If the radius of the Earth contracts by 0.1 % | vedclass

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(B) The weight of a body at the Earth's surface is given by $W = \frac{GMm}{r^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth

Vedclass · Accepted Answer

$0.2 \%$

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(B) The weight of a body at the Earth's surface is given by $W = \frac{GMm}{r^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,$m$ is the mass of the body,and $r$ is the radius of the Earth.Since $G, M,$ and $m$ are constant,$W \propto \frac{1}{r^2}$.Given that the radius contracts by $0.1 \%$,the new radius is $r' = r(1 - 0.001) = 0.999r$.The new weight $W'$ is given by $W' = \frac{GMm}{(0.999r)^2} = \frac{W}{(0.999)^2}$.Using the binomial approximation $(1-x)^{-n} \approx 1+nx$ for small $x$,we have $W' = W(1 - 0.001)^{-2} \approx W(1 + 2 	imes 0.001) = W(1 + 0.002)$.Thus,$W' = W + 0.002W$,which represents an increase of $0.002 	imes 100 = 0.2 \%$.Therefore,the weight increases by $0.2 \%$.

If the radius of the Earth contracts by $0.1 \%$,with its mass remaining the same,then the weight of a body at the Earth's surface will increase by ..............$\%$

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