Question

Calculate the kinetic energy of Earth due to its spinning about its axis. Krot = J Compare your answer with the kinetic energy of the orbital motion of Earth's center of mass about the Sun. Assume Earth to be a homogeneous sphere of mass 6.00 1024 kg and radius 6.40 106 m. The radius of Earth's orbit is 1.50 1011 m.

Answer 1

Answer:

Explanation:

Answer: Let ke = 1/2 IW^2 = 1/2 kMr^2 W^2 be Earth's rotational KE. W = 2pi/24 radians per hour rotation speed and k = 2/5 for a solid sphere M is Earth mass, r = 6.4E6 m.

Then ke = 1/2 2/5 6E24 (6.4E6)^2 (2pi/(24*3600))^2 = ? Joules. You can do the math, note W is converted to radians per second for unit consistency.

Let KE = 1/2 KMR^2 w^2 be Earth's orbital KE. w = 2pi/(365*24) radians per hour K = 1 for a point mass. Note I used 365 days, a more precise number is 365.25 days per year, which is why we have Leap Years.

Find KE/ke = 1/2 KMR^2 w^2//1/2 kMr^2 W^2 = (K/k)(w/W)^2 (R/r)^2 = (5/2) (365)^2 (1.5E11/6.4E6)^2 = 7.81E9 ANS