The time period of revolution of a satellite close to earth is $90$ min. The time period of another satellite in an orbit at a distance of three times the radius of earth from its surface will be

90 \sqrt{8}

min

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360

min

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270 \sqrt{3}

min

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720

min

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Solution

The correct option is B $270 \sqrt{3}$ min
Using $T^{2} \propto R^{3}$
So, $\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{R_{2}^{3}}{R_{1}^{3}}$
Given : $R_{2} = 3 R_{1}$
Or, $\frac{T_{2}^{2}}{90^{2}} = \frac{(3 R_{1})^{3}}{R_{1}^{3}} = 27$
$⟹ T_{2} = 270 \sqrt{3} m i n$

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The time period of revolution of a satellite close to earth is 90 min. The time period of another satellite in an orbit at a distance of three times the radius of earth from its surface will be

The correct option is B 270√3 minUsing T2∝R3So, T22T21=R32R31Given : R2=3R1Or, T22902=(3R1)3R31=27⟹ T2=270√3 min

The time period of revolution of a satellite close to earth is $90$ min. The time period of another satellite in an orbit at a distance of three times the radius of earth from its surface will be

The correct option is B $270 \sqrt{3}$ min
Using $T^{2} \propto R^{3}$
So, $\frac{T_{2}^{2}}{T_{1}^{2}} = \frac{R_{2}^{3}}{R_{1}^{3}}$
Given : $R_{2} = 3 R_{1}$
Or, $\frac{T_{2}^{2}}{90^{2}} = \frac{(3 R_{1})^{3}}{R_{1}^{3}} = 27$
$⟹ T_{2} = 270 \sqrt{3} m i n$