The entropy of systems

Maybe if we start with the force equations.

How does one calculate the force of gravity close to the center of the earth? What is F, m, E, and acceleration of a sample of mass lets say at a distance of 2 Km from the attractor (center of the Earth)?

I think this would have to show the change in momentum is related to the varying energy in that sample of mass as it gets closer to the attractor. Its path is curved around the attractor.

One paper I read had three definitions of force:

1. The Lorentz Force, FL equals the rate of change of momentum dp
dt .
Note: the Lorentz force was derived from Maxwell's equations (energy & force)

2. The Minkowski Force, F, is a 4-vector, and is easy to Lorentz transform.

3. The Newtonian Force, FN = ma = mdv
dt tells how an object will move.

How do we combine these to show the relationship between E, m, and Force of gravity?

GD: 26 Jul 2009 : 14:08: <blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">21How does one calculate the force of gravity close to the center of the earth? What is F, m, E, and acceleration of a sample of mass lets say at a distance of 2 Km from the attractor (center of the Earth)?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">If the standard theories are correct, a hollow sphere of uniform mass can be considered as a point mass at the center for calculating gravity outside the sphere. However, inside the sphere, the mass of the shell contributes zero net gravity. Therefore, the gravity inside a solid sphere of uniform density increases linearly from zero at the center until you reach the surface. Beyond the surface, of course, gravity falls off according to the inverse square law. Since the Earth's density is greater toward the center, the gravity increases at a rate proportional to the density; a graph of field strength vs. radius would rise more steeply near the center and less steeply near the surface. If gravitational shielding exists, I think it would make the curve less steep near the surface, but since we have yet to detect any shielding, it cant be a significant factor.


Fractal Foam Model of Universes: Creator

I would like to show that the force exerted on a sample mass is not the same near the center of the Earth than it is at its surface.

Could we change F = dp/dt with dF = dp/dt ?

The "d" in dp/dt means a small increment of change. For a small increment of time change, you get a small increment of momentum change. dp/dt is the ratio of momentum change to time change for a small increment of time. This is also known as the time rate of change of momentum. If this is new to you, then I recommend that you take a college freshman physics course.

Apparently, my previous post was a bit over your head. The bottom line is, gravity at the center of Earth is zero. If you could have a hollow chamber at the center of Earth, you would float weightlessly inside of it. Viewing gravity as a pulling force, you could say that you have equal masses pulling you in every direction. Viewing gravity as a pushing force, you could say that gravitons are blocked equally in all directions. (Of course, I'm speaking in terms of a reference frame centered on Earth. In another reference frame the gravity at the center of Earth might be different.)

Fractal Foam Model of Universes: Creator

I don't think you understood my question:

Does a change in momentum with time mean a change in force?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Does a change in momentum with time mean a change in force?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
No. A constant force will change momentum at a constant rate. A changing force acting on a constant mass will give acceleration. An acceleration can be expressed as a = dv/dp = d^2p/dt^2. See Wikipedia-acceleration .

Fractal Foam Model of Universes: Creator