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Q. Let g be the acceleration due to gravity at earth's surface and $K$ be the rotational kinetic energy of the earth. Suppose the earth's radius decreased by $2\%$ keeping all other quantities same, then

Rajasthan PETRajasthan PET 2002

Solution:

$ g=\frac{GM}{{{R}^{2}}}\Rightarrow \frac{\Delta g}{g}=-2\frac{\Delta R}{R}=-2(-2%)=+4% $
And $ K=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}{{\omega }^{2}}\times \frac{2}{5}M{{R}^{2}} $
$ \frac{\Delta K}{K}=2\frac{\Delta R}{R}=2\times (-2)%=-4% $