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Question

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axi

Vedclass · Accepted Answer

$48 \sqrt{2}$

Vedclass · Answer

(B) According to Kepler's Third Law of Planetary Motion,the square of the time period $(T)$ is directly proportional to the cube of the semi-major axis $(r)$ of the orbit: $T^2 \propto r^3$.Let the initial orbit radius be $r_1$ and the initial time period be $T_1 = 24 	ext{ hours}$ (since it is a geostationary satellite).Let the new orbit radius be $r_2 = 2r_1$.We need to find the new time period $T_2$.Using the ratio: $\frac{T_2^2}{T_1^2} = \left( \frac{r_2}{r_1} ight)^3$.Substituting the values: $\frac{T_2^2}{24^2} = \left( \frac{2r_1}{r_1} ight)^3 = 2^3 = 8$.$T_2^2 = 8 	imes 24^2$.Taking the square root on both sides: $T_2 = \sqrt{8} 	imes 24 = 2\sqrt{2} 	imes 24 = 48\sqrt{2} 	ext{ hours}$.

$A$ geostationary satellite is taken from one orbit to another orbit,whose distance from the centre of the Earth is $2$ times that of the earlier orbit. The time period in the second orbit is how many hours?

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