Question

A large stone of mass Me/2 is released when centre of mass of the stone is at a height h(h<<Re). Find speed of stone when it is at a height of h/2. Me and Re are mass and radius of earth. Given h=3/20m

Answer 1

mass of large stone, $\frac{M_e}{2}$ is released when centre of mass of the stone is at a height h.

gravitational potential energy between earth and large stone at height h , $P.E_i=-\frac{GM_e\frac{M_e}{2}}{h}$

gravitational potential energy between earth and stone at height h/2 , $P.E_f=-\frac{GM_e\frac{M_e}{2}}{\frac{h}{2}}$

change in potential = change in kinetic energy
$P.E_f-P.E_f=\frac{1}{2}\frac{M_e}{2}v^2$

$\frac{-GM_e^2}{2h}-\frac{-GM_e^2}{h}=\frac{1}{4}M_ev^2$

$\frac{GM_e^2}{2h}=\frac{1}{4}M_ev^2$

$\frac{2GM_e}{h}=v^2$

$v=\sqrt{\frac{2GM_e}{h}}$

hence, speed of stone when it is at a height of h/2 , $v=\sqrt{\frac{2GM_e}{h}}$