If the earth were to suddenly contract to $(1/n)th$ of its present radius without any change in its mass, the duration of the new day will be nearly (in hours) ??

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Answer 1 · 2017-12-30T05:24:19

Duration of the new day $= 24/n^2 hours$

Conservation of angular momentum dictates that the angular momentum before and after the contraction must be the same.

$I_1omega_1=I_2omega_2$

$I = 2/5 Mr^2 larr$ moment of inertia of the earth
$omega = (2pi)/T larr$ angular velocity of the earth = $(2pi)/(day)$ .

Let $r_2 and T_2$ be the new radius and new day duration, and rewrite the conservation equation as:

$(cancel(2/5M)r_1^2)(cancel(2pi)/T_1)=(cancel(2/5M)r_2^2)(cancel(2pi)/T_2)$

$(r_1^2)/T_1=r_2^2/T_2$

$T_2=r_2^2/r_1^2 T_1$

When the earth contracts, $r_2 = 1/nr_1$

$T_2=(1/n r_1)^2/r_1^2 T_1$

$T_2= 1/n^2T_1$

$T_1 =24 hours$

$T_2= 1/n^2T_1 = 24/n^2 (hours)$

If the radius is halved, then the new day is $24/2^2 = 6$ hours long.
If radius is 1/10th, the new day is $24/10^2 = ~ ¼$ hours long

If the earth were to suddenly contract to(1/n)th(1/n)th of its present radius without any change in its mass, the duration of the new day will be nearly (in hours) ??