Question 239: Gravitation

Question

If the earth suddenly shrinks to $$\frac{1}{64}$$th of its original volume with its mass remaining the same, the period of rotation of earth becomes $$\frac{24}{x}$$h. The value of x is __________.

Accepted Answer

From the conservation of angular momentum, we have:
$
\frac{2}{5}MR^2\omega_1 = \frac{2}{5}M\left(\frac{R}{4}\right)^2\omega_2
$
This simplifies to:
$
MR^2\omega_1 = \frac{MR^2}{16}\omega_2
$
From this, we can derive the ratio of the initial and final angular velocities:
$
\frac{\omega_1}{\omega_2} = \frac{1}{16}
$
Since the angular velocity (\omega) is inversely proportional to the period of rotation (T) ((\omega = \frac{2\pi}{T})), we can write:
$
\frac{T_2}{T_1} = \frac{1}{16}
$
We can express this ratio in terms of the variable (x):
$
\frac{T_1}{T_2} = \frac{16}{1} = \frac{24}{x}
$
Solving this equation for (x) gives:
$
x = 16
$
So, if the Earth suddenly shrinks to ( $\frac{1}{64}$ )th of its original volume with its mass remaining the same, the period of rotation of Earth becomes ( $\frac{24}{16}$ )h, or 1.5 hours. Therefore, the value of (x) is 16.

Gravitation

If the earth suddenly shrinks to $\frac{1}{64}th of its original volume with its mass remaining the same, the period of rotation of earth becomes \frac{24}{x}$h. The value of x is __________.

Integer / Numeric Answer Type