A satellite which is geostationary in a parti | vedclass

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(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(r)$: $

Vedclass · Accepted Answer

$48\sqrt{2}$ hours

Vedclass · Answer

(B) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is proportional to the cube of the orbital radius $(r)$: $T^2 \propto r^3$,which implies $T \propto r^{3/2}$.Given that the initial time period $T_1 = 24$ hours (for a geostationary orbit) and the new radius $r_2 = 2r_1$.The new time period $T_2$ is given by: $T_2 = T_1 	imes (r_2/r_1)^{3/2}$.Substituting the values: $T_2 = 24 	imes (2)^{3/2} = 24 	imes 2\sqrt{2} = 48\sqrt{2}$ hours.

$A$ satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of the Earth in the new orbit is $2$ times that of the earlier orbit. The time period in the second orbit is:

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