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