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A satellite is in an elliptic orbit around the earth with an aphelion of 6R and perihelion of 2R where R = 6400 km is the radius of the earth. Find eccentrically of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius 6R?

Answers (1)

$r_{p}= 2R$

$r_{a}= 6R$

$r_{p}= a(1-e)=2R$

$r_{a}= a(1+e)=6R$

$\frac{1-e}{1+e}=\frac{2}{6}$

$3-3e=1+e$

$4e=2\Rightarrow e=\frac{1}{2}$

$L_{1}=L_{2}$

$m_{a}v_{a}r_{a}=m_{p}v_{p}r_{p}$

$m_{a}=m_{p}=m=mass\; of\; satellite$

$\frac{v_{a}}{v_{p}}=\frac{r_{p}}{r_{a}}=\frac{2R}{6R}=\frac{1}{3}$

$v_{p}=3v_{a}$

Applying conservation of energy at apogee and perigee

$\frac{1}{2}mv_{p}^{2}-\frac{GMm}{r_{p}}=\frac{1}{2}mv_{a}^{2}-\frac{GMm}{r_{a}}$

Multiplying $\frac{2}{m}$ to both side and putting $r_{p}=2R$ and $r_{a}=6R$

$v_{p}^{2}-\frac{2GM}{2R}=v_{a}^{2}-\frac{2GM}{6R}$

$v_{a}=\frac{v_{p}}{3}$

$v_{p}^{2}-v_{a}^{2}=\frac{GM}{R}-\frac{1}{3}\frac{GM}{R}$

$v_{p}^{2}-\left ( (\frac{v_{p}}{3})^{2} \right )=\frac{2GM}{3R}$

$v_{p}^{2}\left [ 1-\frac{1}{9} \right ]=\frac{2GM}{3R}$

$\frac{8}{9}v_{p}^{2}=\frac{2GM}{3R}$

$v_{p}^{2}=\frac{3GM}{4R}$

$v_{p}=\sqrt{\frac{3GM}{4R}}=\sqrt{\frac{3\times6.67\times10^{-11}\times6\times10^{24}}{4\times6.6\times10^{6}}}=\sqrt{46.89\times10^{6}}$

$v_{p}=6.85\; km/s$

$v_{a}=\frac{v_{p}}{3}=2.28\; km/s$

$v_{c}=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{6R}}=\sqrt{\frac{6.67\times10^{-11}\times6\times10^{24}}{6\times6.6\times10^{6}}}$

$v_{c}=3.23\; km/s$

Thus, to transfer to a circular orbit at apogee, we have to boost the velocity by

$v_{c}-v_{a}=(3.23-2.28)=0.95\; km/s$

Posted by

infoexpert23

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