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Question

(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

$3$

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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.,$T^2 \propto R^3$.Let $T_1$ be the time period for radius $R_1 = R$ and $T_2$ be the time period for 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 $x$,where $x = 0.02$ and $n = 1.5$:$\frac{T_2}{T_1} \approx 1 + (1.5 	imes 0.02) = 1 + 0.03 = 1.03$.This implies $T_2 \approx 1.03 T_1$.The percentage difference is given by $\frac{T_2 - T_1}{T_1} 	imes 100 = \left( \frac{T_2}{T_1} - 1 ight) 	imes 100$.Substituting the value: $(1.03 - 1) 	imes 100 = 0.03 	imes 100 = 3\%$.

$A$ satellite is launched into a circular orbit of radius $R$ around 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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