Question

If the radius of earth shrinks by $1.5\%$ (mass remaining same), then the value of gravitational acceleration changes by

• A. 2%
• B. -2%
• C. 3%
• D. -3%

Answer 1

Answer: (C)

$ g =\frac{G M}{R^{2}}$
$ g^{\prime} =\frac{G M}{(0.985 R)^{2}} $
$g^{\prime} =(1.0306) \frac{G M}{R^{2}} $
$ \Rightarrow g^{\prime}=1.0306\, g$
$ \Rightarrow $ Acceleration changes by
$ \frac{\Delta g}{g} \times 100=+3 \% $
Alternate method:
$g^{\prime}=\frac{G M}{(R+\Delta R)^{2}}$
$g^{\prime}=G M(R+\Delta R)^{-2} $
$g^{\prime}=\frac{G M}{R^{2}}\left(1+\frac{\Delta R}{R}\right)^{-2}$
for $\frac{\Delta R}{R}<1$, we can use binomial and approximately,
$g^{\prime}=\frac{G M}{R^{2}}\left(1-\frac{2 \Delta R}{R}\right)$
$\Rightarrow g^{\prime}=g-g \frac{2 \Delta R}{R}$
$\Rightarrow \frac{\Delta g}{g}=\frac{-2 \Delta R}{R}$
$=-2 \times\left(\frac{-1.5}{100}\right)$
$=\frac{+3}{100}=3 \% \,\,\left[g^{\prime}-g=\Delta g\right]$