Question

A large stone of mass $\frac{M_e}{2}$ is released when centre of mass of the stone is at a height $h(h<<R_e)$. Find speed of stone when it is at a height of $\frac{h}{2}$. $M_e$ and $R_e$ are mass and radius of earth. Given $h=\frac{3}{20}m.$

Answer 1

Given , $h<<R_e$, Let speed of stone is $v$.

Conserve the $\text{Mechanical Energy}=\text{Potential Energy(PE) + Kinetic Energy(KE)}$

$\Rightarrow P{E}_1+K{E}_1=P{E}_2+K{E}_2$

$\Rightarrow M_{e}gh+0=\frac{M_{e}gh}{2}+\frac{1}{2}{M}_e{v}^2$ $\Rightarrow v=\sqrt{gh}$

$\Rightarrow v=\sqrt{\frac{3g}{20}}$