JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 17)
A geostationary satellite is orbiting around an arbitrary planet 'P' at a height of 11R above the surface of 'P', R being the radius of 'P'. The time period of another satellite in hours at a height of 2R from the surface of 'P' is _________. 'P' has the time period of 24 hours.
3
5
$$6\sqrt 2 $$
$${6 \over {\sqrt 2 }}$$
Explanation
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From Kepler's law
$${T^2} \propto {R^3}$$
$${\left( {{{24} \over T}} \right)^2} = {\left( {{{12R} \over {3R}}} \right)^3}$$
T = 3 sec
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