First slide

Universal law of gravitation

Question

An artificial satellite is moving in a circular orbit around the earth with a speed equal to the half of the escape velocity from the earth of radius r. The height of the satellite above the surface of the earth is

Easy

A

R/2
B

R
C

2R
D

3R

Solution

$\frac{{mv}^{2}}{r} = \frac{GmM}{r^{2}}$
or $V = \sqrt{(\frac{GM}{r})}$ …………(1)
Given that $v = \frac{1}{2} v_{e} = \frac{1}{2} \sqrt{(\frac{2 GM}{R})}$
or $V = \sqrt{(\frac{G M}{2 R})}$ …………(2)
From eqs. (1) and (2), we get
$\sqrt{(\frac{GM}{r})} = \sqrt{(\frac{GM}{2 R})} or r = 2 R$
Height from earth =2 R - R = R

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