First slide

Gravitaional potential energy

Question

An artificial satellite is moving in a circular orbit around the earth at a height fi from the surface of the earth. If the satellite is stopped suddenly in its orbit and allowed to fall freely on the earth, the speed with which it will hit the surface of the earth will be

Difficult

A

$\sqrt{gR}$
B

$\sqrt{(2 gR)}$
C

$\sqrt{(3 gR)}$
D

$2 \sqrt{gR}$

Solution

When the satellite is stopped, its kinetic energy is zero and total energy is only potential energy. This is given by
$E_{1} = - \frac{GMm}{r} = - \frac{GmM}{2 R} (r = R + R)$
Potential energy of the satellite on the earth
$E_{2} = - \frac{GmM}{R}$
loss of potential energy
$\begin{array}{l} = E_{1} - E_{2} = - \frac{GmM}{2 R} - (- \frac{GmM}{R}) \\ = \frac{GmM}{2 R} \end{array}$
This is converted into kinetic energy. Hence
$\frac{1}{2} {mv}^{2} = \frac{GMm}{2 R}$
or $v^{2} = \frac{GM}{R} = gR$
or $v = \sqrt{(gR)}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App