Question

A small block of superdense material has a mass of 3 × 1024kg. It is situated at a height h (much smaller than the earth's radius) from where it falls on the earth's surface. Find its speed when its height from the earth's surface has reduce to to h/2. The mass of the earth is 6 × 1024kg.

Answer 1

Explanation:

Given,  

$h<<<R$

mass of the universe,$m_{1}=6 \times 10^{24} \mathrm{kg}$

Block mass, $\mathrm{m}_{2}=3 \times 10^{24} \mathrm{kg}$

Let earth velocity  be $\mathrm{v} 1$

Let block velocty be $v 2$

The gravitational force of attraction draws two points.  

Two blocks of gravitational energy deposited will be kinetic energy.

$\mathrm{m}_{1} \mathrm{m}^{2}\left[\frac{1}{\left(\mathrm{R}+\left(\frac{\mathrm{h}}{2}\right)\right\}}-\frac{1}{(\mathrm{R}+\mathrm{h})}\right]=\frac{1}{2} \mathrm{m}^{1} \mathrm{x} \mathrm{v}_{1}^{2}+\frac{1}{2} \mathrm{m}_{2} \mathrm{x} \mathrm{v}_{2}^{2}$.........equation 1

As an internal force,

$m_{1} v_{1}=m_{2} v_{2}$$\mathrm{v}_{1}=\frac{\mathrm{m}_{2} \mathrm{v}_{2}}{\mathrm{m}_{1}}$.............................equation 2

Putting value in Equation 1

$m_{1} \mathrm{m}_{2}\left[\frac{2}{(2 \mathrm{R}+\mathrm{h})}-\frac{1}{(\mathrm{R}+\mathrm{h})}\right]=\frac{1}{2} \frac{\mathrm{m}_{1} \mathrm{x} \mathrm{m}_{2}^{2} \mathrm{v}_{2}^{2}}{\mathrm{m}_{1}^{2}} \mathrm{x} \mathrm{v}_{1}^{2}+\frac{1}{2} \mathrm{m}_{2} \mathrm{x} \mathrm{v}_{2}^{2}$

$m_{1} m_{2}\left[\frac{2}{(2 R+h)}-\frac{1}{(R+h)}\right]=\frac{1}{2} x m_{2}^{2} x v_{2}^{2}\left[\frac{m^{2}}{m^{2}}+1\right]$

$m_{1} \mathrm{m}_{2}\left[\frac{(2 \mathrm{R}+2 \mathrm{h}-2 \mathrm{R}-\mathrm{h})}{(2 \mathrm{R}+\mathrm{h})(\mathrm{R}+\mathrm{h})}\right]=\frac{1}{2} \mathrm{x} \mathrm{v}_{2}^{2} \mathrm{x}+\frac{3 \times 10^{24}}{6 \times 10^{24}}+1$

$\left[\frac{\left(m_{1} x m_{2} x h\right)}{\left(2 R^{2}+3 R h+h^{2}\right)}\right]=\frac{1}{2} x v_{2}^{2} x+\frac{3}{2}$

    $h<<<R$ , As mentioned above, it can be ignored    

$\left[\frac{\left(\mathrm{m}_{1} \mathrm{x} \mathrm{m}_{2} \mathrm{x} \mathrm{h}\right)}{\left(2 \mathrm{R}^{2}\right)}\right]=\frac{1}{2} \mathrm{x} \mathrm{v}_{2}^{2} \mathrm{x}+\frac{3}{2}$

$\mathrm{v}_{2}=\sqrt{\frac{2 \mathrm{gh}}{3}}$