Question 8.4: A spacecraft is launched on a mission to Mars starting from ...

1. 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

Table A.2 Gravitational parameter (μ) and sphere of influence (SOI) radius for the sun, the planets, and the moon

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.