Question

a satellite is revolving round the earth with orbital speed v₀ if it stops suddenly,the speed with which it will strike the surface of earth would be-( v_e=escape velocity of particle n earth surface)

Answer 1

$$\begin{gathered} TotalEnergyofSatellite=\frac{-G{M}_{e}m}{R} \\ \frac{m{v}_0^2}{R}=\frac{G{M}_{e}m}{R^2} \\ m{v}_0^2=\frac{G{M}_{e}m}{R} \\ EnergyonsurfaceofEarth=\frac{-G{M}_{e}m}{R}+\frac{1}{2}m{v}^2 \\ \frac{1}{2}m{v}^2-\frac{G{M}_{e}m}{R}=-m{v}_0^2 \\ {v}^2=-2v_0^2+\frac{2G{M}_e}{R} \\ {v}^2=-2v_0^2+v_e^2 \\ v=\sqrt{[v_e^2}-2v_0^2] \end{gathered}$$