Question about gravitational potential

UKBoy1977

Thinker
U
Joined
Dec 2, 2002
Messages
211
I have an assignment to do on this and this question is doing my head in. I know it's basic but I can't seem to get it sorted out.

In a system where other gravitational influences can be discounted, two particles of equal mass, m, are fixed at positions X=0 and X=Xo on the x-axis.

1) Derive an expression for the gravitational potential, Vgrav, at a general position x on the x-axis

2)What is the physical significance of a point where dVgrav/dx = 0?

b) State the coordinates of such a point in this system (You should not require any mathematics for this)


Now I've had a go but it's all a bit of a mess. I've got an answer to part 1 but I'm confused over what terms should be -ve and +ve, I think the answer to 2a) is it's a point where if a particle was placed there it would not move as the gravitational force from each of the two masses are balancing each other out and I'm pretty sure the answer to 2b) is the midpoint between X and Xo, but the whole idea of what Vgrav means and how dVgrav/dx changes as you move along the x-axis is confusing me.

Can anyone give some tips?
 
Last edited:
UKBoy1977 said:
I've got an answer to part 1 but I'm confused over what terms should be -ve and +ve,
Click to expand...
I don't think anything's positive.

The gravitational potential due to a single body is proportional to -1/r, where r is the distance to the body.

The potential due to the pair of bodies is just the sum of the potentials due to each body.

At a point x, the distance to 0 is |x| and the distance to Xo is |x - Xo|, so the potential is proportional to -1/|x| - 1/|x - Xo|. I guess the mass should be in there somewhere, and maybe also the gravitational constant G. But that's the basic idea.


UKBoy1977 said:
I think the answer to 2a) is it's a point where if a particle was placed there it would not move as the gravitational force from each of the two masses are balancing each other out and I'm pretty sure the answer to 2b) is the midpoint between X and Xo,
Click to expand...
That's correct. (Except I assume you meant 0 instead of X.)


UKBoy1977 said:
but the whole idea of what Vgrav means and how dVgrav/dx changes as you move along the x-axis is confusing me.
Click to expand...
dV/dx is the gravitational force. The reason is simply that V is defined to be something for which that's true.

(Actually, the sign is backwards there; the force is really -dV/dx. Gravity pulls things towards lower potential.)
 
UKBoy1977 said:
Now I've had a go but it's all a bit of a mess. I've got an answer to part 1 but I'm confused over what terms should be -ve and +ve, I think the answer to 2a) is it's a point where if a particle was placed there it would not move as the gravitational force from each of the two masses are balancing each other out and I'm pretty sure the answer to 2b) is the midpoint between X and Xo, but the whole idea of what Vgrav means and how dVgrav/dx changes as you move along the x-axis is confusing me. Can anyone give some tips?
Click to expand...
Potentials are, in some respects, kind of abstract things. You can't directly measure a gravitational potential. What you measure is gravitational force. So what's the relationship between gravitational force and gravitational potential? I'm not going to give you the answer, but it's not complex (at least, not once you're comfortable with calculus). And once you understand that relationship, the answers (or at least some of them) should fall right into place.

Edit: 69dodge gave you the answer I was hinting at.
 
UKBoy1977 said:
I have an assignment to do on this and this question is doing my head in. I know it's basic but I can't seem to get it sorted out.

In a system where other gravitational influences can be discounted, two particles of equal mass, m, are fixed at positions X=0 and X=Xo on the x-axis.

1) Derive an expression for the gravitational potential, Vgrav, at a general position x on the x-axis

2)What is the physical significance of a point where dVgrav/dx = 0?

b) State the coordinates of such a point in this system (You should not require any mathematics for this)


Now I've had a go but it's all a bit of a mess. I've got an answer to part 1 but I'm confused over what terms should be -ve and +ve, I think the answer to 2a) is it's a point where if a particle was placed there it would not move as the gravitational force from each of the two masses are balancing each other out and I'm pretty sure the answer to 2b) is the midpoint between X and Xo, but the whole idea of what Vgrav means and how dVgrav/dx changes as you move along the x-axis is confusing me.

Can anyone give some tips?
Click to expand...

We generally define the gravitational potential as the negative of the work we need to do on a unit mass to remove it to infinity against the gravitational pull.

So the gravitational potential at a distance r from a mass m is

V = -Gm/r

The gravitational potential due to a bunch of things is the sum of their individual potentials. Since any conservative force is minus the gradient of the potential energy, there is no force where the derivative of the potential vanishes. If the masses are equal, the point at which the gravitational force of each cancels the other should be obvious.

That is enough information for you to work the problem.
 
Hmm, there are ethical issues involved in helping people with homework. But at this point, most of the answer's been given to you anyway.

UKBoy1977 said:
2)What is the physical significance of a point where dVgrav/dx = 0?
Click to expand...
I think that it's important to note that while derivatives, partial derivatives, and gradients are closely related concepts, they are not the same thing. I don't know if you have studied multidimensional calculus yet, and perhaps your instructor is simply wanting to discuss the physics without getting into the fine points of the math, but I think that it is somewhat dangerous to be using derivatives and gradients interchangeably. If you parameterize a curve with a variable x, define some potential function V(x), and then take the derivative with respect to x, then the answer is the force projected onto the curve that x parameterizes. In this particular case, the force is solely in the x direction, so it doesn't make a difference, but you should realize that in general, if you take a derivative with respect to x, then you're ignoring the y and z components of the force.

State the coordinates of such a point in this system (You should not require any mathematics for this)
Click to expand...
In continuation of the previous point, if your instructor has reduced this problem to a one-dimensional case, then there aren't coordinates; there is only one coordinate: the x-coordinate. I think that it's important to make it clear whether we are discussing a three-dimensional space or a one-dimensional one, and use the appropriate nomenclature.

Cyphermage said:
We generally define the gravitational potential as the negative of the work we need to do on a unit mass to remove it to infinity against the gravitational pull.

So the gravitational potential at a distance r from a mass m is

V = -Gm/r
Click to expand...
A more general definiton, which applies to any conservative force, is that a "potential" is any function whose gradient is equal to the force. This is similar to the indefinite integral, which is defined as a function whose derivative is equal to the function. Just as indefinite integrals can differ by a constant, potentials can differ by constants, as a constant has no effect on the gradient. Thus, the potential at one place is simply the change in potential with respect to some other point. The potential is completely relative; there is no absolute potential. This is actually an important point used to support the law of conservation of charge. If we want to add a charged particle to a system, we should put it at zero potential, otherwise we are adding not just charge but energy. But to do so would require us to decide which potential is "zero" in an absolute sense, which we cannot do.

Thus, the canonical gravitational potential is -Gm/|v| (note that v is a vector, not a scalar), but it can also be given as c-Gm/|v| or -Gm[(1/|v0|)-(1/|v|)] for some reference vector v0.

Since any conservative force is minus the gradient of the potential energy, there is no force where the derivative of the potential vanishes.
Click to expand...
Huh? I don't understand what you're saying here.
 
Last edited:
Art Vandelay said:
Huh? I don't understand what you're saying here.
Click to expand...

The relationship between a conservative force and its associated potential energy is..

F = -grad U

The force is zero when the gradient is zero. In the one dimensional case, the force is zero when the derivative of the potential energy vanishes.
 
69dodge said:
That took me a bit of time to parse too.

Replace "there is no force" with "the force is 0".
Click to expand...
Ah. Reminds me of a joke: which has more legs, three horses, or no horses?
No horses, because three horses have twelve legs, but no horse has twenty legs.


And as long we're editing, I think "negative" would be better than "minus".
 
You must log in or register to reply here.

Back
Top Bottom