Question

If the radius of the earth shrinks by $4\%$ and there is no change in its mass then by how much does the value of $g$ change?
( Given for $x<<1$ , assume $\frac{1}{(1-x{)}^2}\approx 1+2x$)

• A. $2\%$
• B. $4\%$
• C. $16\%$
• D. $8\%$

Answer 1

The correct option is D $8\%$

Let $M_e$ be the mass of earth.
Let $R_1$ be the initial radius of the earth.

Acceleration due to gravity $g_1=\frac{G{M}_e}{{R_1}^2}$

Radius shrinks by $4\%$, and there's no change in mass.

$R_2=R_1\times (1-\frac{4}{100})$
$\Rightarrow R_2=\frac{96}{100}\times R_1$

Now, acceleration due to gravity: $g_2=\frac{G{M}_e}{{R_2}^2}$
$=\frac{G{M}_e}{{(1-\frac{4}{100})}^2{\left(R_1\right)}^2}$
$\approx \frac{G{M}_e}{(R_1{)}^2}\times (1+\frac{2\times 4}{100})$
$=g_1\times (1+\frac{8}{100})$

Thus, the value of $g$ changes by $8\%$.