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)

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