If the radius of the earth shrinks by 1.5% (m | vedclass

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(C) The acceleration due to gravity $g$ on the surface of the earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is th

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(C) The acceleration due to gravity $g$ on the surface of the earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the earth,and $R$ is the radius of the earth.Since the mass $M$ remains constant,we have $g \propto \frac{1}{R^2}$.Taking the logarithmic derivative,we get $\frac{\Delta g}{g} = -2 \frac{\Delta R}{R}$.Given that the radius shrinks by $1.5\%$,we have $\frac{\Delta R}{R} = -1.5\%$.Substituting this value,we get $\frac{\Delta g}{g} = -2 	imes (-1.5\%) = +3\%$.Therefore,the value of acceleration due to gravity increases by $3\%$.

If the radius of the earth shrinks by $1.5\%$ (mass remaining same),then the value of acceleration due to gravity changes by ....... $\%$.

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