Question

Calculate the orbital velocity and period of revolution of an artificial satellite of earth moving at altitude of 200 km given is equal to radius of earth is 6400 km and mass of the earth is 6 into 10 raise to 24 andG is equal to 6.7 into 10 raise to minus 1

Answer 1

Given that,

Altitude, h = 200 km

Radius of Earth, $r=6400\ km$

Mass of earth, $M=6\times 10^{24}\ kg$

To find,

The velocity and period of revolution of an artificial satellite.

Solution,

The orbital velocity of a satellite is given by :

$v=\sqrt{\dfrac{GM}{R}}$

R = r + h

$v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6\times 10^{24}}{6400000+200000}}\\\\v=7786.93\ m/s$

For period of revolution, using Kepler's third law of motion as :

$T^2=\dfrac{4\pi^2}{GM}R^3\\\\T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 6\times 10^{24}}\times (6400000+200000)^3\\\\T=\left(28360537.6002\right)^{\dfrac{1}{2}}\\\\T=5325.46\ s$

Therefore, this is the required explanation.

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Orbital velocity and period of revolution

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