Question 8.10: In Example 8.8, calculate the delta-v required to place the ...
In Example 8.8, calculate the delta-v required to place the spacecraft in an elliptical capture orbit around Mars with a periapsis altitude of 300 km and a period of 48 h. Sketch the approach hyperbola.
From Tables A.1 and A.2, we know that
r_{\text {Mars }}=3380 km
\mu_{ Mars }=42,830 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} | 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 |
The radius to periapsis of the arrival hyperbola is the radius of Mars plus the periapsis of the elliptical capture orbit,
r_p = 3380 + 300 = 3680 km
According to Eq. (8.40) and Eq. (b) of Example 8.8, the speed of the spacecraft at periapsis of the arrival hyperbola is
\left.v_p\right)_{\text {Arrival }}=\sqrt{\left.\left[v_{\infty}\right)_{\text {Arrival }}\right]^2+\frac{2 \mu_{\text {Mars }}}{r_p}}=\sqrt{2.8852^2+\frac{2 \cdot 42,830}{3680}}=5.621 km / s
To find the speed \left.v_p\right)_{\text { ellipse }} at periapsis of the capture ellipse, we use the required period (48 h) to determine the ellipse’s semimajor axis, using Eq. (2.83),
a_{\text {ellipse }}=\left(\frac{T \sqrt{\mu_{\text {Mars }}}}{2 \pi}\right)^{3 / 2}=\left(\frac{48 \cdot 3600 \cdot \sqrt{42,830}}{2 \pi}\right)^{3 / 2}=31,880 km
From Eq. (2.73), we obtain
e_{\text {ellipse }}=1-\frac{r_p}{a_{\text {ellipse }}}=1-\frac{3680}{31,880}=0.8846
Then Eq. (8.59) yields
\left.v_p\right)_{\text {ellipse }}=\sqrt{\frac{\mu_{\text {Mars }}}{r_p}\left(1+e_{\text {ellipse }}\right)}=\sqrt{\frac{42,830}{3680}(1+0.8846)}=4.683 km / s
Hence, the delta-v requirement is
\left.\left.\Delta v=v_p\right)_{\text {Arrival }}-v_p\right)_{\text {ellipse }}=0.9382 km / s
The eccentricity of the approach hyperbola is given by Eq. (8.38),
e=1+\frac{\left.r_p\left(v_{\infty}\right)_{\text {Arival }}\right)^2}{\mu_{\text {Mars }}}=1+\frac{3680 \cdot 2.8852^2}{42,830}=1.715
Assuming that the capture ellipse is a polar orbit of Mars, then the approach hyperbola is as illustrated in Fig. 8.30. Note that Mars’ equatorial plane is inclined 25° to the plane of its orbit around the sun. Furthermore, the vernal equinox of Mars lies at an angle of 85° from that of the earth.
