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(C) According to Kepler's third law,the time period $T$ of a satellite in a circular orbit is related to the orbital radius $r$ by the relation $T \pr

Vedclass · Accepted Answer

$1.5$

Vedclass · Answer

(C) According to Kepler's third law,the time period $T$ of a satellite in a circular orbit is related to the orbital radius $r$ by the relation $T \propto r^{3/2}$.Taking the natural logarithm on both sides: $\ln T = \frac{3}{2} \ln r$.Differentiating both sides,we get the fractional change: $\frac{\Delta T}{T} = \frac{3}{2} \frac{\Delta r}{r}$.Given that the radius increases from $R$ to $(1.01)R$,the percentage change in radius is $\frac{\Delta r}{r} 	imes 100 = 1\%$.Therefore,the percentage change in the time period is $\frac{\Delta T}{T} 	imes 100 = \frac{3}{2} 	imes (1\%) = 1.5\%$.Thus,the period of the second satellite is larger by approximately $1.5\%$.

Column-$I$	Column-$II$
$(1)$ Polar satellite	$(a)$ Telecommunication
$(2)$ Geostationary satellite	$(b)$ Spying
	$(c)$ Weather forecasting

$A$ satellite is launched into a circular orbit of radius $R$ around the earth. $A$ second satellite is launched into an orbit of radius $(1.01)R$. The period of the second satellite is larger than that of the first one by approximately ........... $\%$

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