After a Hohmann transfer from earth to Mars, calculate

(a) the minimum delta-v required to place a spacecraft in orbit with a period of 7 h

(b) the periapsis radius

(c) the aiming radius

(d) the angle between periapsis and Mars’ velocity vector.

Question

After a Hohmann transfer from earth to Mars, calculate

(a) the minimum delta-v required to place a spacecraft in orbit with a period of 7 h

(b) the periapsis radius

(c) the aiming radius

(d) the angle between periapsis and Mars’ velocity vector.

Accepted Answer

The following data are required from Tables A.1 and A.2:

[latex]\mu_{ sun }=1.327\left(10^{11} ight) km ^3 / s ^2[/latex]

[latex]\mu_{ Mars }=42,830 km ^3 / s ^2[/latex]

[latex]R_{ ext {earth }}=149.6\left(10^6 ight) km[/latex]

[latex]R_{ ext {Mars }}=227.9\left(10^6 ight) km[/latex]

[latex]r_{ ext {Mars }}[/latex] = 3396 km

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	[latex]1.989 imes 10^{30}[/latex]	25.38d	7.25°	-	-	-	-
Mercury	2440	[latex]330.2 imes 10^{21}[/latex]	58.56d	0.01°	[latex]57.91 imes 10^{6}[/latex]	0.2056	7.00°	87.97d
Venus	6052	[latex]4.869 imes 10^{24}[/latex]	243[latex]d^{a}[/latex]	177.4°	[latex]108.2 imes 10^{6}[/latex]	0.0067	3.39°	224.7d
Earth	6378	[latex]5.974 imes 10^{24}[/latex]	23.9345h	23.45°	[latex]149.6 times 10^{6}[/latex]	0.0167	0.00°	365.256d
(Moon)	1737	[latex]73.48 imes 10^{21}[/latex]	27.32d	6.68°	[latex]384.4 imes 10^{3}[/latex]	0.0549	5.145°	27.322d
Mars	3396	[latex]641.9 imes 10^{21}[/latex]	24.62h	25.19°	[latex]227.9 imes 10^{6}[/latex]	0.0935	1.850°	1.881y
Jupiter	71,490	[latex]1.899 imes 10^{27}[/latex]	9.925h	3.13°	[latex]778.6 imes 10^{6}[/latex]	0.0489	1.304°	11.86y
Saturn	60,270	[latex]568.5 imes 10^{24}[/latex]	10.66h	26.73°	[latex]1.433 imes 10^{9}[/latex]	0.0565	2.485°	29.46y
Uranus	25,560	[latex]86.83 imes 10^{24}[/latex]	17.24[latex]h^{a}[/latex]	97.77°	[latex]2.872 imes 10^{9}[/latex]	0.0457	0.772°	84.01y
Neptune	24,764	[latex]102.4 imes 10^{24}[/latex]	16.11h	28.32°	[latex]4.495 imes 10^{9}[/latex]	0.0113	1.769°	164.8y
(Pluto)	1187	[latex]13.03 imes 10^{21}[/latex]	6.387[latex]d^{a}[/latex]	122.5°	[latex]5.906 imes 10^{9}[/latex]	0.2488	17.16°	247.9y
[latex]^a[/latex]Retrograde

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

(a) The hyperbolic excess speed is found using Eq. (8.4),

[latex]v_{\infty}=\Delta V_A=\sqrt{\frac{\mu_{ ext {sun }}}{R_{ ext {Mars }}}}\left(1-\sqrt{\frac{2 R_{ ext {earth }}}{R_{ ext {carth }}+R_{ ext {Mars }}}} ight)=\sqrt{\frac{1.327\left(10^{11} ight)}{227.9\left(10^6 ight)}}\left(1-\sqrt{\frac{2 \cdot 149.6\left(10^6 ight)}{149.6\left(10^6 ight)+227.9\left(10^6 ight)}} ight)[/latex]

[latex] herefore v_{\infty}[/latex] = 2.648 km/s

We can use Eq. (2.83) to express the semimajor axis a of the capture orbit in terms of its period T,

[latex]a=\left(\frac{T \sqrt{\mu_{ Mars }}}{2 \pi} ight)^{2 / 3}[/latex]

Substituting T = 7 · 3600 s yields

[latex]a=\left(\frac{25,200 \sqrt{42,830}}{2 \pi} ight)^{2 / 3}=8832 km[/latex]

From Eq. (2.73) we obtain

[latex]a=\frac{r_p}{1-e}[/latex]

On substituting the optimal periapsis radius (Eq. 8.67) this becomes

[latex]a=\frac{2 \mu_{ Mars }}{v_{\infty}^2} \frac{1}{1+e}[/latex]

from which

[latex]e=\frac{2 \mu_{ Mars }}{a v_{\infty}}-1=\frac{2 \cdot 42,830}{8832 \cdot 2.648^2}-1=0.3833[/latex]

Thus, using Eq. (8.70), we find

[latex]\Delta v=v_{\infty} \sqrt{\frac{1-e}{2}}=2.648 \sqrt{\frac{1-03833}{2}}=1.470 km / s[/latex]

(b) From Eq. (8.66), we obtain the periapse radius

[latex]r_p=\frac{2 \mu_{ ext {Mars }}}{v_{\infty}^2} \frac{1-e}{1+e}=\frac{2 \cdot 42,830}{2.648^2} \frac{1-0.3833}{1+0.3833}=5447 km[/latex]

(c) The aiming radius is given by Eq. (8.71),

[latex]\Delta=r_p \sqrt{\frac{2}{1-e}}=5447 \sqrt{\frac{2}{1-0.3833}}=9809 km[/latex]

(d) Using Eq. (8.43), we get the angle to periapsis

[latex]\beta=\cos ^{-1}\left(\frac{1}{1+\frac{r_p v_{\infty}^2}{\mu_{ Mars }}} ight)=\cos ^{-1}\left(\frac{1}{1+\frac{5447 \cdot 2.648^2}{42,830}} ight)=58.09^{\circ}[/latex]

Mars, the approach hyperbola, and the capture orbit are shown to scale in Fig. 8.17. The approach could also be made from the dark side of the planet instead of the sunlit side. The approach hyperbola and capture ellipse would be the mirror image of that shown, as is the case in Fig. 8.12.

Question 8.5: After a Hohmann transfer from earth to Mars, calculate (a) t...

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