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

(A) Given: $r_p = 2R$ and $r_a = 6R$.
Using the properties of an ellipse: $r_a = a(1+e) = 6R$ and $r_p = a(1-e) = 2R$.
Dividing the two equations: $\frac{1+e}{1-e} = \frac{6R}{2R} = 3 \implies 1+e = 3 - 3e \implies 4e = 2 \implies e = 0.5$.
Using conservation of angular momentum: $m v_p r_p = m v_a r_a \implies v_a = v_p \frac{r_p}{r_a} = v_p \frac{2R}{6R} = \frac{v_p}{3}$.
Using conservation of energy: $\frac{1}{2} m v_p^2 - \frac{GMm}{r_p} = \frac{1}{2} m v_a^2 - \frac{GMm}{r_a}$.
Substituting $v_a = v_p/3$: $\frac{1}{2} v_p^2 (1 - 1/9) = GM (\frac{1}{2R} - \frac{1}{6R}) = GM (\frac{2}{6R}) = \frac{GM}{3R}$.
$\frac{4}{9} v_p^2 = \frac{GM}{3R} \implies v_p^2 = \frac{3GM}{4R} \implies v_p = \sqrt{\frac{3 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{4 \times 6400 \times 10^3}} \approx 5.59 \, km/s$.
Then $v_a = v_p/3 \approx 1.86 \, km/s$.
To transfer to a circular orbit of radius $6R$,the satellite must be at apogee $(r=6R)$ and its velocity must be increased to the orbital velocity $v_c = \sqrt{\frac{GM}{6R}} \approx 2.64 \, km/s$.