Gravitation - Position of Stable Equilibrium

[little neutrino]

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!

 

[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 !?

 

[little neutrino]

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!

 

[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).

[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).

 

[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!

 

[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.

 

[BvU]

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

 

[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?

 

[BvU]

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 ?

 

[little neutrino]

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

 

[BvU]

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

 

[little neutrino]

Attached are the screenshots for your reference

 

[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 :)
[edit] 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".

 

[little neutrino]

Ok thanks! :D

 

[little neutrino]

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.

 

[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.

 

[BvU]

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!

 

[little neutrino]

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! :)

 

[little neutrino]

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] 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

 

[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).

 

[little neutrino]

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

 

[BvU]

Still have to convince my colleague homework helpers that there is something flaky in the exercise and the reasoning towards the solution.