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(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) According to Kepler's Third Law of Planetary Motion,the square of the time period $T$ is proportional to the cube of the orbital radius $R$,i.e.,$T^2 \propto R^3$ or $T \propto R^{3/2}$.Taking the logarithmic derivative,we get $\frac{dT}{T} = \frac{3}{2} \frac{dR}{R}$.The percentage change in radius is $\frac{dR}{R} 	imes 100 = \frac{1.02R - R}{R} 	imes 100 = 0.02 	imes 100 = 2\%$.Therefore,the percentage change in the time period is $\frac{dT}{T} 	imes 100 = \frac{3}{2} 	imes (\frac{dR}{R} 	imes 100) = \frac{3}{2} 	imes 2\% = 3\%$.Thus,the percentage difference in the time periods is $3\%$.

$A$ satellite is launched into a circular orbit of radius $R$ around the Earth,while a second satellite is launched into an orbit of radius $1.02 R$. The percentage difference in the time periods of the two satellites is ......... $\%$

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