Gravitation - Position of Stable Equilibrium

In summary: M moves towards its initial position, but not in a straight line since there is a net horizontal force." is incorrect. The graph does not show that M moves towards its initial position, it shows that there is a net downward force pointing to the initial point.
  • #1
little neutrino
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1

Homework Statement


A particle of mass 3m is located 1.00 m from a particle of mass m. You put a third mass M so that the net gravitational force on M due to the two masses is exactly zero. Is the equilibrium of M at this point stable or unstable for points along the line passing through M and perpendicular to the line connecting m and 3m?

Homework Equations


F = GMm/r^2

The Attempt at a Solution


I found that the answer is stable equilibrium by drawing a free body diagram, because there is a net downward/upward force pointing to the initial point. However, will M move in a straight line towards the initial point? How do I derive this mathematically? Thanks!
 
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  • #2
Hello neutrinino, :)

Down and up meaning what ? The way you describe it (pointing to the initial point) one would think the equilibrium is stable !?
 
  • #3
BvU said:
Hello neutrinino, :)

Down and up meaning what ? The way you describe it (pointing to the initial point) one would think the equilibrium is stable !?
Yes I meant stable equilibrium sorry! Edited my post alr!
 
  • #4
So we have a y-axis through M with a particle p1 of mass 3 m on the left and a particle p2of mass m on the right. Force on M is zero, so the x-coordinates of p1 and p2 can be calculated from the relevant equation. This way we know what we are talking about (e.g. "the initial point" is now: "the origin") and we can go into details.


You claim that from a free body diagram you can conclude that there is a net downward/upward force pointing to the initial point (*) and I am very curious how you can derive that. So in that respect I have the same question to you as you have to PF...

I can see that you are approximately right in what you write, but it's only one part of the total, and even that one part may be not exactly right !

To avoid going back and forth (c.q. around in circles :) ) : what are the characteristics of a stable equilibrium in a potential field in two dimensions ? (Two dimensions is enough because of the rotational symmetry of the original configuration around the x-axis).

[edit]: (*) a statement that can be conceived as correct (since it doesn't say that that is the only non-zero component of the force).

I will err on the safe side and I think you are on the safe side too, since you explicitly ask "will it move there in a straight line" (i.e. will it stay on the line x = 0) . It will do so if the net horizontal component of the force is zero.

Your relevant equation isn't good enough in this exercise: you have already made use of the vector character of ##F\;## in your (qualitative?) reasoning. And you need this vector character some more: You will need to work out the net x- and y-components for a quantitative treatment.

Another approach (less work, more insight required) is to look at the ratio of the distances to each of the particles from a point ( 0, y ), knowing that ratio at 0,0 . If that ratio stays the same, the answer to your straight line motion question is yes (because the mass ratio stays the same also).
 
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  • #5
I tried solving it again, my answer is M will move towards its initial position, but not in a straight line. Attached is my working. Is it correct? Thanks!
 

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  • #6
Sub in the values for x, d1 and d2 to obtain 2 expressions with an unknown y.
For the sake of convenience using QuickGraph, change y to x. I.e. y is referred to as x in QuickGraph.
Graph the 2 expressions in QuickGraph.
If the 2 graphs overlap, it means all values of y will give F1x = F2x and M will move to initial position in a straight line.
However, as shown in attached graph, the 2 graphs only overlap at x = 0 (read y = 0), thus M will move towards its initial position, but not in a straight line since there is a net horizontal force.
 

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  • #7
Once it starts moving off the line, what happens to the forces ?
 
  • #8
There is still a net vertical force pointing towards y = 0, and a net horizontal force pointing towards 3m (since F1x > F2x for y > 0, M moves off the line towards 3m, so distance between M and 3m decreases, increasing the gravitational force between them, and vice versa for m). This means that M moves increasingly close to 3m in a curved path? But how is this stable equilibrium?
 
  • #9
Where was it established that there is a stable equilibrium at all ?

[edit] Your statement "However, as shown in attached graph, the 2 graphs only overlap at x = 0 (read y = 0), thus M will move towards its initial position, but not in a straight line since there is a net horizontal force." doesn't sound healthy at all: the net force does not point to (0,0) , so why would it move there at all ?
 
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  • #10
Oh this is a qn from University Physics textbook and the answer key says stable equilibrium .-.
 
  • #11
Is that an argument ? Even universities are known to have made mistakes. Was the problem statement in post #1 correct and complete ?
 
  • #12
Attached are the screenshots for your reference
 

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  • #13
I see what you mean. Looks pretty authoritative. But...

Apparently you've already done the stability under disturbances in the x-direction for y=0 and concluded: unstable. For good reasons: (0,0) is a saddle point.

Now I come to a question I have witheld so far (until we had established the criteria asked for in post #4) : what about stability under disturbances in the x-direction for y##\ne##0 ? You've already found a net horizontal force (I haven't checked the calculations -- I'm too lazy -- but I trust your result because it agrees with my more intuitive approach (*) ). From there we both conclude: unstable
(it shoots off toward the bigger mass and forms a black hole there -- at least mathematically :)
[edit] o:)nonsense. In mechanics we have conservation of angular momentum, so it will go into orbit around 3m or something. Never mind. Shouldn't have brought that up. Don't bother teacher with that ...).


The only way I can curb this into stable is to assume that the third mass is constrained only to move on the line x=0. Like a bead on a wire. But there is no mentioning of that whatsoever in 13.9.

My conclusion: with all respect for the book answer, unstable on x=0 for all y. And stationary (which is also unstable) only in (0,0).


(*) when on y=0 the distances squared have a ratio 3 (needed for net force = 0), for y ##\ne## 0 that ratio becomes ##\displaystyle {3x_1^2 +y^2\over x_1^2+y^2}\; <\;3## , so the 3m "wins".
 
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  • #14
Ok thanks! :D
 
  • #15
BvU said:
Where was it established that there is a stable equilibrium at all ?

[edit] Your statement "However, as shown in attached graph, the 2 graphs only overlap at x = 0 (read y = 0), thus M will move towards its initial position, but not in a straight line since there is a net horizontal force." doesn't sound healthy at all: the net force does not point to (0,0) , so why would it move there at all ?

Sorry I meant M will move towards y = 0, but not in a straight line.
 
  • #16
Ah, perhaps I can attribute my deviating answer to the way the exercise is formulated:
it says (you rendered it just fine):
(a) Where should you put a third mass M so that the net gravitational force on M is 0 ?
(b) Is the equilibrium of M at this point stable or unstable
(i) for points along the line connecting m and 3 m
(ii) for points along the line passing through M ##\perp## the line connecting m and 3 m​

And what I read in post #1 (which you also rendered just fine) was something else (namely -- rephrased -- does M return to (0,0) if released from (y,0) when y ##\ne## 0)
( (0,0) being the point found in (a) )

Basically I have an issue with the formulation ... equilibrium of M at this point for points ...
which I (non-native english speaker) find meaningless; I would have preferred
... equilibrium of M at this point wrt disturbances in the ... direction (... = x in (i), y in (ii) )

I think you, little ##\nu##, have really placed your finger on a sore spot in this exercise: when you go in depth it becomes flaky. All the authors intended was this simple "net force opposite displacement ##\Rightarrow## stable". And the more I think about it, the more I think it's wrong. The net force is approximately opposite the displacement. The approximation is fantastically good for small displacements, but it's never exact.
 
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  • #17
little neutrino said:
Sorry I meant M will move towards y = 0, but not in a straight line.
Yes, and you have already seen in (a) that it's unstable there!
 
  • #18
BvU said:
Ah, perhaps I can attribute my deviating answer to the way the exercise is formulated:
it says (you rendered it just fine):
(a) Where should you put a third mass M so that the net gravitational force on M is 0 ?
(b) Is the equilibrium of M at this point stable or unstable
(i) for points along the line connecting m and 3 m
(ii) for points along the line passing through M ##\perp## the line connecting m and 3 m​

And what I read in post #1 (which you also rendered just fine) was something else (namely -- rephrased -- does M return to (0,0) if released from (y,0) when y ##\ne## 0)
( (0,0) being the point found in (a) )

Basically I have an issue with the formulation ... equilibrium of M at this point for points ...
which I (non-native english speaker) find meaningless; I would have preferred
... equilibrium of M at this point wrt disturbances in the ... direction (... = x in (i), y in (ii) )

I think you, little ##\nu##, have really placed your finger on a sore spot in this exercise: when you go in depth it becomes flaky. All the authors intended was this simple "net force opposite displacement ##\Rightarrow## stable". And the more I think about it, the more I think it's wrong. The net force is approximately opposite the displacement. The approximation is fantastically good for small displacements, but it's never exact.

Ahh yes I agree! Thanks again! :)
 
  • #19
BvU said:
I see what you mean. Looks pretty authoritative. But...

Apparently you've already done the stability under disturbances in the x-direction for y=0 and concluded: unstable. For good reasons: (0,0) is a saddle point.

Now I come to a question I have witheld so far (until we had established the criteria asked for in post #4) : what about stability under disturbances in the x-direction for y##\ne##0 ? You've already found a net horizontal force (I haven't checked the calculations -- I'm too lazy -- but I trust your result because it agrees with my more intuitive approach (*) ). From there we both conclude: unstable
(it shoots off toward the bigger mass and forms a black hole there -- at least mathematically :)
[edit] o:)nonsense. In mechanics we have conservation of angular momentum, so it will go into orbit around 3m or something. Never mind. Shouldn't have brought that up. Don't bother teacher with that ...).
The only way I can curb this into stable is to assume that the third mass is constrained only to move on the line x=0. Like a bead on a wire. But there is no mentioning of that whatsoever in 13.9.

My conclusion: with all respect for the book answer, unstable on x=0 for all y. And stationary (which is also unstable) only in (0,0).(*) when on y=0 the distances squared have a ratio 3 (needed for net force = 0), for y ##\ne## 0 that ratio becomes ##\displaystyle {3x_1^2 +y^2\over x_1^2+y^2}\; <\;3## , so the 3m "wins".

Haha don't worry I took the forming a black hole as a figure of speech :-p
 
  • #20
I posted this on the homework helper hideout (asking the others to check us).
If no one shows we're wrong I might ask the author (Young died in 2013, but Geller is at UCSB).
 
  • #21
BvU said:
I posted this on the homework helper hideout (asking the others to check us).
If no one shows we're wrong I might ask the author (Young died in 2013, but Geller is at UCSB).

Ok! :w
 
  • #22
Still have to convince my colleague homework helpers that there is something flaky in the exercise and the reasoning towards the solution.
 

1. What is the definition of "gravitation"?

Gravitation is a natural phenomenon by which physical bodies with mass are attracted to each other.

2. What is the position of stable equilibrium in relation to gravitation?

The position of stable equilibrium in relation to gravitation is the point where an object is in a state of balance and will not move unless an external force is applied.

3. How is the position of stable equilibrium affected by the mass of an object?

The position of stable equilibrium is affected by the mass of an object because the greater the mass, the stronger the gravitational force between objects. This can result in a shift in the position of stable equilibrium.

4. Can the position of stable equilibrium change over time?

Yes, the position of stable equilibrium can change over time due to various factors such as changes in mass, distance, and external forces acting on the objects.

5. How is the position of stable equilibrium used in practical applications?

The position of stable equilibrium is used in various practical applications, such as designing stable structures and determining the stability of objects in space or on land. It is also important in understanding the motion of celestial bodies and the behavior of particles in physics experiments.

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