Question

Prove that the orbital velocity of a satellite at a height

h

from the earth's surface is equal to

R_{e} \sqrt{\frac{g}{(R_{e} + h)}},

where

R_{e}

is the radius of earth and

g

is the acceleration due to gravity. Find out the time period of revolution of the satellite.

Accepted Answer

If the mass of earth is $M_{e}$ . The gravitational force exerted by the earth on the satellite of mass $m$ is $F = G \frac{M_{e} m}{r^{2}}$ where $G$ is the gravitational constant.

The gravitational force provides the required centripetal force, we have

$G \frac{M_{e} m}{r^{2}} = \frac{m {v_{o}}^{2}}{r} {GM}_{e} = {v_{o}}^{2} r {v_{o}}^{2} = \frac{{GM}_{e}}{r} v_{o} = \sqrt{\frac{{GM}_{e}}{r}} . . . . . (1)$

Its $R_{e}$ be the radius of earth and $h$ the height of the satellite above the earth's surface. So the distance of the satellite from the center of the earth will be

$r = R_{e} + h$

Substituting the value of $r$ in above equation, we have $v_{o} = \sqrt{G \frac{M_{e}}{R_{e} + h}} . . . . . . (2)$

If the acceleration due to gravity on the earth's surface $g$ , then

$g = G \frac{M_{e}}{{R_{e}}^{2}} or {GM}_{e} = {gR}_{e}^{2}$

Substituting the value of ${GM}_{e}$ in equation (1) we get

$v_{o} = R_{e} \sqrt{\frac{g}{R_{e} + h}} .$

It is the evident from the above equations that the speed of satellite depend only upon its height above the earth.

The time period of the revolving satellite is given by:

$T = \frac{2 πr}{V} = \frac{2 πr}{\sqrt{\frac{G M}{r}}} = \frac{2 {πr}^{\frac{3}{2}}}{\sqrt{G M}} = \frac{2 π (R_{e} + h)^{\frac{3}{2}}}{\sqrt{G M}}$ using equation $1$ here the letters have usual meaning

G L Mittal and TARUN MITTAL Solutions for Chapter: Gravitation : Planets and Satellites, Exercise 1: QUESTIONS

G L Mittal Physics Solutions for Exercise - G L Mittal and TARUN MITTAL Solutions for Chapter: Gravitation : Planets and Satellites, Exercise 1: QUESTIONS

Questions from G L Mittal and TARUN MITTAL Solutions for Chapter: Gravitation : Planets and Satellites, Exercise 1: QUESTIONS with Hints & Solutions

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