Question

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.

Answer #1

Given,

Initial time period, T_{i} = 1 day = 24*60*60 = 86400
s

Change time period, T = 5.9 s

Final time period, T_{f} = 86400 - 5.9 = 86394.1 s

Thus,

Initial angular frequency, w_{o} = 2*pi/T_{i} =
2*3.14 / 86400 = 726851.85 *10^{-10} s^{-1}

Final angular frequency, w_{f} =
2pi/T_{f} = 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,

w_{f} - w_{i} =
*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*R^{2} =
0.4*5.98*10^{24}*(6.380*10^{6})^{2}

I = 9.73*10^{37} kg.m^{2}

Now, Torque
= I*
= 9.73*10^{37} *5.74*10^{-14} = 55.89
*10^{23} kg*m^{2}/s^{2}

Let the force be F

I* = F*R

55.89 *10^{23} = F*6.38*10^{6}

F = 8.76*10^{17} N

F = 0.876 *10^{18} N

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