A planet (P_1) is moving around a star of mas | vedclass

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Question

(B) According to Kepler's Third Law, the time period $T$ of a planet orbiting a star of mass $M$ at a distance $R$ is given by $T^2 = \frac{4\pi^2 R^3

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) According to Kepler's Third Law, the time period $T$ of a planet orbiting a star of mass $M$ at a distance $R$ is given by $T^2 = \frac{4\pi^2 R^3}{GM}$.Thus, $T \propto \sqrt{\frac{R^3}{M}}$.For planet $P_1$: $T_1 \propto \sqrt{\frac{R^3}{2M}}$.For planet $P_2$: $T_2 \propto \sqrt{\frac{(2R)^3}{4M}} = \sqrt{\frac{8R^3}{4M}} = \sqrt{\frac{2R^3}{M}}$.Taking the ratio $\frac{T_2}{T_1}$:$\frac{T_2}{T_1} = \frac{\sqrt{2R^3/M}}{\sqrt{R^3/2M}} = \sqrt{\frac{2R^3}{M} \cdot \frac{2M}{R^3}} = \sqrt{4} = 2$.Therefore, the ratio of the time periods of revolution of $P_2$ and $P_1$ is $2$.

$A$ planet $(P_1)$ is moving around a star of mass $2M$ in an orbit of radius $R$. Another planet $(P_2)$ is moving around another star of mass $4M$ in an orbit of radius $2R$. The ratio of the time periods of revolution of $P_2$ and $P_1$ is . . . . . . .

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