A satellite is launched into a circular orbit | vedclass

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Question

(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

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(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 $R$: $T^2 \propto R^3$.Taking the logarithmic derivative of both sides,we get: $2 \frac{\Delta T}{T} = 3 \frac{\Delta R}{R}$.Given the change in radius $\Delta R = 1.03 R - R = 0.03 R$,so $\frac{\Delta R}{R} = 0.03$.Substituting this into the derivative equation: $2 \frac{\Delta T}{T} = 3 	imes 0.03$.Therefore,the percentage change in the time period is: $\frac{\Delta T}{T} 	imes 100 = \frac{3 	imes 0.03}{2} 	imes 100 = 4.5 \%$.

$A$ satellite is launched into a circular orbit of radius $R$ around the Earth. $A$ second satellite is launched into an orbit of radius $1.03 R$. The time period of revolution of the second satellite is larger than the first one approximately by: (in $\%$)

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