Calculate the ratio of the radii of the orbits

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In summary, two objects of masses move in circular orbits that have the same center. The force that gives rise to this motion is a force of attraction between the two objects acting along the line joining them. Conservation of momentum and N's 2nd Law allow us to calculate the ratio of the radii of the orbits.
  • #1
StephenPrivitera
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Two objects of masses m and M move in circular orbits that have the same center. The force that gives rise to this motion is a force of attraction between the two objects acting along the line joining them. Use conservation of momentum and N's 2nd Law to show that they move with the same angular speed. Calculate the ratio of the radii of the orbits.

My attempt:
Conservation of Momentum
mv1+Mv2=C

N's 2nd Law
Fnet=ma

For uniform (I assume, because I know no better, rather than show that the motion is uniform - feel free to give me a lesson on that topic) circular motion: F=mv2/r

s=[the]r
ds/dt=v=d[the]/dt*r=wr

So, F=mrw2, for uniform circular motion

But since v=wr
mwr + MWR = K

dP/dt= mrw2 + MRW2=F1+F2=0

dP/dt=0=mr(dw/dt)+MR(dW/dt)

mr(dw/dt)+MR(dW/dt)= mrw2 + MRW2=0
mr(dw/dt-w^2)=MR(W^2-dW/dt)=0
dw/dt=w^2
dW/dt=W^2
And you may have thought I was going somewhere with that. But I'm not. No matter what I do to these equations I can't get w=W!
A little hint would be nice. Thank you.
 
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  • #2
Seems no one likes my problem :( - Well, here's another anyway. A billard ball collides head on with a second billard ball that is touching a third billard ball (I'm guessing this means they form a straight line). Assume all collisions are elastic. Describe the motion of the balls.

Not much I can do with this problem. Since they all have approximately the same mass, p conservation yields
v1i= v1f+v2f+v3f
energy conservation yields
v1i2= v1f2+v2f2+v3f2
There are three unknowns (I consider v1i a known), but two equations. So at best I can give some sort of relation between the variables, but there appears to be so many possible outcomes that this would be of little use.
Anyway, I came up with
v1f*v1i+v2f*v3f=v1f
through various calculations. Can anyone do any better?
 
  • #3
Originally posted by StephenPrivitera
So, F=mrw2, for uniform circular motion

But since v=wr
mwr + MWR = K
Correct. Now since the attraction is mutual, you know F1 = -F2. Now you use a reference frame where the center-of-mass is at rest. This means K=0. And there you are.

As for the 2nd problem:
Do you really not know what happens in this case ? If so, you should try a little experiment - it's very instructive. Well, I can help you with the math...
 
  • #4
You mean that since
mwr + MWR=0
that Psystem=0=mVsystem?
I can agree with that if that means the CM has v=0. I guess I didn't consider that the constant is arbitrary depending on your reference frame.
But the idea is to show that the angular velocities are equal (ie. w=W).
And given that K=0 doesn't help that matter.
w/W = -(MR)/(mr)
Depending on which direction I define to be negative, either r or R will be negative. If w=W, then w/W=1. Or, |(MR)/mr|=1, which is only true if MR=mr. We don't know that to be true.
?
As for the second question, I'm sorry I don't know what happens. I could speculate, but that's not physics. I thought I'd do some math instead. I would appreciate some help with that matter, anyway. If you can describe what happens that would be nice, but I like to see the math behind it all. Actually, I was playing pool today (ironically), but it didn't occur to me do that experiment. I was busy paying attention to whether they really depart at right angles or not. I don't know. I guess they do. I was wondering how one would treat a problem when the velocity of the billard ball is not collinear with the line joining the point of contact and the CM... Still wondering.
 
  • #5
Fine. Please show me a more direct way to get to this result. I've done it, but it's excessively complicated I think. Say there are two balls, and one is initially at rest.
v1i=v1f+ v2f
v1i2= v1f2+ v2f2
= (v1f+ v2f)2
v1f2+ 2v1f*v2f + v2f2= v1f2+ v2f2
v1f*v2f =0
so either v_1f=0, v_2f=0 or both. Clearly it can't be both, since that would violate conservation of momentum.
Using the first two equations,
v1i-v1f=v2f
v1i2-v1f2= v2f2
Divide each by each to get,
v_1i+v_1f=v_2f
multiply by v_2f on each side
v_1i=v_2f
thus, v_1f=0
Alternatively, I could have multiplied by v_1f on each side. I would have found,
v_1f=-v_1i
and v_2f=0
Unlikely. Strange that the math supports that possibility. My guess is the complicated nature of the solution made the last one feasible. There must be a way to show that v_2f[x=]0. But anyway, that's only the solution for two balls. Could I find the solution where two are at rest by considering the collision which immediately occurs when the second ball begins to move. It would (ideally) be the same as the first collision with v_2f=v_1i, and thus I would expect to see the last ball move with v_3f=v_1i, and the other two balls at rest, in contact. But again, at this point, pure speculation.
 
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  • #6
Originally posted by StephenPrivitera
But the idea is to show that the angular velocities are equal (ie. w=W).
1) actio = reactio =>
F1 + F2 = 0
mrw^2 + MRW^2 = 0.
2) conservation of momentum =>
mrw + MRW = 0

2) => mrw = -MRW
2) into 1) -MRWw + MRW^2 = 0
=> -w + W = 0
=> W = w
========

into 1) mr + MR = 0
=> mr = -MR
=> r/R = -M/m
============
 
  • #7
Originally posted by StephenPrivitera
so either v_1f=0, v_2f=0 or both. Clearly it can't be both, since that would violate conservation of momentum.
Right. And that's all that math can do for you here. Now, we know that a collison happens, and that means, ball #2 gets some momentum, and from the math we know that it gets all. And that's it.
I would expect to see the last ball move with v_3f=v_1i, and the other two balls at rest, in contact.
Correct.
 

1. What is the formula for calculating the ratio of the radii of the orbits?

The formula for calculating the ratio of the radii of the orbits is R2/R1 = (T2/T1)2/3, where R2 and R1 are the radii of the orbits, and T2 and T1 are the orbital periods of the two objects.

2. How do you determine the orbital periods of two objects?

The orbital periods of two objects can be determined by measuring the time it takes for each object to complete one full orbit around the central body, such as a planet or a star. This can be done using observations or calculations based on the objects' masses and distances from the central body.

3. What is the significance of calculating the ratio of the radii of the orbits?

Calculating the ratio of the radii of the orbits allows us to understand the relative sizes of the orbits of two objects in a system. This can provide insights into the dynamics and stability of the system, as well as the masses and distances of the objects.

4. Can the ratio of the radii of the orbits change over time?

Yes, the ratio of the radii of the orbits can change over time due to various factors such as gravitational interactions with other objects, tidal forces, and the objects' own orbital evolution. However, in stable systems, this ratio tends to remain relatively constant.

5. How is the ratio of the radii of the orbits used in astrophysics?

The ratio of the radii of the orbits is used in astrophysics to study various celestial objects and systems, such as planets and their moons, binary star systems, and galaxies. It can also be used to estimate the masses and distances of objects in these systems, and to make predictions about their future evolution.

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