A satellite is in an elliptic orbit around the earth with aphelion of $6R$ and perihelion of $2R$ where $R = 6400 \,km$ is the radius of the earth. Find eccentricity 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 $6R$ ? $($ $G = 6.67 \times 10^{-11}\,SI$ $\rm {unit}$ and $M = 6 \times 10^{24}\,kg$$)$

Given, $r_{p}=$ radius of perihelion $=2 \mathrm{R}$

$r_{a}=$ radius of aphelion $=6 \mathrm{R}$

Hence, we can write,

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

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

Solving equ. $(i)$ and $(ii)$, we get

eccentricity, $e=\frac{1}{2}$

Angular momentum remains unchanged.

$\therefore m v_{p} r_{p}=m v_{a} r_{a}$

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

Energy is same at perigee and apogee,

$\frac{1}{2} m v_{p}^{2}-\frac{\mathrm{GM} m}{r_{p}}=\frac{1}{2} m v_{a}^{2}-\frac{\mathrm{GM} m}{r_{a}}$

$\therefore v_{p}^{2}\left(1-\frac{1}{9}\right)=-2 \mathrm{GM}\left(\frac{1}{r_{a}}-\frac{1}{r_{p}}\right)=2 \mathrm{GM}\left(\frac{1}{r_{p}}-\frac{1}{r_{a}}\right)\left(\right.$ By putting $\left.v_{a}=\frac{v_{p}}{3}\right)$

${p}=\frac{\left[2 \mathrm{GM}\left(\frac{1}{r_{p}}-\frac{1}{r_{a}}\right)\right]^{\frac{1}{2}}}{\left[1-\left(\frac{v_{a}}{v_{p}}\right)^{2}\right]^{\frac{1}{2}}}=\left[\frac{\frac{2 \mathrm{GM}}{\mathrm{R}}\left(\frac{1}{2}-\frac{1}{6}\right)}{\left(1-\frac{1}{9}\right)}\right]^{\frac{1}{2}}$ $=\left(\frac{\frac{2}{3} \mathrm{GM}}{\frac{8}{9} \mathrm{R}}\right)^{\frac{1}{2}}=\sqrt{\frac{3}{4} \frac{\mathrm{GM}}{\mathrm{R}}}=6.85 \mathrm{~km} / \mathrm{s}$ 

$v_{p}=6.85 \mathrm{~km} / \mathrm{s}, v_{a}=2.28 \mathrm{~km} / \mathrm{s}$