Answer to Question #105126 in Classical Mechanics for Ayatollah

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:

$$E_e=m\bigg(G\frac{M}{r}+\frac{v^2}{2}\bigg),$$

where the velocity is

$$v=\frac{2\pi r}{T}:$$

$$E_e=m\bigg(G\frac{M}{r}+\frac{2\pi^2 r^2}{T^2}\bigg).$$

Compare the two energies:

$$\xi=\bigg(\frac{E_e}{E_p}-1\bigg)\cdot100\%=\frac{2\pi^2 r^3}{GMT^2}\cdot100\%=0.17\%.$$

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

On the pole the energy is

$$E_p=G\frac{mM}{r}=62\text{ MJ/kg}.$$