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:

- 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:

- The 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

- The 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

- The energy E required is :

                           E = P / t

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

                           E = 2.06301 * 10^13 J

the answer is b i think

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.