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

In summary: The two masses in Newton's formula are "source" masses, not "test" masses. And the formula for the gravitational field of a point mass is not the same as the formula for the force between two point masses. The first formula involves only one mass and is an expression for the gravitational field created by that mass. The second formula involves two masses and is an expression for the force experienced by one mass due to the gravitational field created by the other mass. So there is no contradiction between the two formulas - they are describing different things.
  • #1
Ebenshap
9
0
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
 
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  • #2
Ebenshap said:
I want a better visual model of what Newton's gravitational force formula represents:
Have you tried field lines? Note that Coulomb force between opposite charges is analogous to Newtonian gravity,
 
  • #3
Ebenshap said:
I want a better visual model of what Newton's gravitational force formula represents:

$$-{G m_1 m_2\over d^2}$$

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
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.

There really is no contradiction.

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.
 
  • #4
I am not sure how to interpret your question. It seems to be confusing two different questions (1) the gravitational field of a point source and (2) the force between two point sources.
 

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

What is the formula for gravitational force between two point sources?

The formula for gravitational force between two point sources is F = G * m1 * m2 / d2, where G is the gravitational constant, m1 and m2 are the masses of the two point sources, and d is the distance between them.

What is the relationship between mass and gravitational force?

The greater the mass of an object, the greater the gravitational force it exerts. This is because mass is directly proportional to gravitational force in the gravitational force formula.

How does distance affect gravitational force?

The gravitational force between two point sources is inversely proportional to the square of the distance between them. This means that as the distance between two objects increases, the gravitational force between them decreases.

What is the unit of measurement for gravitational force?

The SI unit for gravitational force is Newtons (N). This is derived from the gravitational constant, which has a value of 6.674 x 10^-11 N*m^2/kg^2.

Can the gravitational force formula be used to calculate the intensity of gravitational force?

Yes, the intensity of gravitational force can be calculated using the gravitational force formula. The intensity of gravitational force is the force per unit mass, and it is directly proportional to the gravitational force in the formula.

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