Question

Let g be the acceleration due to gravity at the earth’s surface and K the rotational kinetic energy of the earth. Supposing that the earth’s radius decreases by 2% (while keeping all other quantities constant), then:
A. g increases by 2% and K increases by 2%
B. g increases by 4% and K increases by 4%
C. g decreases by 4% and K decreases by 2%
D. g decreases by 2% and K decreases by 4%

Answer 1

Hint: As a first step, you could find the expressions for acceleration due to gravity g and rotational kinetic energy K in terms of the radius. Then, we could find the percentage change using that expression. Also do note that the other physical quantities related to earth such as its mass, angular momentum, etc is said to be constant.
Formula used:
Acceleration due to gravity,
$g=\dfrac{GM}{{{R}^{2}}}$
Rotational kinetic energy,
$K=\dfrac{1}{2}\dfrac{{{L}^{2}}}{I}$

Complete answer:
Here in the question, we have the acceleration due to gravity g and rotational kinetic energy K of the earth. We are said that the earth’s radius is being reduced by 2% with all the other quantities constant and we are supposed to find which of the following statements are correct.
We are supposed to find the increase or decrease that happens in the acceleration due to gravity and rotational kinetic energy as the result of Earth’s radius being reduced by 2%.
We have the expression for acceleration due to gravity in terms of radius of earth given by,
$g=\dfrac{GM}{{{R}^{2}}}$
$\Rightarrow \dfrac{\Delta g}{g}=\left( -2 \right)\dfrac{\Delta R}{R}$
$\Rightarrow \dfrac{\Delta g}{g}\times 100=\left( -2 \right)\left( \dfrac{\Delta R}{R}\times 100 \right)$
Substituting the percentage decrease in radius,
$\therefore \dfrac{\Delta g}{g}\times 100=\left( -2 \right)\left( -2 \right)=+4%$ …………………….. (1)
Now, rotational kinetic energy,
$K=\dfrac{1}{2}\dfrac{{{L}^{2}}}{I}$
But, $I=\dfrac{2}{5}M{{R}^{2}}$
$\Rightarrow K=\dfrac{5}{4}\dfrac{{{L}^{2}}}{M{{R}^{2}}}$
$\Rightarrow \dfrac{\Delta K}{K}\times 100=\left( -2 \right)\left( \dfrac{\Delta R}{R}\times 100 \right)$
$\therefore \dfrac{\Delta K}{K}\times 100=\left( -2 \right)\left( -2 \right)=+4%$…………………………………………. (2)
From (1) and (2) we find that both the acceleration due to gravity and rotational kinetic energy increases by 4%.

Hence, option B is the correct answer.

Note:
We are clearly given in the question that all the other quantities remain constant other than the radius of earth. So, we could conclude that earth’s mass and angular momentum remains constant. Thereby, we have avoided these constants along with the gravitational constant while finding the percentage change.