- Home
- Standard 11
- Physics
If the radius of earth shrinks by $1.5 \%$ (mass remaining same), then the value of gravitational acceleration changes by ......... $\%$
$2$
$-2$
$3$
$-3$
Solution
(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 $\frac{\Delta R}{R} \ll 1$, we can use binomial and approximately,
$g^{\prime}=\frac{G M}{R^2}\left(1-\frac{2 \Delta R}{R}\right)$
$\Rightarrow g^{\prime}=g-g \frac{2 \Delta R}{R}$
$\Rightarrow \frac{\Delta g}{g}=\frac{-2 \Delta R}{R}=-2 \times\left(\frac{-1.5}{100}\right)=\frac{+3}{100}=3 \% \quad\left[g^{\prime}-g=\Delta g\right]$
