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You know the angular momentum equation that makes keplers second law work

L=rmv where L is angular momentum is r radius, m is mass and v is velocity.

Well according to that equation if we decrease the radius then either the mass or the velocity would have to increase (sort of how keplers second law works) right? Well then if we decreased the radius to Planck length or something extremely small, wouldn't that mean that the object would have to increase its velocity (since it can't increase its mass )so it could keep angular momentum constant?... Pls explain this to me (try to use equations) but any help would be helpful...Thanx in advance

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    $\begingroup$ Right, Newtonian mechanics get spoiled by that. But Quantum mechanics kicks in. At a small scale, everything is "smeared" and not located at one single point. Don't ask me, I'm myself working on a deeper understanding. $\endgroup$ – Gyro Gearloose Sep 26 '20 at 16:20
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You are right, given a point mass and some way to pull the mass toward its center of revolution. But it would take an infinite amount of energy to reduce that radius of revolution to zero.

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  • $\begingroup$ thanx for the answer but why would it take a lot of energy to do that? I mean its as simple as pulling a string $\endgroup$ – alienare 4422 Sep 27 '20 at 4:17
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    $\begingroup$ See if you can calculate the force required to pull the string, as a function of radius- remembering that angular momentum must remain constant. $\endgroup$ – S. McGrew Sep 27 '20 at 5:21
  • $\begingroup$ well... it wasn't easy but I gave it a go: $\endgroup$ – alienare 4422 Sep 27 '20 at 10:33
  • $\begingroup$ centripetal force = ((angular momentum)^2)/mr^(3) (you can prove the equation your self if you substitute angular momentum equation to the above equation) so that would mean just halving the radius would have an affect of increasing the centripetal force by a factor of 8....which would only affect the tension of the string right? So I don't see why you need a lot of force to pull the string...I mean there should be zero force since theres a reaction force balancing centripetal force $\endgroup$ – alienare 4422 Sep 27 '20 at 10:41
  • $\begingroup$ Also it would be helpful if you could relate that force to some type of energy so that I could get a sense of how much energy will be needed...thanx in advance $\endgroup$ – alienare 4422 Sep 27 '20 at 10:46

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