Question

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

• A. $2\%$
• B. $-2\%$
• C. $3\%$
• D. $-3\%$

Answer 1

The correct option is C $3\%$

The accelaration due to gravity on the surface of earth is given by$$g=\frac{GM}{R^2}...\left(1\right)$$ Where, $R$ is the radius of earth and $M$ of mass of the earth.

Let, the radius of the earth be $R^{\prime }$ when its shrinks by $1.5\%$.

So, $R^{\prime }=R-\frac{1.5R}{100}$

$\Rightarrow R^{\prime }=\frac{98.5R}{100}$

$\Rightarrow \frac{R^{\prime }}{R}=\frac{98.5}{100}$

Using, $\left(1\right)$ we can write,

$\frac{g^{\prime }}{g}={\left(\frac{R}{R^{\prime }}\right)}^2$

Substituting values,

$\Rightarrow \frac{g^{\prime }}{g}={\left(\frac{100}{98.5}\right)}^2$

$\Rightarrow \frac{g^{\prime }}{g}=1.03069\approx 1.03$

So, the percentage change in acceleration is, $$\left(\frac{g^{\prime }-g}{g}\right)\times 100$$$=(\frac{g^{\prime }}{g}-1)\times 100$

Substituting the data obtained,

$\%$ change in $g$ $=(1.03-1)\times 100$

$=3\%$

$\therefore$ There is a $3\%$ increase in acceleration due to gravity.

OR

The accelaration due to gravity on the surface of earth is given by$$g=\frac{GM}{R^2}...\left(1\right)$$ Where, $R$ is the radius of earth and $M$ of mass of the earth.
Differentiating equation(1), we get,
$\frac{dg}{g}=-2\frac{dR}{R}$

Given that, earth shrinks by $1.5\%$.
$\Rightarrow \frac{dR}{R}=1.5\%$

Hence, $\frac{dg}{g}=-2\left(1.5\%\right)=-3\%$

$\text{Key Concept}:$ Accelaration due to gravity on the surface of a planet is given by$$g=GM/R^2$$$\text{Why the question:}$To make students understand the quantities on which the accelaration due to gravity depends.