Question

If the radius of Earth becomes half of its present value, with its mass remaining the same, the duration of one day will become:

• A. 8h
• B. 6h
• C. 12h
• D. 18h

Hint:

This problem illustrates the principle of conservation of angular momentum and how changes in a body’s structure affect its rotational dynamics. It’s a fundamental concept in understanding planetary physics.

Answer 1

The Correct Option is B

The duration of one day is determined by the rotational period of the Earth. If the radius is halved while the mass remains constant, the Earth's moment of inertia \(I\) will change according to:
\[I = k M R^2\]
where \(k\) is a constant, \(M\) is mass, and \(R\) is radius.
Reducing \(R\) by half means:
\[I_{\text{new}} = k M \left(\frac{R}{2}\right)^2 = \frac{1}{4} I_{\text{old}}\]
With a smaller moment of inertia and the same angular momentum (since angular momentum is conserved if no external torques are applied), the Earth would spin faster. The new rotational period \(T_{\text{new}}\) can be estimated by considering the conservation of angular momentum:
\[T_{\text{new}} = \frac{T_{\text{old}}}{2}\]
Given the current period is approximately 24 hours, the new period would be:
\[T_{\text{new}} = \frac{24 \, \text{h}}{2} = 12 \, \text{h}\]
However, because the moment of inertia is reduced by a factor of 4, the day will shorten further, leading to a period of \(6 \, \text{h}\).