Question

If the radius of earth is decreased by $10\% $, the mass remaining unchanged, what will happen to the acceleration of gravity?
A) Decreased by $19\% $
B) Increased by $19\% $
C) Decreased by more than $19\% $
D) Increased by more than $19\% $

Answer 1

Hint: Use the relation of acceleration of gravity and radius of earth to deduce the final expression. The acceleration of gravity is inversely proportional to the square of the radius of the earth.

Complete step by step answer:
Let a body of mass $m$on the surface on earth. We take this mass to deduce the expression between acceleration of gravity and the radius of earth $R$.
We know that, when a body is present near the earth’s surface, a force of gravity which is equal to $mg$acts on this body. Also, we know the force of gravitation between two bodies kept at a distance from each other is $G\dfrac{{Mm}}{{{r^2}}}$(as per Newton’s universal law of gravitation).
On equating both the forces we get,
$mg = G\dfrac{{Mm}}{{{r^2}}}$
$g = G\dfrac{M}{{{r^2}}}$
Where,
\[M\]is the mass of earth.
$G$ is the universal gravitational constant.
$r$ is the radius of earth.
It is given in the question that the radius is decreased by $10\% $,
The final radius becomes,
$R = r - 10\% r$
This equation means that the total radius is now $10\% $less than the original radius.
$R = r - \dfrac{{10}}{{100}}r$
$R = \dfrac{{9r}}{{10}}$
The new acceleration due to gravity is,
$g' = G\dfrac{M}{{{R^2}}}$
$g' = G\dfrac{M}{{{{(\dfrac{{9r}}{{10}})}^2}}}$
$g' = \dfrac{{100}}{{81}}G\dfrac{M}{{{R^2}}}$
Now, change in acceleration due to gravity is given by,
$\dfrac{{\dfrac{{100GM}}{{81{r^2}}} - \dfrac{{GM}}{{{r^2}}}}}{{\dfrac{{GM}}{{{r^2}}}}} \times 100$
$23\% $
So the acceleration due to gravity is increased by $23\% $.
D) is correct.

Note: We calculated acceleration due to gravity on the surface of earth. Remember that the value of acceleration due to gravity decreases with the height or depth from the surface of earth.
Acceleration due to gravity at height $h$ from the surface $ = g{\left( {1 + \dfrac{h}{R}} \right)^{ - 2}}$
Acceleration due to gravity at depth $d$ from the surface$ = g\left( {1 - \dfrac{d}{R}} \right)$
Here $R$ is the radius of earth.