Question

If an artificial satellite is moving in a circular orbit around the earth with a speed equal to half the
magnitude of the escape velocity from the earth, the height of the satellite above the surface of the earth is

Easy

Solution

We know that, $v_{o} = \frac{1}{2} \times v_{e}$ …(i)
$∵ v_{o} = \sqrt{\frac{G M_{e}}{R_{e} + h^{x}}} and v_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}}$
Putting these values in Eq. (i), we get
$\sqrt{\frac{G M_{e}}{R_{e} + h}} = \frac{1}{2} \sqrt{\frac{2 G M_{e}}{R_{e}}}$
$\Rightarrow \frac{G M_{e}}{R_{e} + h} = \frac{1}{2} \times \frac{1}{2} \times \frac{2 G M_{e}}{R_{e}}$
$\Rightarrow R_{e} + h = 2 R_{e}$
$\Rightarrow h = R_{e} or R$

Get Instant Solutions

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

Recieve an sms with download link

SCAN ME