Question

If the earth shrinks to 1/8 of its original volume without changing its mass, what will be the new duration of the day?

Answer 1

Given:

The earth shrinks to 1/8 of its original volume without changing its mass.

To find:

New duration of day ?

Calculation:

• Let initial radius of earth be r .

Now, the new radius will be :

$\dfrac{4}{3} \pi {R}^{3} = \dfrac{1}{8} \times \dfrac{4}{3} \pi {r}^{3}$

$\implies {R}^{3} = \dfrac{1}{8} \times {r}^{3}$

$\implies R = \dfrac{r}{2}$

• Now, we will apply CONSERVATION OF ANGULAR MOMENTUM.

$I_{1}\omega_{1} = I_{2}\omega_{2}$

$\implies \dfrac{2}{5} m {r}^{2} \omega_{1} = \dfrac{2}{5}m {( \dfrac{r}{2} )}^{2} \omega_{2}$

$\implies {r}^{2} \omega_{1} = \dfrac{ {r}^{2} }{4} \omega_{2}$

• Let initial angular velocity be $\omega$.

$\implies {r}^{2} \omega= \dfrac{ {r}^{2} }{4} \omega_{2}$

$\implies \omega_{2} = 4 \omega$

• So, duration of day will be :

$T = \dfrac{2\pi}{ \omega_{2}}$

$\implies T = \dfrac{2\pi}{ 4\omega}$

$\implies T = \dfrac{1}{4} \times \dfrac{2\pi}{ \omega}$

• Now , $2\pi/\omega$ is 24 hours.

$\implies T = \dfrac{24}{4}$

$\implies T =6 \: hours$

So, new duration of day is 6 hours.