A geostationary satellite is orbiting earth at a height of 6,R above surface of earth&nbs

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7.Gravitation

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A geostationary satellite is orbiting the earth at a height of $6\,R$ above the surface of earth ($R$ is the radius of earth). The time period of another satellite at a height of $2.5\,R$ from the surface of the earth is :-

A

$3 \sqrt 2 \,hour$

B

$6 \sqrt 2\, hour$

C

$6\, hour$

D

$72\, hour$

Solution

$\frac{\mathrm{T}_{1}^{2}}{\mathrm{r}_{1}^{3}}=\frac{\mathrm{T}_{2}^{2}}{\mathrm{r}_{2}^{3}}$

$\frac{(24)^{2}}{(7 R)^{3}}=\frac{T_{2}^{2}}{(3.5 R)^{3}}$

$\mathrm{T}_{2}=6 \sqrt{2}$

Standard 11

Physics

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