Question

Earth shrinks to $\frac{1}{64}$ times of its initial volume. Time period of Earth rotation is found to be $\frac{24}{x}$ hrs. Find the value of x.

Answer 1

Initially,
Volume of the earth, V1 = \frac{4}{3}𝜋R13$$\frac{4}{3}{\pi}R_1^{3}
Where, R1 is the radius of the earth
Time period of Earth rotation is T1 = 24 hrs
Angular frequency, $\omega_1$ = $\frac{2\pi}{T_1}$= $\frac{2\pi}{24}$ = $\frac{\pi}{12}$
Moment of inertia of the earth about its diameter, I1 = $\frac{2}{5}MR_1^{2}$
Finally,
Volume of the earth, V2 = $\frac{4}{3}{\pi}R_2^{3}$
Given, V2 = $\frac{V_1}{64}$ ⇒ $\frac{4}{3}{\pi}R_2^{3}$ = $\frac{\frac{4}{3}{\pi}R_1^{3}}{64}$
⇒ R23 = $\frac{R_1^{3}}{64}$ ⇒ R23 = $(\frac{R_1}{64})$3 ⇒ R2 = $\frac{R_1}{4}$
Time period of Earth rotation is T2 = $\frac{24}{x}$ hrs
Angular frequency, $\omega_2$ = $\frac{2\pi}{T_2}$ = $\frac{2\pi}{\frac{24}{x}}$ = $\frac{{\pi}X}{12}$
Moment of inertia of the earth about its diameter, I2 = $\frac{2}{5}MR_2^{2}$
According to conservation of angular momentum,
I1$\omega_1$ = I2$\omega_2$
⇒ $\frac{2}{5}MR_1^{2} * \frac{{\pi}}{12}$ = $\frac{2}{5}MR_2^{2} * \frac{{\pi}x}{12}$
⇒ R12 = R22x ⇒ x = $\frac{R_1^{2}}{R_2^{2}}$ = $\frac{4R_2^{2}}{R_2^{2}}$ =16
Answer. 16