A geostationary satellite is orbiting the ear | vedclass

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Question

According to Kepler's third law $T \propto {r^{3/2}}$

$	herefore \frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} ight)^{3/2

Vedclass · Accepted Answer

$6\sqrt 2 \,\,hr$

Vedclass · Answer

According to Kepler's third law $T \propto {r^{3/2}}$

$	herefore \frac{{{T_2}}}{{{T_1}}} = {\left( {\frac{{{r_2}}}{{{r_1}}}} ight)^{3/2}} = {\left( {\frac{{R + 2R}}{{R + 5R}}} ight)^{3/2}} = \frac{1}{{{2^{3/2}}}}$

Since ${T_1} = 24\,hours$

$So,\,\frac{{{T_2}}}{{24}} = \frac{1}{{{2^{3/2}}}}\,\,or\,\,{T_2} = \frac{{24}}{{{2^{3/2}}}} = \frac{{24}}{{2\sqrt 2 }} = 6\sqrt 2 \,hours$

A geostationary satellite is orbiting the earth at a height $5R$ above the surface of the earth , $R$ being the radius of the earth. The time period of another satellite in hours at a height of $2R$ from the surface of the earth is

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