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## Q. 6.14

Orbit 1 has angular momentum h and eccentricity e. The direction of motion is shown. Calculate the Δv required to rotate the orbit 90° about its latus rectum BC without changing h and e. The required direction of motion in orbit 2 is shown in Figure 6.35.

## Verified Solution

By symmetry, the required maneuver may occur at either B or C, and it involves a rigid body rotation of the ellipse, so that $v_{r}$ and $v_{⊥}$ remain unaltered. Because of the directions of motion shown, the true anomalies of B on the two orbits are

$\theta _{B_{1}}=-90^{◦}$ $\theta _{B_{2}}=+90^{◦}$

The radial coordinate of B is

$r_{B}=\frac{h^{2}}{\mu }\frac{1}{1+e\cos (\pm 90)}=\frac{h^{2}}{\mu }$

For the velocity components at B, we have

$v_{\bot _{B}})_{1}= v_{\bot _{B}})_{2}=\frac{h}{r_{B}}=\frac{\mu }{h}$

$v_{rB})_{1}=\frac{\mu }{h}e\sin (\theta _{B_{1}})=-\frac{\mu e}{h}$

$v_{rB})_{2}=\frac{\mu }{h}e\sin (\theta _{B_{2}})=\frac{\mu e}{h}$

Substituting these into Equation 6.19, yields

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

$\Delta v_{B}=\sqrt{[v_{r_{B}})_{2} – v_{r_{B}})_{1}]^{2}+v_{\bot_{B}})^{2}_{1}+v_{\bot_{B}})^{2}_{2}-2v_{\bot_{B}})_{1}v_{\bot _{B}})_{2}\cos 90^{◦} }$
$=\sqrt{ [\frac{\mu e}{h}-(-\frac{\mu e}{h}) ]^{2}+(\frac{\mu }{h} ) ^{2}+(\frac{\mu }{h} ) ^{2}-2(\frac{\mu }{h} )(\frac{\mu }{h} )· 0}$

$=\sqrt{4\frac{\mu ^{2}}{h^{2}}e^{2}+2\frac{\mu^{2} }{h^{2}} }$

so that

$\Delta v_{B}=\frac{\sqrt{2}\mu}{h}\sqrt{1+2e^{2}}$           (a)

If the motion on ellipse 2 were opposite to that shown in Figure 6.35, then the radial velocity components at B (and C) would be in the same rather than in the opposite direction on both ellipses, so that instead of (a) we would find a smaller velocity increment,

$\Delta v_{B}=\frac{\sqrt{2}\mu}{h}$