Explanation:
\(g = \frac{{GM}}{{{R^2}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\)
\(\dfrac{d g}{d R}=\dfrac{-2 G M}{R^{3}}\) putting \(d R=h\), we obtain
\( \Rightarrow \frac{{dg}}{h} = \frac{{ - 2GM}}{{{R^2}}} \cdot \frac{1}{R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\)
From (1) and (2) \(\Rightarrow \dfrac{d g}{g}=-2\left(\dfrac{h}{R}\right)\)
\(\Rightarrow\) Change is \(-v e\). That means \(g\) decreases.