# Gravitation independnet of motion/momentum?

1. Jan 24, 2013

### bahamagreen

Scenario 1: A small mass is moving away from a large mass, slows down, reverses direction such that at t=0 it has an instantaneous velocity of 0 wrt height r.

Scenario 2: An identical small mass is moving directly toward the large mass such that it arrives at height r with a velocity relative to r at t=0.

So at height r at t=0 a small mass will be present, in scenario 1 it will be stationary for an instant, in scenario 2 it will have an instantaneous velocity.

I'm wondering about the difference in the small mass momentum between the two scenarios... does a difference in momentum change the magnitude of the "gross" gravitational influence at r?
In other words, does motion radially through the gravitational field effect the magnitude of influence? How are momentum variations accounted for if not?
Does it work out that rate changes in radial momentum just balance out with the "gross" force of position to be the same "net" force?

Is it improper to think of masses moving forward or backward through the gravitational field encountering different magnitudes of "gross" force compensated by momentum changes?

Is it like the velocity of light being invariant - net gravitational acceleration experienced by a mass is always independent of its motion or momentum?

2. Jan 24, 2013

### Vorde

I'm a bit confused by your post. But classically the acceleration from a massive body is $\frac{gm_0}{r^2}$ where $m_0$ is the mass of the heavy body and r is the distance. There is no place for velocity or momentum there, so it does not affect the acceleration due to gravity.

3. Jan 25, 2013

### bahamagreen

Sorry for the confusion... let me ask it another way:

An object in relative motion wrt a light source may measure aberration of the light; is there no similar effect with motion and momentum of masses wrt to gravitational force?

4. Jan 25, 2013

### A.T.

Light is not a field. It is a propagating disturbance of the EM-field. The E-field itself has no aberration. Charges in uniform linear relative motion are pulled exactly towards each other. Same with gravity.

5. Jan 26, 2013

### bahamagreen

I'll try again...

But let me ask a question, first.

Concerning the force applied at a particular time on a small mass some radius from a large mass;

Is this force, as it is locally applied at the small mass, instantaneous?

This is not asking if the speed of gravitational interaction between the large and small masses is instantaneous.
Just like the small mass has an instantaneous speed and position at a specific radius and time, it is asking if the small mass is also subject to an instantaneous magnitude of force at that instantaneous place and time.

6. Jan 26, 2013

### Staff: Mentor

Yes. At a particular moment t the mass is at a particular location r(t), and the force on it is written (as Vorde said above) as a function of that distance r but not the velocity of the object at that moment. If you want to see the equation for the force written out with all the time dependencies, we'd say
$$F(t)=\frac{Gm_1m_2}{r(t)^2}$$

Now r(t) is a bit complicated as t changes... but for any given value of t, F(t) depends on the value of r(t) at that time but not on the velocity, nor the more or less complicated ways in which r has changed in the past or will change in the future.

7. Jan 27, 2013

### bahamagreen

OK, that gets right to the heart of the matter that I'm pondering.

Imagine I have a special machine that swings a golf club head so that at "r" the club face has reached a specific speed. All golf balls are the same mass, but sometimes when the club swings, the ball is at rest at "r" before contact, other times the ball is rolling through "r" toward the pin so that contact occurs at "r" while the ball has an instantaneous velocity at "r".

Is the club presenting the same instantaneous force at "r" in both cases?
The ball at rest at "r" is going to get a harder smack than the ball rolling through "r" in the direction of the club's movement.

Where does this analogy fail to relate to the gravitational situation?

8. Jan 27, 2013

### A.T.

The question is rather: Where does your analogy even start to resemble gravity?

Aside from the trivial fact that some forces are involved, I see no reason why your collision due to relative motion should have anything to do with a static gravitational field that was postulated, without stating any underlying mechanism, based solely on observation .

9. Jan 27, 2013

### bahamagreen

I admit I'm having trouble with it. Maybe I'm thinking that the force would act differently because of the difference in kinetic energy... I'll sleep on it.

10. Jan 27, 2013

### HallsofIvy

You seem to be under the impression that changing the wording of the question will change the answer. You have been told repeatedly that the motion of a gravitating object relative to another object has not affect on the gravitational force between them. There is nothing more to be said.

11. Jan 27, 2013

### bahamagreen

I'm not doubting it, just having trouble presenting what I'm asking about.

Let me try asking another a simple question.

When gravitational force applied at radius "r" accelerates a mass at "r", does that mass react back against the immediate local instantaneous gravitational force at "r"?

I'm led to think it does because of the definition of inertial mass and Newton's second law.

If so, what characterizes the measurement of the local gravitational force's opposite and equal reaction to accelerating the mass at "r"?

12. Jan 27, 2013

### Vorde

The "equal and opposite reaction" you are talking about is the gravitational force that the object at radius r applies to the massive body that is drawing the object in.

Put otherwise, if object 1 exerts a gravitational pull on object 2 which is r distance away, then the gravitational pull that object 2 exerts on object 1 is 'equal and opposite'

13. Jan 28, 2013

### Khashishi

In Newtonian gravity, gravity is not dependent on momentum--only on mass.

In the more accurate general relativity, gravity depends on mass and momentum of the source. The answer depends on which coordinate system you adopt. In the frame of the small mass, the large mass is moving, and there will be a slight frame dragging effect on the small mass. In the frame of the large mass, the acceleration at t=0 on the small mass is the same regardless of its motion.

14. Jan 29, 2013

### bahamagreen

Thanks for the help so far.
Alright, I think I'm see where my problem is; I'm trying to use a couple of principles that I think are correct, but when I put them together it's not correct. Or maybe it is correct but I'm missing a link somewhere.
We should be able to find where I'm going of course.

The three things are:

Newton's third law
The point of action of a force
The distinction that a force has magnitude, direction, and a point of action:
but a vector only has magnitude and direction

I think it has been confirmed from responses that the point of action of the gravitational force on a mass at radius "r" is applied at "r".

And it has been stated that the small mass' opposite and equal reaction to the force applied at "r" is not applied at "r" but at the COM of the large mass.

The vectors can be positioned to show this because a vector does not have a point of action... but the vectors represent forces which do have points of action.

The force on the small mass at "r" is being applied at "r", and inertial mass is measured by its resistance to acceleration which is also at "r". If that resistance to acceleration is the reaction of Newton's third law, then the point of action of this equal and opposite reaction force looks like it should occur at "r" rather than the COM of the large object.

The vectors representing these forces can be relocated to find the net force, but how is it that the point of action of a force (part of its definition) is allowed to be moved like a vector? How does the reaction force at "r" appear at the COM of the large mass?

It seems like the gravity equation is describing the vectors, relocated and standing in for the forces they represent without points of action, to get the right answer, but when the points of action are assigned to the forces themselves, the correct vector answer is just assigned to the force points of action as "action at a distance"... and this agrees with observation and measurement, is that what is happening?

Last edited: Jan 29, 2013
15. Jan 29, 2013

### A.T.

You do realize that the two equal but opposite forces in Newtons 3rd are always acting on two different objects? So in the cases of Newtonian gravity they obviously cannot act at the same point.

16. Jan 29, 2013

### bahamagreen

It looks to me if the small mass was reacting against a force at "r" radius from the large mass, and the large mass was reacting against a force at its center of mass, then there would be two pairs of unbalanced forces the net of each individual force pair resulting in acceleration, the net resultant of the two unbalanced force pairs being taken together equal and opposite.

I think I'm assuming:

that the point of action for the gravitational force is in the field.
And that the third law applies at this point of action.
And that the fields of the two masses share the common radius "r".
And that at each end of this distance "r" is a pair of unbalanced forces causing acceleration.
And that the resultant net force and acceleration (I think?) of the center of mass of the whole affair is 0.

And if these forces with their points of action are treated a vectors without points of action, then the two force pairs in reaction collapse down to one pair (the two masses only) and these become the two different objects presenting the equal and opposite, and observable reaction.

Is the mistake in assigning the point of action to the field?
Or a misunderstanding of the difference between forces and vectors?
Or a misapplication of N' 3rd Law?

17. Jan 29, 2013

### HallsofIvy

Yes, that is why what you are saying is wrong. The force "acts" on the center of gravity of both objects and r is measured from the their center of masses.

What do you mean by "in the field"?

There are no "unbalanced forces". The force that each of two masses, m and M, apply to each other is $GmM/r^2$. It is only the directions of the forces that are different.

Yes, there is no "outside" force and so the center of mass does not move.

Treating the two vectors without points of action, you would see that they have equal size in opposite directions and so cancel.

I am not clear what you mean by that.

18. Jan 29, 2013

### A.T.

What two pairs? There are only two objects, with one interaction. So one pair, with two forces.
The two masses are accelerating. There is one unbalanced force acting on each of them.

19. Jan 29, 2013

### Khashishi

bahamagreen, there isn't such a thing as a point of action of the gravitational field. The gravitational field exists at every point. In fact, the term "field" usually means a value at every point in space. The total gravitational field from a solid object with geometrical extent is the sum of the field generated by each point that makes up the mass. However, for a spherically symmetric mass, the effect of the shape cancels out, and we can treat the mass as all stored in a single point at the center of the sphere.

Each point in the small mass feels a slightly different acceleration because the field is weaker for points farther away from the large mass (and vice versa). But usually we can approximate the mass as all at the center of mass. In this case, the relevant distance is r12, the distance between the centers of masses of the two objects.

20. Jan 29, 2013

### bahamagreen

I'll name and locate the forces so we can identify whether some of them are incorrect or not being handled correctly...
I'll go step by step to see if we can find the problem.

l is the large mass
s is the small mass

COMl is center of mass of large mass
COMs is center of mass of small mass

"r" is mutual distance between COMl and COMs

[COMl]----------"r"----------[COMs]

Fgsl is gravitational force applied at COMl due to field of small mass at COMl
Fgls is gravitational force applied at COMs due to field of large mass at COMs

[Fgsl]->[COMl]----------"r"----------[COMs]<-[Fgls]

Adding 3rd law... (locally at each mass' point of action)

When the large mass' field acts at COMs, the small mass reacts 3rd law against it locally at COMs
Likewise for the small mass' field acting at COMl.

Frl is the opposite and equal reaction force to application of Fgsl applied locally at COMl
Frs is the opposite and equal reaction force to application of Fgls applied locally at COMs

Locally at COMl

[Fgsl]->[COMl]<-[Frl]

and locally at COMs

[Frs]->[COMs]<-[Fgls]

The force pair for the large mass [Fgsl]->[COMl]<-[Frl] is unbalanced and yields a net acceleration ->
The force pair for the small mass [Frs]->[COMs]<-[Fgls] is unbalanced and yields a net acceleration <-

All together The two pairs of forces diagram like this:

[Fgsl]->[COMl]<-[Frl]----------"r"----------[Frs]->[COMs]<-[Fgls]

with net accellerations like this:

->[COMl]----------"r"----------[COMs]<---

The point of action for the gravitational force is "in the field" in the sense that each mass is in the other mass' field at radius "r" wrt each other. So the point of action is at the COM of each other's mass.

I'll pause at this point to hear if all this is correct so far...