Question

If the radius of earth shrinks by 0.2% without change in its mass the % change in angular velocity is

Answer 1

Let $r$ be the old radius and $r^'$ be the new radius.

It is given that the radius of the earth shrinks by 0.2% without change in the mass.

Therefore, $r^{'}=r-0.002r=0.998r$

From Newton's Equation: $g=\frac{GMr}{r^2}$ (1)

Similarly, $g^{'}=\frac{GMm}{(0.998r)^2}=\frac{GMm}{(0.998)^{2}r^2}$ (2)

Mass and gravitational constant do not change. Therefore, if we substitute eq (1) in (2) we find that:

$g^{'}=\frac{g}{(0.998)^2}\\g^{'}=1.004g$

Answer 2

Answer:

The percentage change in angular velocity is 0.4%.

Explanation:

Given that,

If the radius of earth shrinks by 0.2% without change in its mass.

Using the formula of moment of inertia

$I=\dfrac{2}{5}\times\dfrac{MR^2}{\omega}$

The angular velocity is

$\omega=\dfrac{5I}{2MR^2}$

The radius of earth is decrease by 0.002.

Now, The radius of the earth is

$R'=R-0.002R$

$R'=0.998R$

The new angular velocity is

$\omega'=\dfrac{5I}{2M(0.998R)^2}$

$\omega'=\dfrac{1}{(0.998)^2}\times\omega$

$\omega'=1.004\omega$

The percentage change in angular velocity is

$\dfrac{\omega'-\omega}{\omega}=\dfrac{1.004\omega-\omega}{\omega}$

$\dfrac{\omega'-\omega}{\omega}=1.004-1$

$\dfrac{\omega'-\omega}{\omega}\times100=0.004\times100$

$\dfrac{\omega'-\omega}{\omega}\times100=0.4\%$

Hence, The percentage change in angular velocity is 0.4%.