- #1
spacetimedude
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Homework Statement
A particle of mass m moves under the influence of a central force
F(r)=-mk[(3/r^2)-2a/r^3]rhat
Show that if the particle is moving in a circular orbit of radius a, then its angular momentum is L=mh=m√(ka)
Homework Equations
L=mvr for circular orbit
The Attempt at a Solution
From the equation, we extract the second derivative of the position vector r''=-k[(3/r^2)-2a/r^3]rhat. We integrate this in order to find the velocity to use in L=mvr.
(At this point, can I dot product both sides by rhat to get rid of the vector?)
r''=-k[(3/r^2)-2a/r^3]
Integrating:
r'=-k[-3/r+a/r^2]=-k[(-3r+a)/r^2]
Using the angular momentum equation L=mvr=m*r'*r:
L=m*-k[(-3r+a)/r^2]*r.
We have r=a, so
L=m*(-k[(-3a+a)/a^2]*a)=m2k.
So obviously, the integration is wrong because I haven't included the integrating constant and also the answer we get is wrong. And intuitively by substituting in r=a, we are saying that r doesn't change.
How could I correctly approach this problem?
Thank you!