Inverse Cube Force Law

  • Thread starter Spanky1996
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Homework Statement


I'm given a force law is [tex] F = \frac{-k}{r^3} [/tex] and that initially, the particle is in a circular orbit the particle is given an impulse parallel and in the opposite direction to its velocity find the distance from the center for the particle as a function of time.

Homework Equations



I started with the first integral of motion. We know the energy is conserved because this is central force motion (also know angular momentum is conserved).

The Attempt at a Solution



After taking the time derivative of the first integral of motion I get: [tex] r^3*\frac{dr^2}{dt^2} = \frac{L^2-mk}{m} [/tex]

I'd then have to solve that and I'd technically have my answer I think? I guess I'd have to analyze its form and figure out what it's doing when it's going slower than a certain threshold to find it's motion after the impulse is given? Is there possibly a simpler way to approach this problem? Thanks.
 
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Answers and Replies

  • #2
TSny
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Hello and welcome to PF!
After taking the time derivative of the first integral of motion I get: [tex] r^3*\frac{dr^2}{dt^2} = \frac{L^2-mk}{m} [/tex]
Can you show the steps you used to arrive at this? Note that the units don't match for the left and right sides of the equation. Does ##L## represent angular momentum? Did you mean to write ##\frac{d^2r}{dt^2}## instead of ##\frac{dr^2}{dt^2}##?
 
  • #3
Oops yes, that is the 2nd time derivative of r(t)

[tex] E = \frac{1}{2}m(\frac{dr}{dt})^2+\frac{L^2}{2mr^2}-\frac{k}{2r^2} [/tex]

take the time derivative of this equation and that is how I got the expression from my previous post. L is the angular momentum, yes.
 
  • #4
haruspex
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the energy is conserved
So what equation does that give you?
After taking the time derivative
Why would you do that? That's going in the wrong direction. You want to integrate again.
 
  • #5
TSny
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Oops yes, that is the 2nd time derivative of r(t)

[tex] E = \frac{1}{2}m(\frac{dr}{dt})^2+\frac{L^2}{2mr^2}-\frac{k}{2r^2} [/tex]

take the time derivative of this equation and that is how I got the expression from my previous post. L is the angular momentum, yes.
OK. If you take the time derivative of this, you will not quite get what you wrote for the right hand side of your equation in the first post. L2/m does not have the correct units.

But, rather than taking the time derivative of the energy equation, try working with the equation itself. You should be able to manipulate it so that you can integrate it. [as pointed out by haruspex]
 
  • #6
The equation it gives me is the one just above your post I believe.

I suppose you're right I maybe should not take the time derivative. I guess E can be found with the initial conditions. [tex] E = \frac{L^2}{2mr_o^2}-\frac{k}{2r_o^2} [/tex]

from there I have a first-order equation I suppose.
 
  • #7
TSny
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I suppose you're right I maybe should not take the time derivative. I guess E can be found with the initial conditions. [tex] E = \frac{L^2}{2mr_o^2}-\frac{k}{2r_o^2} [/tex]

from there I have a first-order equation I suppose.
Yes
 
  • #8
So, TSny and Haruspex I think I've got this one figured out. What if the radial velocity at time = 0 isn't 0? What if the particle is given some type of kick outward or inward? Would I then want to turn it into a 2nd order D.E. so I could take that initial condition into account? Or is there a better way?
 
  • #9
TSny
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So, TSny and Haruspex I think I've got this one figured out. What if the radial velocity at time = 0 isn't 0? What if the particle is given some type of kick outward or inward? Would I then want to turn it into a 2nd order D.E. so I could take that initial condition into account? Or is there a better way?
I don't think it would change things much. You would just need to account for an adjustment in the angular momentum. Unless I'm overlooking something.
 
  • #10
I don't think it would change things much. You would just need to account for an adjustment in the angular momentum. Unless I'm overlooking something.
If the impulse was purely radial the angular momentum wouldn't change right? Shoot, have I done something wrong with the original problem since the angular momentum would change? I plugged it into the integral and treated as a constant. I'm not sure if there is a mistake there or not now...
 
  • #11
haruspex
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have I done something wrong with the original problem since the angular momentum would change?
Hard to say without seeing your work. You are not given the size of the impulse.
Note that it says opposite to velocity direction. It ought to add that the result is reduced speed. This is important in the nature of the solution.
 

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