A satellite S_1 of mass m is moving in an orb | vedclass

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(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $(T)$ of a satellite is directly proportional to the cube of the

Vedclass · Accepted Answer

$2\sqrt{2}:1$

Vedclass · Answer

(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $(T)$ of a satellite is directly proportional to the cube of the radius $(r)$ of its orbit:$T^2 \propto r^3$Given for satellite $S_1$: radius = $r$,time period = $T_1$.Given for satellite $S_2$: radius = $2r$,time period = $T_2$.Using the proportionality:$\frac{T_2}{T_1} = \left( \frac{r_2}{r_1} \right)^{3/2}$Substituting the values:$\frac{T_2}{T_1} = \left( \frac{2r}{r} \right)^{3/2} = (2)^{3/2} = 2^{1} \cdot 2^{1/2} = 2\sqrt{2}$Thus,the ratio is $2\sqrt{2}:1$.

$A$ satellite $S_1$ of mass $m$ is moving in an orbit of radius $r$. Another satellite $S_2$ of mass $2m$ is moving in an orbit of radius $2r$. The ratio of the time period of satellite $S_2$ to that of $S_1$ is

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