Question 8.6: A spacecraft departs earth with a velocity perpendicular to ...

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

\mu_{ sun }=1.327\left(10^{11}\right)  km ^3 / s ^2

\mu_{ Venus }=324,900  km ^3 / s ^2

R_{\text {earth }}=149.6\left(10^6\right)  km

R_{\text {Venus}}=108.2\left(10^6\right)  km

r_{\text {Venus}} = 6052 km

Table A.1 Astronomical data for the sun, the planets, and the moon

Table A.2 Gravitational parameter (μ) and sphere of influence (SOI) radius for the sun, the planets, and the moon

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

R_{\text {earth }}=\frac{h_1^2}{\mu_{\text {sun }}} \frac{1}{1-e_1}

Thus,

h_1^2=\mu_{\text {sun }} R_{\text {earth }}\left(1-e_1\right)                               (a)

At intercept,

R_{\text {Venus }}=\frac{h_1^2}{\mu_{\text {sun }}} \frac{1}{1+e_1 \cos \left(\theta_1\right)}

Substituting Eq. (a) and \theta_1 = – 30° into this expression and solving the resulting expression for e_1leads to

e_1=\frac{R_{\text {earth }}-R_{\text {Venus }}}{R_{\text {earth }}+R_{\text {Venus }} \cos \left(\theta_1\right)}=\frac{149.6\left(10^6\right)-108.2\left(10^6\right)}{149.6\left(10^6\right)+108.2\left(10^6\right) \cos \left(-30^{\circ}\right)}=0.1702

With this result, and Eq. (a) yields

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

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

V_{\perp_1}^{(v)}=\frac{h_1}{R_{\text {Venus }}}=\frac{4.059\left(10^9\right)}{108.2\left(10^6\right)}=37.51  km / s

V_{r_1}^{(v)}=\frac{\mu_{ sun }}{h_1} e_1 \sin \left(\theta_1\right)=\frac{1.327\left(10^{11}\right)}{4.059\left(10^9\right)} \cdot 0.1702 \cdot \sin \left(-30^{\circ}\right)=-2.782  km / s

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

\gamma_1=\tan ^{-1} \frac{V_{r_1}^{(v)}}{V_{\perp_1}^{(v)}}=\tan ^{-1}\left(\frac{-2.782}{37.51}\right)=-4.241^{\circ}

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

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

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

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

\left.\left. V _1^{(v)}=V_1^{(v)}\right)_V \hat{ u }_V+V_1^{(v)}\right)_S \hat{ u }_S                         (8.75)

\left.\left.V_1^{(v)}\right)_V=V_{\perp_1}^{(v)} \quad V_1^{(v)}\right)_S=-V_{r_1}^{(v)}                              (8.77)

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

Hence

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

It follows that

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

The periapsis radius is

r_p=r_{\text {Venus }}+300=6352  km

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

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

e=1+\frac{r_p v_{\infty}{ }^2}{\mu_{\text {Venus }}}=1+\frac{6352 \cdot 3.733^2}{324,900}=1.272

The turn angle and true anomaly of the asymptote are

\delta=2 \sin ^{-1}\left(\frac{1}{e}\right)=2 \sin ^{-1}\left(\frac{1}{1.272}\right)=103.6^{\circ}

\theta_{\infty}=\cos ^{-1}\left(-\frac{1}{e}\right)=\cos ^{-1}\left(-\frac{1}{1.272}\right)=141.8^{\circ}

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

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

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

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

\Delta=a \sqrt{e^2-1}                                      (2.107)

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

\phi_1=\tan ^{-1} \frac{\left.v_{\infty_1}\right)_S}{\left.v_{\infty_1}\right)_V}=\tan ^{-1} \frac{2.782}{2.490}=48.17^{\circ}                                   (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 v_{\infty} and V _{\text {Venus }} at the outbound crossing is

\phi_2=\phi_1+\delta=48.17^{\circ}+103.6^{\circ}=151.8^{\circ}

Hence, by Eq. (8.86),

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

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

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

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

It follows from Eq. (8.89) that

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

\left.\left.V_{\perp_2}^{(v)}=V_2^{(v)}\right)_V \quad V_{r_2}^{(v)}=-V_2^{(v)}\right)_S

The speed of the spacecraft at the outbound crossing is

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

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)

h_2=R_{\text {venus }} V_{\perp_2}^{(v)}=108.2(10)^6 \cdot 31.73=3.434\left(10^9\right)  \left( km ^2 / s \right)                        (h)

From Eq. (8.91),

e \cos \theta_2=\frac{h_2^2}{\mu_{\text {sun }} R_{ V \text { enus }}}-1=\frac{\left[3.434\left(10^9\right)\right]^2}{1.327\left(10^{11}\right) \cdot 108.2\left(10^6\right)}-1=-0.1790                    (i)

and from Eq. (8.92)

e \sin \theta_2=\frac{V_{r_2}^{(v)} h_2}{\mu_{\text {sun }}}=\frac{-1.766 \cdot 3.434(10)^9}{1.327\left(10^{11}\right)}=-0.04569                              (j)

Thus

\tan \theta_2=\frac{e \sin \theta_2}{e \cos \theta_2}=\frac{-0.04569}{-0.1790}=0.2553                            (k)

which means

\theta_2 = 14.32°   or    194.32°                      (l)

But \theta_2 must lie in the third quadrant since, according to Eqs. (i) and (j), both the sine and cosine are negative.

\theta_2 = 194.32°                                     (m)

With this value of \theta_2, we can use either Eq. (i) or Eq. (j) to calculate the eccentricity,

e_2 = 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),

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

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

\phi_2=\phi_1-\delta=48.17^{\circ}-103.6^{\circ}=-55.44^{\circ}

Therefore,

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

The spacecraft’s heliocentric velocity at the outbound crossing is

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

which means

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

The speed of the spacecraft at the outbound crossing is

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

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.

h_2=R_{\text {Venus }} V_{\perp_2}^{(v)}=108.2\left(10^6\right) \cdot 37.14=4.019\left(10^9\right)  \left( km ^2 / s \right)

e \cos \theta_2=\frac{h_2^2}{\mu_{ sun } R_{ Venus }}-1=\frac{\left[4.019\left(10^9\right)\right]^2}{1.327\left(10^{11}\right) \cdot 108.2\left(10^6\right)}-1=0.1246                    (o)

e \sin \theta_2=\frac{V_{r_2}^{(v)} h_2}{\mu_{\operatorname{sun}}}=\frac{3.074 \cdot 4.019\left(10^9\right)}{1.327\left(10^{11}\right)}=0.09309                      (p)

\tan \theta_2=\frac{e \sin \theta_2}{e \cos \theta_2}=\frac{0.09309}{0.1246}=0.7469 \Rightarrow \theta_2=36.76^{\circ} \text { or } 216.76^{\circ}

\theta_2 must lie in the first quadrant since both the sine and cosine are positive. Hence,

\theta_2 = 36.76°                             (q)

With this value of \theta_2, we can use either Eq. (o) or (p) to calculate the eccentricity,

e_2 = 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

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

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