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

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(A) 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 ax

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) 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 or the orbital radius $(r)$: $T^2 \propto r^3$.Given that the initial time period $T_1 = 24 	ext{ hours}$ and the initial radius is $r_1$.The new radius is $r_2 = \frac{1}{4} r_1$.Using the relation $\frac{T_2^2}{T_1^2} = \left( \frac{r_2}{r_1} ight)^3$, we get:$\frac{T_2^2}{T_1^2} = \left( \frac{1}{4} ight)^3 = \frac{1}{64}$.Taking the square root on both sides:$\frac{T_2}{T_1} = \sqrt{\frac{1}{64}} = \frac{1}{8}$.Therefore, $T_2 = \frac{T_1}{8} = \frac{24}{8} = 3 	ext{ hours}$.

The time period of a satellite of Earth is $24 \text{ hours}$. 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: (in $\text{ hours}$)

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