Question 8.5: After a Hohmann transfer from earth to Mars, calculate (a) t...
After a Hohmann transfer from earth to Mars, calculate
(a) the minimum delta-v required to place a spacecraft in orbit with a period of 7 h
(b) the periapsis radius
(c) the aiming radius
(d) the angle between periapsis and Mars’ velocity vector.
The following data are required from Tables A.1 and A.2:
\mu_{ sun }=1.327\left(10^{11}\right) km ^3 / s ^2
\mu_{ Mars }=42,830 km ^3 / s ^2
R_{\text {earth }}=149.6\left(10^6\right) km
R_{\text {Mars }}=227.9\left(10^6\right) km
r_{\text {Mars }} = 3396 km
Table A.1 Astronomical data for the sun, the planets, and the moon
| Object | Radius (km) |
Mass (kg) | Sidereal rotation period |
Inclination of equator to orbit plane |
Semimajor axis of orbit (km) |
Orbit eccentricity |
Inclination of orbit to the ecliptic plane |
Orbit sidereal period |
| Sun | 696000 | 1.989 \times 10^{30} | 25.38d | 7.25° | – | – | – | – |
| Mercury | 2440 | 330.2 \times 10^{21} | 58.56d | 0.01° | 57.91 \times 10^{6} | 0.2056 | 7.00° | 87.97d |
| Venus | 6052 | 4.869 \times 10^{24} | 243d^{a} | 177.4° | 108.2 \times 10^{6} | 0.0067 | 3.39° | 224.7d |
| Earth | 6378 | 5.974 \times 10^{24} | 23.9345h | 23.45° | 149.6 times 10^{6} | 0.0167 | 0.00° | 365.256d |
| (Moon) | 1737 | 73.48 \times 10^{21} | 27.32d | 6.68° | 384.4 \times 10^{3} | 0.0549 | 5.145° | 27.322d |
| Mars | 3396 | 641.9 \times 10^{21} | 24.62h | 25.19° | 227.9 \times 10^{6} | 0.0935 | 1.850° | 1.881y |
| Jupiter | 71,490 | 1.899 \times 10^{27} | 9.925h | 3.13° | 778.6 \times 10^{6} | 0.0489 | 1.304° | 11.86y |
| Saturn | 60,270 | 568.5 \times 10^{24} | 10.66h | 26.73° | 1.433 \times 10^{9} | 0.0565 | 2.485° | 29.46y |
| Uranus | 25,560 | 86.83 \times 10^{24} | 17.24h^{a} | 97.77° | 2.872 \times 10^{9} | 0.0457 | 0.772° | 84.01y |
| Neptune | 24,764 | 102.4 \times 10^{24} | 16.11h | 28.32° | 4.495 \times 10^{9} | 0.0113 | 1.769° | 164.8y |
| (Pluto) | 1187 | 13.03 \times 10^{21} | 6.387d^{a} | 122.5° | 5.906 \times 10^{9} | 0.2488 | 17.16° | 247.9y |
| ^aRetrograde | ||||||||
Table A.2 Gravitational parameter (μ) and sphere of influence (SOI) radius for the sun, the planets, and the moon
| Celestial body | μ (km³/s²) | SOI radius (km) |
| Sun | 132,712,440,018 | – |
| Mercury | 22,032 | 112,000 |
| Venus | 324,859 | 616,000 |
| Earth | 398,600 | 925,000 |
| Earth’s moon | 4905 | 66,100 |
| Mars | 42,828 | 577,000 |
| Jupiter | 126,686,534 | 48,200,000 |
| Saturn | 37,931,187 | 54,800,000 |
| Uranus | 5,793,939 | 51,800,000 |
| Neptune | 6,836,529 | 86,600,000 |
| Pluto | 871 | 3,080,000 |
(a) The hyperbolic excess speed is found using Eq. (8.4),
v_{\infty}=\Delta V_A=\sqrt{\frac{\mu_{\text {sun }}}{R_{\text {Mars }}}}\left(1-\sqrt{\frac{2 R_{\text {earth }}}{R_{\text {carth }}+R_{\text {Mars }}}}\right)=\sqrt{\frac{1.327\left(10^{11}\right)}{227.9\left(10^6\right)}}\left(1-\sqrt{\frac{2 \cdot 149.6\left(10^6\right)}{149.6\left(10^6\right)+227.9\left(10^6\right)}}\right)
\therefore v_{\infty} = 2.648 km/s
We can use Eq. (2.83) to express the semimajor axis a of the capture orbit in terms of its period T,
a=\left(\frac{T \sqrt{\mu_{ Mars }}}{2 \pi}\right)^{2 / 3}
Substituting T = 7 · 3600 s yields
a=\left(\frac{25,200 \sqrt{42,830}}{2 \pi}\right)^{2 / 3}=8832 km
From Eq. (2.73) we obtain
a=\frac{r_p}{1-e}
On substituting the optimal periapsis radius (Eq. 8.67) this becomes
a=\frac{2 \mu_{ Mars }}{v_{\infty}^2} \frac{1}{1+e}
from which
e=\frac{2 \mu_{ Mars }}{a v_{\infty}}-1=\frac{2 \cdot 42,830}{8832 \cdot 2.648^2}-1=0.3833
Thus, using Eq. (8.70), we find
\Delta v=v_{\infty} \sqrt{\frac{1-e}{2}}=2.648 \sqrt{\frac{1-03833}{2}}=1.470 km / s
(b) From Eq. (8.66), we obtain the periapse radius
r_p=\frac{2 \mu_{\text {Mars }}}{v_{\infty}^2} \frac{1-e}{1+e}=\frac{2 \cdot 42,830}{2.648^2} \frac{1-0.3833}{1+0.3833}=5447 km
(c) The aiming radius is given by Eq. (8.71),
\Delta=r_p \sqrt{\frac{2}{1-e}}=5447 \sqrt{\frac{2}{1-0.3833}}=9809 km
(d) Using Eq. (8.43), we get the angle to periapsis
\beta=\cos ^{-1}\left(\frac{1}{1+\frac{r_p v_{\infty}^2}{\mu_{ Mars }}}\right)=\cos ^{-1}\left(\frac{1}{1+\frac{5447 \cdot 2.648^2}{42,830}}\right)=58.09^{\circ}
Mars, the approach hyperbola, and the capture orbit are shown to scale in Fig. 8.17. The approach could also be made from the dark side of the planet instead of the sunlit side. The approach hyperbola and capture ellipse would be the mirror image of that shown, as is the case in Fig. 8.12.
