Question 8.8: A spacecraft departs earth’s sphere of influence on November......
A spacecraft departs earth’s sphere of influence on November 7, 1996 (0 h UT), on a prograde coasting flight to Mars, arriving at Mars’ sphere of influence on September, 12, 1997 (0 h UT). Use Algorithm 8.2 to determine the trajectory and then compute the hyperbolic excess velocities at departure and arrival.
Step 1:
Algorithm 8.1 yields the state vectors for earth and Mars.
Step 2:
The position vector R_1 of the spacecraft at crossing the earth’s sphere of influence is just that of the earth,
On arrival at Mars’ sphere of influence, the spacecraft’s position vector is
R _2= R _{\text {Mars }}=-2.0858 \times 10^7 \hat{ I }-2.1842 \times 10^8 \hat{J }-4.06244 \times 10^6 \hat{ K } ( km )According to Eqns (5.47) and (5.48)
JD =J_0+\cfrac{ UT }{24} (5.47)
J_0=367 y-\operatorname{INT}\left\{\cfrac{7\left[y+\operatorname{INT}\left(\cfrac{m+9}{12}\right)\right]}{4}\right\}+\operatorname{INT}\left(\cfrac{275 m}{9}\right)+d+1,721,013.5 (5.48)
\begin{aligned}& J D_{\text {Departure }}=2,450,394.5 \\& J D_{\text {Arrival }}=2,450,703.5\end{aligned}Hence, the time of flight is
t_{12}=2,450,703.5-2,450,394.5=309 \text { days }Entering R_1, R_2, and t_{12} into Algorithm 5.2 yields
Using the state vector R_1,V^{(v)}_ D , we employ Algorithm 4.2 to find the orbital elements of the transfer trajectory.
\boxed{\begin{aligned}& h=4.8456 \times 10^6 km ^2 / s \\& e=0.20581 \\& Ω=44.898^{\circ} \\& i=1.6622^{\circ} \\& \omega=19.973^{\circ} \\& \theta_1=340.04^{\circ} \\& a=1.8475 \times 10^8 km \end{aligned}}Step 3:
At departure, the hyperbolic excess velocity is
Therefore, the hyperbolic excess speed is
\left.\left. v _{\infty}\right)_{\text {Departure }}=\| v _{\infty}\right)_{\text {Departure }} \|=\boxed{3.1656 km / s} (a)
Likewise, at arrival
\left. v _{\infty}\right)_{\text {Arrival }}= V _A^{(v)}- V _{\text {Mars }}=-2.8805 \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 }} \|=\boxed{2.8852 km / s} (b)