The time period of a satellite of earth is 24 | vedclass

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Question

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

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) According to Kepler's third law of planetary motion,the square of the time period $(T)$ is directly proportional to the cube of the orbital radius $(R)$: $T^2 \propto R^3$.Given the initial time period $T_1 = 24 \ hours$ and initial radius $R_1 = R$.The new radius is $R_2 = \frac{R}{4}$.Using the ratio formula: $\frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$.Substituting the values: $\left(\frac{24}{T_2}\right)^2 = \left(\frac{R}{R/4}\right)^3 = (4)^3 = 64$.Taking the square root on both sides: $\frac{24}{T_2} = \sqrt{64} = 8$.Therefore,$T_2 = \frac{24}{8} = 3 \ hours$.

The time period of a satellite of earth is $24 \ hours$. If the separation between the earth and the satellite is decreased to one-fourth of the previous value,then its new time period will become $....... \ hours$.

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