Question

If the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth, find the speedwith which it hits the surface of earth

Answer 1

Increase in kinetic energy=decrease in potential energy
$\therefore$ $\frac{1}{2}m{v}^2=\frac{mgh}{1+\frac{h}{R}}$
$\therefore$ $v=\sqrt{\frac{2hg}{1+\frac{h}{R}}}$
Substituting the values we have,
$v=\sqrt{\frac{2\times 9.81times6400\times {10}^3}{1+\frac{R}{R}}}$
$=7924$ m/s$=7.92$ km/s