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 5.2 seconds? The Earth has mass 5.98x1024kg and radius 6380 km. Answer:

Answer 1

Given,

Initial time period, T_i = 23 hours 56 min

                                  = 86160 s

Change time period, T = 5.2 s

Final time period, T_f = 86160 - 5.2 = 86154.8 s

Thus,

Initial angular frequency, ₀ = 2/T_i = 2*3.14 / 86160

                                               = 728876 * 10^-10 s^-1

Final angular frequency, _f = 2/T_f = 2*3.14 / 86154.8

                                             = 728920 * 10^-10 s^-1

Now,

Let the angular acceleration be

Time for which force is applied, t = 1 day = 86160 s

Now,

_f - ₀ = *t

=> (728920 * 10^-10) - (728876 * 10^-10) = *86160

=> 44 * 10^-10 = *86160

=> = (44 * 10^-10) / 86160

          = 5.10678 * 10^-14 s^-2

Now,

Given,

mass of earth, m = 5.98*10²⁴ kg

radius of earth, R = 6380 km = 6.380*10⁶ m

Since earth is solid sphere,

Moment of inertia of earth, I = 2/5*m*R²

                                               = 0.4*5.98*10²⁴*(6.380*10⁶)²

                                               = 9.73649 * 10³⁷ kg.m²

Now,

Torque, = I

Let the force be F

=> I = F*R

=> (9.73649 * 10³⁷)*(5.10678 * 10^-14) = F * (6.380*10⁶)

=> F * (6.380*10⁶) = 49.7221 * 10²³

=> F = (49.7221 * 10²³ ) / (6.380*10⁶)

=> F = 7.7934 * 10¹⁷ N

Applied force is 7.7934 * 10¹⁷ N