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(B) 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 $R$ $(T^2

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$3$

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(B) 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 $R$ $(T^2 \propto R^3)$.Let $T_1$ be the period of the first satellite with radius $R_1 = R$,and $T_2$ be the period of the second satellite with radius $R_2 = 1.02 R$.Then,$\frac{T_2}{T_1} = \left( \frac{R_2}{R_1} ight)^{3/2} = (1.02)^{3/2}$.Using the binomial approximation $(1+x)^n \approx 1+nx$ for small $x$:$\frac{T_2}{T_1} \approx 1 + \frac{3}{2}(0.02) = 1 + 0.03 = 1.03$.Therefore,$T_2 \approx 1.03 T_1$.The percentage increase in the time period is $\frac{T_2 - T_1}{T_1} 	imes 100 = (1.03 - 1) 	imes 100 = 3 \%$.

$A$ satellite is launched into a circular orbit of radius $R$ around the Earth. $A$ second satellite is launched into an orbit of radius $1.02 R$. The period of the second satellite is larger than the first one by approximately ........ $\%$

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