The fractional change in the value of free-fall acceleration ‘g’ for a particle when it is lifted from the surface to an elevation h (h<<R) is

$-\frac{2 h}{R}$

$g=\frac{G M}{R^2}$     . . .(i)

$\frac{dg}{dR}=\frac{-2 GM}{R^3}$   putting dR = h we obtain

$\Rightarrow \frac{dg}{h}=\frac{-2 GM}{R^2} . \frac{1}{R}$     . . . (ii)

From (i) and (ii)

$\Rightarrow \frac{dg}{g}=-2\left(\frac{h}{R}\right)$

⇒ Change is –ve. That means g decreases