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
r_{\text {earth }}=6378 km
\mu_{\text {earth }}=398,600 km ^3 / s ^2
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 km
Substituting this and Eq. (a) from Example 8.8 into Eq. (8.40) we get the speed of the spacecraft at periapsis of the departure hyperbola,
\left.ν_p\right)_{\text {Departure }}=\sqrt{\left.\left[ν_{\infty}\right)_{\text {Departure }}\right]^2+\frac{2 \mu_{\text {earth }}}{r_p}}=\sqrt{3.1651^2+\frac{2.398,600}{6558}}=11.47 km / s
The speed of the spacecraft in its circular parking orbit is
ν_c=\sqrt{\frac{\mu_{ earth }}{r_p}}=\sqrt{\frac{398,600}{6558}}=7.796 km / s
Hence, the delta-v requirement is
\left.\Delta ν=ν_p\right)_{\text {Departure }}-ν_c=3.674 km / s
The eccentricity of the hyperbola is given by Eq. (8.38),
e=1+\frac{\left.r_p\left[ ν_{\infty}\right)_{\text {Departure }}\right]^2}{\mu_{\text {earth }}}=1+\frac{6558 \cdot 3.1656^2}{398,600}=1.165
If 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 Fig. 8.29.
