What is the ratio of the time periods of a sa | 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

$1 : 2\sqrt{2}$

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 $(r)$: $T^2 \propto r^3$.Therefore,the ratio of the time periods is given by $\frac{T_A}{T_B} = \left( \frac{r_A}{r_B} \right)^{3/2}$.Given $r_A = r$ and $r_B = 2r$,we substitute these values into the formula:$\frac{T_A}{T_B} = \left( \frac{r}{2r} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{2\sqrt{2}}$.Thus,the ratio of the time periods is $1 : 2\sqrt{2}$.

What is the ratio of the time periods of a satellite $A$ of mass $m$ at a distance $r$ from the center of the Earth and a satellite $B$ of mass $2m$ at a distance $2r$ from the center of the Earth?

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