Solve the following :

Question:

A small block of superdense material has a mass of $3 \times 10^{24} \mathrm{~kg}$. It is situated at a height $h$ (much smaller than the earth's surface) from where it falls on the earth's surface. Find its speed when its height from the earth's surface has reduced to $\mathrm{h} / 2$. The mass of the earth is $3 \times 10^{24} \mathrm{~kg}$.

Solution:

b-block

e-earth

Use C.O.L.M

$M_{e} V_{e}=m_{b} V_{b}$

$M_{e} V_{e}=m_{b} V_{b}$

$V_{e}=\frac{m_{b} V_{B}}{M}$

Use C.O.E.L

$G M_{e} m_{b}\left[\frac{1}{R+\left(\frac{h}{2}\right)}-\frac{1}{R+h}\right]=\frac{1}{2} M_{e} V_{e}^{2}+\frac{1}{2} m_{b} V_{b}^{2}$

$\Rightarrow G M_{e} m_{b}\left[\frac{2}{2 R+h}-\frac{1}{R+h}\right]=\frac{1}{2} M_{e} \frac{m_{b}^{2} V_{b}^{2}}{M_{e}^{2}}+\frac{1}{2} m_{b} V_{b}^{2}$

$=\frac{1}{2} m_{b} V_{b}^{2}\left(\frac{m_{b}}{M_{e}}+1\right)$

$\Rightarrow G M_{e}\left[\frac{h}{2 R^{2}+3 R h+h^{2}}\right]=\frac{1}{2} V_{b}^{2}\left(\frac{3}{2}\right)$

$\mathrm{h}<<\mathrm{R}$

$\Rightarrow \frac{G M_{e} h}{2 R^{2}}=\frac{1}{2} V_{b}^{2}\left(\frac{3}{2}\right) \Rightarrow V_{B}=\sqrt{\frac{2 g h}{3}}$

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