Question 253: Gravitation

Question

If the radius of earth is reduced to three-fourth of its present value without change in its mass then value of duration of the day of earth will be ________ hours 30 minutes.

Accepted Answer

Given that the radius of the Earth is decreased to three-fourths of its current value while its mass remains unchanged, we need to determine the new duration of the Earth's day.
First, using the principle of conservation of angular momentum, we have:
$ 	au_{	ext{ext}} = 0 \implies 	ext{Angular momentum is conserved} $
Therefore,
$ \frac{2}{5} M R^2 \cdot \omega_i = \frac{2}{5} M \left(\frac{3R}{4}ight)^2 \cdot \omega_f $
Simplifying this equation, we find:
$ \omega_f = \frac{16}{9} \omega $
Since the period $ T $ of rotation is given by:
$ T = \frac{2\pi}{\omega} $
For the new period $ T_1 $, we have:
$ T_1 = \frac{2\pi}{\omega_f} = \frac{2\pi}{\frac{16}{9}\omega} = \frac{9}{16} 	imes T $
Given that the initial period $ T $ is 24 hours:
$ T_1 = \frac{9}{16} 	imes 24 	ext{ hours} $
$ T_1 = 13 	ext{ hours} ~30 	ext{ minutes} $
Thus, the new duration of the Earth's day would be 13 hours and 30 minutes.

Gravitation

If the radius of earth is reduced to three-fourth of its present value without change in its mass then value of duration of the day of earth will be ________ hours 30 minutes.

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