A geostationary satellite is orbiting the ear | vedclass

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Question

$T = 2\pi \sqrt {\frac{{{r^3}}}{{GM}}} $

$	herefore {\left( {\frac{{{T_1}}}{{{T_2}}}} ight)^2} = {\left( {\frac{{{r_1}}}{{{r_2}}}} ight)^3} =

Vedclass · Accepted Answer

$6\sqrt 2 \,hours$

Vedclass · Answer

$T = 2\pi \sqrt {\frac{{{r^3}}}{{GM}}} $

$	herefore {\left( {\frac{{{T_1}}}{{{T_2}}}} ight)^2} = {\left( {\frac{{{r_1}}}{{{r_2}}}} ight)^3} = {\left( {\frac{{6R + R}}{{2.5R + r}}} ight)^3} = 8$

${T_2} = \frac{{{T_1}}}{{\sqrt 8 }} = \frac{{24}}{{\sqrt 8 }} = 6\sqrt 2 h\,r$

A geostationary satellite is orbiting the earth at a height of $6\, R$ from the earth’s surface ($R$ is the earth’s radius ). What is the period of rotation of another satellite at a height of $2.5\, R$ from the earth’s surface

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