Question

What force, applied tangentially to the Earth along 45° of latitude in the direction of rotation...

What force, applied tangentially to the Earth along 45° of latitude in the direction of rotation for 1 day would result in a new day length that is shorter by 5.9 seconds? The Earth has mass 5.98x1024kg and radius 6380 km. The answer should be a value times 10^18N!

Thank you.

Given,

Initial time period, Ti = 1 day = 24*60*60 = 86400 s

Change time period, T = 5.9 s

Final time period, Tf = 86400 - 5.9 = 86394.1 s

Thus,

Initial angular frequency, wo = 2*pi/Ti = 2*3.14 / 86400 = 726851.85 *10-10 s-1

Final angular frequency,  wf = 2pi/Tf = 2*3.14 / 86394.1 = 726901.48*10-10 s-1

Now,

Let the angular acceleration be

Time for which force is applied, t = 1 day = 86400 s

Now,

wf - wi = *t

726901.48*10-10 -   726851.85 *10-10 = *86400

= 5.74*10-14 s-2

Now,

Moment of inertia of earth, I = 2/5*M*R2 = 0.4*5.98*1024*(6.380*106)2

I = 9.73*1037 kg.m2

Now, Torque   = I* = 9.73*1037 *5.74*10-14 = 55.89 *1023 kg*m2/s2

Let the force be F

I* = F*R

55.89 *1023 = F*6.38*106

F = 8.76*1017 N

F = 0.876 *1018 N

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