Mission: Find a Hohmann transfer from a 300 km altitude initial circular orbit about the Earth to a circular orbit about the Earth at the same distance as the Moon’s orbit.
Details to include:
Figure 3.1 shows the steps used.
\begin{align*} r_p &= r_{earth} + \text{alt} \\ &= 6378 + 300 \\ &= 6678\, \text{km} \end{align*}
The apogee distance \(r_a\) is the moon’s distance from center of earth given by \(r_a= 384400\) km. Therefore the semi-major axis \(a\) is \begin{align*} a &= \frac{r_a+r_p}{2}\\ &= \frac{384400+6678}{2}\\ &= \boxed{195539\, \text{km}} \end{align*}
The eccentricity \(e\) is \begin{align*} e &= \frac{r_a-r_p}{r_a-r_p}\\ &= \frac{384400+6678}{384400-6678}\\ &= \boxed{0.96585} \end{align*}
\(V_1\) is the spacecraft velocity in LEO and is given by \begin{align*} V_1 &= \sqrt{\frac{\mu _{earth}}{r_p}}\\ &= \sqrt{\frac{3.986 \times 10^5}{6678}}\\ &= \boxed{ 7.7258\, \text{km per second}} \end{align*}
The spacecraft required speed at perigee of the Hohmann transfer orbit \(V_p\) is \begin{align*} V_p &= \sqrt{\mu _{earth} \left ( \frac{2}{r_p}-\frac{1}{a} \right ) }\\ &= \sqrt{3.986 \times 10^5 \left ( \frac{2}{6678}-\frac{1}{195539} \right ) }\\ &= \boxed{ 10.8323\, \text{km per second}} \end{align*}
Since the moon is inside the sphere of influence of the earth, the difference of the above two speeds is all that is needed to send the spacecraft to the moon using a Hohmann orbit. Therefore \begin{align*} \Delta V_1 &= V_p - V_1\\ &= 10.8323 - 7.7251\\ &= \boxed{ 3.1065\, \text{km per second} } \end{align*}
When the spacecraft reaches the apogee of the Hohmann orbit, its speed \(V_a\) will be \begin{align*} V_a &= \sqrt{\mu _{earth} \left ( \frac{2}{r_a}-\frac{1}{a} \right ) }\\ &= \sqrt{3.986 \times 10^5 \left ( \frac{2}{384400}-\frac{1}{195539} \right ) }\\ &= \boxed{ 0.1882\, \text{km per second} } \end{align*}
The required speed \(V_2\) to put the satellite in the moon’s circular orbit is \begin{align*} V_2 &= \sqrt{\frac{\mu _{earth}}{r_a}}\\ &= \sqrt{\frac{3.986 \times 10^5}{3844008}}\\ &= \boxed{ 1.0183\, \text{ km per second} } \end{align*}
Therefore the impulse needed is \begin{align*} \Delta V_2 &= 1.0183 - 0.1882\\ &= \boxed{ 0.83\, \text{km per second} } \end{align*}
The total \(\Delta V\) is found from \begin{align*} \Delta V &= |\Delta V_1| + |\Delta V_2| \\ &= 3.1065 + 0.83 \\ &= \boxed{ 3.937\, \text{km per second} } \end{align*}
The transfer time \(\Delta T\) in seconds from the earth’s LEO orbit to the moon’s circular orbit is half the period of the Hohmann ellipse. Therefore \begin{align*} \Delta T &= \pi \sqrt{\frac{a^3}{\mu _{earth}}}\\ &= \pi \sqrt{\frac{195539^3}{3.986 \times 10^5}}\\ &= 4.3026e5\, \text{second} \\ &= \boxed{ 4.9798\, \text{day} } \end{align*}
Figure 3.2 shows the final orbit which is to scale and was generated from STK.
The following parameters are used in the calculations that follows \begin{align*} \mu _{earth} &= 3.986e5\, \text{km}^3\,\text{per second squared} \\ \mu _{moon} &= 4902.8\, \text{km}^3\,\text{per second squared} \\ r_a &= 384400\, \text{km}\\ r_{earth} &= 6378\, \text{km} \\ r_{moon} &= 1737.4\, \text{km} \\ r_{bo} &= 1760\, \text{km} \\ SOI_{moon} &= 6.61e4\, \text{km} \end{align*}
Figure 3.3 shows a more detailed Hohmann transfer orbit used as a guide in the calculations that follows. This diagram is not drawn to scale. A diagram drawn to scale is given at the end of this section.
The velocity of the spacecraft at the apogee of the Hohmann transfer was found in part (I) as \(V_a=0.188184\, \text{km per second}\). The speed of the moon relative to earth is \(V_{moon}=\sqrt{\frac{\mu _{earth}}{r_a}}=1.0183\, \text{km per second}\), therefore the speed of the spacecraft relative to the moon at the entry of the moon’s sphere of influence is \[ \boxed{V_\infty =V_{moon}-V_a= 0.830119\, \text{km per second}} \] Using the energy equation we can solve for the burn out speed \(V_{bo}\), which is the speed of the spacecraft at \(r_{bo}\), the closest distance from the moon surface \begin{align*} \frac{V_{bo}^2}{2} - \frac{\mu _{moon}}{r_{bo}} &= \frac{V_{\infty }^2}{2} - \frac{\mu _{moon}}{SOI_{moon}} \\ \frac{V_{bo}^2}{2} - \frac{4902.8}{1760} &= \frac{0.830119^2}{2} - \frac{4902.8}{6.6 \times 10^4} \end{align*}
Solving gives \[ \boxed{V_{bo} = 2.47222\, \text{km per second}} \] The impact distance \(b\) is found by solving \begin{align*} b \, V_\infty &= r_{bo} \, V_{bo} \\ b (0.830119) &= (1760)(2.47222) \end{align*}
Giving \[ \boxed{b= 5241.56\, \text{km}} \]
Figure 3.4 drawn to scale shows a moon centered fly-by of the spacecraft.
The eccentricity of the flyby hyperbolic orbit is found as follows \begin{align*} e &= 1+ \frac{r_{bo} V_\infty ^2}{\mu _{moon}} \\ &= 1+ \frac{ (1760) (0.830119)}{4902.8} \\ &= \boxed{ 1.24737 } \end{align*}
Hence \begin{align*} \eta &= \arccos \left ({\frac{-1}{e}} \right )\\ &= \arccos \left ({\frac{-1}{ 1.24737}} \right )\\ &= 2.50665\, \text{radian} \\ &= \boxed{ 143.621\, \text{degree} } \end{align*}
Therefore \begin{align*} \theta &= 2 \eta - 180\, \text{degree} \\ &= 2 (143.621) - 180\, \text{degree} \\ &= \boxed{ 107.241\, \text{degree} } \end{align*}
We now calculate \(V_D\), the departure speed relative to earth, using figure 3.5 that shows the change in speed and direction of the spacecraft as it enters and exists the moon’s sphere of influence.
Hence \[ \boxed{V_D= 1.49236\, \text{km per second} } \] The angle \(\gamma _d\) is found from the law of sines \begin{align*} \frac{V_D}{\sin{\theta }} &= \frac{V_\infty }{\sin{\gamma _d}}\\ \sin{\gamma _d} &= \frac{V_\infty \sin{\theta }}{V_D} \\ &= \frac{(0.830119) \sin{(107.241\, \text{degree})}}{1.49236} \end{align*}
Hence \(\sin{\gamma _d}=0.531252\) and \[ \boxed{\gamma _d = 32.0901\, \text{degree} } \]
The semi-major axis of the new orbit \(a_{new}\) is found from \begin{align*} V_D &= \sqrt{\mu _{earth} \left ( \frac{2}{r_a} - \frac{1}{a_{new}} \right ) }\\ 1.49236 &= \sqrt{3.986\times 10^5 \left ( \frac{2}{384400} - \frac{1}{a_{new}} \right ) } \end{align*}
Solving numerically for \(a_{new}\) gives \[ \boxed{a_{new} = -2.60104e6\, \text{km}} \] The new eccentricity is found from \begin{align*} \cos{\gamma _d} &= \sqrt{ \frac{a_{new}^2 (1-e^2)}{r_a(2 a_{new} - r_a)} }\\ \cos{( 32.0901\, \text{degree})} &= \sqrt{ \frac{(-2.60104\times 10^6)^2 (1-e^2)}{384400(2 (-2.60104\times 10^6) - 384400)} } \end{align*}
Solving numerically for the new \(e\) and taking the positive root gives \[ \boxed{e =1.10808} \] Therefore, the new trajectory is hyperbolic when the spacecraft exits the moon’s sphere of influence.
Since the new trajectory is hyperbolic, the true anomaly \(f\) can be found using the hyperbolic equation \begin{align*} r_1 &= \frac{a_{new}(e^2-1)}{1+e\cos{f}} \\ 384400 &= \frac{2.60104\times 10^6(1.10808^2-1)}{1+(1.10808)\cos{f}} \end{align*}
Solving for \(f\) and taking the positive value since the spacecraft is in the positive half plane gives \begin{align*} f &= 1.06009\, \text{radian}\\ &= 60.7387\, \text{degree} \end{align*}
This value of the true anomaly is used to locate the new perigee of the post flyby orbit. The \(r_p\) of the hyperbola is first found from \begin{align*} r_p &= a_{new}(e-1)\\ &= 2.60104\times 10^6(1.10808-1)\\ &= 281109\, \text{km} \end{align*}
Figure 3.6 shows the new post flyby hyperbolic trajectory
Figure 3.7 shows the velocity vector diagram
Figure 3.8 was generated from STK showing the LEO and small part of the Hohmann transfer orbit with the moon orbit at a distance. This is to scale.
The above results are summarized in table 3.1
| variable | pre flyby | post flyby |
| orbit type | elliptical | hyperbolic |
| \(e\) | 0.96585 | 1.10808 |
| semi-major axis \(a\) | 195539 km | -2.60104e6 km |
| true anomaly \(f\) | 180 degree | 60.7387 degree |
| \(r_p\) | 6678 km | 281109 km |
The above results for the flyby hyperbolic trajectory are summarized in table 3.2
| parameter | value |
| \(e\) | 1.24737 |
| \(V_A\) | 0.188184 km per second |
| \(V_D\) | 1.49236 km per second |
| \(\gamma _A\) | 77.37 degree |
| \(\gamma _D\) | 32.09 degree |
| \(b\) | 5241.56 km |
| \(V_{\infty }\) | 0.83 km per second |
| \(\eta \) | 143.621 degree |
| \(\theta \) | 107.241 degree |
The following parameters are used in the calculations that follows \begin{align*} \mu _{earth} &= 3.986e5 \, \text{km}^3\,\text{per second squared} \\ \mu _{moon} &= 4902.8 \, \text{km}^3\,\text{per second squared} \\ r_{earth} &= 6378\, \text{km} \\ r_{moon} &= 1737.4\, \text{km} \\ SOI_{moon} &= 6.61e4\, \text{km} \\ r_p &= r_{earth}+300\\ &= 6378+300 \\ &= 6678\, \text{km}\\ v_{moon} &= \sqrt{\frac{mu_{earth}}{r_{moon}}}\\ &= \sqrt{\frac{3.986\times 10^5}{1737.4}}\\ &= 1.0183\, \text{km per second} \, \text{(velocity of moon relative to earth)}\\ r_1 &= 384400\, \text{km} \, \text{(distance from earth to the moon)}\\ a &= 300000\, \text{km}\,\text{(semi-major axis of the Hohmann transfer ellipse)}\\ \end{align*}
Figure 3.10 gives a general view of the initial phase of the orbit showing the non-Tangential approach to the moon’s circular orbit using the initial Hohmann transfer ellipse. This is not scale.
part 1 The given \(a\) is used to solve for \(r_a\). Since \(a=\frac{r_p + r_a}{2}\) hence \begin{align*} r_a &= 2a-r_p\\ &= (2)(300000)-6678\\ &= \boxed{ 593322\, \text{km}} \end{align*}
The eccentricity of the Hohmann ellipse is now found as follows \begin{align*} e &= \frac{r_a-r_p}{r_a+r_p}\\ &= \frac{593322-6678}{593322+6678}\\ &= \boxed{0.97774} \end{align*}
The speed of the spacecraft at the location where Hohmann orbit intersects the the moon’s circular orbit is called \(V_A\) and found as follows \begin{align*} V_A &= \sqrt{\mu _{earth} \left ( \frac{2}{r_1} -\frac{1}{a} \right ) }\\ &= \sqrt{3.986\times 10^5 \left ( \frac{2}{384400} -\frac{1}{300000} \right ) }\\ &= 0.863258\, \text{km} \end{align*}
\(\gamma _A\) is the angle between the path of the spacecraft and the moon’s velocity vector direction \begin{align*} \cos{\gamma _A} &= \sqrt{\frac{a^2(1-e^2)}{r_1(2 a- r_1)}}\\ &= \sqrt{\frac{300000^2(1-0.97774^2)}{384400 (2 (300000)- 384400)}}\\ &= 0.218651 \end{align*}
Hence \begin{align*} \gamma _A &= 1.35036\, \text{radian}\\ &= \boxed{ 77.3702\, \text{degree}} \end{align*}
The true anomaly \(f\) of the pre flyby Hohmann transfer at the above location can now be found \begin{align*} \tan{\gamma _A} &= \frac{e \sin{f}}{1+e \cos{f}} \\ \tan{( 77.3702\, \text{degree})} &= \frac{(0.97774) \sin{f}}{1+(0.97774) \cos{f}} \end{align*}
Solving for \(f\) gives \begin{align*} f &={2.8582\, \text{radian}} \\ &= \boxed{{ 163.763\, \text{degree}} } \end{align*}
Relative to the moon, and at the entry to the moon’s sphere of influence, the velocity of the spacecraft is given by \(V_{\infty _a}\) as shown in figure 3.11
And the angle \(\beta \) is \begin{align*} \frac{V_A}{\sin{\beta }} &= \frac{V_{\infty _a}}{\sin{\gamma _A}} \\ \frac{0.863258}{\sin{\beta }} &= \frac{1.18226}{\sin{1.35036}} \end{align*}
Solving for \(\beta \) gives \begin{align*} \beta &= 1.21158\, \text{radian}\\ &= \boxed{ 45.439\, \text{degree}} \end{align*}
The eccentricity of the flyby hyperbolic trajectory \(e_{flyby}\) inside the moon’s sphere of influence can be found from the energy equation, using the burn out distance \(r_{bo}= 1760\, \text{km}\)1 . \begin{align*} \frac{V_{\infty _a}^2}{2} - \frac{\mu _{moon}}{SOI_{moon}} &= \frac{V_{bo}^2}{2} - \frac{\mu _{moon}}{r_{bo}}\\ \frac{1.18226^2}{2} - \frac{4902.8}{6.61\times 10^4} &= \frac{V_{bo}^2}{2} - \frac{4902.8}{1760} \end{align*}
Solving for \(V_{bo}\) gives \[ \boxed{V_{bo} = 2.6116\, \text{km per second}} \] Therefore \begin{align*} e_{flyby} &= \sqrt{ 1 + \frac{ V_{bo}^2 V_{\infty _a}^2 r_{bo}^2}{\mu _{moon}^2} }\\ &= \sqrt{ 1 + \frac{ (2.6116^2) (1.18226)^2 (1760)^2}{4902.8^2} }\\ &= \boxed{1.4928} \end{align*}
The angle \(\eta \) is \begin{align*} \eta &= \arccos{\left ( \frac{-1}{e_{flyby}} \right ) } \\ &= \arccos{\left ( \frac{-1}{1.4928} \right ) } \\ &= 2.3048\, \text{radian} \\ &= \boxed{132.05\, \text{degree} } \end{align*}
The turning angle of the asymptotic is \(\theta \) as shown in figure 3.12.
The departure speed of the spacecraft \(V_D\) relative to earth is found from the velocity vector in figure 3.15 as follows
The semi-major axis \(a_{new}\) of the post flyby orbit is found from \begin{align*} V_D &= \sqrt{\mu _{earth} \left ( \frac{2}{r_1} - \frac{1}{a_{new}} \right ) }\\ 1.99197 &= \sqrt{398600 \left ( \frac{2}{384400} - \frac{1}{a_{new}} \right ) } \end{align*}
Solving for \(a_{new}\) gives \[ \boxed{a_{new} = -2.10444e5\, \text{km}} \] Therefore the departure angle \(\gamma _D\) is \[ \frac{V_D}{\sin{(\beta +\theta )}} = \frac{V_{\infty _d}}{\sin{\gamma _D}} \] Solving for \(\gamma _D\) \begin{align*} \sin{\gamma _D} &= \frac{v_{\infty _d} \sin{(\beta +\theta )}}{V_D}\\ &= \frac{(1.18226) \sin{(45.439\, \text{degree}+84.114\, \text{degree})}}{1.99197} \end{align*}
Hence \begin{align*} \gamma _D &= 0.4753\, \text{radian}\\ &=\boxed{ 27.233\, \text{degree}} \end{align*}
The eccentricity of the post flyby orbit is \begin{align*} \cos{\gamma _D} &= \sqrt{\frac{a_{new}^2(1-e^2)}{r_1(2 a_{new}- r_1)}}\\ 0.8891 &= \sqrt{\frac{(-2.10444\times 10^5)^2(1-e^2)}{384400 (2 (-2.10444\times 10^5)- 384400)}} \end{align*}
Solving for \(e\) gives \[ \boxed{e=2.5546} \]
part 2 Since the new trajectory after the flyby is found to be a hyperbola, then the hyperbolic equation is used to obtain the true anomaly \(f\) \begin{align*} r_1 &= \frac{a_{new}(e^2-1)}{1+e\cos{f}} \\ 384400 &= \frac{(2.10444\times 10^5)(2.5546^2-1)}{1+(2.5546)\cos{f}} \end{align*}
Solving for \(f\) and taking the positive value since the spacecraft is in the positive half plane gives \begin{align*} f &={0.665415\, \text{radian}} \\ &= \boxed{{37.552\, \text{degree}} } \end{align*}
This value of the true anomaly is used to locate the new value of perigee of the post flyby orbit. The \(r_p\) of the hyperbola is found from \begin{align*} r_p &= a_{new}(e-1)\\ &= (2.10444\times 10^5)(2.55546-1)\\ &= 3.27157e5\, \text{km} \end{align*}
Figure 3.14 shows the pre flyby and the new post flyby changes to the orbit. The effect of the flyby is to produce an instantaneous \(\Delta V\) that comes from the change of energy of the spacecraft due to its going and leaving the moon’s sphere of influence.
part 3 Figure 3.15 shows the velocity triangles of the flyby trajectory.
summary for non-tangential flyby. Behind the moon case The above results for the pre and post flyby trajectories are summarized in table 3.5
| variable | pre flyby | post flyby |
| orbit type | elliptical | hyperbolic |
| \(e\) | 0.97774 | 2.5546 |
| semi-major axis \(a\) | 300000 km | -2.10444e5 km |
| true anomaly \(f\) | 163.76 degree | 37.552 degree |
| \(r_p\) | 6678 km | 3.27157e5 km |
The results for the flyby hyperbolic trajectory are summarized in table 3.6
| parameter | value |
| \(e\) | 1.4928 |
| \(V_A\) | 0.863 km per second |
| \(V_D\) | 1.99197 km per second |
| \(\gamma _A\) | 77.37 degree |
| \(\gamma _D\) | 27.233 degree |
| \(V_{\infty }\) | 1.18226 km per second |
| \(\beta \) | 45.439 degree |
| \(\eta \) | 132.05 degree |
| \(\theta \) | 84.11 degree |
When the spacecraft flies by the moon from behind it gains energy and the new speed relative to earth \(V_D\) is larger than the arrival speed \(V_A\) relative to earth. The reverse happens when the spacecraft flies in front of the moon. Its new velocity \(V_D\) will be smaller than \(V_A\).
The computation for this part follows closely what was done for the case of flying behind the moon. The difference is in how the velocity vector diagram is constructed to make sure the correct angles are used. This results in a velocity of the spacecraft \(V_D\) after leaving the moon sphere of influence slower than the above case.
The computation that follows starts from the new velocity vector diagram as follows.
The turning angle of the asymptotic \(\theta \) is shown in figure 3.16.
The departure speed of the spacecraft \(V_D\) relative to earth is found from the velocity vector in figure 3.19
The semi-major axis \(a_{new}\) of the post flyby orbit is \begin{align*} V_D &= \sqrt{\mu _{earth} \left ( \frac{2}{r_1} - \frac{1}{a_{new}} \right ) }\\ 0.74492 &= \sqrt{3.986\times 10^5 \left ( \frac{2}{384400} - \frac{1}{a_{new}} \right ) } \end{align*}
Solving for \(a_{new}\) gives \[ \boxed{a_{new} = 2.62413e5\, \text{km}} \] The departure angle \(\gamma _D\) is found from \begin{align*} \sin{\gamma _D} &= \frac{v_\infty \sin{(\beta -\theta )}}{V_D}\\ &= \frac{(1.18226) \sin{(45.439\, \text{degree}-(84.11\, \text{degree})}}{1.99197} \end{align*}
Solving for \(\gamma _D\) gives \begin{align*} \gamma _D &=-1.4425\, \text{radian}\\ &=\boxed{ -82.649\, \text{degree}} \end{align*}
The eccentricity of the post flyby orbit is found from \begin{align*} \cos{\gamma _D} &= \sqrt{\frac{a_{new}^2(1-e^2)}{r_1(2 a_{new}- r_1)}}\\ 0.1279 &= \sqrt{\frac{(2.62413\times 10^5)^2(1-e^2)}{384400 (2 (2.62413\times 10^5)- 384400)}} \end{align*}
Solving for \(e\) gives \[ \boxed{e=0.9935} \]
part 2 Since the new trajectory after the flyby is elliptic in this case, the elliptic equation is used to obtain the new true anomaly \(f\) \begin{align*} r_1 &= \frac{a_{new}(1-e^2)}{1+e\cos{f}} \\ 384400 &= \frac{(2.62413\times 10^5)(1-0.9935^2)}{1+(0.9935)\cos{f}} \end{align*}
Solving for \(f\) gives \begin{align*} f &= -3.0728\, \text{radian} \\ &= -176.06\, \text{degree} \end{align*}
Since \(\gamma _D<0\) then the post flyby true anomaly is between \(180\) and \(360\) degrees. Therefore, \begin{align*} f &= 3.21\, \text{radian}\\ &= \boxed{{183.94\, \text{degree}} } \end{align*}
This value of the true anomaly is now used to locate the new value of perigee of the post flyby orbit. The \(r_p\) of the new ellipse is found from \begin{align*} r_p &= a_{new}(1-e)\\ &= (2.62413\times 10^5)(1-0.9935)\\ &= 1689.13\, \text{km} \end{align*}
Figure 3.18 shows the pre flyby and the post flyby changes to the orbit.
part 3 Figure 3.19 shows the velocity triangle of the flyby.
summary of non-tangential flyby. Front of the moon case The above results for the pre and post flyby trajectories are summarized in table 3.5
| variable | pre flyby | post flyby |
| orbit type | elliptical | elliptical |
| \(e\) | 0.97774 | 0.9935 |
| semi-major axis \(a\) | 300000 km | 262413 km |
| true anomaly \(f\) | 163.76 degree | 183.94 degree |
| \(r_p\) | 6678 km | 1689 km |
The above results for the flyby hyperbolic trajectory are summarized in table 3.6
| parameter | value |
| \(e\) | 1.4928 |
| \(V_A\) | 0.863 km per second |
| \(V_D\) | 0.7449 km per second |
| \(\gamma _A\) | 77.37 degree |
| \(\gamma _D\) | -82.649 degree |
| \(V_{\infty }\) | 1.18226 km per second |
| \(\beta \) | 45.439 degree |
| \(\eta \) | 132.05 degree |
| \(\theta \) | 84.11 degree |
Since new \(r_p\) is smaller than \(r_{earth}\), the spacecraft will hit earth on way back on way back on the new post flyby trajectory.
When the spacecraft flies by the moon from front, it losses energy and the new speed relative to earth \(V_D\) is smaller than the arrival speed \(V_A\) relative to earth.
The trajectory that was selected for the pre flyby part is to send the spacecraft to front of the moon. The reason is because the post flyby velocity of the spacecraft \(V_D\) in this case will be smaller that the approach velocity \(V_D\) and the new flight path angle \(\gamma _D\) will be negative and the post flyby trajectory being an ellipse. This insures the the spacecraft will return back to earth.
It is assumed that the spacecraft will rendezvous with the moon when it reaches the moon’s orbit. Timing considerations are discussed in part (IV).
Since the original altitude above earth of the spacecraft was fixed by the project requirement to be in LEO at 300 km, the other free variable that can be used to adjust the trajectory is the semi-major axis \(a\) of the pre flyby orbit.
Changing \(a\) is the same as changing the initial \(\Delta V_1\). The lunar burn out radius \(r_{bo}\) was also fixed by project requirement to be 1760 km.
A program was written to make it easier to change the semi-major axis \(a\) using a flyby in front of the moon approach. The program calculates all the parameters of the new post flyby trajectory.
The resulting post flyby ellipse was checked after each simulation run to see if it meets the requirement of having a return altitude on earth of between 300 km and 500 km. In addition, The selected trajectory was required to have its velocity at perigee (closest point to earth) to be below 12 km per second to ensure safety of the spacecraft as it enters earth.
The selected trajectory had a new \(V_p\) of 10.8079 km per second. This is faster than the initial elliptical orbit \(V_p\) which was 7.725 km per second but it is still a safe entry velocity back to earth.
Figure 3.20 shows the user interface of the program with the final selected trajectory.
Figure 3.21 shows the final result of the trajectory selected. This table was generated by the simulation program written for this project. The semi-major axis of the initial orbit is
\(\displaystyle a=415000 \text{km}\) and the eccentricity is
\(\displaystyle e=0.983908\) and
\(\displaystyle \Delta V_1=3.1561 \text{km per second}\)
The altitude at the perigee of the post flyby ellipse is
\(\displaystyle \text{ALT}= 347.028 \text{km}\) which meets the requirements
The program developed for this project plots the final selected trajectory to scale. It shows both the pre flyby ellipse and the post flyby ellipse.
Figure 3.22 shows the selected trajectory generated by the simulation program (to scale).
Figure 3.25 shows a zoomed version of the selected trajectory near earth. The new perigee is 6725.03 km which represents an altitude of 347.028 km.
It was found during simulation that finding the return ellipse with the required final altitude was very sensitive to small changes in value of the semi-major axis \(a\) for the initial orbit. There was a small range of values of \(a\) which generated an acceptable free return trajectories. Using a simulation program helped in finding this small range of values of \(a\) easier.
The time to reach the moon is given by \(\Delta T=\sqrt{\frac{a^3}{\mu _{earth}}}\left ( E-e\sin E \right )\) where \(E\) is the eccentric anomaly of the pre flyby trajectory.
\(E\) is found by solving \(r=a(1-e\cos E)\) where \(r\) here is the distance between earth and the moon and \(e\) is the eccentricity of the pre flyby orbit. Using the result of the selected trajectory of part (4) \begin{align*} r&=a(1-e\cos E) \\ 384400 &= 415000(1- (0.9839)\cos E) \end{align*}
Solving gives \(E=1.4957\, \text{radian}\) or
\(\displaystyle E=85.7 \text{degree}\) . Therefore the time to reach the moon is \begin{align*} \Delta T&=\sqrt{\frac{a^3}{\mu _{earth}}}\left ( E-e\sin E \right )\\ &=\sqrt{\frac{(415000)^3}{3.986\times 10^5}}\left ( 1.4957-(0.9839)\sin{(1.4957)} \right )\\ &=217926\, \text{second}\\ &=60.535\, \text{hour}\\ &=\boxed{2.523\, \text{day}} \end{align*}
The angular velocity of the moon in its orbit around earth is given by \(\omega =\sqrt{\frac{\mu _{earth}}{r^3}}\) where \(r\) is the distance from earth to the moon. During the \(\Delta T\) found in part (1), the moon will travel \begin{align*} \theta &= \omega (\Delta T)\\ &= \sqrt{\frac{\mu _{earth}}{r^3}} (\Delta T)\\ &= \sqrt{\frac{3.986\times 10^5}{384400}} (217926)\\ &= 0.5773\, \text{radian}\\ &= \boxed{ 33.077\, \text{degree}} \end{align*}
Since the true anomaly was found in part (1) for the pre flyby to be 168.9 degree, therefore the moon has to be at angle \(\theta _0=168.9-33.077\) or \[ \theta _0=135.833\, \text{degree} \] In front of the spacecraft initial position as shown in figure 3.26
The angular velocity of the spacecraft around earth \begin{align*} \omega _1 &= \sqrt{\frac{\mu _{earth}}{r_p}} \\ &= \sqrt{\frac{3.986\times 10^5}{6678}} \\ &= 0.00115691\, \text{radian per second} \end{align*}
The angular velocity of the moon around earth is \begin{align*} \omega _{moon} &= \sqrt{\frac{\mu _{earth}}{r_1}} \\ &= \sqrt{\frac{3.986\times 10^5}{384400}} \\ &= 2.64907e(-6)\, \text{radian per second} \end{align*}
The synodic period of the moon relative to the spacecraft is how often the space craft and the moon have the correct alignment, which is given by \begin{align*} \tau _s &= \frac{2 \pi }{|\omega _1-\omega _{moon}|} \\ &= \frac{2 \pi }{|0.00115691-2.64907\times 10^{-6}|} \\ &= \boxed{ 1.51208\, \text{hour} } \end{align*}
Figure 3.27 shows the flyby hyperbola.
The eccentric anomaly \(F\) for the hyperbolic trajectory is found from \begin{align*} r_{bo} &= a(e \cosh{F}-1)\\ 1760 &= 2823.51(1.62334 \cosh{F}-1) \end{align*}
Solving gives \begin{align*} F &=3.40112\, \text{radian}\\ &=194.87\, \text{degree} \end{align*}
Therefore the time for the overall flyby, which is the time that the spacecraft is inside the moon’s sphere of influence is \begin{align*} \Delta T &=2 \sqrt{ \frac{a^3}{\mu _{moon}} }(e \sinh{F}-F)\\ &=2 \sqrt{ \frac{2823.51^3}{4902.8} }\left ((1.62334) \sinh{(3.40112)}-3.40112\right )\\ &=24.9029\, \text{hour} \end{align*}
The time to fly back to earth from the moon after the flyby phase is complete is found from the elliptical equation for the return flight solution found above. \begin{align*} r &= a(1-e\cos E) \\ 384400 &= 230327(1- (0.970802)\cos E) \end{align*}
Solving gives \(E=2.33098\, \text{radian}\) or
\(\displaystyle E=133.55\, \text{degree}\) . Therefore the time is \begin{align*} \Delta T_2&=\sqrt{\frac{a^3}{\mu _{earth}}}\left ( E-e\sin E \right )\\ &=\sqrt{\frac{(230327)^3}{3.986\times 10^5}}\left ( 2.33098-(0.97080)\sin{(2.33098)} \right )\\ &=217926\, \text{second}\\ &=79.149\, \text{hour}\\ &=\boxed{3.298\, \text{day}} \end{align*}
Using the time to flyby the moon found earlier in part (1) above, the total flight time for the whole journey is therefore \begin{align*} T &= 60.535 + 24.9029 + 79.149 \\ &= \boxed{164.587\, \text{hour}} \end{align*}
Hence, the percentage of time in flyby around the moon is \(\frac{24.9029}{164.587}\) or
\(\displaystyle 15.13\%\) .
During the time the spacecraft is inside the moon’s sphere of influence, the moon will have traveled \begin{align*} \Delta \theta &= \omega _{moon}\, (\text{flyby time}) \\ &= \sqrt{\frac{\mu _{earth}}{r_1^3}} \times (24.9029\, \text{hour})\\ &= \sqrt{\frac{3.986\times 10^5}{384400^3}} \times (24.9029\, \text{hour})\\ &= \boxed{13.607\, \text{degree}} \end{align*}
This is \(\frac{13.607}{360}=\fbox{3.78\%}\) of the full orbit of the moon around the earth. This shows that the change of speed \(\Delta V\) that occurs due to the flyby is not instantaneous and takes about \(3.8\%\) of the period of the moon around the earth.
Therefore, the conic method can be considered only to be a first order approximation, and therefore, for practical spacecraft trajectory design, numerical methods can be used to obtain a more accurate solutions.