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(D) According to Kepler's third law,the time period $T$ of a satellite is related to the orbital radius $r$ by the relation $T \propto r^{3/2}$.Taking

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$3.0$

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(D) According to Kepler's third law,the time period $T$ of a satellite is related to the orbital radius $r$ by the relation $T \propto r^{3/2}$.Taking the logarithmic derivative of both sides,we get $\frac{\Delta T}{T} = \frac{3}{2} \frac{\Delta r}{r}$.Given that the radius increases from $R$ to $1.02 R$,the change in radius is $\Delta r = 1.02 R - R = 0.02 R$.Therefore,the fractional change in radius is $\frac{\Delta r}{r} = \frac{0.02 R}{R} = 0.02$ or $2\%$.Substituting this into the derivative equation: $\frac{\Delta T}{T} = \frac{3}{2} 	imes 2\% = 3\%$.Thus,the time period increases by $3\%$.

$A$ satellite orbits the Earth in a circular orbit of radius $R$. Another satellite orbits in an orbit of radius $1.02 R$. What is the percentage increase in the time period of the second satellite compared to the first (in $\%$)?

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