Question

A small block of super dense material has a mass of $3\times {10}^{24}$ kg.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 reduced to $\frac{h}{2}$.The mass of the earth is $6\times {10}^{24}$ kg.

Answer 1

Given h<<R

G mass=$6\times {10}^{24}$ kg

mb=$3\times {10}^{24}$ kg

Let $V_e\to$ Velocity of earth

$V_b\to$ Velocity of the block

The two blocks are attracted by gravitational force of attraction.The gravitational potential energy stored will be the K.E. of two blocks.

$G{M}_e{m}_b\frac{1}{R+\left(\frac{h}{2}\right)}-\frac{1}{R+h}$

$=\left(\frac{1}{2}\right)m_e\times v_e^2+\left(\frac{1}{2}\right)M_b\times V_b^2$

Again As the internal force acts

$M_e{V}_e=M_b{V}_b$

$\Rightarrow V_e=\frac{M_b{V}_b}{M_e}...\left(ii\right)$

Putting in equation (i),

$G{m}_e\times m_b[\frac{2}{2R+h}-\frac{1}{R+h}]$

$=\left(\frac{1}{2}\right)\times M_e\times \frac{M_b^2{V}_b^2}{M_e^2}+\left(\frac{1}{2}\right)\times M_b\times V_b^2$

$=\left(\frac{1}{2}\right)\times V_b^2\frac{M_b}{M_e}+\frac{1}{2}\times M_b\times v_b^2$

$\Rightarrow G{M}_e\frac{2r+2h-2R-h}{(2R+h)(R+h)}$

$=\left(\frac{1}{2}\right)\times V_b^2\times (\frac{3\times {10}^{24}}{6\times {10}^{24}}+1)$

$\Rightarrow \left[\frac{GM\times h}{2R^2}\right]=\left(\frac{1}{2}\right)\times V_b^2\times \left(\frac{3}{2}\right)$

$\Rightarrow gh=V_b^2\times \left(\frac{3}{2}\right)$

$\Rightarrow V_b=\frac{2gh}{3}$