A spacecraft departs earth with a velocity perpendicular to the sun line on a flyby mission to Venus. Encounter occurs at a true anomaly in the approach trajectory of – 30°. Periapsis altitude is to be 300 km.

(a) For an approach from the dark side of the planet, show that the postflyby orbit is as illustrated in Fig. 8.20.

(b) For an approach from the sunlit side of the planet, show that the postflyby orbit is as illustrated in Fig. 8.21.

Question

A spacecraft departs earth with a velocity perpendicular to the sun line on a flyby mission to Venus. Encounter occurs at a true anomaly in the approach trajectory of - 30°. Periapsis altitude is to be 300 km.

(a) For an approach from the dark side of the planet, show that the postflyby orbit is as illustrated in Fig. 8.20.

(b) For an approach from the sunlit side of the planet, show that the postflyby orbit is as illustrated in Fig. 8.21.

Accepted Answer

The following data are found in Tables A.1 and A.2:

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

[latex]\mu_{ Venus }=324,900 km ^3 / s ^2[/latex]

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

[latex]R_{ ext {Venus}}=108.2\left(10^6 ight) km[/latex]

[latex]r_{ ext {Venus}}[/latex] = 6052 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 × 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

Preflyby ellipse (orbit 1)
Evaluating the orbit formula (Eq. 2.45) at aphelion of orbit 1 yields

[latex]R_{ ext {earth }}=\frac{h_1^2}{\mu_{ ext {sun }}} \frac{1}{1-e_1}[/latex]

Thus,

[latex]h_1^2=\mu_{ ext {sun }} R_{ ext {earth }}\left(1-e_1 ight)[/latex] (a)

At intercept,

[latex]R_{ ext {Venus }}=\frac{h_1^2}{\mu_{ ext {sun }}} \frac{1}{1+e_1 \cos \left( heta_1 ight)}[/latex]

Substituting Eq. (a) and [latex] heta_1[/latex] = - 30° into this expression and solving the resulting expression for [latex]e_1[/latex]leads to

[latex]e_1=\frac{R_{ ext {earth }}-R_{ ext {Venus }}}{R_{ ext {earth }}+R_{ ext {Venus }} \cos \left( heta_1 ight)}=\frac{149.6\left(10^6 ight)-108.2\left(10^6 ight)}{149.6\left(10^6 ight)+108.2\left(10^6 ight) \cos \left(-30^{\circ} ight)}=0.1702[/latex]

With this result, and Eq. (a) yields

[latex]h_1=\sqrt{1.327\left(10^{11} ight) \cdot 149.6\left(10^6 ight)(1-0.1702)}=4.059\left(10^9 ight) km ^2 / s[/latex]

Now we can use Eqs. (2.31) and (2.49) to calculate the radial and transverse components of the spacecraft’s heliocentric velocity at the inbound crossing of Venus’ sphere of influence

[latex]V_{\perp_1}^{(v)}=\frac{h_1}{R_{ ext {Venus }}}=\frac{4.059\left(10^9 ight)}{108.2\left(10^6 ight)}=37.51 km / s[/latex]

[latex]V_{r_1}^{(v)}=\frac{\mu_{ sun }}{h_1} e_1 \sin \left( heta_1 ight)=\frac{1.327\left(10^{11} ight)}{4.059\left(10^9 ight)} \cdot 0.1702 \cdot \sin \left(-30^{\circ} ight)=-2.782 km / s[/latex]

The flight path angle, from Eq. (2.51), is

[latex]\gamma_1= an ^{-1} \frac{V_{r_1}^{(v)}}{V_{\perp_1}^{(v)}}= an ^{-1}\left(\frac{-2.782}{37.51} ight)=-4.241^{\circ}[/latex]

The negative sign is consistent with the fact that the spacecraft is flying toward perihelion of the preflyby elliptical trajectory (orbit 1).
The speed of the space vehicle at the inbound crossing is

[latex]V_1^{(v)}=\sqrt{\left(V_{r_1}^{(v)} ight)^2+\left(V_{\perp_1}^{(v)} ight)^2}=\sqrt{(-2.782)^2+37.51^2}=37.62 km / s[/latex] (b)

Flyby hyperbola
From Eqs. (8.75) and (8.77), we obtain

[latex]V _1^{(v)}=37.51 \hat{ u }_V+2.782 \hat{ u }_S ( km / s )[/latex]

[latex]\left.\left. V _1^{(v)}=V_1^{(v)} ight)_V \hat{ u }_V+V_1^{(v)} ight)_S \hat{ u }_S[/latex] (8.75)

[latex]\left.\left.V_1^{(v)} ight)_V=V_{\perp_1}^{(v)} \quad V_1^{(v)} ight)_S=-V_{r_1}^{(v)}[/latex] (8.77)

[latex]V =\sqrt{\frac{\mu_{ ext {sun }}}{R_{ ext {Venus }}}} \hat{ u }_V=\sqrt{\frac{1.327(10)^{11}}{108.2(10)^6}} \hat{ u }_V=35.02 \widehat{ u }_V ( km / s )[/latex] (c)

Hence

[latex]v _{\infty_1}= V _1^{(v)}- V =\left(37.51 \hat{ u }_V+2.782 \hat{ u }_S ight)-35.02 \hat{ u }_V=2.490 \hat{ u }_V+2.782 \hat{ u }_S ( km / s )[/latex] (d)

It follows that

[latex]v_{\infty}=\sqrt{ v _{\infty_1} \cdot v _{\infty_1}}=3.733 km / s[/latex]

The periapsis radius is

[latex]r_p=r_{ ext {Venus }}+300=6352 km[/latex]

Eqs. (8.38) and (8.39) are used to compute the angular momentum and eccentricity of the planetocentric hyperbola.

[latex]h=6352 \sqrt{v_{\infty}{ }^2+\frac{2 \mu_{ ext {Venus }}}{6352}}=6352 \sqrt{3.733^2+\frac{2 \cdot 324,900}{6352}}=68,480 km ^2 / s[/latex]

[latex]e=1+\frac{r_p v_{\infty}{ }^2}{\mu_{ ext {Venus }}}=1+\frac{6352 \cdot 3.733^2}{324,900}=1.272[/latex]

The turn angle and true anomaly of the asymptote are

[latex]\delta=2 \sin ^{-1}\left(\frac{1}{e} ight)=2 \sin ^{-1}\left(\frac{1}{1.272} ight)=103.6^{\circ}[/latex]

[latex] heta_{\infty}=\cos ^{-1}\left(-\frac{1}{e} ight)=\cos ^{-1}\left(-\frac{1}{1.272} ight)=141.8^{\circ}[/latex]

From Eqs. (2.50), (2.103), and (2.107), the aiming radius is

[latex]\Delta=r_p \sqrt{\frac{e+1}{e-1}}=6352 \sqrt{\frac{1.272+1}{1.272-1}}=18,340 km[/latex] (e)

[latex]r_p=\frac{h^2}{\mu} \frac{1}{1+e}[/latex] (2.50)

[latex]a=\frac{h^2}{\mu} \frac{1}{e^2-1}[/latex] (2.103)

[latex]\Delta=a \sqrt{e^2-1}[/latex] (2.107)

Finally, from Eqs. (8.84) and (d) we obtain the angle between [latex]V_{\infty_1}[/latex] and V,

[latex]\phi_1= an ^{-1} \frac{\left.v_{\infty_1} ight)_S}{\left.v_{\infty_1} ight)_V}= an ^{-1} \frac{2.782}{2.490}=48.17^{\circ}[/latex] (f)

There are two flyby approaches, as shown in Fig. 8.22. In the dark-side approach, the turn angle is counterclockwise (+102.9°), whereas for the sunlit-side, approach it is clockwise (- 102.9°).

(a) Dark-side approach
According to Eq. (8.85), the angle between [latex]v_{\infty}[/latex] and [latex]V _{ ext {Venus }}[/latex] at the outbound crossing is

[latex]\phi_2=\phi_1+\delta=48.17^{\circ}+103.6^{\circ}=151.8^{\circ}[/latex]

Hence, by Eq. (8.86),

[latex]v _{\infty_2}=3.733\left(\cos 151.8^{\circ} \hat{ u }_V+\sin 151.8^{\circ} \hat{ u }_S ight)=-3.289 \hat{ u }_V+1.766 \hat{ u }_S ( km / s )[/latex]

[latex]v _{\infty_2}=v_{\infty} \cos \phi_2 \hat{ u }_V+v_{\infty} \sin \phi_2 \hat{ u }_S[/latex] (8.86)

Using this and Eq. (c), we compute the spacecraft’s heliocentric velocity at the outbound crossing.

[latex]V _2^{(v)}= V + v _{\infty_2}=31.73 \hat{ u }_V+1.766 \hat{ u }_S ( km / s )[/latex]

It follows from Eq. (8.89) that

[latex]V_{\perp_2}^{(v)}=31.73 km / s \quad V_{r_2}^{(v)}=-1.766 km / s[/latex] (g)

[latex]\left.\left.V_{\perp_2}^{(v)}=V_2^{(v)} ight)_V \quad V_{r_2}^{(v)}=-V_2^{(v)} ight)_S[/latex]

The speed of the spacecraft at the outbound crossing is

[latex]V_2^{(v)}=\sqrt{\left(V_{r_2}^{(v)} ight)^2+\left(V_{\perp_2}^{(v)} ight)^2}=\sqrt{(-1.766)^2+31.73^2}=31.78 km / s[/latex]

This is 5.83 km/s less than the inbound speed.

Postflyby ellipse (orbit 2) for the dark-side approach
For the heliocentric postflyby trajectory, labeled orbit 2 in Fig. 8.20, the angular momentum is found using Eq. (8.90)

[latex]h_2=R_{ ext {venus }} V_{\perp_2}^{(v)}=108.2(10)^6 \cdot 31.73=3.434\left(10^9 ight) \left( km ^2 / s ight)[/latex] (h)

From Eq. (8.91),

[latex]e \cos heta_2=\frac{h_2^2}{\mu_{ ext {sun }} R_{ V ext { enus }}}-1=\frac{\left[3.434\left(10^9 ight) ight]^2}{1.327\left(10^{11} ight) \cdot 108.2\left(10^6 ight)}-1=-0.1790[/latex] (i)

and from Eq. (8.92)

[latex]e \sin heta_2=\frac{V_{r_2}^{(v)} h_2}{\mu_{ ext {sun }}}=\frac{-1.766 \cdot 3.434(10)^9}{1.327\left(10^{11} ight)}=-0.04569[/latex] (j)

Thus

[latex] an heta_2=\frac{e \sin heta_2}{e \cos heta_2}=\frac{-0.04569}{-0.1790}=0.2553[/latex] (k)

which means

[latex] heta_2[/latex] = 14.32° or 194.32° (l)

But [latex] heta_2[/latex] must lie in the third quadrant since, according to Eqs. (i) and (j), both the sine and cosine are negative.

[latex] heta_2[/latex] = 194.32° (m)

With this value of [latex] heta_2[/latex], we can use either Eq. (i) or Eq. (j) to calculate the eccentricity,

[latex]e_2[/latex] = 0.1847 (n)

Perihelion of the departure orbit lies 194.32° clockwise from the encounter point (so that aphelion is 14.32° therefrom), as illustrated in Fig. 8.20. The perihelion radius is given by Eq. (2.50),

[latex]R_{ ext {perihelion }}=\frac{h_2^2}{\mu_{ ext {sun }}} \frac{1}{1+e_2}=\frac{\left[3.434\left(10^9 ight) ight]^2}{1.327\left(10^{11} ight)} \frac{1}{1+0.1847}=74.98\left(10^6 ight) km[/latex]

which is well within the orbit of Venus.
(b) Sunlit-side approach
In this case, the angle between [latex]v_{\infty}[/latex] and [latex]V _{ ext {Venus }}[/latex] at the outbound crossing is

[latex]\phi_2=\phi_1-\delta=48.17^{\circ}-103.6^{\circ}=-55.44^{\circ}[/latex]

Therefore,

[latex]v _{\infty_2}=3.733\left[\cos \left(-55.44^{\circ} ight) \hat{ u }_V+\sin \left(-55.44^{\circ} ight) \hat{ u }_S ight]=2.118 \hat{ u }_V-3.074 \hat{ u }_S ( km / s )[/latex]

The spacecraft’s heliocentric velocity at the outbound crossing is

[latex]V _2^{(v)}= V _{ ext {Venus }}+ v _{\infty_2}=37.14 \hat{ u }_V-3.074 \hat{ u }_S ( km / s )[/latex]

which means

[latex]V_{\perp_2}^{(v)}=37.14 km / s \quad V_{r_2}^{(v)}=3.074 km / s[/latex]

The speed of the spacecraft at the outbound crossing is

[latex]V_2^{(v)}=\sqrt{\left(V_{r_2}^{(v)} ight)^2+\left(V_{\perp_2}^{(v)} ight)^2}=\sqrt{3.050^2+37.14^2}=37.27 km / s[/latex]

This speed is just 0.348 km/s less than the inbound crossing speed. The relatively small speed change is due to the fact that the apse line of this hyperbola is nearly perpendicular to Venus’ orbital track, as shown in Fig. 8.23. Nevertheless, the periapses of both hyperbolas are on the leading side of the planet.

Postflyby ellipse (orbit 2) for the sunlit-side approach
To determine the heliocentric postflyby trajectory, labeled orbit 2 in Fig. 8.21, we repeat Steps (h) through (n) above.

[latex]h_2=R_{ ext {Venus }} V_{\perp_2}^{(v)}=108.2\left(10^6 ight) \cdot 37.14=4.019\left(10^9 ight) \left( km ^2 / s ight)[/latex]

[latex]e \cos heta_2=\frac{h_2^2}{\mu_{ sun } R_{ Venus }}-1=\frac{\left[4.019\left(10^9 ight) ight]^2}{1.327\left(10^{11} ight) \cdot 108.2\left(10^6 ight)}-1=0.1246[/latex] (o)

[latex]e \sin heta_2=\frac{V_{r_2}^{(v)} h_2}{\mu_{\operatorname{sun}}}=\frac{3.074 \cdot 4.019\left(10^9 ight)}{1.327\left(10^{11} ight)}=0.09309[/latex] (p)

[latex] an heta_2=\frac{e \sin heta_2}{e \cos heta_2}=\frac{0.09309}{0.1246}=0.7469 \Rightarrow heta_2=36.76^{\circ} ext { or } 216.76^{\circ}[/latex]

[latex] heta_2[/latex] must lie in the first quadrant since both the sine and cosine are positive. Hence,

[latex] heta_2[/latex] = 36.76° (q)

With this value of [latex] heta_2[/latex], we can use either Eq. (o) or (p) to calculate the eccentricity,

[latex]e_2[/latex] = 0.1556

Perihelion of the departure orbit lies 36.76° clockwise from the encounter point as illustrated in Fig. 8.21. The perihelion radius is

[latex]R_{ ext {perihelion }}=\frac{h_2^2}{\mu_{ ext {sum }}} \frac{1}{1+e_2}=\frac{\left[4.019\left(10^9 ight) ight]^2}{1.327\left(10^{11} ight)} \frac{1}{1+0.1556}=105.3\left(10^6 ight) km[/latex]

which is just within the orbit of Venus. Aphelion lies between the orbits of earth and Venus.

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