1) Calculate the torque required to accelerate the Earth in 5 days from rest to its present angular speed about its axis. 2) Calculate the energy required. 3) Calculate the average power required.

a) τ = 4.47746 * 10^25 N-m

b) E = 2.06301 * 10^13 J

c) P = 3.25511*10^21 W

Explanation:

Given that,

The radius of earth r = 6.3781×10^6 m

The angular speed of earth w = 7.27*10^-5 rad/s

The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s

The mass of earth m = 5.972 × 10^24 kg

The inertia of sphere I = 2/5 * m* r^2

Solution:

angular acceleration of the earth from rest to w is given by α:

                               α = w / t

                               α = (7.27*10^-5) / (1.57784760 * 10^8)

                               α = 4.60754*10^-13 rad/s^2

The required torque τ is given by:

                               τ = I*α

                               τ = 2/5 * m* r^2 * α

 τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)

 τ = 4.47746 * 10^25 N-m

Power required P to turn the earth to the speed w is:

                          P = τ*w

                          P = (4.47746 * 10^25)*(7.27*10^-5)

                          P = 3.25511*10^21 W

Energy E required is :

                          E = P / t

                          E = (3.25511*10^21) / (1.57784760 * 10^8)

                          E = 2.06301 * 10^13 J

ι вєℓιєνє ιт ιѕ тнє fιяѕт ¢нσι¢є(α.)

нє ѕнσυℓ∂ ѕмιℓє ρσℓιтℓу αи∂ ѕρєαк тσ тнє ρєяѕσи ωнσ ιѕ ιитєяяυριиg αfтєя тнє ∂ιѕ¢υѕѕισи

нσρє ι ¢συℓ∂ нєℓρ.

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge