Answer to Question #167875 in Mechanics | Relativity for Akshay

Question

A satellite going around Earth in an elliptic orbit has a speed of 10 km s−1 at the perigee which is at a distance of 227 km from the surface of the earth. Calculate the apogee distance and its speed at that point.

Answer 1

$\frac{v^2_a}{2}-\frac{GM}{r_a}=\frac{v^2_p}{2}-\frac{GM}{r_p}\\ v_ar_a=v_pr_p\\ \frac{v^2_a}{2}-\frac{GM}{r_pv_p}v_a=\frac{v^2_p}{2}-\frac{GM}{r_p}\\$

$\frac{v^2_a}{2}-\frac{(6.67\cdot10^{-11})(5.97\cdot10^{24})}{227\cdot10^3(10^4)}v_a\\=\frac{(10^4)^2}{2}-\frac{(6.67\cdot10^{-11})(5.97\cdot10^{24})}{227\cdot10^3}\\$

$v_a=10\frac{km}{s}$

$r_a=227\ km$