If we consider the Earth to be a perfect sphere, then the acceleration due to gravity at its surface is given by \[g = G\frac{M}{R^2}\]Here, M is the mass of Earth; R is the radius of the Earth and G is universal gravitational constant. If the radius of the earth is decreased by 1%, then the new radius becomes \[R' = R - \frac{R}{100} = \frac{99}{100}R\] \[ \Rightarrow R' = 0 . 99R\] New acceleration due to gravity will be given by \[g' = G\frac{M}{R '^2} = G\frac{M}{(0 . 99R )^2}\] \[ \Rightarrow g' = 1 . 02 \times \left( G\frac{M}{R^2} \right) = 1 . 02g\] Hence, the value of the acceleration due to gravity increases when the radius is decreased. \[\frac{g' - g}{g} \times 100\] \[ = \frac{0 . 02g}{g} \times 100\] = 2 % |