A satellite moves in a circular orbit around the earth at height $R / 2$ from earth's surface where $R_{e}$ is the radius of the earth. Calculate its period of revolution. Given $R_{e} = 6.38 \times 10^{6}$ m.

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Solution

$\frac{{m v}^{2}}{r} = \frac{G M m}{R^{2}} v = \sqrt{\frac{G M}{R}}$
At surface
$T = \frac{2 π}{v} = \frac{2 π \sqrt{R}}{\sqrt{G M}} = 24 h$
At distance $\frac{3}{2} R$
$T = 2 π \sqrt{\frac{3 R}{2 G M}} = \sqrt{\frac{3}{2}} \times 24 h = 29.39 h$

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A satellite moves in a circular orbit around the earth at height R/2 from earth's surface where Re is the radius of the earth. Calculate its period of revolution. Given Re=6.38×106m.

mv2r=GMmR2v=√GMRAt surfaceT=2πv=2π√R√GM=24hAt distance 32RT=2π√3R2GM=√32×24h=29.39h

A satellite moves in a circular orbit around the earth at height $R / 2$ from earth's surface where $R_{e}$ is the radius of the earth. Calculate its period of revolution. Given $R_{e} = 6.38 \times 10^{6}$ m.

$\frac{{m v}^{2}}{r} = \frac{G M m}{R^{2}} v = \sqrt{\frac{G M}{R}}$
At surface
$T = \frac{2 π}{v} = \frac{2 π \sqrt{R}}{\sqrt{G M}} = 24 h$
At distance $\frac{3}{2} R$
$T = 2 π \sqrt{\frac{3 R}{2 G M}} = \sqrt{\frac{3}{2}} \times 24 h = 29.39 h$