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## Q. 6.8

An earth satellite in orbit 1 of Figure 6.19 undergoes the indicated delta-v maneuver at its perigee. Determine the rotation η of its apse line.

## Verified Solution

From the figure

$r_{A_{1}}=17000$ Km                      $r_{p_{1}}=7000 Km$

The eccentricity of orbit 1 is

$e_{1}=\frac{r_{A_{1}}-r_{p_{1}}}{r_{A_{1}}+r_{p_{1}}}=0.41667$                   (a)

As usual, we use the orbit equation to find the angular momentum,

$r_{p_{1}}=\frac{h^{2}_{1} }{\mu }\frac{1}{1+e_{1}\cos (0)} \Rightarrow 7000=\frac{h^{2}_{1}}{398600}\frac{1}{1+0.41667}\Rightarrow h_{1}=62871 Km^{2}/s$

At the maneuver point $P_{1}$, the angular momentum formula and the fact that $P_{1}$ is perigee of orbit 1$(θ_{1} = 0)$ imply that

$v_{\bot _{1}}=\frac{h_{1}}{r_{p_{1}}}=\frac{62 871}{7000}=8.9816 km/s$

$v_{r_{1}}=0$               (b)

From Figure 6.18 it is clear that

$\Delta v_{\bot _{1}}=\Delta v\cos 60^{◦}=1$ Km/s

$\Delta v_{r}=\Delta v\sin 60^{◦} =1.7321$ Km/s                 (c)

To compute $θ_{2}$, we use Equation 6.18b together with (a), (b) and (c):

$\tan \theta _{2}=\frac{(v_{\bot _{1}}+\Delta v_{\bot })(v_{r_{1}}+\Delta v_{r})}{(v_{\bot _{1}}+\Delta v_{\bot })^{2}e_{1}\cos \theta _{1}+(2v_{\bot _{1}}+\Delta v_{\bot })\Delta v_{\bot } } \frac{v^{2}_{\bot _{1}} }{\mu /r}$              (6.18b)

$\tan \theta_{2}=\frac{\left(v_{\perp_{1}}+\Delta v_{\perp}\right)\left(v_{r_{1}}+\Delta v_{r}\right)}{\left(v_{\perp_{1}}+\Delta v_{\perp}\right)^{2} e_{1} \cos \theta_{1}+\left(2 v_{\perp_{1}}+\Delta v_{\perp}\right) \Delta v_{\perp}} \frac{v_{\perp_{1}}^{2}}{\left(\mu / r_{P_{1}}\right)}$

$= \frac{(8.9816 + 1)(0 + 1.7321)}{(8.9816 + 1)^{2} · 0.41667·\cos (0)+(2 · 8.9816 + 1)· 1}·\frac{8.9816^{2} }{(398 600/7000)} =0.4050$

It follows that $θ_{2} = 22.047^{◦}$, so that Equation 6.12 yields

$η = θ_{1} − θ_{2}$              (6.12)

$\eta =- 22.05^{◦}$

which means the rotation of the apse line is clockwise, as indicated in Figure 6.19.