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## Q. 6.13

A spacecraft is in a 500 km by 10 000 km altitude geocentric orbit which intersects the equatorial plane at a true anomaly of 120º (see Figure 6.34). If the inclination to the equatorial plane is 15º, what is the minimum velocity increment required to make this an equatorial orbit?

## Verified Solution

The orbital parameters are

$e=\frac{r_{A}-r_{P}}{r_{A}+r_{P}}=\frac{(6378 + 10 000) − (6378 + 500)}{(6378 + 10 000) + (6378 + 500)}= 0.4085$

$r_{P}=\frac{h^{2}}{\mu }\frac{1}{1+e\cos (0)}\Rightarrow 6878 =\frac{h^{2}}{398600}\frac{1}{1+0.4085}\Rightarrow h = 62 141 km/s$

The radial coordinate and velocity components at points B and C, on the line of intersection with the equatorial plane, are

$r_{B}=\frac{h^{2}}{\mu }\frac{1}{1+e\cos \theta _{B}}=\frac{62141^{2}}{398600}\frac{1}{1+0.4085\cdot \cos120^{\circ}}=12 174 km$

$v_{\bot _{B}}=\frac{h}{r_{B}}= \frac{62141}{12174}= 5.1043 km/s$

$v_{r _{B}}=\frac{\mu }{h}e\sin \theta _{B}=\frac{398600}{62141}· 0.4085 ·\sin 120^{\circ}= 2.2692 km/s$

and

$r_{C}=\frac{h^{2}}{\mu }\frac{1}{1+e\cos \theta _{C}}=\frac{62141^{2}}{398600}\frac{1}{1+0.4085\cdot \cos300^{◦}}=8044.6 km$

$v_{\bot _{C}}=\frac{h}{r_{C}}= \frac{62141}{8044.6}= 7.7246 km/s$

$v_{r _{C}}=\frac{\mu }{h} e\sin \theta _{C}=\frac{398600}{62141}· 0.4085 ·\sin 300^{◦} = −2.2692 km/s$

All we wish to do here is rotate the plane of the orbit rigidly around the node line BC. The impulsive maneuver must occur at either B or C. Equation 6.19 applies, and since the radial and perpendicular velocity components remain fixed, it reduces to

$\Delta v=\sqrt{(v_{r_{2}}-v_{r_{1}})^{2}+v^{2}_{\bot _{1}} +v^{2}_{\bot _{2}} -2v_{\bot _{1}}v_{\bot _{2}}\cos δ}$ (6.19)

$\Delta v=v_{\bot }\sqrt{2(1-\cosδ )}$

where δ = 15°. For the minimum Δv, the maneuver must be done where $v_{⊥}$ is smallest, which is at B, the point farthest from the center of attraction F. Thus,

$\Delta v=5.1043\sqrt{2(1-\cos 15^{◦})}= 1.3325 km/s$