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
Answer & explanation
Correct answer: option 2
$g=\frac{G M}{R^2}$ . . .(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