Question

If the radius of the earth decreases by 1% without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease? If so, by what per cent?

Answer 1

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

$$\begin{gathered} R\text{'}=R-\frac{R}{100}=\frac{99}{100}R \\ \Rightarrow R\text{'}=0.99R \end{gathered}$$

New acceleration due to gravity will be given by
$$\begin{gathered} g\text{'}=G\frac{M}{R{\text{'}}^2}=G\frac{M}{(0.99R{)}^2} \\ \Rightarrow g\text{'}=1.02\times \left(G\frac{M}{R^2}\right)=1.02g \end{gathered}$$
Hence, the value of the acceleration due to gravity increases when the radius is decreased.

Percentage increase in the acceleration due to gravity is given by
$$\begin{gathered} \frac{g\text{'}-g}{g}\times 100 \\ =\frac{0.02g}{g}\times 100 \\ =2\% \end{gathered}$$