Explanation

(c) $g = \frac{GM}{r^{2}}$
$\therefore\log g = \log{G + \log{M - 2\log r}}$
Differentiating both sides w.r.t. $t$
$\frac{1}{g} = \frac{dg}{dt} = 0 - 2 \times \frac{1}{r}\frac{dr}{dt}\left( \frac{dr}{dt} \times 100 = - 1 \right)$
$\Rightarrow \ \frac{1}{g}\left( \frac{dg}{dt} \times 100 \right) = - 2 \times \frac{1}{r}\left( \frac{dr}{dt} \times 100 \right)$
$\Rightarrow \ \frac{dg}{dt} \times 100 = - 2 \times ( - 1) = 2$
$\therefore g$ increasing by 2%