Particle between two masses

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Homework Statement


Statement of the problem (quoting from my assignment):
a) write equations of motion
b) try to solve analytically

Given: m1, m2 - two masses
R - distance between two masses

Homework Equations


V=-G(m1/r + m2/(R-r))
F=-dV/dr

The Attempt at a Solution


a) Equations of motion: v = dr/dt, a=dv/dt
b) Solution
dV/dt = G(m1/r2 - m2/(R-r)2)

Separating variables
dV=Gm1*1/r2dr+Gm2*1/(R-r)2dr

Integrating both sides I basically get what I started with
V=-G(m1/r +m2/(R-r))+C

So, I know what I am doing is not right. I know that somehow the equations of motions need to come into play, but don't understand the relation. My system is in equilibrium, so both velocity and acceleration are equal to 0.
I am not a physics student (I am in Math), but I'm working on a little summer project in Physics.
 
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Answers and Replies

  • #2
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My system is in equilibrium, so both velocity and acceleration are equal to 0.
Then you don't have to solve any differential equations. Just set the acceleration to zero and solve for r.
That doesn't seem to agree with your problem statement, however. Solving the equations of motion for the equilibrium point only is trivial.
 
  • #3
Ray Vickson
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Homework Statement


Statement of the problem (quoting from my assignment):
a) write equations of motion
b) try to solve analytically

Given: m1, m2 - two masses
R - distance between two masses

Homework Equations


V=-G(m1/r - m2/(R-r))
F=-dV/dr

The Attempt at a Solution


a) Equations of motion: v = dr/dt, a=dv/dt
b) Solution
dV/dt = G(m1/r2 - m2/(R-r)2)

Separating variables
dV=Gm1*1/r2dr-Gm2*1/(R-r)2dr

Integrating both sides I basically get what I started with
V=-G(m1/r - m2/(R-r))+C

So, I know what I am doing is not right. I know that somehow the equations of motions need to come into play, but don't understand the relation. My system is in equilibrium, so both velocity and acceleration are equal to 0.
I am not a physics student (I am in Math), but I'm working on a little summer project in Physics.
(1) I don't see anywhere in the problem statement the stipulation that your system is in equilibrium. In fact, it specifically asks for equations of motion.
(2) The derivative ##dV/dt## has no relevance in this problem; the derivative ##dV/dr## is what matters here.
(3) The concept of "equilibrium" is suspect in this case. If you happen to be at the equilibrium point and with zero velocity, then even the tiniest fluctuation to one side or the other will put you into a location that will cause the particle to crash into either ##0## or ##R##. So, even if you were not quite at equilibrium, but were off a by a bit in the 200th decimal place, your system would move away and eventually crash.
 
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  • #4
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Ray,mfb,
Thanks for the input. I changed the equation for potential. And well, yes you both are probably right, let's assume that the system is not in equilibrium. Then the equations of motion would be
F1 = m1d2r/dt2 and F2=m2d2(R-r)/dt2 ?
 
  • #5
vela
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It's not clear to me what the situation is. Are you asked to write the equation of motion for an object between the two masses, which is what the potential in your original post suggests, or are you being asked to write the equations of motion for the two masses, which is what your last post suggests?
 
  • #6
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Well, yes it wasn't clear to me either. But, I think now it is. So, I need the equation of motion for the test particle, and I hope it will look like this
$$F=m_0\ddot r (t),$$
where ##m_0## is the mass of the test particle.
Then the differential equation will be
$$m_0\ddot r=-G \frac {m_1} {r^2} + G\frac {m_2} {(R-r)^2}$$
The only thing I am not sure is the signs. And also how to solve it. It's separable and I probably need to reduce the order by some smart substitution. But, Ray, you mentioned that the equation can be simplified by using the fact that the system's energy is conserved. How would you go about it?
 
  • #7
vela
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The righthand side is missing a factor of ##m_0##.

You seem to be assuming that the test particle lies directly between the two masses. Was this your intent?
 
  • #8
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Vela,
I am not sure what you mean by the right hand side missing a factor of ##m_0##. Yes, the particle lies between the two masses.
 
  • #9
vela
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What's Newton's law of gravity?
 
  • #10
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Aha! So, ##m_0##'s cancel, and we have:

$$\ddot r=-G \frac {m_1} {r^2} + G\frac {m_2} {(R-r)^2}$$
I hope, this time it's correct.
 
  • #11
haruspex
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Aha! So, ##m_0##'s cancel, and we have:

$$\ddot r=-G \frac {m_1} {r^2} + G\frac {m_2} {(R-r)^2}$$
I hope, this time it's correct.
Yes, but that is what you had already in the line following
b) Solution
Where you went wrong is here:
Integrating both sides
You integrated the left hand side wrt t but the right hand side wrt r. That is invalid.
As Ray hinted in post #3, you need to express the acceleration in terms of dv/dr. This is a standard trick which you've probably encountered.
 
  • #12
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Thanks haruspex,
So, we have:
$$\frac {dv} {dt} = -G\frac {M_1} {r^2} + G\frac {M_2} {(R-r)^2}$$
And using the standard trick:
$$v\frac {dv} {dr} = -G\frac {M_1} {r^2} + G\frac {M_2} {(R-r)^2}$$
Now, separate the variables and integrate:
$$\int v \,dv=-\int G \frac {M_1} {(R-r)^2}\,dr + \int G \frac {M_2} {r^2}\,dr$$
$$v^2=-2G\frac {M_1} {(R-r)} + 2G \frac {M_2} {r}$$
$$\frac {dr} {dt}=\pm\sqrt {-2G\frac {M_1} {(R-r)} + 2G \frac {M_2} {r}}$$

What do I do now? I guess, I could separate the variables again, but integrating it is not going to be easy. Any hints appreciated. By the way, I'd like to thank everybody chipping in, it's been fun working this out, so far.
 
  • #13
haruspex
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Thanks haruspex,
So, we have:
$$\frac {dv} {dt} = -G\frac {M_1} {r^2} + G\frac {M_2} {(R-r)^2}$$
And using the standard trick:
$$v\frac {dv} {dr} = -G\frac {M_1} {r^2} + G\frac {M_2} {(R-r)^2}$$
Now, separate the variables and integrate:
$$\int v \,dv=-\int G \frac {M_1} {(R-r)^2}\,dr + \int G \frac {M_2} {r^2}\,dr$$
$$v^2=-2G\frac {M_1} {(R-r)} + 2G \frac {M_2} {r}$$
$$\frac {dr} {dt}=\pm\sqrt {-2G\frac {M_1} {(R-r)} + 2G \frac {M_2} {r}}$$

What do I do now? I guess, I could separate the variables again, but integrating it is not going to be easy. Any hints appreciated. By the way, I'd like to thank everybody chipping in, it's been fun working this out, so far.
Right. And you might notice the first integration gives you the energy equation but with mass cancelled out.
 
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