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*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