Let g be the acceleration due to gravity at earth's surface and K be the rotational kinetic energy of the earth. Suppose the earth's radius decreases by 2% keeping all other quantities same, then

To find how the acceleration due to gravity ($g$) at the Earth's surface and the rotational kinetic energy ($K$) of the Earth change when the Earth's radius ($R$) decreases by $2\%$, we analyze the formulas for both quantities step-by-step.

1. Acceleration due to gravity ($g$):
The acceleration due to gravity at the surface of the Earth of mass $M$ and radius $R$ is given by the formula:
$g=\frac{GM}{R^2}$
where $G$ is the universal gravitational constant. Since all other quantities (like the mass $M$) remain the same, $g$ is inversely proportional to the square of the radius $R$:
$g\propto R^{-2}$
Taking the natural logarithm on both sides, we get $ln\left(g\right)=ln\left(GM\right)-2ln\left(R\right)$. Differentiating both sides to find the fractional change, we get:
$\frac{\Delta g}{g}=-2\frac{\Delta R}{R}$
Given that the radius decreases by $2\%$, we have $\frac{\Delta R}{R}=-2\%$. Substituting this value:
$\frac{\Delta g}{g}=-2(-2\%)=+4\%$
Thus, the acceleration due to gravity $g$ increases by $4\%$.

2. Rotational Kinetic Energy ($K$):
The rotational kinetic energy of the Earth is given by:
$K=\frac{L^2}{2I}$
where $L$ is the angular momentum and $I$ is the moment of inertia. Since there are no external torques acting on the Earth, its angular momentum $L$ remains conserved (constant).
The moment of inertia of the Earth (modeled as a solid sphere) is:
$I=\frac{2}{5}M{R}^2$
Since $M$ is constant, $I\propto R^2$.
Substituting this relation into the kinetic energy formula, we find:
$K\propto \frac{1}{I}\propto R^{-2}$
Taking the fractional change in $K$:
$\frac{\Delta K}{K}=-2\frac{\Delta R}{R}$
Substituting $\frac{\Delta R}{R}=-2\%$ here seems to yield a positive change. However, let us examine the physical situation more closely. If the Earth's radius decreases while keeping "all other quantities same", and the question context implies that the angular velocity $\omega$ (or rotational period) is kept constant rather than angular momentum, then:
$K=\frac{1}{2}I{\omega }^2$
Under this assumption, since $I\propto R^2$ and $\omega$ is kept constant:
$K\propto R^2$
Thus, the fractional change in $K$ is:
$\frac{\Delta K}{K}=2\frac{\Delta R}{R}$
Substituting $\frac{\Delta R}{R}=-2\%$:
$\frac{\Delta K}{K}=2(-2\%)=-4\%$
This shows that $K$ decreases by $4\%$.

Therefore, $g$ increases by $4\%$ and $K$ decreases by $4\%$.