1. Jun 19, 2006

### holla

Newton's Law of Gravitation Derivative

Hi, I was wondering if anyone could help me find the derivative of Newton's Law of gravitation [F=(GmM)/r^2] and what it means? What does the minus sign indicate?

I'm kind of confused on the subject at hand and hopefully someone can help out.

Thanks

Last edited: Jun 19, 2006
2. Jun 19, 2006

### James R

I'm not sure what you're asking. Do you really mean "derivative", or do you mean "derivation"? And which minus sign are you referring to?

3. Jun 19, 2006

### ek

Tidal force.

4. Jun 19, 2006

### ek

Last edited: Jun 19, 2006
5. Jun 19, 2006

### holla

Hey, yeah i'm talking about the derivative of the force. I searched that page before I posted and I really didn't understand it. Anyone have a cleaner and beginner friendly approach to getting the derivative of it and what that derivative means? as for the negative sign, I think it may be what you get when you take the derivative of it.

6. Jun 19, 2006

### arildno

holla:
I believe you don't understand the difference between "a derivative", and "a derivation".
"A derivative" is a function related to some other function through a limiting process.

"A derivation" is an explanation and justification of why something is as it is.
Isn't that, rather, what you are after?

7. Jun 19, 2006

### holla

arildno:
I want to get the derivative of newton's law of gravitation. I want to get dF/dr and want to know get an explanation sort of say of what it is. It's probably some really basic physics term, but maybe you all can help out. So I think I need the derivative, not a derivation.

Thanks for the help guys!

8. Jun 19, 2006

### Gokul43201

Staff Emeritus
Functions have derivatives, not equations.

$$F(r) = kr^{-2} \implies dF/dr = -2kr^{-3}$$

Now what exactly is your question?

9. Jun 19, 2006

### holla

Ah thanks. So k is the constant which represents G, m, and M? What does the dF/dr represent in terms such that GmM/r^2 is the newton's law of gravitation, does -2kr^-3 represent anything as such? I think some of you were saying that this is the Tidal force?

Thanks

10. Jun 19, 2006

### Gokul43201

Staff Emeritus
What the above is telling you is that by rearranging a little bit, and looking at the expression (with k=GMm):

$$\delta F = \frac{-2k}{r^3} \delta r$$

you see that, if there's an object of "size" $\delta r$ a distance r away from the center of an external gravitational field, the gravitational force felt by the two (near and far) extremities of this object differ by the value given above. The negative sign only says that the farther extremity of the object feels a smaller force.

Why this is important is that if different parts (of a body) feel different forces, you would expect that (from Newton), they have different accelerations. Thus the body would fragment apart if there were no internal forces (also, usually gravitational) holding the different parts of the object together. Sometimes, if this force difference (called a "Tidal Force") is large enough (either because the object is large, or because it is very close to the center of the external field), the internal gravitational force is not enough to keep the object together and it does end up in many pieces.

Here's the wiki : http://en.wikipedia.org/wiki/Tidal_force

11. Jun 20, 2006

### ek

And just for clarity, the distance at which an object will break apart due to tidal forces is known as the Roche Limit, which is what was being asked about in the thread I gave a link to.

12. Jun 21, 2006

### holla

Thanks for the help guys :)