# Gravitational force formula: mass 2 point sources, intensity like 1?

1. Sep 1, 2015

### Ebenshap

I want a better visual model of what Newton's gravitational force formula represents:

(G⋅m1⋅m2)/d2

But there are two contradictory things that I'm having trouble reconciling:

Multiplying the two masses shows a relationship between the two point sources, but using the area of the sphere to come up with how the intensity lessens over distance is best represented visually with one point source. For example, a point of light. When gravity involves two point sources, how can one justify dividing the results by a formula that involves one point source? It's almost as if the pull from the two point sources is represented as a single virtual point source that sends gravity out in all directions, but this is very abstract and it's hard to say if that's actually what's going on.

Also light goes out in all directions. But if gravity is the relationship between masses, then that would imply that the gravity of two masses only goes in the direction where the two would meet.

Does anyone know of any historical debate that may have arisen at the time that this information was published? Is there some kind of rationale that can explain away the contradictions above?

Thank you,

Eben

2. Sep 1, 2015

### A.T.

Have you tried field lines? Note that Coulomb force between opposite charges is analogous to Newtonian gravity,

3. Sep 1, 2015

### BvU

How about Newton's third law? Gravitational field strength (N/kg) of $m_1$ at $m_2$ is $-{G m_1\over d^2}$ so force on $m_2$ is $-{G\; m_1 \over d^2}\;{\bf m_2} = -{G\; m_1\; m_2 \over d^2}$.

And gravitational field strength (N/kg) of $m_2$ at $m_1$ is $-{G m_2\over d^2}$ so force is $-{G \; m_2 \over d^2} \;{\bf m_1} = -{G\; m_1\; m_2 \over d^2}$.

If you do it decently there is a direction vector in the field and out comes $\vec F_{12} = -\vec F_{21}$

Earth gravitational field at surface is $-{G \; m_{\rm earth}\over r_{\rm earth}^2}$ which is usually designated $g$, so my weight is $-{G \; m_{\rm earth} \; m_{\rm me} \over r_{\rm earth}^2} = -m_{\rm me} \; g$. The minus sign tells us it's pointing down (fortunately).

electric charges can be attracting or repelling, mass can only attract.

And the pull from the two point sources is approximately represented by the field from a single "virtual" point at distances r >> d but then you have a gravitational field strength $-{G \;(m_1 + m_2)\over r^2}$. And closer by you have the vector sum of two field strengths.