackrel{.fwdarw.}{u}& \left(5.19\right)\end{array}$

(276) The second term on the right-hand side is the thrust generated in vector form where {right arrow over (u)} is the unit vector in the direction of the flight path relative to the rocket, which is opposite to the rocket velocity vector . Rearranging (5.19) gives,

(277) 0$\begin{array}{cc}m\frac{d\stackrel{.fwdarw.}{v}}{\mathrm{dt}}={\stackrel{.fwdarw.}{F}}_{\mathrm{EXT}}-v_e\frac{dm}{\mathrm{dt}}\stackrel{.fwdarw.}{u}& \left(5.2\right)\end{array}$

(278) The constant force {right arrow over (F)}.sub.EXT generated by a constant acceleration g opposing the direction of the thrust can be introduced into (5.20), and ignoring air drag gives the one-dimensional scalar rocket equation of motion as,

(279) $\begin{array}{cc}m\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{mg}-v_e\frac{\mathrm{dm}}{\mathrm{dt}}& \left(5.21\right)\end{array}$

(280) Multiplying both sides of (5.21) with dt as done before gives,
md=mgdt.sub.edm(5.22)

(281) Dividing both sides of (5.22) by m gives,

(282) $\begin{array}{cc}\mathrm{dv}=-\mathrm{gdt}-v_e\frac{\mathrm{dm}}{m}& \left(5.23\right)\end{array}$

(283) (5.23) can be analytically integrated as,

(284) $\begin{array}{cc}{}_{v_0}^{v_f}\mathrm{dv}=-{}_{t_0}^{t_f}\mathrm{gdt}-v_e{}_{m_0}^{m_f}\frac{\mathrm{dm}}{m}& \left(5.24\right)\end{array}$

(285) Giving,
f.sub.0=g(t.sub.ft.sub.0).sub.e[ln(m.sub.f)ln(m.sub.0)](5.25)

(286) On the other hand, (t.sub.ft.sub.0) in (5.25) is the burn time T.sub.B for constant burn rate b.sub.r calculated as,

(287) $\begin{array}{cc}{T}_B=\left(t_f-t_0\right)=\frac{m_{\mathrm{prop}}}{b_r}& \left(5.26\right)\end{array}$

(288) Arranging (5.25) and substituting (5.26) in it gives,

(289) $\begin{array}{cc}{v}_f=v_0+v_e\ln \left(\frac{m_0}{m_f}\right)-g\frac{m_{\mathrm{prop}}}{b_r}& \left(5.27\right)\end{array}$

(290) Equation (5.27) is the corresponding equation (5.11) for the case of no external forces. In this work the rocket in the launch stage moves in the opposite direction of gravitational force, but during the descent or capture stage the rocket moves in the same direction as gravitational force. If the gravitational acceleration is in the direction of the thrust, the sign of the last term in (5.27) changes to a positive sign (+). Covering both cases (5.27) can be written as,

(291) $\begin{array}{cc}v=v_f-v_0=v_e\ln \left(\frac{m_0}{m_f}\right)g\frac{m_{\mathrm{prop}}}{b_r}& \left(5.28\right)\end{array}$ where the + sign corresponds to gravitational acceleration if it is in the same direction as thrust, the case where it is used in the descent stage of the system described herein. The velocity difference (+g), which is defined as the rocket velocities going opposite and in the same direction of gravitational acceleration, has the same mass parameters,

(292) $\begin{array}{cc}v\left(g\right)=2.Math.g\frac{m_{\mathrm{prop}}}{b_r}=2{\mathrm{gT}}_B& \left(5.29\right)\end{array}$

(293) Equation (5.29) shows that the same velocity can be gained with a significantly smaller mass of propellant. Rocket assisted descent is important and requires a more detailed analysis as is given in Section 5.2. Coming back to lift-off case, since the rocket should be able to lift-off the ground with full initial mass m.sub.0, the thrust F.sub.T must satisfy,

(294) $\begin{array}{cc}{m}_0=\frac{F_T}{T2W.Math.g}=\frac{v_e{b}_r}{T2W.Math.g}\mathrm{where}T2W>1& \left(5.3\right)\end{array}$

(295) The parameter T2W is the thrust-to-weight ratio of the rocket at the launch pad, which must be greater than 1 for a successful launch. Due to safety of the launch the typically T2W at the launch is set to a number greater than 1.5.

(296) Relation (5.27) and its more general form (5.28) can be represented better by introducing a parameter (t) as,

(297) $\begin{array}{cc}\left(t\right)=\frac{m\left(t\right)}{m_0}\mathrm{where}0<1& \left(5.31\right)\end{array}$

(298) The minimum value of (t) is reached when all the propellant is consumed. Since there is always a payload involved with any launch m.sub.f>0, the final value of .sub.f is greater than zero as given in (5.31). The inverse of (t) can be written as,

(299) 0$\begin{array}{cc}\frac{1}{{}_f}=\frac{m_0}{m_f}=\frac{m_0}{m_0-m_{\mathrm{prop}}}& \left(5.32\right)\end{array}$

(300) Since,
m.sub.prop=m.sub.0m.sub.f(5.33)

(301) Time dependency of (t) in terms of burn rate can be written explicitly as,

(302) $\begin{array}{cc}\left(t\right)=\frac{m\left(t\right)}{m_0}=\frac{m_0-b_{r}t}{m_0}& \left(5.34\right)\end{array}$

(303) Substituting t=0 and t=T.sub.B in (5.34), the upper (final) and lower (initial) limits of .sub.f in powered flight can be written as,

(304) $\begin{array}{cc}\left(0\right)={}_0=1\mathrm{and}\left(T_B\right)={}_f=\frac{m_0-m_{\mathrm{prop}}}{m_0}& \left(5.35\right)\end{array}$

(305) Some arithmetic performed on the second part of (5.27) and (5.28) gives,

(306) $\begin{array}{cc}g\frac{m_{\mathrm{prop}}}{b_r}=g\frac{m_0-m_f}{b_r}=g\frac{m_0\left(1-\frac{m_f}{m_0}\right)}{b_r}& \left(5.36\right)\end{array}$

(307) Multiplying dominator and the denominator of (5.36) with .sub.E and T2W gives,

(308) $\begin{array}{cc}\frac{\left(v_{E}T2W\right){\mathrm{gm}}_0\left(1-\frac{m_f}{m_0}\right)}{\left(v_{E}T2W\right)b_r}=\frac{v_E\left({\mathrm{gm}}_{0}T2W\right)\left(1-\right)}{\left(v_E{b}_r\right)T2W}& \left(5.37\right)\end{array}$

(309) On the other hand, writing thrust in terms of thrust-to-weight ratio T2W gives,
F.sub.T=gm.sub.0T2W=.sub.Eb.sub.r(5.38)

(310) Substituting (5.38) in (5.37) gives,

(311) $\begin{array}{cc}\frac{v_E{\mathrm{gm}}_{0}T2W\left(1-f\right)}{v_E{b}_{r}T2W}=\frac{v_E{F}_T\left(1-f\right)}{F_T.Math.T2W}=\frac{v_E\left(1-f\right)}{T2W}& \left(5.39\right)\end{array}$

(312) The equation (5.29) which gives the velocity of the rocket accelerating in the opposite direction of the uniform gravitational acceleration g after burning all its propellant m.sub.prop becomes expressed in a very compact form with very simple rocket related variables pf and T2W as,

(313) $\begin{array}{cc}v\left({}_f,T2W\right)=v_B=v_E\left[\ln \left(\frac{1}{{}_f}\right)-\frac{\left(1-{}_f\right)}{T2W}\right]& \left(5.4\right)\end{array}$

(314) Since the initial velocity of the rocket when it is standing on the launch pad is zero, using .sub.B, where the subscript B representing Burnt, instead of (.sub.f, T2W), is a more meaningful and more convenient in the following math. Close examination of (5.40) gives,
for f.fwdarw.0.sub.B(0).fwdarw.(5.41)

(315) This result is clearly non-physical and needs correction which is explained in Section 6 below. And,
for .sub.f=1.sub.B(1)=0(5.42)

(316) FIG. 4A depicts .sub.B(.sub.f, T2W) for thrust-to-weight ratio T2W=1.1 and 8 with a constant acceleration g. The asymptote at which makes .sub.B.fwdarw. at .sub.f=0 is marked with dotted vertical line is clearly shown. One of the early questions to be explored in rocketry was to determine if rocket propulsion could generate velocities sufficient to put an object into orbit, or sufficient to create an escape velocity .sub.ESCAPE from the Earth. To illustrate, escape velocity .sub.ESCAPE is superimposed on the .sub.B(.sub.f, T2W) curves. As can be seen, for .sub.f values very close to zero, .sub.B(.sub.f, T2W).sub.ESCAPE, where the rocket escapes the Earth's gravitational pull and reaches an infinite altitude h.fwdarw.+. The linear scale of .sub.f in FIG. 4A does not clearly show how close the .sub.f value must become for .sub.B(.sub.f, T2W).sub.ESCAPE or for any hypersonic velocities to put a satellite into a desired orbit as shown in FIGS. 1A and 1B. This could be achieved by presenting the same data in FIG. 4A where the horizontal axis is drawn in logarithmic scale.

(317) FIG. 4B depicts .sub.B(.sub.f, T2W) for thrust-to-weight ratio T2W=1.1 and 8 with a constant acceleration g, where the horizontal axis is in logarithmic scale. The required .sub.f values for achieving .sub.B(.sub.f, T2W).sub.ESCAPE or for any hypersonic velocities needed to put a satellite into a desired orbit have to be below 0.1, meaning that majority of the rocket mass has to be propellant.

(318) Writing everything in scalar form for simplicity, if z represents the distance traveled up, or in the opposite direction of the gravitational acceleration, the velocity along the same direction becomes the derivative of Z traveled expressed as,

(319) $\begin{array}{cc}{v}_B=\frac{\mathrm{dz}}{\mathrm{dt}}& \left(5.43\right)\end{array}$

(320) Integrating (5.43) as,

(321) $\begin{array}{cc}{}_{z=0}^{z=h_B}\mathrm{dz}=v_E{}_{t=0}^{t=T_B}\left[\ln \left(\frac{1}{}\right)-\frac{\left(1-\right)}{T2W}\right]\mathrm{dt}& \left(5.44\right)\end{array}$

(322) The integral (5.44) can be evaluated with a variable transformation applying Leibnitz's chain rule [20-24] given as,

(323) $\begin{array}{cc}{v}_B=\frac{\mathrm{dz}}{\mathrm{dt}}=\frac{\mathrm{dz}}{d}\frac{d}{\mathrm{dt}}=-\frac{b_r}{m_0}\frac{\mathrm{dz}}{d}& \left(5.45\right)\end{array}$

(324) Differentiating (5.34) with respect to t gives,

(325) 0$\begin{array}{cc}\frac{d}{\mathrm{dt}}=-\frac{m_0}{b_r}& \left(5.46\right)\end{array}$

(326) Substituting (5.46) in (5.45) becomes,

(327) $\begin{array}{cc}\mathrm{dz}=-\frac{m_0}{b_r}v\left(\right)d& \left(5.47\right)\end{array}$

(328) Giving the integral with the help of the lower and upper limits of given in (5.35),

(329) $\begin{array}{cc}{}_{z=0}^{z=h_B}\mathrm{dz}=-\frac{m_0}{b_r}{v}_E{}_{={}_0}^{={}_f}\left[\ln \left(\frac{1}{}\right)-\frac{\left(1-\right)}{T2W}\right]d& \left(5.48\right)\end{array}$

(330) The upper limit h.sub.B in the integral on the left-hand side of the integral equation (5.48) is the altitude that the rocket reaches after ejecting all its propellant, or at t=T.sub.B.

(331) To integrate the first term in the right hand-side of the integral equation (5.48),

(332) $\begin{array}{cc}{}_{={}_0=1}^{={}_f}\ln \left(\frac{1}{}\right)d& \left(5.49\right)\end{array}$

(333) With the variable transformation,

(334) $\begin{array}{cc}x=\frac{1}{}\mathrm{giving}d=-\frac{1}{x^2}\mathrm{dx}& \left(5.5\right)\end{array}$ resulting in the integral which has an open form integral expression [22-24] as,

(335) $\begin{array}{cc}\ln \left(\frac{1}{}\right)d=-\frac{\ln \left(x\right)}{x^2}\mathrm{dx}=\frac{\ln \left(\right)}{}+\frac{1}{}& \left(5.51\right)\end{array}$

(336) Substituting .sub.f in (5.51) for calculating the integral value at the upper integration limit I.sub.2 of (5.49) becomes,

(337) $\begin{array}{cc}{I}_2={}_f\ln \left(\frac{1}{{}_f}\right)+{}_f& \left(5.52\right)\end{array}$

(338) Substituting the lower integration limit .sub.0=1 in (5.51) for calculating the integral value at the lower limit I.sub.1 of (5.49) gives,

(339) $\begin{array}{cc}{I}_1={}_0\ln \left(\frac{1}{{}_0}\right)+{}_0=1.Math.\ln \left(\frac{1}{1}\right)+1=1& \left(5.53\right)\end{array}$

(340) The resulting integral value I.sub.2I.sub.1 of the first part in (5.49) becomes,

(341) $\begin{array}{cc}{}_{=1}^{={}_f}\ln \left(\frac{1}{}\right)d={}_f\ln \left(\frac{1}{{}_f}\right)+{}_f-1& \left(5.54\right)\end{array}$

(342) Integration of the second term in right hand-side of (5.48) is straightforward giving [22-24],

(343) $\begin{array}{cc}\frac{\left(1-\right)}{T2W}d=-\frac{1}{2}\frac{{\left(1-\right)}^2}{T2W}& \left(5.55\right)\end{array}$

(344) Applying the integration limits at (5.48) to (5.55) gives

(345) 00$\begin{array}{cc}-{}_{=1}^{={}_f}\frac{\left(1-\right)}{T2W}d=\frac{1}{2}\frac{{\left(1-{}_f\right)}^2}{T2W}& \left(5.56\right)\end{array}$

(346) Substituting (5.56) in (5.48) the integral (5.48) finally becomes,

(347) 01$\begin{array}{cc}{h}_B=v_E\frac{m_0}{b_r}\left[1-{}_f\ln \left(\frac{1}{{}_f}\right)-{}_f-\frac{1}{2}\frac{{\left(1-{}_f\right)}^2}{T2W}\right]& \left(5.57\right)\end{array}$

(348) The multiplier in front of (5.57) can be simplified further by multiplying denominator and the dominator with (g.Math..sub.E) giving,

(349) 02$\begin{array}{cc}\frac{v_E{m}_0\left(v_{E}g\right)}{b_r\left(v_{E}g\right)}=\frac{v_E^2{m}_{0}g}{F_{T}g}=\frac{v_E^2}{g.Math.T2W}& \left(5.58\right)\end{array}$

(350) Giving the height h.sub.B that the rocket reaches in powered flight at the time of T.sub.B, in other words when it runs out of propellant, with a vertical velocity in opposing direction of constant acceleration g as,

(351) 03$\begin{array}{cc}{h}_B\left({}_f,T2W\right)=\frac{v_E^2}{g.Math.T2W}\left[1-{}_f\ln \left(\frac{1}{{}_f}\right)-{}_f-\frac{1}{2}\frac{{\left(1-{}_f\right)}^2}{T2W}\right]& \left(5.59\right)\end{array}$

(352) At that point the rocket has a velocity .sub.B as given in (5.40) and it keeps gaining altitude. In other words, it coasts until it reaches its final altitude h also known as apogee in the rocket literature where its velocity becomes zero.

(353) The limits of h.sub.B(.sub.f, T2W) at .sub.f=0 and .sub.f=1 are worth mentioning giving,

(354) 04$\begin{array}{cc}\mathrm{for}{}_f.fwdarw.0h_B\left(0,T2W\right).fwdarw.\frac{{}_E^2}{g.Math.T2W}\left(1-\frac{1}{2T2W}\right)& \left(5.6\right)\end{array}$
for .sub.f.fwdarw.1h.sub.B(1,T2W)=0(5.61)

(355) FIG. 5A depicts h.sub.B(.sub.f, T2W) for thrust-to-weight ratio T2W=1.1, 2, 4, and 8 with a constant acceleration g.

(356) FIG. 5B depicts h.sub.B(.sub.f, T2W) for thrust-to-weight ratio T2W=1.1 and 8 where the horizontal axis is in logarithmic scale.

(357) What is interesting to observe is that h.sub.B(0, T2W), as given in (5.60) and shown in FIGS. 5A and 5B, has a finite value that is less than 600 [km] for any practical thrust to weight ratio of T2W. As shown in FIGS. 2A, 2B, and 2C, the g(h) variation for altitudes of 600 [km], which can be considered to be the maximum burn height h.sub.BMAX, is negligible, meaning that the constant acceleration g assumption for any condition is an excellent approximation, at least for the burn stage of any rocket launched from Earth. FIG. 2C depicts the gravitational acceleration normalized to its sea level value as a function of altitude g(h) with the maximum burn altitude h.sub.B(.sub.f=0) for T2W=1.1, 2, 4 and 8 marked on the curve, which graphically displays this important fact for a wide range of T2W values. As can be seen at the highest burn altitude the normalized gravitational acceleration does not go below 0.85 meaning that constant acceleration g assumption is a good approximation for any rocket launched from the Earth for its h.sub.B(0, T2W) calculation. This property is used for the derivation of the inverse square law gravitational case in as described in Section 6.

(358) The horizontal lines superimposed on FIGS. 5A and 5B show h.sub.GIVEN=40 and 400 [km]. Their intersection points with the h.sub.B(.sub.f, T2W) curves give the corresponding pf values which is the graphical solution to the non-linear problem given as the equation h.sub.B If, T2W)h.sub.GIVEN=0. FIG. 5A shows that for h.sub.B(.sub.f, T2W)=40 [km], .sub.f should be roughly in the interval of 0.3.sub.f0.6 for all the ranges of T2W, which demonstrates the graphical solution method to this non-linear problem related to rocket design altitude. FIG. 5A does clearly show the intersection points of h.sub.B(.sub.f, T2W) curves and h.sub.GIVEN=400 [km], but FIG. 5B, having a logarithmic horizontal axis, gives roughly .sub.f=0.06, illustrating some of the difficulties in reaching higher altitudes.

(359) The final altitude h can be solved by applying the energy conservation law [21] as,
KE(h.sub.B)+PE(h.sub.B)=KE(h)+PE(h)(5.62)

(360) Where KE(h.sub.B) and PE(h.sub.B) are the kinetic and potential energies at the altitude h.sub.B, where the velocity of the rocket is the known value of .sub.B. Similarly, KE (h) and PE(h) are the kinetic and potential energies at the final altitude h, where the rocket velocity .sub.h.sup.2 is zero, giving zero kinetic energy at apogee, which can be written explicitly as,

(361) 05$\begin{array}{cc}\frac{1}{2}{m}_f{v}_B^2+m_{f}g{h}_B=\frac{1}{2}{m}_f{v}_h^2+m_{f}gh& \left(5.63\right)\end{array}$

(362) Solving h from (5.59) for .sub.h.sup.2=0 gives,

(363) 06$\begin{array}{cc}h\left({}_f,T2W\right)=\left[\frac{v_B^2\left({}_f,T2W\right)}{2g}+h_B\left({}_f,T2W\right)\right]& \left(5.64\right)\end{array}$

(364) FIG. 6A depicts the climb altitude h(.sub.f, T2W for the constant gravitational acceleration of g=9.8 [m/s.sup.2] with thrust-to-weight ratio T2W=1.1 and 8 where the vertical axis is in logarithmic scale. Since the .sub.B(.sub.f, T2W) term in (5.64) becomes+ for .sub.f=0, h(.sub.f, T2W) also becomes+, as shown with an asymptote (dotted vertical line) in FIG. 6A at .sub.f=0.

(365) The horizontal lines superimposed on h.sub.GIVEN=40, 400, and 400,000 [km], and their intersection points with the h(.sub.f, T2W) curves give the corresponding if values that is the graphical solution to the non-linear problem h(.sub.f, T2W)h.sub.GIVEN=0. FIG. 6A shows that for h(.sub.f, T2W)=40 [km], .sub.f should be roughly in the interval of 0.6.sub.f0.75 for all the ranges of T2W, which is an easily achievable task. This demonstrates the graphical solution method of solving this non-linear problem for rocket design altitude. FIG. 6A also clearly shows that the intersection points of the h(.sub.f, T2W) curves and h.sub.GIVEN=400 [km] are roughly in the interval of 0.275.sub.f0.45. FIG. 6A shows the difficulty in reaching extreme altitudes of 400,000 [km] which is the approximate distance to the Moon, requiring very small .sub.f values. FIG. 6A graphically demonstrates that reaching LEO orbital heights is only achievable with multi-stage rockets.

(366) The first term in the brackets of (5.64) is the distance that the rocket coasts after depleting all its propellant under constant acceleration g and is represented as,

(367) 07$\begin{array}{cc}{h}_C\left({}_f,T2W\right)=\frac{v_B^2\left({}_f,T2W\right)}{2g}& \left(5.65\right)\end{array}$

(368) Naming this distance h.sub.C, as the coast distance, writing (5.64) in terms of the sum of h.sub.B and h.sub.C becomes handy in evaluating the derivatives and limits with respect to pf giving,
h(.sub.f,T2W)=h.sub.C(.sub.f,T2W)+h.sub.B(.sub.f,T2W)(5.66)

(369) Since (5.41) holds the limit of h.sub.C(.sub.f, T2W) at .sub.f=0 becomes,
.sub.f.fwdarw.0.sub.B.sup.2(0,T2W).fwdarw.h.sub.C(0,T2W).fwdarw.(5.67)

(370) The limit of h.sub.C(.sub.f, T2W) at .sub.f=1 gives,
.sub.f.fwdarw.1.sub.B.sup.2(1,T2W).fwdarw.0h.sub.C(1,T2W).fwdarw.0(5.68)

(371) Due to (5.66), the limits of h at at .sub.f=0 and .sub.f=1 give the same values as (5.67) and (5.68). Due to (5.67), .sub.f.fwdarw.0 becomes the asymptotes for the h.sub.C(.sub.f, T2W) and h(.sub.f, T2W) curves, as shown in FIGS. 6A and 6B.

(372) FIG. 6B depicts burn height h.sub.B(.sub.f, T2W) and coast height h.sub.C(.sub.f, T2W) for the constant gravitational acceleration of g=9.8 [m/s.sup.2] with thrust-to-weight ratio T2W=1.1 and 8 where the vertical axis is in logarithmic scale. The asymptote for h.sub.C (.sub.f, T2W) is shown at .sub.f.fwdarw.0 with the dotted vertical line while h.sub.B(.sub.f, T2W) has a finite value less than 600 [km].

(373) 5.2 Solution of Tsiolkovsky's Rocket Equation for Constant Gravitational Acceleration

(374) When the Rocket Thrust is in the Same Direction of the Gravitational Force

(375) The modification to accommodate the rocket thrust being in the same direction of the gravitational force is straightforward. The burn velocity .sub.B after consuming all the rocket propellant as given in initially at (5.28), (5.29) and finally in (5.40) can be written for gravitational force is in the opposite direction of the rocket thrust as,

(376) 08$\begin{array}{cc}{v}_B^{-g}=v_E\left[\ln \left(\frac{1}{{}_f}\right)-\frac{\left(1-{}_f\right)}{T2W}\right]& \left(5.69\right)\end{array}$

(377) Superscript g is not an exponent, it is a superscript indicating it is the velocity gained when the rocket thrust is in the opposite direction of the gravitational force symbolized as g. After some simple math, the velocity gained when the rocket thrust is in the same direction of gravitational force can be written as,

(378) 09$\begin{array}{cc}{v}_B^{+g}=v_E\left[\ln \left(\frac{1}{{}_f}\right)+\frac{\left(1-{}_f\right)}{T2W}\right]& \left(5.7\right)\end{array}$

(379) Similar notation can be used for the h.sub.B, derived in (5.59) for the distance covered when the rocket thrust is in the opposite direction of the gravitational force symbolized as gas,

(380) 0$\begin{array}{cc}{h}_B^{-g}\left({}_f,T2W\right)=\frac{v_E^2}{g.Math.T2W}\left[1-{}_f\ln \left(\frac{1}{{}_f}\right)-{}_f-\frac{1}{2}\frac{{\left(1-{}_f\right)}^2}{T2W}\right]& \left(5.71\right)\end{array}$

(381) Again, after some simple math, the velocity gained when the rocket thrust is in the same direction of gravitational force can be written as,

(382) $\begin{array}{cc}{h}_B^{+g}\left({}_f,T2W\right)=\frac{v_E^2}{g.Math.T2W}\left[1-{}_f\ln \left(\frac{1}{{}_f}\right)-{}_f+\frac{1}{2}\frac{{\left(1-{}_f\right)}^2}{T2W}\right]& \left(5.72\right)\end{array}$

(383) FIG. 3H depicts the rocket thrust assisted velocity reached .sub.B, having rocket thrust being in the same direction of the constant gravitational acceleration shown as .sub.B.sup.+g (in rocket powered vertical descent) and rocket thrust being in the opposite direction of the constant gravitational acceleration shown as .sub.B.sup.g (in rocket powered vertical ascent) using the solution of modified Tsiolkovsky's rocket equation for T2W=1.1 and 4 versus .sub.f. In both cases velocities are after all the propellant is consumed. Escape velocity and orbital velocities for 400 [km] altitude is also shown for reference.

(384) FIG. 3I depicts the rocket thrust assisted distance covered having rocket thrust being in the same direction of the constant gravitational acceleration shown as h.sub.B.sup.+g (in rocket powered vertical descent) and rocket thrust being in the opposite direction of the constant gravitational acceleration shown as h.sub.B.sup.9 (in rocket powered vertical ascent) using the solution of modified Tsiolkovsky's rocket equation for T2W=1.1 and 4 versus .sub.f. In both cases distances are after all the propellant is consumed. The distances of 400 and 40 [km] are also shown for reference.

(385) FIG. 3J is same data depicted in FIG. 3I in logarithmic x axes to show the covered distances clearer for larger .sub.f values.

(386) FIG. 3K depicts the ratios between the velocities reached .sub.VB(.sub.f)=.sub.B.sup.+g/.sub.B.sup.9 and ratios of the distances covered .sub.HB(.sub.f)=h.sub.B.sup.+g(.sub.f, T2W)/h.sub.B.sup.g(.sub.f, T2W) for T2W=1.1.

(387) FIG. 3L depicts the ratios between the velocities reached .sub.VB(.sub.f)=.sub.B.sup.+g/.sub.B.sup.g and ratios of the distances covered .sub.HB(f)=h.sub.B.sup.+g(.sub.f, T2W)/h.sub.B.sup.9(.sub.f, T2W) for T2W=4.

(388) As can be seen the only differences are the sign change in the last terms in (5.69) and (5.71) but it has a very significant effect on the velocities gained and distances covered between the vertical rocket assisted descent described in this invention and prior art ascend.

(389) With all the notations given above FIG. 3H-FIG. 3L displays the comparative analysis between (5-69) and (5.70) for velocities gained graphically for T2W=1.1 and 4. As can be seen in in FIG. 3H for T2W=1.1 the same payload can gain orbital velocity for orbital height of

(390) h=400 [km] (ISS orbital altitude) with .sub.f=0.2 in rocket assisted decent, while the same velocity can be gained with .sub.f=0.05 giving a .sub.f(T2W=1.1)=0.15 resulting in very significant propellant savings and reduction in rocket thrust for the same payload in descent versus ascent. FIG. 3H also shows the similar fact with less savings for T2W=4. Precise numerical values of .sub.f(.sub.f, T2W) are given in the third column of Table 3B for all six orbital altitudes listed in Table 3A for T2W=1.1, 2, and 8.

(391) FIG. 3I depicts the rocket thrust assisted distance covered having rocket thrust being in the same direction of the constant gravitational acceleration shown as h.sub.B.sup.+g (in rocket powered vertical descent) and rocket thrust being in the opposite direction of the constant gravitational acceleration shown as h.sub.B.sup.g (in rocket powered vertical ascent) using the solution of modified Tsiolkovsky's rocket equation for T2W=1.1 and 4 versus .sub.f. In both cases distances are after all the propellant is consumed. The distances of 400 and 40 [km] are also shown for reference.

(392) FIG. 3J is same data depicted in FIG. 3I in logarithmic x axes to show the covered distances clearer for larger .sub.f values.

(393) FIG. 3K depicts the ratios between the velocities reached .sub.VB(.sub.f)=.sub.B.sup.+g/.sub.B.sup.g and ratios of the distances covered .sub.HB(.sub.f)=h.sub.B.sup.+g(.sub.f, T2W)/h.sub.B.sup.g(.sub.f, T2W) for T2W=1.1.

(394) FIG. 3L depicts the ratios between the velocities reached .sub.VB(.sub.f)=.sub.B.sup.+g/.sub.B.sup.g and ratios of the distances covered .sub.HB(.sub.f)=h.sub.B.sup.+g (.sub.f, T2W)/h.sub.B.sup.g (.sub.f, T2W) for T2W=4.

(395) As a conclusion, the main reason of achieving orbital velocities by going into smaller radiuses then initially calculated r.sub.0 given as,
r.sub.0=2.Math.r.sub.ORBIT(5.73) using energy conservation principle with the same mass as in (1.1). This results in defining a third radius r.sub.h with a scaling factor s.sub.f as given in (1.2) as,
r.sub.h=s.sub.f.Math.r.sub.ORBIT where 1.1<s.sub.f<2(5.74)

(396) It is important to point out that all these calculations are done for vertical descent and ascent using rocket propulsion based on mass ejection for constant gravitational force. FIG. 3H and FIG. 3L clearly show that the thrust to weight ratio (T2W) plays an important part in the results. From the curves it is evident that as the T2W increases, the curves get piled on top of each other and are difficult to distinguish from each other. Table 3B gives a clearer view of the analysis done for satellites at different orbital altitudes showing the T2W effect on the results and the logic of coming up with a third radius r.sub.h and its relation to r.sub.0 through the scaling factor s.sub.f.

(397) This is done in two steps:

(398) a. Find the .sub.f.sup.+g which will give the orbital velocity .sub.ORBIT(h) when initially stationary at the altitude h by solving the equation given in (5.70) as,

(399) $\begin{array}{cc}{v}_{\mathrm{ORBIT}}\left(h\right)=v_E\left[\ln \left(\frac{1}{{}_f^{+g}}\right)+\frac{\left(1-{}_f^{+g}\right)}{T2W}\right]& \left(5.75\right)\end{array}$

(400) ii) By substituting .sub.f.sup.+g for p in (5.72) h.sub.B.sup.+g(.sub.f, T2W) is obtained, which is the fall distance covered when the rocket thrust is in the same direction of gravitational acceleration while reaching .sub.ORBIT(h) from an initial velocity of zero and is written on the 5.sup.th column of Table 3B for all 6 orbital altitudes and for 4 different values of T2W. The arithmetic for calculating s.sub.f is straightforward and is listed in column 7 of Table 3B. Column 6 gives the h.sub.B.sup.+g(.sub.f, T2W)/h ratio. As can be seen all the way to orbital altitudes of 781 [km], the results are consistent, but the inaccuracy caused by assuming constant acceleration throughout the entire descent path becomes apparent.

(401) TABLE-US-00005 Orbital Altitude h [km] T2W .sub.B.sup.+g .sub.B.sup.g .sub.f(h,T2W) h.sub.B.sup.g [km] h.sub.B.sup.+g [km] $\frac{h_B^{+g}}{h}$ $sf=\frac{r_h^{+g}}{r_{\mathrm{ORBIT}}}$ $\frac{m_{\mathrm{PROP}}^{-g}-m_{\mathrm{PROP}}^{+g}}{\mathrm{payload}}$ 255 1.1 0.167 207.35 865.90 3.396 1.131 18.56 255 2 0.092 239.97 467.23 1.832 1.070 9.198 255 4 0.046 164.37 225.13 0.833 1.034 4.440 255 8 0.023 93.93 109.57 0.430 1.017 2.200 400 1.1 0.170 212.12 892.20 2.231 1.132 18.077 400 2 0.094 247.17 482.47 1.206 1.071 8.951 400 4 0.047 169.77 232.77 0.582 1.034 4.320 400 8 0.023 97.13 113,36 0.283 1.017 2.141 540 1.1 0.173 216.69 917.70 1.699 1.133 17.629 540 2 0.096 254.15 497.28 0.921 1.072 8.726 540 4 0.048 175.03 240.21 0.445 1.035 4.211 540 8 0.024 100.24 117.05 0.217 1.017 2.087 781 1.1 0.178 224.50 961.83 1.232 1.134 16.910 781 2 0.099 266.21 522.97 0.67 1.073 8.365 781 4 0.049 184.14 253.15 0.32 1.035 4.036 781 8 0.025 105.66 123.47 0.158 1.017 2.000 20,200 1.1 0.356 727.84 4,910 0.243 1.185 4.792 20,200 2 0.210 1,309.7 2,994 0.148 1.113 2.277 20,200 4 0.108 1,069 1,571 0.078 1.059 1.081 20,200 8 0.054 662.6 798 0.04 1.030 0.533 35,786 1.1 0.394 1,055 8,271 0.231 1,196 3.437 35,786 2 0.233 2,169 5,192 0.145 1,123 1,596 35,786 4 0.119 1,861 2,791 0.078 1,066 0.751 35,786 8 0.060 1,181 1,439 0.04 1,034 0.369

(402) The remaining columns of Table 3B are for comparative analysis between the rocket thrust against gravity and in the same direction as the gravity.

(403) Solving (5.69) gives .sub.f.sup.g, which gives the orbital velocity .sub.ORBIT(h) when initially stationary at the altitude h when the rocket thrust is against gravity as,

(404) $\begin{array}{cc}{v}_{\mathrm{ORBIT}}\left(h\right)=v_E\left[\ln \left(\frac{1}{{}_f^{-g}}\right)-\frac{\left(1-{}_f^{-g}\right)}{T2W}\right]& \left(5.76\right)\end{array}$

(405) The third column of Table 3B gives the differences between the solutions of (5.75) and (5.76) as,
.sub.f(h,T2W)=.sub.f.sup.+g.sub.f.sup.g(5.77)

(406) By substituting .sub.f.sup.+g for .sub.f in (5.71) h.sub.B.sup.g(.sub.f, T2W) is obtained, which is the height from which the payload is launched if rocket thrust is opposite to gravitational force given at fourth column of Table 3B.

(407) On the other hand, since the whole objective is to put a satellite into orbit, the final masses in both cases are equal and it is equal to the satellite mass or referred to as payload m.sub.PAY in earlier calculations. Therefore written as,

(408) $\begin{array}{cc}\frac{1}{{}_f^{-g}}-\frac{1}{{}_f^{+g}}=\frac{m_0^{-g}-m_0^{+g}}{m_f}=\frac{m_0^{-g}-m_0^{+g}}{m_{PAY}}& \left(5.78\right)\end{array}$

(409) Since most of the mass of a rocket is propellant (5.78) can be approximated as,

(410) $\begin{array}{cc}\frac{m_0^{-g}-m_0^{+g}}{m_{\mathrm{PAY}}}\frac{m_{\mathrm{PROP}}^{-g}-m_{\mathrm{PROP}}^{+g}}{m_{\mathrm{PAY}}}& \left(5.79\right)\end{array}$

(411) Which gives the eighth (last) column in Table 3B. Having a large ratio shows how much propellant is saved for the same payload with rocket thrust in the same direction of gravitation versus against it.

(412) Real calculations and trajectory optimizations are only possible with numerical solution of three-dimensional equation of motion, and the r.sub.h and related s.sub.f calculation is payload and orbital altitude dependent, but it is clearly shown in Table 3B that s.sub.f values can be even as low as 1.1.

(413) It was earlier proven that the total energy of an object in a circular orbit at a radius r.sub.ORBIT is equal to the total energy of the same object, which is stationary at the radius of r.sub.0=2.Math.r.sub.ORBIT. Then when the same object is launched to a smaller radius, r.sub.h then 2r.sub.ORBIT, as given in (1.2) and (5.74), it will have less gravitational potential energy and since it is stationary its total energy, compared to the total energy when orbiting at a radius of r.sub.ORBIT.

(414) FIG. 13I demonstrates the methodology employed for two different values of s.sub.f=1.5 and 2 along with the related mass versus time plots of FIG. 13J and FIG. 13K. Similar graphical demonstration is given for 20,200 [km] (GPS) orbital launch in FIG. 13M with the related mass versus time plots FIG. 13N and FIG. 13O.

(415) Modification of Tsiolkovsky's Rocket Equation for Inverse Square Law Gravitational Field.

(416) If the launch altitude is in the order of, or larger than the Earth's radius, Tsiolkovsky's Rocket Equation for constant gravitational acceleration must be modified for the inverse square law gravitational field. In this work this modification is done by applying the energy conservation law for inverse square law gravitational fields which gives much better correlation to reality and numerical solution of the equations of motion. In this approach, the non-linearities in the equation relating the final and initial rocket masses to final velocity and altitude get more complex and solving it with Newton's method becomes more challenging, requiring more care and is explained in detail below.

(417) The Inverse Square Law Gravitational Field Relation can be incorporated into the conservation of energy formulation for calculating a more accurate final altitude h. If the final altitude h, calculated by (5.61) is comparable to or larger than the radius of Earth r.sub.EARTH, the potential energy expression for constant gravitational acceleration g becomes inaccurate. FIG. 2B clearly shows the ratio of the inverse square law calculated acceleration versus constant acceleration value on the surface of the Earth as a function of radius measured from the center of the Earth and at sea level. The marked altitude on the curve is h.sub.BMAX, which shows that highest altitude that any rocket can reach during its burn phase giving 85% of the gravitation acceleration of the surface value, giving a much smaller h than the gravitational potential energy derived for the inverse square law gravitational field as shown in FIG. 2C and explained above. This problem can be fixed by replacing the potential energy with the formulation for the inverse square law gravitational field giving (6.1),

(418) $\begin{array}{cc}\frac{1}{2}{m}_f{v}_B^2-G\frac{m_f{m}_{\mathrm{EARTH}}}{r_B}=\frac{1}{2}{m}_f{v}_{\mathrm{rhINV}2}^2-G\frac{m_f{m}_{\mathrm{EARTH}}}{r_{\mathrm{hINV}2}}& \left(6.1\right)\end{array}$

(419) Subscripts INV2 represent the inverse square law gravitational field formulation quantities, where r.sub.B and r.sub.hINV2 are radiuses corresponding to the altitudes h.sub.B and h.sub.INV2 given as,
r.sub.B=h.sub.B+r.sub.EARTH and r.sub.hINV2=h.sub.INV2+r.sub.EARTH(6.2)

(420) Since r.sub.EARTH is a constant, the following derivative relations are also valid,

(421) 0$\begin{array}{cc}\frac{{\mathrm{dr}}_B}{d{}_f}=\frac{{\mathrm{dh}}_B}{d{}_f}\mathrm{and}\frac{{\mathrm{dr}}_{\mathrm{hINV}2}}{d{}_f}=\frac{{\mathrm{dh}}_{\mathrm{INV}2}}{d{}_f}& \left(6.3\right)\end{array}$

(422) Solving 13D from (6.1), gives,

(423) $\begin{array}{cc}{r}_{\mathrm{hINV}2}=\frac{-2{\mathrm{Gr}}_B{m}_{\mathrm{EARTH}}}{-2{\mathrm{Gm}}_{\mathrm{EARTH}}+r_B\left(v_B^2-v_{\mathrm{rhINV}2}^2\right)}& \left(6.4\right)\end{array}$

(424) Multiplying denominator and dominator of (6.4) with 1 gives,

(425) $\begin{array}{cc}{r}_{\mathrm{hINV}2}=2{\mathrm{Gm}}_{\mathrm{EARTH}}\frac{r_B}{2{\mathrm{Gm}}_{\mathrm{EARTH}}-r_B\left(v_B^2-v_{\mathrm{rhINV}2}^2\right)}& \left(6.5\right)\end{array}$

(426) Substituting r.sub.hINV2.sup.2=0 in (6.5), which is the case of interest gives,

(427) $\begin{array}{cc}{r}_{\mathrm{hINV}2}=2{\mathrm{Gm}}_{\mathrm{EARTH}}\frac{r_B}{2{\mathrm{gm}}_{\mathrm{EARTH}}-r_B{v}_B^2}& \left(6.6\right)\end{array}$

(428) Since a proper (r.sub.B, .sub.B) combination can make the denominator of (6.6) zero, an infinite value for both r.sub.hINV2 and h.sub.INV2 is possible with a finite .sub.B. The equation gives the finite .sub.B. The equation which gives the finite .sub.B, resulting with an infinite value for both r.sub.hINV2 and h.sub.INV2 can be solved by solving,
2Gm.sub.EARTHr.sub.B.sub.B.sup.2=0(6.7)

(429) Giving,

(430) $\begin{array}{cc}{v}_{\mathrm{BASY}}=\sqrt{\frac{2{\mathrm{Gm}}_{\mathrm{EARTH}}}{r_B}}=v_{\mathrm{ESCAPE}}\left(r_B,m_{\mathrm{EARTH}}\right)& \left(6.8\right)\end{array}$

(431) This corresponds to the escape velocity from a spherical boundary with a radius of r.sub.B enclosing the mass of m.sub.EARTH in it. The escape velocity from Earth vs. altitude is very clearly shown in FIG. 1D along with markings of the orbital altitudes given in Table 2. This means that the rocket reaches infinite radius or altitude if it has a velocity of .sub.BASY at r.sub.B or at the altitude h.sub.B. As can be seen this makes perfect physical sense for the defined escape velocity, which the solution of the rocket equation under constant g formulation does not give. Writing (6.8) in terms of the previously calculated altitude h.sub.B and velocity .sub.B gives,

(432) $\begin{array}{cc}{h}_{\mathrm{INV}2}=2{\mathrm{Gm}}_{\mathrm{EARTH}}\frac{h_B+r_{\mathrm{EARTH}}}{2{\mathrm{Gm}}_{\mathrm{EARTH}}-r_B{v}_B^2}-r_{\mathrm{EARTH}}& \left(6.9\right)\end{array}$

(433) The corresponding coast height h.sub.CINV2 becomes,
h.sub.CINV2=h.sub.INV2h.sub.B(6.10)

(434) FIG. 7A depicts asymptote x intercept .sub.ASYM(T2W) vs. T2W and .sub.MIN(T2W) by solving the non-linear equation (6,7). As can be seen, the asymptotes for any T2W are very close to zero, in the range of 0.016<.sub.f<0.036, and getting closer to zero with decreasing thrust-to-weight ratio T2W. As given by (6.6) r.sub.hINV2 becomes+ along with the altitude h.sub.INV2 due to the linear relation given in relation (6.2), which causes plotting and interpolation to be very difficult, if not impossible for .sub.f=.sub.ASYM(T2W). To avoid this problem, .sub.MIN(T2W) is defined by a small value greater than the .sub.ASYM(T2W) with very large altitude values in the family of altitude vs .sub.f curves in FIG. 7B through 7E.

(435) FIG. 7B depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field with asymptotes where vertical axis is in logarithmic scale for thrust-to-weight ratio T2W=1.1 and 8.

(436) FIG. 7C depicts burn height h.sub.BINV2(.sub.f, T2W) and coast height h.sub.CINV2(.sub.f, T2W) at T2W for the inverse square gravitational field with asymptotes where the vertical axis is in logarithmic scale.

(437) FIG. 7D depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field with asymptotes and h(.sub.f, T2W) for the constant gravitational acceleration of g=9.8 [m/s.sup.2] where the vertical axis is in logarithmic scale for thrust-to-weight ratio T2W=1.1 and 8. Although the differences are self-explanatory, they can be better seen in logarithmic horizontal axes.

(438) FIG. 7E depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field with asymptotes and h(.sub.f, T2W) for the constant gravitational acceleration of g=9.8 [m/s.sup.2] where the vertical and horizontal axes are in logarithmic scale for thrust-to-weight ratio T2W=1.1 and 8. Important to note is that the inverse square gravitational field assumption has to be used for solving pf for altitudes higher than 400 [km]. The constant gravitational acceleration assumption can be safely used for calculating pf as well as altitudes less than 400 [km].

(439) FIG. 7F depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field with asymptotes for <.sub.ASYM(T2W) with thrust-to-weight ratio T2W=1.1, 2, 4, and 8.

(440) As can be seen very clearly in FIG. 7F there are 2 asymptotes for each .sub.INV2(.sub.f, T2W) curve. The first one is at .sub.f=0 and all the h.sub.INV2(.sub.f, T2W) functions approach, which is non-physical. The second asymptotes of every h.sub.INV2(.sub.f, T2W) curve has an asymptote where their x axes intercept, as shown in FIG. 7A, at .sub.f=.sub.ASYM(T2W), where the h.sub.INV2(.sub.f, T2W) functions approach+. This asymptote brings also a jump type discontinuity where h.sub.INV2(.sub.f, T2W) jumps from + to zero when .sub.f crosses the asymptote as can be seen clearly in FIG. 7E and FIG. 7F, This can be worded as in the interval of 0.sub.f.sub.ASYM(T2W) the h.sub.INV2(.sub.f, T2W) function becomes non-physical.

(441) FIG. 8 depicts climb altitude h.sub.INV2(.sub.f, T2W) for the inverse square gravitational field and climb altitude h(.sub.f, T2W) with the constant gravitational acceleration of g=9.8 [m/s.sup.2] vs. T2W, where .sub.f=0.1, 0.2, 0.3, and 0.4. At .sub.f>0.3 the h(.sub.f, T2W) and h.sub.INV2(.sub.f, T2W) curves become indistinguishable. FIG. 8 is another way of showing that the constant gravitational acceleration assumption can be safely used for calculating .sub.f, by simply looking at the calculated .sub.f value. If .sub.f>0.3, the constant gravitational acceleration assumption is accurate enough to use, with no need to employ the inverse square gravitational field formulation or to question the formulation used in the solution, both give very close results.

(442) Introducing Propellant Tank to Propellant Mass Ratio k.

(443) The propellant tank mass is related to the mass of the propellant that it stores, by introducing a parameter k as,

(444) $\begin{array}{cc}k=\frac{m_{\mathrm{tank}}}{m_{\mathrm{prop}}}& \left(7.1\right)\end{array}$

(445) The initial mass m.sub.0 of the rocket can be written as,
m.sub.0=m.sub.prop(1+k)+m.sub.RE+m.sub.PAYLOAD(7.2) where m.sub.prop, m.sub.RE and m.sub.PAYLOAD are the mass of the propellant, rocket engine and payload, and where the parameter k is called the propellant tank-to-propellant mass ratio. Typical values of k should be a small number like 0.05 to 0.2, with a smaller ratio being more advantageous. The chemistry and the resulting k parameter for liquid and solid fuel rockets are different. Here the analysis for a very simplified formulation applicable of a liquid rocket having a propellant tank being made from only a single cylinder is presented. A similar approach for k parameter calculations for the solid rockets can also be derived.

(446) In a liquid fuel rocket the m.sub.prop consists of oxidizer plus the fuel masses given as,
m.sub.prop=m.sub.fuel+m.sub.oxi(7.3) which are typically stored in two different cylindrical tanks having two semi-spherical caps at the top and bottom. Due to the chemistry of burning the fuel with the oxidizer, their masses, densities, and their resulting volumes are not necessarily equal and can be calculated with their reaction chemistry. The simplest example uses hydrogen; H.sub.2 as fuel and oxygen; O.sub.2 for the oxidizer giving the chemical reaction of,

(447) $\begin{array}{cc}{H}_2+\frac{1}{2}{O}_2.fwdarw.H_{2}O& \left(7.4\right)\end{array}$

(448) Applying stoichiometry analysis to the chemical reaction given at (7.4) shows that 2 kg of hydrogen reacting with 16 kg of oxygen ( mass ratio) gives 18 kg of steam. This 18 kg of hot steam is ejected from the rocket nozzle with a velocity of .sub.E. Assuming each are stored in liquid form, hydrogen has a density of 71 kg/m.sup.3 at 20.28 K (252.87 C.) and liquid oxygen has a density of 1,141 kg/m.sup.3 at 90.19 K (297.33 C.). As can be seen liquid oxygen is denser than water and approximately 16 times denser than liquid H.sub.2. The question becomes how to calculate the volumetric ratios of the liquid H.sub.2 and O.sub.2 satisfying the calculated mass ratio. Using this example, 2 kg of hydrogen volume V(H.sub.2)=2/71=0.02817 m.sup.3, reacts with an oxygen volume of V(O.sub.2)=16/1,141=0.01402 m.sup.3. This shows that the H.sub.2 tank must be volumetrically 2.009 times larger than the O.sub.2 tank to satisfy the calculated mass ratio.

(449) In rocket design there are some other factors that are considered for maximizing thrust and cooling issues of the rocket engine. For H.sub.2/O.sub.2 rockets the highest impulse power, I.sub.SP is achieved when the H.sub.2/O.sub.2 mass ratio is (leaving half of the H.sub.2 unburnt), not when it is corresponding to full burn of H.sub.2. In practice the mass ratio is kept as for other reasons. As an example, the space shuttle liquid O.sub.2 tank is 19,563 cubic feet (553.96 m.sup.3) and H.sub.2 tank is 53,518 cubic feet (1,515.46 m.sup.3), having volumetric ratio of 2.7357. The full load oxygen and hydrogen mass that can be stored in these tanks are 632.068 and 107.597 tons, giving the mass ratio, as given earlier. As can be seen, calculating optimal mass and volumetric propellant/oxidant ratios is not that simple.

(450) In a liquid propellant rocket the fuel and oxidizer are stored in two separate tanks, with piping and some additional essential parts like turbo pumps, controls, etc. To simplify all the calculations, it is assumed that the oxidizer and propellant tanks are two cylinders with same radiuses and spherical caps both having a uniform skin thickness t.sub.skin. Once the oxidizer and fuel masses and volumes for the mission are calculated as shown in the H.sub.2/O.sub.2 example above, calculating the parameter k simply becomes a trivial volume and mass calculation. The goal here is to quantify the significant dependency of the parameter k to the geometrical parameters of the rocket, like its height and radius under these assumptions. This can be achieved by doing the analysis for only one tank, named the propellant tank. The parameter k as given in (7.1) is defined as the mass ratio between the propellant tank and the propellant stored in it, k can be simply calculated by the propellant tank skin area S.sub.skin, skin material density P.sub.skin, propellant volume contained in the propellant tank, V.sub.prop, and its density, P.sub.prop. A good estimate of k can be given as,

(451) $\begin{array}{cc}k\frac{m_{\mathrm{skin}}}{m_{\mathrm{prop}}}\frac{S_{\mathrm{skin}}{t}_{\mathrm{skin}}}{V_{\mathrm{prop}}}\frac{{}_{\mathrm{skin}}}{{}_{\mathrm{prop}}}& \left(7.5\right)\end{array}$

(452) Ignoring the masses of both end caps of the propellant tank gives k as,

(453) $\begin{array}{cc}k\frac{r^2-{\left(r-t_{\mathrm{skin}}\right)}^2}{{\left(r-t_{\mathrm{skin}}\right)}^2}\frac{{}_{\mathrm{skin}}}{{}_{\mathrm{prop}}}=\frac{{}_{\mathrm{skin}}}{{}_{\mathrm{prop}}}\frac{\left(2r{t}_{\mathrm{skin}}+t_{\mathrm{skin}}^2\right)}{{\left(r-t_{\mathrm{skin}}\right)}^2}& \left(7.6\right)\end{array}$

(454) As can be seen in (7.6) k is a decreasing function of the rocket radius r, closer to a function inversely proportional to the radius r of the rocket. A better estimate of k can be obtained by adding the masses of the top and bottom caps of the propellant tank. Assuming the caps are semi-spherical and has the same skin thickness t.sub.skin, the skin volume V.sub.sksph of the two semi-spherical shell regions on the top and the bottom of the propellant tank is,

(455) 0$\begin{array}{cc}{V}_{\mathrm{sksph}}=\frac{4}{3}\left[r^3-{\left(r-t_{\mathrm{skin}}\right)}^3\right]& \left(7.7\right)\end{array}$

(456) The volume of the cylindrical shell region V.sub.skcyl with a height of h.sub.cyl and a skin thickness of t.sub.skin is,
V.sub.skcyl=h.sub.cyl[r.sup.2t.sub.skin).sup.2](7.8)

(457) The empty mass of the tank becomes,
m.sub.tank=P.sub.skin(V.sub.sksph+V.sub.skcyl(7.9)

(458) Assuming the propellant tank is filled completely prior to launch gives the internal volume of the propellant tank as,

(459) $\begin{array}{cc}{V}_{\mathrm{int}}\left(h\right)=\left[{h_{\mathrm{cyl}}\left(r-t_{\mathrm{skin}}\right)}^2+\frac{3}{8}{\left(r-t_{\mathrm{skin}}\right)}^3\right]& \left(7.1\right)\end{array}$

(460) The propellant mass m.sub.prop becomes,
m.sub.prop=p.sub.oxiV.sub.int(h.sub.oxi)+P.sub.fuelV.sub.int(h.sub.fuel)(7.11)

(461) FIG. 11 depicts the skin of a propellant tank.

(462) FIG. 12 depicts the mass of the propellant tank m=f(r, t.sub.skin) for the skin thickness of t.sub.skin=3, 6, and 10 cm carrying 20 tons of propellant mass. It is clear from FIG. 12 that the mass of the propellant tank m is inversely proportional to the radius of the propellant tank for a desired volume or mass.

(463) FIG. 13A depicts k=f(r, t.sub.skin) for an aluminum skin thickness of t.sub.skin=3, 6, and 10 cm when carrying 20 tons of propellant mass. It is clear from FIG. 13A that k is inversely proportional to the radius of the propellant tank for a desired volume or mass. Section 8 below describes how k determines the maximum altitude that a rocket can reach.

(464) Thrust, F.sub.T and the Remaining Rocket Parameters Calculated from .sub.f=m.sub.f/m.sub.0 and k=m.sub.tank/m.sub.prop.

(465) The introduction of the parameter k leads into very elegant solutions for all the rocket parameters and very interesting design relations. Propellant with a mass of m.sub.prop in a rocket is stored in a cylindrical tank with a mass of m.sub.tank. Since most of the rocket mass is propellant mass, most of the volume of the rocket will also be the volume of the propellant tank. Since the density of the propellant is known, it is straightforward to calculate the cylindrical volume of the propellant tank for a given diameter. The mass of the tank can be calculated with a given skin thickness and density of the tank.

(466) $\begin{array}{cc}{}_f=\frac{m_f}{m_0}=\frac{m_0-m_{\mathrm{prop}}}{m_0}=1-\frac{m_{\mathrm{prop}}}{m_0}& \left(8.1\right)\end{array}$

(467) Solving m.sub.prop from (8.1) gives,

(468) $\begin{array}{cc}\frac{m_{\mathrm{prop}}}{m_0}=1-{}_f& \left(8.2\right)\end{array}$

(469) The initial mass m.sub.0 is related to the thrust F.sub.T given as,

(470) $\begin{array}{cc}{m}_0=\frac{F_T}{g.Math.T2W}& \left(8.3\right)\end{array}$

(471) Solving m.sub.prop from (8.2) gives,
m.sub.prop=m.sub.0(1.sub.f)(8.4)

(472) Resulting in,

(473) $\begin{array}{cc}{m}_{\mathrm{prop}}=\frac{F_T}{g.Math.T2W}\left(1-{}_f\right)& \left(8.5\right)\end{array}$

(474) With the introduction of the parameter k, the propellant tank mass can be simply related to the propellant mass. Employing the parameter k the initial rocket mass m.sub.0 becomes,
m.sub.0=m.sub.prop(1+k)+m.sub.RE+m.sub.PAY(8.6)

(475) As can be seen m.sub.prop(1+k) term in (8.6) gives the sum of the propellant mass and the mass of the tank with which it is stored. In general, propellant mass also is the main structure of the rocket, everything is basically attached to it, which means that the parameter k determines a significant portion of the rocket mass as a function of the propellant mass which is in the rocket equation.

(476) Another parameter in equation (8.6) is m.sub.RE, which is the mass of the rocket engine that includes the mass of additional components such as pumps, electronics, guidance, etc. m.sub.PAY is the mass of the payload and associated additional mass related to its housing. Substituting (8.5) into (8.6) gives the thrust equation,

(477) $\begin{array}{cc}\frac{F_T}{g.Math.T2W}=\frac{F_T}{g.Math.T2W}\left(1-{}_f\right)\left(1+k\right)+m_{\mathrm{RE}}+m_{\mathrm{PAY}}& \left(8.7\right)\end{array}$

(478) Both sides of the equation are equal to the total initial mass m.sub.0 and becomes the thrust equation. The thrust F.sub.T, required to put a payload m.sub.PAY into an altitude of h, equation (8.7) can be solved giving,

(479) $\begin{array}{cc}{F}_T=\frac{g.Math.T2W\left(m_{\mathrm{RE}}+m_{\mathrm{PAY}}\right)}{1-\left(1-{}_f\right)\left(1+k\right)}& \left(8.8\right)\end{array}$

(480) Relation (8.8) assumes that the rocket has a single rocket engine. Simplifying the denominator gives,

(481) $\begin{array}{cc}{F}_T=\frac{g.Math.T2W\left(m_{\mathrm{RE}}+m_{\mathrm{PAY}}\right)}{{}_f\left(1+k\right)-k}& \left(8.9\right)\end{array}$

(482) Since there is only one rocket engine and its thrust and mass are already specified as F.sub.T and m.sub.RE, solving its thrust does not make much sense, as it is already known. Instead, one can calculate the payload m.sub.PAY that the rocket can put to an altitude h with that thrust,

(483) $\begin{array}{cc}{m}_{\mathrm{PAY}}=\frac{F_T\left[{}_f\left(1+k\right)-k\right]}{g.Math.T2W}-m_{\mathrm{RE}}& \left(8.1\right)\end{array}$

(484) Since F.sub.T, m.sub.PAY>0 and must be finite, the denominator of relation (8.9) puts some important restrictions between .sub.f and k as,

(485) 0$\begin{array}{cc}{}_f>\frac{k}{\left(1+k\right)}& \left(8.11\right)\end{array}$

(486) $\begin{array}{cc}k<\frac{{}_f}{\left(1-{}_f\right)}& \left(8.12\right)\end{array}$

(487) FIG. 13B depicts the plot of relation (8.12), which gives the minimum value of k=k.sub.MAX that needs to be satisfied to give a desired value of .sub.f. Since a given value of pf also gives the height h.sub.INV2(.sub.f, T2W) that the rocket reaches, a very important relationship exists between the height that the rocket can reach and the k.sub.MAX value, as shown in FIG. 13C.

(488) FIG. 13C depicts the height h.sub.INV2(k.sub.MAX, T2W) function, an important relationship between the height that the rocket can reach and the k.sub.MAX value that the rocket design must satisfy, graphically permitting the determination of k.sub.MAX for h.sub.GIVEN 400 km, 20,200 km, and 400,000 km. As can be seen, reaching a given altitude requires a minimum value of k.sub.MAX to be satisfied. The higher the desired altitude, the smaller the value of k.sub.MAX that needs to be satisfied. If kk.sub.MAX, the rocket cannot reach the given altitude, h.sub.GIVEN h.sub.INV2(k.sub.MAX, T21W).

(489) Relations (8.11) and (8.12) give the necessary conditions to be satisfied between them. The solution of the rocket equation for the rocket to reach the desired altitude h for a given T2W gives .sub.f(h, T2W) and (8.12) gives the minimum value of k needed. If this number is not a realizable quantity, then the launch will not be successful, leaving a multi-stage rocket as the only option to employ for a smaller k, which is an original relationship derived herein. If k is already given, then (8.11) gives the maximum value of .sub.f and moreover the maximum altitude the rocket can reach. Since FIG. 13A gives k=f(r, t.sub.skin), and as shown in FIG. 13C there is a functional relation between k.sub.MIN and h.sub.INV2(k.sub.MIN, T2W), the altitude that the rocket can reach, a functional relationship h.sub.INV2(MIN, T2W), the radius of the rocket can be derived which is of interest. In other words, to reach a desired altitude there is a minimum rocket radius r.sub.MIN that enables it, which is a function of burn chemistry, density of the oxidizer and fuel, their mix ratios, thickness of skin, its density, and machining. Ideally a value can be given to the denominator of (8.9) which satisfies (8.11) and (8.12).

(490) FIG. 13D depicts the SpaceX Starship Super Heavy (loaded with 3,400 tons of propellant) and the ULA Atlas first stage (loaded with 284 tons of propellant) booster heights with a t.sub.skin thickness of 3 [cm] with respective diameters of 9 [m] and 3.81 [m]. The relationship between the booster radius and the booster height is shown in FIG. 13D for the SpaceX Starship Super Heavy and the ULA (United Launch Alliance) first stage of Atlas V for known parameters such as their actual propellant mass and tank parameters. Both curves in FIG. 13D show an excellent fit for rocket booster heights. For the SpaceX Super Heavy the curve gives 71 [m] of booster height with the actual diameter of 9 [m] loaded with 3,400,000 [kg] of propellant mass. For ULA's first stage of Atlas V the curve gives 32.5 [m] of booster height at the actual diameter of 3.81 [m] loaded with 284,089 [kg] of propellant mass.

(491) FIG. 13E depicts h.sub.INV2(r, T2W), where r is the rocket diameter of the SpaceX Starship Super Heavy loaded with 3,400 Tons of propellant having t.sub.SKIN=3 [cm]. FIG. 13E shows the relationship between the SpaceX Starship Super Heavy maximum altitude as a function of its radius loaded with 3,400,000 [kg] of propellant mass and same tank parameters for T2W=1.1, 2, 4, and 8. It is clear that having a larger diameter gives a lower k, resulting in a higher maximum altitude that can be reached. The lowest k can be obtained with a spherical tank geometry giving the smallest area enclosing a given propellant volume.

(492) If the single rocket engine cannot give the desired thrust F.sub.T, then several rocket engines must be deployed. For this case the thrust relation (8.9) must be modified. Assuming all the rocket engines have the same mass m.sub.PRE (mass Per Rocket Engine) and they all generate the same thrust per engine F.sub.TPRE, the number of rocket engines n.sub.RE can be calculated from the calculated F.sub.T. The equivalent rocket engine mass m.sub.RE and the total thrust in (8.6) and (8.7) become related to the number of rocket engines that gives the necessity of including this effect. As an example, for cases where the needed thrust F.sub.T is larger than F.sub.TPRE, the number of rocket engines n.sub.RE that is needed becomes,

(493) $\begin{array}{cc}{n}_{\mathrm{RE}}=\frac{F_T}{F_{\mathrm{TPRE}}}1& \left(8.13\right)\end{array}$ where, F.sub.TPRE is the thrust per rocket engine. There can only be an integer number of rocket engines and this number also has to be greater than 1. The number n.sub.RE as calculated by relation (8.13) does not necessarily give an integer number. It can even be a number smaller than 1 if the calculated thrust F.sub.T is less than the selected rocket engine thrust F.sub.TPRE. Therefore, n.sub.RE is named the engine thrust scaling factor to represent its non-integer value. The engine thrust scaling factor n.sub.RE is used to calculate the total equivalent rocket engine mass to match the needed thrust F.sub.T. Using F.sub.TPRE, the corresponding equivalent total rocket engine mass m.sub.RE, which appears in the F.sub.T thrust relation (8.9) becomes,

(494) $\begin{array}{cc}{m}_{\mathrm{RE}}=n_{\mathrm{RE}}.Math.m_{\mathrm{PRE}}=\frac{F_T}{F_{\mathrm{TPRE}}}{m}_{\mathrm{PRE}}& \left(8.14\right)\end{array}$

(495) A more widely used rocket engine parameter instead of mass per rocket engine m.sub.PRE is the thrust-to-weight ratio of the rocket engine T2W.sub.PRE, a number like 100 to 200, a much larger number than any jet engine. As an example, the jet engine with the highest thrust-to-weight ratio is 8 for the GE J85 powering many airplanes like the F-5 and T-38, giving 13.1 [kN] with a mass of 140 kg (afterburner versions give 22 [kN] with a mass of 230 [kg]). The F-1 rocket engine that powered the Saturn-5 had a thrust-to-weight ratio of 94. This is an important parameter in making the rocket engine selection. Using the T2W.sub.PRE parameter the mass per rocket engine can be calculated as,

(496) $\begin{array}{cc}{m}_{\mathrm{PRE}}=\frac{F_{\mathrm{TPRE}}}{g.Math.T2W_{\mathrm{PRE}}}& \left(8.15\right)\end{array}$

(497) Substituting (8.15) in (8.14) gives,

(498) $\begin{array}{cc}{m}_{\mathrm{RE}}=\frac{F_T}{F_{\mathrm{TPRE}}}.Math.\frac{F_{\mathrm{TPRE}}}{g.Math.T2W_{\mathrm{PRE}}}=\frac{F_T}{g.Math.T2W_{\mathrm{PRE}}}& \left(8.16\right)\end{array}$

(499) Substituting (8.16) in (8.7) gives another term with F.sub.T dependency on the right-hand side as,

(500) $\begin{array}{cc}\frac{F_T}{g.Math.T2W}=\frac{F_T}{g.Math.T2W}\left(1-{}_f\right)\left(1+k\right)+\frac{F_T}{g.Math.T2W_{\mathrm{PRE}}}+m_{\mathrm{PAY}}& \left(8.17\right)\end{array}$