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

Question

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.

Accepted Answer

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

[latex]R _{ ext {earth }}=1.0499\left(10^8 ight) \hat{ I }+1.0465\left(10^8 ight) \hat{ J }+716.93 \hat{ K } ( km ) \quad\left(R_{ ext {earth }}=1.4824\left(10^8 ight) km ight)[/latex]

[latex]V _{ ext {earth }}=-21.515 \hat{ I }+20.958 \hat{ J }+0.00014376 \hat{ K } ( km / s ) \quad\left(V_{ ext {earth }}=30.055 km / s ight)[/latex]

[latex]R _{ ext {Mars }}=-2.0858\left(10^7 ight) \hat{ I }-2.1842\left(10^8 ight) \hat{ J }+4.06244\left(10^6 ight) \hat{ K } ( km ) \quad\left(R_{ ext {Mars }}=2.1945\left(10^8 ight) km ight)[/latex]

[latex]V _{ ext {Mars }}=25.037 \hat{ I }+0.22311 \hat{ J }-0.62018 \hat{ K } ( km / s ) \quad\left(V_{ ext {Mars }}=25.046 km / s ight)[/latex]

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

[latex]R _1= R _{ ext {earth }}=1.0499\left(10^8 ight) \hat{ I }+1.0465\left(10^8 ight) \hat{ J }+716.93 \hat{ K } ( km )[/latex]

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

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

According to Eqs. (5.47) and (5.48)

[latex]J D_{ ext {Departure }}=2,450,394.5[/latex]

[latex]J D_{ ext {Arrival }}=2,450,703.5[/latex]

[latex]J D=J_0+\frac{U T}{24}[/latex] (5.47)

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

Hence, the time of flight is

[latex]t_{12}[/latex] = 2,450,703.5 - 2,450,394.5 = 309 days

Entering [latex]R_1[/latex], [latex]R_2[/latex], and [latex]t_{12}[/latex] into Algorithm 5.2 yields

[latex]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 ight)[/latex]

[latex]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 ight)[/latex]

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

h = 4.8456 [latex]\left(10^6 ight)[/latex] km²/s
e = 0.20581
Ω = 44.898°
i = 1.6622°
ω = 19.973°
[latex] heta_1[/latex] = 340.04°
a = 1.8475 [latex]\left(10^8 ight)[/latex] km

Step 3:
At departure, the hyperbolic excess velocity is

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

Therefore, the hyperbolic excess speed is

[latex]\left.\left.v_{\infty} ight)_{ ext {Departure }}=\| v _{\infty} ight)_{ ext {Departure }} \|=3.1656 km / s[/latex] (a)

Likewise, at arrival

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

so that

[latex]\left.\left. v _{\infty} ight)_{ ext {Arrival }}=\| v _{\infty} ight)_{ ext {Arrival }} \|=2.8852 km / s[/latex] (b)

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