
#1
Jan1806, 03:18 PM

P: 1,445

the trajectory of a particle moving in a cnetral potential V(r) is given by [itex] r = a + b \sin(\eta \phi) [/itex] where a b and eta are constants
Compute the potential V(r) in which the particel moves (for arbitrary a, b, and eta) and sketch [itex] V_{eff} (r) = V(r) + \frac{L^2}{2mr^2} [/itex] for cases a =b sketching is not hte problem here.. i just need to find out V(r) well i know that i need to compute the force f first and then integrate f w.r.t. r to get V(r) find find [tex] u = \frac{1}{r} = \frac{1}{a + b \sin(\eta \phi)} [/tex] then using [tex] f = \frac{L^2 u^2}{m} (u'' + u) [/tex] and [tex] u '' = \frac{2b^2 \eta^2 \cos^2 (\eta \phi)}{r^3} + \frac{b \eta^2 \sin(\eta \phi)}{r^2} [/tex] (phew!) once i sub expressions for u'' and u i get [tex] f(r) =  \frac{L^2}{mr^2} \left( \frac{2b^2 \eta^2 \cos^2 (\eta \phi)}{r^3} + \frac{b \eta^2 \sin(\eta \phi)}{r^2} \right)+ \frac{1}{r} [/tex] is that fine? Can i simplify that any more in terms of r? are the steps correct? Any problems with the derivative? Just need to know if i can simplify any further. Your help is greatly appreciated! Thank you 


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