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Gravitation independnet of motion/momentum?

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Khashishi
#19
Jan29-13, 11:24 AM
P: 887
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.
bahamagreen
#20
Jan29-13, 07:22 PM
P: 543
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]

Adding gravity...

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...
Vorde
#21
Jan29-13, 08:15 PM
Vorde's Avatar
P: 784
The thing you're missing is that the your Fgsl is equal to your Fgls. The equation has been posted in this thread already so I won't repeat it but look at the equation for universal gravitation and think about it for a while and you'll come to that conclusion.

Once you realize that, then you'll realize that you don't need 4 'forces' because the Fgsl = (and opposite) to Fgls and therefore they are each other's equal and opposite forces.
TurtleMeister
#22
Jan29-13, 08:28 PM
TurtleMeister's Avatar
P: 757
You're making this much more difficult than it is. Think of a balance scale with a fulcrum between the two masses, the large mass on one side, and the small mass on the other side. The fulcrum points to the COM of the two bodies. This COM should be the inertial frame of reference for all of your calculations. If the two bodies (M1 and M2) are interacting gravitationally with no other forces involved then according to Newton's second law; F1 = M1A1 and F2 = M2A2, where A1 and A2 are the accelerations of M1 and M2 respectively. And according to Newton's third law; F1 = -F2. It's a balanced system. The forces are always equal in magnitude and opposite in direction, and the momentums are always equal in magnitude and opposite in direction. And this is true regardless of the velocity of the bodies (as long as you take the COM as your frame of reference).

Edit: Of course you'll need the universal law of gravitation to determine F, then you can use A = F/M to get the accelerations. The above is to illustrate how the system is balanced.
bahamagreen
#23
Jan30-13, 04:05 AM
P: 543
Let's step back just a moment to make sure of something...

If a single mass is accelerated by some force, the point of action of that force is the COM of the single mass, and the point of action of the 3rd law equal and opposite reaction is also the COM of this single mass. So, both force and reaction forces points of action are coincident in the COM of this single mass.

Please indicate if this is correct for a single mass being accelerated by a force. If this is incorrect, please indicate where the points of action for the force and reaction should be assigned.
A.T.
#24
Jan30-13, 05:14 AM
P: 4,064
Quote Quote by bahamagreen View Post
If a single mass is accelerated by some force,
What force? Action at a distance or direct interaction with another mass?

Quote Quote by bahamagreen View Post
the point of action of that force is the COM of the single mass, and the point of action of the 3rd law equal and opposite reaction is also the COM of this single mass.
Not if it is action at a distance, like Newtons gravity.

Quote Quote by bahamagreen View Post
So, both force and reaction forces points of action are coincident in the COM of this single mass.
Let me repeat my question to you:

Do you 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, can they?
xAxis
#25
Jan30-13, 05:35 AM
P: 210
Quote Quote by bahamagreen View Post
Let's step back just a moment to make sure of something...

If a single mass is accelerated by some force, the point of action of that force is the COM of the single mass, and the point of action of the 3rd law equal and opposite reaction is also the COM of this single mass. So, both force and reaction forces points of action are coincident in the COM of this single mass.

Please indicate if this is correct for a single mass being accelerated by a force. If this is incorrect, please indicate where the points of action for the force and reaction should be assigned.
That's incorect because there are no two forces acting on COM. The reaction force of each body is the action force of the other. So there is only one pair of forces.
bahamagreen
#26
Jan30-13, 07:46 AM
P: 543
If the gravitational field acts locally against the small mass, doesn't the small mass react locally against the field?

If not the latter, how the former?
Vorde
#27
Jan30-13, 09:24 AM
Vorde's Avatar
P: 784
You're making this much more confusing than you need to. Newton's third law doesn't stipulate that the reverse force has to occur at the same spot, just that one occurs. With mechanical forces the spot is the same, with forces at a distance like gravity this is not the case.
A.T.
#28
Jan31-13, 03:43 AM
P: 4,064
Quote Quote by bahamagreen View Post
If the gravitational field acts locally against the small mass, doesn't the small mass react locally against the field?
Not in Newtonian gravity.
Quote Quote by bahamagreen View Post
If not the latter, how the former?
Newtonian gravity doesn't propose any underlying mechanism. It just quantifies the effect.
xAxis
#29
Feb1-13, 10:42 AM
P: 210
Quote Quote by bahamagreen View Post
If the gravitational field acts locally against the small mass, doesn't the small mass react locally against the field?

If not the latter, how the former?
Just to add to A.T.s explanation, Newton proposed that every part of matter gravitationaly attracts every other part of matter. If you for instance add vectorialy all gravitational forces that the sun exerts on earth, the resultant force will act on COM of the earth, and will point to the COM of the sun.
The idea of gravitational field came after Newton.
bahamagreen
#30
Feb1-13, 04:37 PM
P: 543
I looked here to see if it would clear up my misunderstanding.

Feynmann's Lectures on Physics, Vol 1
Chapter 12 Characteristics of Force
Section 12-4 Fundamental forces, Fields

He discusses the electric force and the gravitational force. He introduces concepts with the electric force:

He states: "...the formula gives the force between two objects only when the objects are standing still."

He says, ""To analyze this force as a field concept... ... we say that the force F on q2 at R can be written in two parts."

Then, "One part says the something produces a field. The other part says that something is acted on by the field." (his italics)

In the next paragraph he turns to gravity; his first sentence is, "In the case of gravitation, we can do exactly the same thing."

-----

Now the inadequacy of the formula for electric force when the objects are moving sounds like my original question, but I'm not sure how much of the electrical force discussion is meant to be extended with the "exactly like gravity" statement.

The use of two forces for each object when using the field concept does look like what I was asking about.

Maybe someone can look at these two pages and tell me something about it.
A.T.
#31
Feb1-13, 05:23 PM
P: 4,064
If you want to avoid action at a distance, but still want momentum conservation, the you have treat the field as an object that carries momentum. But the force is still independent of the motion.
bahamagreen
#32
Feb1-13, 05:46 PM
P: 543
OK, so assigning the reaction forces to the field momentum... then at the two objects' locations there are two forces, the gravitational force applied by the field to the object, and the object's inertial resistance to acceleration applied back to the field.

So that disposes of bare "action at a distance", but it does that by putting in its place "maintenance of a field at a distance"...

So the field is being "pushed" by the object's reaction, and the field is being pushed oppositely and equally by the other object's reaction, and since the field is a sum of the two objects' fields, the net is balanced... the 3rd law.

Is that it?


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