View Full Version : Question about stress energy of the Earth

2004-Aug-19, 02:59 PM
For a body in orbit around the Earth the forces of gravity and centrifugal motion are balanced:

(1) mv^2/r = GMm/r^2 ,

where M is the Earth, m is the body and r is the distance from the Earth's centre.

We then have

(2) mv^2 = GMm/r ,

and the kinetic energy of the body is equal to the gravitational potential energy (except for a factor of 2).

Now if the same body is lying on the Earth's surface, the gravitational force on it is still GMm/r^2. To prevent the body from falling into the Earth's centre, there must be an equal outward force to balance this. This force must be related to the stress energy of the Earth that was built up during the Earth's formation. As each piece of mass was added to the young Earth, it caused a compression of the existing Earth body. If this is true, then would it be correct to say that the total stress energy of the Earth is equal to the total gravitational potential energy of the Earth? The latter is about - 10^32 J.

2004-Aug-19, 07:39 PM
Thanks for the PM, ExpErdMann.

Now if the same body is lying on the Earth's surface, its gravitational potential energy is still GMm/r. To prevent the body from falling into the Earth's centre, there must be an outward force equal to this.
You've mixed a couple quantities there, since force and energy cannot be equal.

Also, potential energy of an object is said to be zero at infinity, to balance the books. That is, if it is infinitely far from any object, it has no potential. As it gets closer to another mass, it loses potential energy. Thus, at a finite distance r it will have a negative potential energy, and the closer it gets to the Earth, the less it will have. All other things considered, that loss of potential energy is converted to kinetic energy, so at distance r its velocity would be just the escape velocity at that distance--the velocity that would be needed to "reverse" the process.

So, you would have 1/2 mv^2 + GMm/r = 0, where v is the escape velocity.

Orbital velocity is 1/sqrt(2) of the escape velocity. Put another way, its kinetic energy during orbit is a balance (an average) between zero (which would mean it would fall into the Earth) and escape.

2004-Aug-19, 08:18 PM
Thanks, milli360. I've tidied my OP up a bit. Basically, I'm suggesting that there are analogous equations to (1) and (2) to account for a mass not falling into the Earth's centre. There is a 'stress force' playing the role of centrifugal force which prevents this fall. I imagine I'm not saying anything new here, but was wondering if anyone had estimated the total amount of stress energy in the Earth. Is it just the total GPE of the Earth?