Question

An artificial satellite of mass $m$ is moving in circular orbit at a height equal to the radius $R$ of the earth. Suddenly due to internal explosion the satellite breaks into two parts of equal pieces. One part of the satellite stops just after the explosion. The increase in the mechanical energy of the system due to explosion will be (Given: acceleration due to gravity on the surface of earth is $g$)

• A. $mgR$
• B. $\frac{mgR}{2}$
• C. $\frac{mgR}{4}$
• D. $\frac{3mgR}{4}$

Answer 1

The correct option is C $\frac{mgR}{4}$

Just before collision , $v_0=\sqrt{\frac{GM}{r}}=\sqrt{\frac{GM}{R+R}}$ where M be the mass of earth.

So, $v_0=\sqrt{\frac{GM}{2R}}$

Now after collision using conservation of linear momentum, $m{v}_0=\left(m/2\right)0+\left(m/2\right)v$

$v=2v_0$

Thus the increase in mechanical energy of the system

$=K{E}_f-K{E}_i=\left(1/2\right)\left(m/2\right)v^2-\left(1/2\right)m{v}_0^2$

$=\left(1/2\right)m[\frac{4v_0^2}{2}-v_0^2]$

$=\frac{1}{2}m{v}_0^2$

$=\frac{1}{2}m\times \frac{GM}{2R}$

$=\frac{MgR}{4}$ where earth surface gravity, $g=\frac{GM}{R^2}$