Question 8.8: A spacecraft departs earth’s sphere of influence on November...

Step 1:
Algorithm 8.1 yields the state vectors for earth and Mars

R _{\text {earth }}=1.0499\left(10^8\right) \hat{ I }+1.0465\left(10^8\right) \hat{ J }+716.93 \hat{ K }  ( km ) \quad\left(R_{\text {earth }}=1.4824\left(10^8\right)  km \right)

V _{\text {earth }}=-21.515 \hat{ I }+20.958 \hat{ J }+0.00014376 \hat{ K }  ( km / s ) \quad\left(V_{\text {earth }}=30.055  km / s \right)

R _{\text {Mars }}=-2.0858\left(10^7\right) \hat{ I }-2.1842\left(10^8\right) \hat{ J }+4.06244\left(10^6\right) \hat{ K }  ( km ) \quad\left(R_{\text {Mars }}=2.1945\left(10^8\right)  km \right)

V _{\text {Mars }}=25.037 \hat{ I }+0.22311 \hat{ J }-0.62018 \hat{ K }  ( km / s ) \quad\left(V_{\text {Mars }}=25.046  km / s \right)

Step 2:
The position vector R_1 of the spacecraft at crossing the earth’s sphere of influence is just that of the earth,

R _1= R _{\text {earth }}=1.0499\left(10^8\right) \hat{ I }+1.0465\left(10^8\right) \hat{ J }+716.93 \hat{ K }  ( km )

On arrival at Mars’ sphere of influence, the spacecraft’s position vector is

R _2= R _{ mars }=-2.0858\left(10^7\right) \hat{ I }-2.1842\left(10^8\right) \hat{ J }-4.06244\left(10^6\right) \hat{ K }  ( km )

According to Eqs. (5.47) and (5.48)

J D_{\text {Departure }}=2,450,394.5

J D_{\text {Arrival }}=2,450,703.5

J D=J_0+\frac{U T}{24}                  (5.47)

J_0=367 y-\operatorname{INT}\left\{\frac{7\left[y+\operatorname{INT}\left(\frac{m+9}{12}\right)\right]}{4}\right\}+\operatorname{INT}\left(\frac{275 m}{9}\right)+d+1,721,013.5                            (5.48)

Hence, the time of flight is

t_{12} = 2,450,703.5 – 2,450,394.5 = 309 days

Entering R_1, R_2, and t_{12} into Algorithm 5.2 yields

V _D^{(v)}=-24.429 \hat{ I }+21.782 \hat{ J }+0.94810 \hat{ K }  ( km / s ) \quad\left(V_D^{(v)}=32.743  km / s \right)

V _A^{(v)}=22.157 \hat{ I }+0.19959 \hat{ J }+0.45793 \hat{ K }  ( km / s ) \quad\left(V_A^{(v)}=22.162  km / s \right)

Using the state vector \left( R _1, V_D^{(v)}\right), we employ Algorithm 4.2 to find the orbital elements of the heliocentric transfer trajectory.

h = 4.8456 \left(10^6\right) km²/s
e = 0.20581
Ω = 44.898°
i = 1.6622°
ω = 19.973°
\theta_1 = 340.04°
a = 1.8475 \left(10^8\right) km

Step 3:
At departure, the hyperbolic excess velocity is

\left. v _{\infty}\right)_{\text {Departure }}= V _D^{(v)}- V _{\text {earth }}=-2.9138 \hat{ I }+0.79525 \hat{ J }+0.94796 \hat{ K }  ( km / s )

Therefore, the hyperbolic excess speed is

\left.\left.v_{\infty}\right)_{\text {Departure }}=\| v _{\infty}\right)_{\text {Departure }} \|=3.1656  km / s                     (a)

Likewise, at arrival

\left. v _{\infty}\right)_{\text {Arrival }}= V _A^{(v)}- V _{\text {Mars }}=-2.58805 \hat{ I }+0.023514 \hat{ J }+0.16254 \hat{ K }  ( km / s )

so that

\left.\left. v _{\infty}\right)_{\text {Arrival }}=\| v _{\infty}\right)_{\text {Arrival }} \|=2.8852  km / s                           (b)