# Energy conservation for a point at the Earth's surface

1. Jul 19, 2012

### Kamikaze_951

Hi everyone,

This is my first post here and I am really sorry for that question, but I have found the answer nowhere.

Consider a mass at the Earth's equator that is static in the Earth's referential during an entire day. Put the Earth at one of its equinoxes to simplify the problem. Then, at noon, the potential energy due to the sun of a mass m at that point is :

E_p noon = -GMm/(distance Sun-Earth - radius of Earth)

At midnight,

E_p midnight = -GMm/(distance Sun-Earth + radius of Earth)

Clearly, there is a difference in potential energy. By energy conservation, it should be balanced (by another kind of periodic energy variation). However, if the mass stand on solid ground, its rotation speed is the Earth rotation speed and it should be constant during the day (not going up and down) and therefore, the difference in kinetic energy is null.

My questions : What balances this difference in potential energy? Is there a mistake in my reasoning?

2. Jul 19, 2012

### Simon Bridge

You have only got half the rotation of course ... when you do a whole rotation the net energy change is zero.

Not happy? Well that really just means that, wherever the energy went, we got it back.

As a mass "falls" from the dark side of the Earth towards the light side (towards the sun) is gains energy which goes into "lifting" the equivalent mass on the light side into the dark side. Since the masses are balanced, and they are connected by the rest of the Earth, there is no gain or loss.

You get the same for any rotation in gravity ... or any conservative force field.

3. Jul 19, 2012

### Kamikaze_951

Your post just made me realize that I considered the wrong system (Sun+mass rather than Sun+Earth) and forgot to take into account the energy transfer between two masses of the same system.

4. Jul 19, 2012

### Simon Bridge

No worries - it's actually a very common mistake.

5. Jul 19, 2012

### Staff: Mentor

If you look at force differences of this size, keep in mind that the earth orbits the sun, which leads (in the reference frame of earth) to a centrifugal force outwards - "down" at day and "up" at night.

Note that the effect of variable gravitational forces is bigger if you use the moon. Tides exist due to these differences, and there are power plants which use those tides.

6. Jul 19, 2012

### Staff: Mentor

The OP's question assumes an unbalanced earth, which is wrong.