Answer to Question #106057 in Classical Mechanics for Jerico

Question

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)

Answer 1

At the pole the total mechanical energy is $E _ p ​ =G \frac{mM} r​ .$

At the equator the body has also kinetic energy:

$E_ e ​ =m(\frac{G M}r ​ + v ^ 2 ​ ),$

where the velocity is

$v= \frac{ 2πr} {T}$

$E _ e ​ =m(\frac {G M}r ​ + \frac{2π ^2 r ^2 }{T^2} ​ ).$

Compare the two energies:

$ξ=( \frac{E _p}{ E_ e ​} ​ −1)⋅100\%= \frac{ 2π ^ 2 r ^ 3}{GMT^2} ​ ⋅100\%=0.17\%.$

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

On the pole the energy is

$E_ p ​ =G \frac{ mM}r ​ =62 MJ/kg.$