Question 8.4: A spacecraft is launched on a mission to Mars starting from ...
A spacecraft is launched on a mission to Mars starting from a 300-km circular parking orbit. Calculate (a) the delta-v required, (b) the location of perigee of the departure hyperbola, and (c) the amount of propellant required as a percentage of the spacecraft mass before the delta-v burn, assuming a specific impulse of 300 s.
- From Tables A.1 and A.2, we obtain the gravitational parameters for the sun and the earth,
\mu_{ sun }=1.327\left(10^{11}\right) km ^3 / s ^2
\mu_{\text {earth }}=398,600 km ^3 / s ^2
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} | 243da | 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} | 17.24ha | 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} | 6.387da | 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 |
and the orbital radii of the earth and Mars,
R_{\text {earth }}=149.6\left(10^6\right) km
R_{\text {Mars }}=227.9\left(10^6\right) km
(a) According to Eq. (8.35), the hyperbolic excess speed is
v_{\infty}=\sqrt{\frac{\mu_{\text {sun }}}{R_{\text {earth }}}}\left(\sqrt{\frac{2 R_{ Mars }}{R_{\text {earth }}+R_{ Mars }}}-1\right)=\sqrt{\frac{1.327\left(10^{11}\right)}{149.6\left(10^6\right)}}\left(\sqrt{\frac{2 \cdot 227.9\left(10^6\right)}{149.6\left(10^6\right)+227.9\left(10^6\right)}-1}\right)
from which
v_{\infty}=2.943 km / s
The speed of the spacecraft in its 300-km circular parking orbit is given by Eq. (8.41),
v_c=\sqrt{\frac{\mu_{\text {earth }}}{r_{\text {earth }}+300}}=\sqrt{\frac{398,600}{6678}}=7.726 km / s
Finally, we use Eq. (8.42) to calculate the delta-v required to step up to the departure hyperbola
\Delta v=v_c\left(\sqrt{2+\left(\frac{v_{\infty}}{v_c}\right)^2}-1\right)=7.726\left(\sqrt{2+\left(\frac{2.943}{7.726}\right)^2}-1\right)
Δv = 3.590 km/s
(b) Perigee of the departure hyperbola, relative to the earth’s orbital velocity vector, is found using Eq. (8.43),
\beta=\cos ^{-1}\left(\frac{1}{1+\frac{r_p v_{\infty}{ }^2}{\mu_{\text {earth }}}}\right)=\cos ^{-1}\left(\frac{1}{1+\frac{6678 \cdot 2.943^2}{368,600}}\right)
β = 29.16°
Fig. 8.12 shows that the perigee can be located on either the sunlit or the dark side of the earth. It is likely that the parking orbit would be a prograde orbit (west to east), which would place the burnout point on the dark side.
(c) From Eq. (6.1), we have
\frac{\Delta m}{m}=1-\exp \left(-\frac{\Delta v}{l_{\text {sp }} g_0}\right)
Substituting \Delta v=3.590 km / s , I_{ sq }=300 s \text {, and } g_0=9.81\left(10^{-3}\right) km / s ^2, this yields
\frac{\Delta m}{m}=0.705
That is, prior to the delta-v maneuver over, 70% of the spacecraft mass must be propellant.
