A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth; R being the radius of the earth. What will be the time period of another satellite at a height 2.5 R from the surface of the earth?

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A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth, where R is the radius of the earth. The time period of another satellite at a height of 2.5R from the surface of the earth is ______________ hours.

The correct option is C 8.5T1=24 hrs R1=6R+R=7R R2=2.5R+R=3.5R Using Kepler's third law, As, T2∝R3 T21T22=R31R32 242T22=(7R)3(3.5R)3 T2=8.5 hrs

A geostationary satellite is orbiting the earth at a height of $6 R$ above the surface of the earth, where $R$ is the radius of the earth. The time period of another satellite at a height of $2.5 R$ from the surface of the earth is ______________ hours.

The correct option is C $8.5$
$T_{1} = 24 h r s$
$R_{1} = 6 R + R = 7 R$
$R_{2} = 2.5 R + R = 3.5 R$
Using Kepler's third law,
As, $T^{2} \propto R^{3}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{R_{1}^{3}}{R_{2}^{3}}$
$\frac{24^{2}}{T_{2}^{2}} = \frac{(7 R)^{3}}{(3.5 R)^{3}}$
$T_{2} = 8.5$ $h r s$