# Gravitation - Position of Stable Equilibrium

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1. Jan 26, 2015

### little neutrino

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
F = GMm/r^2

3. 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!

Last edited: Jan 26, 2015
2. Jan 26, 2015

### BvU

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. Jan 26, 2015

### little neutrino

Yes I meant stable equilibrium sorry! Edited my post alr!

4. Jan 26, 2015

### BvU

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).

: (*) 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).

Last edited: Jan 26, 2015
5. Jan 26, 2015

### little neutrino

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!

#### Attached Files:

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6. Jan 26, 2015

### little neutrino

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. Jan 27, 2015

### BvU

Once it starts moving off the line, what happens to the forces ?

8. Jan 27, 2015

### little neutrino

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. Jan 27, 2015

### BvU

Where was it established that there is a stable equilibrium at all ?

 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 ?

Last edited: Jan 27, 2015
10. Jan 27, 2015

### little neutrino

Oh this is a qn from University Physics textbook and the answer key says stable equilibrium .-.

11. Jan 27, 2015

### BvU

Is that an argument ? Even universities are known to have made mistakes. Was the problem statement in post #1 correct and complete ?

12. Jan 27, 2015

### little neutrino

Attached are the screenshots for your reference

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13. Jan 27, 2015

### BvU

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 :)
 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".

14. Jan 27, 2015

### little neutrino

Ok thanks! :D

15. Jan 27, 2015

### little neutrino

Sorry I meant M will move towards y = 0, but not in a straight line.

16. Jan 27, 2015

### BvU

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.

17. Jan 27, 2015

### BvU

Yes, and you have already seen in (a) that it's unstable there!

18. Jan 27, 2015

### little neutrino

Ahh yes I agree! Thanks again! :)

19. Jan 27, 2015

### little neutrino

Haha don't worry I took the forming a black hole as a figure of speech

20. Jan 27, 2015

### BvU

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).