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

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

$\begin{aligned}\mu_{\text {sun }} & =1.3271 \times 10^{11}  km ^3 / s ^2 \\\mu_{\text {Venus }} & =324,900  km ^3 / s ^2 \\R_{\text {earth }} & =149.6 \times 10^6  km \\R_{\text {Venus }} & =108.2 \times 10^6  km \\r_{\text {Venus }} & =6052  km\end{aligned}$

Preflyby ellipse (orbit 1)
Evaluating the orbit formula, Eqn (2.45), at aphelion of orbit 1 yields

$\boxed{r=\cfrac{h^2}{\mu} \cfrac{1}{1+e \cos \theta}}$                           (2.45)

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

Thus,

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

At intercept,

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

Substituting Eqn (a) and $\theta_1=-30^{\circ}$ and solving the resulting expression for $e_1$ leads to

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

With this result, Eqn (a) yields

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

Now we can use Eqns (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.

$\boxed{h=r ν_{\perp}}$                            (2.31)

$\boxed{ν_r=\cfrac{\mu}{h} e \sin \theta}$                             (2.49)

The flight path angle, from Eqn (2.51), is

$\tan \gamma=\cfrac{ν_r}{ν_{\perp}}$                          (2.51)

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

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

$V_1^{(v)}=\sqrt{v_{n_1}^2+v_{\perp_1}^2}=\sqrt{(-2.782)^2+37.51^2}=37.62  km / s$                  (b)

Flyby hyperbola
From Eqns (8.75) and (8.77), we obtain

$V _1^{(ν)}=\left[V_1^{(ν)}\right]_V \hat{ u }_V+\left[V_1^{(ν)}\right]_S \hat{ u }_S$                                               (8.75)

$\left[V_1^{(ν)}\right]_V=V_{\perp_1}\left[V_1^{(ν)}\right]_S=-V_{r_1}$                             (8.77)

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

The velocity of Venus in its presumed circular orbit around the sun is

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

$r_p=r_{\text {Venus }}+300=6352  km$

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

$e=1+\cfrac{r_p ν_{\infty}^2}{\mu_1}$                              (8.38)

 

$h=r_p \sqrt{ν_{\infty}^2+\cfrac{2 \mu_1}{r_p}}$                               (8.39)

 

The turn angle and true anomaly of the asymptote are

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

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

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

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

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

Finally, from Eqn (d) we obtain the angle between $v _{\infty_1} \text { and } V \text {, }$

$\phi_1=\tan ^{-1} \cfrac{2.782}{2.490}=48.17^{\circ}$                  (f)

There are two flyby approaches, as shown in Figure 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°).
Dark side approach
According to Eqn (8.85), the angle between $v _{\infty} \text { and } V _{\text {Venus }}$ at the outbound crossing is

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

Hence, by Eqn (8.86),

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

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

Using this and Eqn (c) above, 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 Eqn (8.89) that

$V_{\perp_2}=\left[V_2^{(ν)}\right]_V \quad V_{r_2}=-\left[V_2^{(ν)}\right]_S$                              (8.89)

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

The speed of the spacecraft at the outbound crossing is

$v_2^{(v)}=\sqrt{v_{t_2}^2+v_{\perp_2}^2}=\sqrt{(-1.766)^2+31.73^2}=31.78  km / s$

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 Figure 8.20, the angular momentum is found using Eqn (8.90)

$h_2=R V_{\perp_2}$                             (8.90)

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

From Eqn (8.91),

$R=\cfrac{h_2^2}{\mu_{\text {sun }}} \cfrac{1}{1+e_2 \cos \theta_2}$                                (8.91)

and from Eqn (8.92)

$V_{r_2}=\cfrac{\mu_{\text {sun }}}{h_2} e_2 \sin \theta_2$                            (8.92)

$e \sin \theta_2=\cfrac{V_{r_2} h_2}{\mu_{\text {sun }}}=\cfrac{-1.766 \cdot 3.434 \times 10^9}{1.327 \times 10^{11}}=-0.04569$                                           (j)

Thus

$\tan \theta_2=\cfrac{e \sin \theta_2}{e \cos \theta_2}=\cfrac{-0.045,69}{-0.1790}=0.2553$                            (k)

which means

$\theta_2=14.32^{\circ} \text { or } 194.32^{\circ}$                     (l)

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

$\theta_2=194.32^{\circ}$                               (m)

With this value of $\theta_2$, we can use either Eqn (i) or Eqn (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 Figure 8.20. The perihelion radius is given by Eqn (2.50),

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

which is well within the orbit of Venus.
Sunlit side approach
In this case, the angle between $v _{\infty} \text { and } V _{\text {Venus }}$ 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

$V_{\perp_2}=37.14  km / s \quad V_{r_2}=3.074  km / s$

The speed of the spacecraft at the outbound crossing is

$V_2^{(v)}=\sqrt{3.074^2+v_{\perp 2}^2}=\sqrt{3.050^2+37.14^2}=37.27  km / s$

This speed is just 0.348 km/s less than 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 Figure 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 orbit2in Figure 8.21, werepeat Steps (h) through (n) above.

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

$\theta_2=36.76^{\circ}$                       (q)

With this value of $\theta_2$, we can use either Eqn (o) or Eqn (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 Figure 8.21. The perihelion radius is

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