Relative velocity of two gravitationally attracting spheres

[etotheipi]

Homework Statement

Two masses $$m_{A}, m_{B}$$ of radii $$r_{A}, r_{B}$$ are released from rest at infinity. Find their final relative velocity at the point of collision.

Relevant Equations

$$GPE = -\frac{Gm_{A}m_{B}}{r}$$

There have been some other threads on similar problems but none address one specific point I'm confused about.

The change in GPE of a body is the negative of the work done on that body by a gravitational field between two points; by this logic, since the same (but opposite) gravitational forces have acted on both A and B from infinity to a final separation of r_A + r_B, I assumed that both undergo a change in GPE of $$\Delta GPE = -\frac{Gm_{A}m_{B}}{r_{A} + r_{B}}$$ the negative of which should be then equal to the work done on each body and consequently also the change in kinetic energy.

So I thought that the final relative velocity of the masses should be $$v_{rel}= |v_{A}| + |v_{B}| = \sqrt{\frac{2Gm_{B}}{r_{A}+r_{B}}} + \sqrt{\frac{2Gm_{A}}{r_{A}+r_{B}}}$$ However, the solution given equates the sum of the final kinetic energies to one GPE, as in the following $$\frac{Gm_{A}m_{B}}{r_{A} + r_{B}} = \frac{1}{2}mv_{A}^{2} + \frac{1}{2}mv_{B}^{2}$$ which when followed through by considering momentum (that part of the argument is fine) yields a different relative velocity of $$\sqrt{\frac{2G(m_{A}+m_{B})}{r_{A}+r_{B}}}$$I don't understand how we can equate the drop in GPE of only one body to the gain in KE of both bodies, since surely this same amount of work is done on each separately and is not shared - I was wondering if someone could help me out. Thanks a bunch!

 

[PeroK]

The GPE equation you quote is valid only when the other mass is assumed to be so large that it does not move. Try deriving the correct equation for GPE of a system where both masses move.

 

[etotheipi]

The GPE equation you quote is valid only when the other mass is assumed to be so large that it does not move. Try deriving the correct equation for GPE of a system where both masses move.

I think I've gotten really confused somewhere. Here's where I've got to,

Considering the case of a 'source' mass assumed stationary, we can calculate the work done on another mass by the source's gravitational field as usual and equate this to the change in GPE: is this the GPE of only the moving body, or the GPE of the system of both masses?

Similarly, if both masses are now allowed to move then both masses do work on each other and undergo some decrease in potential energy with respect to the other's field. Do these individual decreases sum to the overall decrease in GPE of the system?

I wonder whether this confusion comes down to the definition of GPE I am using. By the negative of the work done by gravity, does this mean the negative of the work done on one body only by other gravitational fields (which would imply the GPE is a property of a single body) or the work done by all objects in a system by gravitational forces?

Sorry about this!

 

[PeroK]

I think I've gotten really confused somewhere. Here's where I've got to,

Considering the case of a 'source' mass assumed stationary, we can calculate the work done on another mass by the source's gravitational field as usual and equate this to the change in GPE: is this the GPE of only the moving body, or the GPE of the system of both masses?

Similarly, if both masses are now allowed to move then both masses do work on each other and undergo some decrease in potential energy with respect to the other's field. Do these individual decreases sum to the overall decrease in GPE of the system?

I wonder whether this confusion comes down to the definition of GPE I am using. By the negative of the work done by gravity, does this mean the negative of the work done on one body only by other gravitational fields (which would imply the GPE is a property of a single body) or the work done by all objects in a system by gravitational forces?

Sorry about this!

Do the maths, as they say. Both masses represent accelerating reference frames, so best to use the initial rest frame.

 

[etotheipi]

Do the maths, as they say. Both masses represent accelerating reference frames, so best to use the initial rest frame.

It seems like there is a really straightforward solution but I've been trying all morning and just can't figure out where to start! I've confused myself to the point where I don't even know what the GPE actually corresponds to.

For instance, say we only have 2 bodies. Lots of places say that when we compute the work done by the gravitational force of the other we deduce the potential energy of that body, whilst others say that is the potential energy of the 2 body system.

Going back to basics, if one body is stationary and we bring another mass from infinity we deduce the work done to be$$\int_{infinity}^{x} - \frac{GMm}{r^{2}} dr$$which, when made negative, gives us our good old GPE formula. I think this corresponds to the GPE of this body with respect field associated with the stationary one. However, many places quote that this potential energy accounts for the entire system. I guess this makes conceptual sense, since then when the moving body has reached the larger one, the larger one has accelerated so little (COM argument) - which can be assumed to be negligible as you mentioned earlier 0 - that all the kinetic energy has been transferred to the moving body.

Now if both bodies are able to move, I don't really know what's going on. For instance, if one starts at x = 0 and the other at x = infinity, they will fall toward some point in the middle as both do work on the other one, decreasing the potential energy of each in the other's fields. I can only assume that the sum of the individual decreases in potential energy would equal the original GPE formula, but don't know how I would set this up.

 

[hutchphd]

Perhaps this will help:
If you are talking about the system, then you are tacitly making all of your measurements in the Center of Mass frame of that system. Then the ΔPE and KE balance.
Clearly kinetic energy of individual particles will change according to frame choice and the (non-relativistic) kinetic energy must include the C of M energy if you choose to change your frame of reference.
If one of the particles is infinitely massive then this fixes the C of M frame as static.

 

[PeroK]

Try integrating using the force on each body and the infinitesimal time and displacement. This will give you increase in KE by reduction in separation. From this you can solve the problem directly and generate the true equation for the GPE of the system.

 

[etotheipi]

Try integrating using the force on each body and the infinitesimal time and displacement. This will give you increase in KE by reduction in separation. From this you can solve the problem directly and generate the true equation for the GPE of the system.

I really am sorry about this, I've been staring at it literally all day but it just won't click. I set body A to be at x=0 and body B to be at x=infinity.

Just for reference (so if what I do below makes no physics sense at all) I would derive the body-A-is-stationary case like so:

$$-\Delta GPE_{system} = W_{total} = W_{on A} + W_{on B} = 0 + \int_{infinity}^{x_{1}} \frac{-GMm}{x^{2}} dx$$

Since body A can no longer be considered stationary, work is done on A as well as B so

$$-\Delta GPE_{system} = W_{total} = W_{on A} + W_{on B} = \int_{0}^{x_{1}} F_{a} dx + \int_{infinity}^{x_{2}} F_{b} dx$$

where F_A and F_B are functions of the distances between the x coordinates. I don't know how to represent this distance because the distances moved by both bodies will depend on their respective masses.

 

[PeroK]

The first observation or calculation is that the force on each body is the same, hence the acceleration of each body is related to the mass of the other. As this is the true at all times, the velocity and displacement of each body are also similarly related. Formally you can integrate wrt to get this.

Then you have the relationship between the total separation and the displacement of each mass. It's the separation distance that you really want to focus on.

Finally you have the relationship between KE and force times displacement. From that you get a formula for the total incremental increase in KE in terms of the incremental change in separation distance. You can then integrate that.

That's how I'd do it.

 

[etotheipi]

The first observation or calculation is that the force on each body is the same, hence the acceleration of each body is related to the mass of the other. As this is the true at all times, the velocity and displacement of each body are also similarly related. Formally you can integrate wrt to get this.

Then you have the relationship between the total separation and the displacement of each mass. It's the separation distance that you really want to focus on.

Finally you have the relationship between KE and force times displacement. From that you get a formula for the total incremental increase in KE in terms of the incremental change in separation distance. You can then integrate that.

That's how I'd do it.

Thank you for all of your help; I'm going to put this a rest for tonight because I'm not really thinking properly at the moment (:D) and try this tomorrow.

 

[ehild]

The change in GPE of a body is the negative of the work done on that body by a gravitational field between two points; by this logic, since the same (but opposite) gravitational forces have acted on both A and B from infinity to a final separation of r_A + r_B, I assumed that both undergo a change in GPE of $$\Delta GPE = -\frac{Gm_{A}m_{B}}{r_{A} + r_{B}}$$ the negative of which should be then equal to the work done on each body and consequently also the change in kinetic energy.

The PE of the system will be equal to the negative of the work done on the bodies by the force of gravity between them.
The bodies are in rest at infinite distance between them. You can choose their centre of mass as the origin of the frame of reference. Imagine you move body A to the position x=-r_A. Is there any work done? The other body is infinitely far away...
Imagine body A is kept at place x=-r_A and body B is brought to x=r_B. The work done by gravity is $W = \frac{Gm_{A}m_{B}}{r_A+r_B}$, so the gravitational potential energy of the system is $GPE = -\frac{Gm_{A}m_{B}}{r_A+r_B}$

 

[PeroK]

The KE of the system will be equal to the negative of the work done on the bodies by the force of gravity between them.
The bodies are in rest at infinite distance between them. You can choose their centre of mass as the origin of the frame of reference. Imagine you move body A to the position x=-r_A. Is there any work done? The other body is infinitely far away...
Imagine body A is kept at place x=-r_A and body B is brought to x=r_B. The work done by gravity is $W = \frac{Gm_{A}m_{B}}{r_A+r_B}$, so the gravitational potential energy of the system is $GPE = -\frac{Gm_{A}m_{B}}{r_A+r_B}$

That can't be right. That's effectively the solution where only one body moves.

 

[hutchphd]

That can't be right. That's effectively the solution where only one body moves.

Usually your arguments are very clear to me but I don't understand this one.

If neither body is constrained then the net change in gravitational PE all goes to total KE of both bodies apportioned according to momentum conservation. The work is done by both (reciprocally but not equally!) and equals change in PE
If one body is constrained then all the (same) PE now goes to the KE of the nonstationary ball. At collision obviously this C of M frame is not the initial C of M frame because of the external constraint required. The work is done by the stationery ball and equals the change in PE..

I don't see what you are arguing here...what am I missing?

 

[ehild]

That can't be right. That's effectively the solution where only one body moves.

Starting with both bodies at infinity and arriving to the situation when body A is at position xA and body B is at xB is accompanied with the same work independent of the way how does it happen, as the force of interaction is conservative - either both move. or first one moves, then the other one. The gravitational potential energy of two bodies A and B is $GPE=-\frac {Gm_Am_B}{d}$ where d is the distance between the bodies. In the same way, the electric potential energy between point charges Q1 and Q2 is $U=\frac {kQ_1Q_2}{d}$

 

[etotheipi]

It certainly appears to be the case that the gravitational potential energy of a system as viewed from a fixed reference frame must be the same no matter how it is set up.

In the case where one body is assumed stationary (large mass), the work done by gravitational forces on the system reduces to the work done on only the moving body (no work is done on the stationary body) so the change in GPE of the system is derived as normal.

If now we consider both bodies as capable of moving, the work done by gravitational forces on the system includes the work done on both bodies. I believe PeroK was alluding to the more complex but essentially equivalent derivation, though it does seem that this would yield the same formula for GPE.

 

[PeroK]

Starting with both bodies at infinity and arriving to the situation when body A is at position xA and body B is at xB is accompanied with the same work independent of the way how does it happen, as the force of interaction is conservative - either both move. or first one moves, then the other one. The gravitational potential energy of two bodies A and B is $GPE=-\frac G{m_AM_B}{d}$ where d is the distance between the bodies. In the same way, the electric potential energy between point charges Q1 and Q2 is $U=\frac {kQ_1Q_2}{d}$

Yes, sorry. I wasn't thinking. That's the quick way.

 

[PeroK]

It certainly appears to be the case that the gravitational potential energy of a system as viewed from a fixed reference frame must be the same no matter how it is set up.

In the case where one body is assumed stationary (large mass), the work done by gravitational forces on the system reduces to the work done on only the moving body (no work is done on the stationary body) so the change in GPE of the system is derived as normal.

If now we consider both bodies as capable of moving, the work done by gravitational forces on the system includes the work done on both bodies. I believe PeroK was alluding to the more complex but essentially equivalent derivation, though it does seem that this would yield the same formula for GPE.

No. Perok just got it wrong!

 

[ehild]

The kinetic energy of the relative motion can be derived from Newton second law, by integration. without the concept of potential energy.
There are two interacting point-like bodies, with no external forces, body A at position xA, body B at position xB. The force between them is along the line connecting the bodies. Body B exerts force f on A, body A exerts force -f on B.
From Newton's second law:
$m_A\ddot x_A = f$
$m_B\ddot x_B = -f$.

$f=G\frac{m_Am_B}{(x_B-x_A)^2}$

Divide the first equation with m_A, the second one by m_B and subtact them:
$\ddot x_A - \ddot x_B = f(\frac{1}{m_A}+\frac{1}{m_B})$
The relative position of B with respect to A is $X = x_B-x_A$.
$\frac{1}{m_A}+\frac{1}{m_B}=1/\mu$ (reciprocal of the reduced mass).
$\mu=\frac{m_A m_B}{m_A+m_B}$.
So we got the equation
$\mu\ddot X=-f$,
with
$f=G\frac{\mu(m_A+m_B)}{X^2}$
as if there would be a single body with mass mu in the gravitation field of a mass $M=m_A+m_B$.

The relative velocity is $V=\dot X= \dot x_B-\dot x_A$
$\ddot X =\frac{dV}{dX}v$
$\frac{dV}{dX}v=-G\frac{M}{X^2}$
Integrate with respect X between infinity (where V=0) and $X=r_A+r_B$.