The time period of a satellite of the Earth i | vedclass

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(D) 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

$40$

Vedclass · Answer

(D) 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 the initial time period $T_1 = 5$ hours.Let the initial separation be $R_1$ and the new separation be $R_2 = 4R_1$.The new time period $T_2$ is given by the ratio: $\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} ight)^{3/2}$.Substituting the values: $\frac{T_2}{5} = \left( \frac{4R_1}{R_1} ight)^{3/2} = (4)^{3/2}$.Calculating the exponent: $(4)^{3/2} = (2^2)^{3/2} = 2^3 = 8$.Therefore,$T_2 = 5 	imes 8 = 40$ hours.

The time period of a satellite of the Earth is $5$ hours. If the separation between the Earth and the satellite is increased to four times the previous value,the new time period will become ......... $hours$.

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