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Question

(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 its o

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 $T$ of a satellite is directly proportional to the cube of its orbital radius $r$,i.e.,$T^2 \propto r^3$.This relationship is independent of the mass of the satellite.Given for satellite $A$: $r_A = r$.Given for satellite $B$: $r_B = 2r$.The ratio of their time periods is given by:$\frac{T_A}{T_B} = \left( \frac{r_A}{r_B} \right)^{3/2} = \left( \frac{r}{2r} \right)^{3/2} = \left( \frac{1}{2} \right)^{3/2} = \frac{1}{2^{3/2}} = \frac{1}{2 \cdot 2^{1/2}} = \frac{1}{2\sqrt{2}}$.Thus,the ratio is $1:2\sqrt{2}$.

$A$ satellite $A$ of mass $m$ is at a distance of $r$ from the centre of the earth. Another satellite $B$ of mass $2m$ is at a distance of $2r$ from the earth's centre. Their time periods are in the ratio of:

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