A body (mass $\mathrm{m}$ ) starts its motion from rest from a point distant $R_0\left(R_0>R\right)$ from…
A body (mass $\mathrm{m}$ ) starts its motion from rest from a point distant $R_0\left(R_0>R\right)$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of earth will be ( $\mathrm{G}=$ universal constant of gravitation, $\mathrm{M}=$ mass of earth, $\mathrm{R}$ = radius of earth)
According to law of conservation of energy,
$\begin{aligned}
& \quad \frac{1}{2} \mathrm{mv}^2=-\frac{\mathrm{GMm}}{\mathrm{R}_0}-\left(\frac{\mathrm{GMm}}{\mathrm{R}}\right)=\mathrm{GMm}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right) \\
& \therefore \quad \mathrm{v}^2=2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)
\end{aligned}$
$\therefore \quad$ The velocity acquired by the body when it reaches the surface of earth is:
$\mathrm{V}=\left[2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}$