First slide

Motion of planets and satellites

Question

If an artificial satellite is moving in a circular orbit around earth with speed equal to one fourth of $V_{e}$ from earth, then height of the satellite above the surface of the earth is

Easy

A

7R
B

4R
C

3R
D

R

Solution

$v = {\frac{v_{e}}{4}}_{0} \Rightarrow \sqrt{\frac{G M}{R + h}} = \frac{\sqrt{\frac{2 G M}{R}}}{4} S q u a r i n g o n b o t h s i d e s w e g e t \Rightarrow \frac{G M}{R + h} = \frac{\frac{2 G M}{R}}{16} \Rightarrow \frac{1}{R + h} = \frac{2}{16 R} \Rightarrow 8 R = R + h \Rightarrow h = 7 R$

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