How Do Orbital Dynamics Change Under Different Force Laws?

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Suppose a particle is under the influence of a "1/r^3" force. The particle travels in a circular orbit, but its nudged so that its energy and angular momentum changes slightly. What would happen?

What if the force varies with distance as 1/r^2 (like Earth)? (Earth's orbit is almost circular) What would happen?:confused:
 
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Answer to the second question: a nudge off inverse-square central force orbiting body from circular orbit would make its orbit elliptical. (Of course a circle is an ellipse, but I mean an ellipse with some eccentricity.) I used to know the answer to the first question, but have forgotten. Someone here will know.
 
Janitor said:
Answer to the second question: a nudge off inverse-square central force orbiting body from circular orbit would make its orbit elliptical. (Of course a circle is an ellipse, but I mean an ellipse with some eccentricity.) I used to know the answer to the first question, but have forgotten. Someone here will know.

Thank you Janitor. :smile:
 
Odyssey said:
Suppose a particle is under the influence of a "1/r^3" force. The particle travels in a circular orbit, but its nudged so that its energy and angular momentum changes slightly. What would happen?

What if the force varies with distance as 1/r^2 (like Earth)? (Earth's orbit is almost circular) What would happen?:confused:

Look for the other thread (two threads) on the inverse cube force law.

https://www.physicsforums.com/showthread.php?t=45277

https://www.physicsforums.com/showthread.php?t=45330

The effective potential for an particle in a circular orbit is flat (zero). So the behavior of a particle given a small radial nudge will be the same as a particle on a flat line given a nudge, i.e. dr/dt will be positive or negative and constant. I'd assume this continues on blissfully until the particle impacts the surface, at which point it will be moving quite rapidly in because of the conservation of angular momentum.

If you give the particle a nudge that increases its angular momentum, the effective potential won't be zero anymore, making the problem a little more difficult, so I'll refer you to the first thread above.
 
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