Physics - Gravitation Question with Solution | TestHub

from kepler's law,

$T^2\propto R^3$$\Rightarrow T\propto R^{\frac{3}{2}}$

$\Rightarrow T=k{R}^{\frac{3}{2}};k$k is some constant . for radius$=R$

$T=k{R}^{\frac{3}{2}}$

$\frac{dT}{dR}=\frac{3}{2}k{R}^{\frac{1}{2}}$

when radius$$\begin{gathered} =1.02R; \\ dR=1.02R-R=0.02R \end{gathered}$$

$\frac{dT}{dR}=\frac{3}{2}k{R}^{\frac{1}{2}}$

$\Rightarrow dT=\frac{3}{2}k{R}^{\frac{1}{2}}\times dR$

$\Rightarrow dT=1.5k{R}^{\frac{1}{2}}\times 0.02R$

$\%\text{change =}\frac{dT}{T}\times 100=\frac{0.03T}{T}\times 100=3$