# Derivation of Newton's Law of Universal Gravitation

1. Aug 13, 2013

### WK95

$F = G \frac{ m_{1} m_{2}}{ r^{2} }$

Where does the formula come from? And why does it work that way?

How would it relate to Newton's Second Law?
$F = ma$
Using Newton's Second Law, is it possible to get the Law of Universal Gravitation?

2. Aug 13, 2013

### TurtleMeister

It comes from Issac Newton. No one knows "why" it works.

No, I don't think so. Newton's second law and the law of universal gravitation involve two different properties of mass (inertial and gravitational).

3. Aug 13, 2013

### Astrum

Newton's law of universal gravitation relates to $\vec{F} = m \vec{a}$ in the following way:

$$\vec{F}=\frac{GM_1 m_2}{r^2}\hat{r} = \frac{GM_1}{r^2}m_2 \hat{r}=m_2 \vec{a}$$ where $a=g=\frac{GM_1}{r^2}$

Why? it is what it is.

4. Aug 14, 2013

### MikeGomez

$G$ is a constant of proportionality, and can therefore be made to be 1 under the correct choice of units.

$F = m{1}m{2}/r{2}$

The force of gravity falling off as the inverse of the distance squared is due the generally spherical shape of the source (usually planets and stars and such). If the gravitational source is a cylinder, then the strength of gravity would fall off as the inverse of the distance, not the inverse of the distance squared. If the gravitational source were an infinite plane, then (perhaps surprisingly) the strength of the force of gravitation does not fall off with distance.

$F = m{1}m{2}$

That leaves the force of gravity as simply the product of the two masses. As to the exact nature of why matter attracts gravitationally, scientists are still working on that.

5. Aug 14, 2013

### WK95

Since everything with mass has gravity, how come if the gravitational source is a cylinder, the gravity's strength would fall of as the inverse of the distance? So lets say I have a cylinder. At two points one distance x and another distance 2x from the cylinder, the strength at the latter point is only half the strength at the former?

What if the object in question were another shape such as a square pyramid? How could one find the gravitational strength of such an object? Or what about irregularly shaped objects?

With a infinite plane, I get how if one were to move parallel to the plane, the gravitational strength would not fall off but I don't get why it wouldn't fall if a given point were to move away in a directional perpendicular to the plane.

6. Aug 14, 2013

### technician

It might help if you can picture gravitational lines of force. Lines of force are commonly used to illustrate electric and magnetic fields.
The strength of the field is indicated by the number of lines per unit area.
This should help you to 'see' the inverse square law for a point mass.
Can you now 'see' the cylindrical situation?

7. Aug 14, 2013

### robphy

Superposition of vector-fields.
Use Gauss' Law (as one does in electromagnetism) for highly-symmetrical situations.

8. Aug 14, 2013

### Staff: Mentor

When you're "distance x from the cylinder", you're at distance x from only one small part of the cylinder, the part that's directly "opposite" you. When you look "up" or "down" at other parts of the cylinder, they're further away from you, at different distances. Each part of the cylinder produces a gravitational field at your location, whose strength depends on how far that part is from you. To get the total gravitational field at your location, you have to add up the contributions from all the pieces, taking into account that they're in different directions. This is an exercise in vector integration.

Again you add up the contributions to the field from each small part of the object, by vector integration.

One way to look at this is that a featureless infinite plane "looks the same" to you no matter how far you are from it.

9. Aug 14, 2013

### WK95

How would unit area work when the distance from a given point to a gravitational source is simply an imaginary straight line?

With a point mass, lines of force would just be lines pointing towards the point.

With a cylinder at the flat surfaces, the liens of force would be point in perpendicular to the flat face and at the rounded part, the lines would be pointed into towards the cylinder's center line.

10. Aug 14, 2013

### D H

Staff Emeritus
I think you meant inverse of the square of the distance, and it doesn't, at least not close to the cylinder. At very great distances from the cylinder it does, but that's because every compact mass distribution looks like a point mass at a large enough distance from the object in question.

11. Aug 14, 2013

### technician

Have a look at electric fields and magnetic fields to see how lines of force are used.
Faraday came up with the idea of lines of force to represent force fields (magnetism in his case) because he was not so good at mathematics.

12. Aug 14, 2013

### WK95

I was quoting Mikegomez on that part.
" If the gravitational source is a cylinder, then the strength of gravity would fall off as the inverse of the distance, not the inverse of the distance squared." - MikeGomez

13. Aug 14, 2013

### robphy

There is an implicit assumption that the cylinder is infinitely-long.

14. Aug 14, 2013

### WK95

A genius scientist who isn't that good at mathematics? Is that even possible?!

15. Aug 14, 2013

### technician

This is a different question !...start a new thread if you want to raise this question and you may get appropriate responses.
The genius is in the idea of lines of force. They link nicely with the mathematics so you have 2 techniques to get to grips with force fields.

16. Aug 14, 2013

### MikeGomez

I like Dr. Ramachandran's explanation in this video. He is discussing the electric field, but it applies to the strength of the gravitational field as well. At 55:10 he summarizes the three cases for a sphere, line (cylinder), and infinite plane. Look a little earlier for his proofs (infinite plane is at 47:00), or better yet watch the entire video.

Last edited by a moderator: Sep 25, 2014
17. Aug 14, 2013

### robphy

18. Aug 14, 2013