Question 8.9: In Example 8.8, calculate the delta-v required to launch the......
In Example 8.8, calculate the delta-v required to launch the spacecraft onto its cruise trajectory from a 180 km circular parking orbit. Sketch the departure trajectory.
Recall that
\begin{aligned}& r_{\text {earth }}=6378 km \\& \mu_{\text {earth }}=398,600 km ^3 / s ^2\end{aligned}The radius to periapsis of the departure hyperbola is the radius of the earth plus the altitude of the parking orbit,
r_p=6378+180=6558 kmSubstituting this and Eqn (a) from Example 8.8 into Eqn (8.40) we get the speed of the spacecraft at periapsis of the departure hyperbola,
ν_p=\cfrac{h}{r_p}=\sqrt{ν_{\infty}^2+\cfrac{2 \mu_1}{r_p}} (8.40)
v_p=\sqrt{\left.\left[v_{\infty}\right)_{\text {Departure }}\right]^2+\cfrac{2 \mu_{\text {earth }}}{r_p}}=\sqrt{3.1651^2+\cfrac{2 \cdot 398,600}{6558}}=11.47 km / sThe speed of the spacecraft in its circular parking orbit is
v_c=\sqrt{\cfrac{\mu_{ earth }}{r_p}}=\sqrt{\cfrac{398,600}{6558}}=7.796 km / sHence, the delta-v requirement is
\Delta v=v_p-v_c=\boxed{3.674 km / s}The eccentricity of the hyperbola is given by Eqn (8.38),
e=1+\cfrac{r_p ν_{\infty}^2}{\mu_1} (8.38)
e=1+\cfrac{r_p v_{\infty}{ }^2}{\mu_{\text {earth }}}=1+\cfrac{6558 \cdot 3.1656^2}{398,600}=1.165If we assume that the spacecraft is launched from a parking orbit of 28° inclination, then the departure appears as shown in the three-dimensional sketch in Figure 8.29.
