Answer to Question #106057 in Classical Mechanics for Jerico

Question #106057

1. Consider the mechanical energy of a body at rest on the ground at the Earth's equator, at
re = 6400 km.

Consider the mechanical energy of the same body at rest on the ground at the South pole, at re = 6400 km. For this problem, we consider the Earth to be spherical.

(Remember, the object at the equator traces a circular path, the object at the Pole does not.)

G = 6.67×10^−11 Nm^2 kg^−2, and the mass of the Earth is M = 5.97×10^24 kg.

a. How much more mechanical energy per kilogram does an object on the ground at the Equator than on the ground at the Pole?

b. E= ___ MJ.kg^−1. (2 sig figs, do not use scientific notation)

Expert's answer

At the pole the total mechanical energy is "E _\np\n\u200b\t\n =G \n\n\\frac{mM}\nr\u200b\t\n ."

At the equator the body has also kinetic energy:

"E_ \ne\n\u200b\t\n =m(\\frac{G \n\nM}r\n\u200b\t\n + \n\nv ^\n2\n \n\u200b\t\n ),"

where the velocity is

"v= \n\\frac{\n2\u03c0r}\n{T}"

"E _\ne\n\u200b\t\n =m(\\frac {G \n\nM}r\n\u200b\t\n + \n\n \n\\frac{2\u03c0 \n^2\n r \n^2\n }{T^2}\n\u200b\t\n )."

Compare the two energies:

"\u03be=( \\frac{E \n_p}{\t\n \nE_ \ne\n\u200b}\t\n \n\u200b\t\n \u22121)\u22c5100\\%= \n\n\n \\frac{\n2\u03c0 ^\n2\n r ^\n3}{GMT^2}\n \n\u200b\t\n \u22c5100\\%=0.17\\%."

Here T is the length of a day (24 hours converted to seconds).

On the pole the energy is

"E_ \np\n\u200b\t\n =G \n\\frac{\nmM}r\n\u200b\t\n =62 MJ\/kg."