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(C) 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

$3.0$

Vedclass · Answer

(C) 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.,$T^2 \propto R^3$.Taking the logarithm and differentiating both sides,we get $2 \frac{dT}{T} = 3 \frac{dR}{R}$,which simplifies to $\frac{dT}{T} = \frac{3}{2} \frac{dR}{R}$.Given that the change in radius is $\Delta R = 1.02 R - R = 0.02 R$,the fractional change in radius is $\frac{\Delta R}{R} = 0.02$.Substituting this into the expression for the fractional change in time period: $\frac{\Delta T}{T} = \frac{3}{2} 	imes 0.02 = 0.03$.To find the percentage difference,multiply by $100$: $\frac{\Delta T}{T} 	imes 100 = 0.03 	imes 100 = 3\%$.Thus,the percentage difference in the time periods is $3\%$.

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

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