Consider the following potential energy in a two-body system: U(r) = -A/r^n Wher
ID: 1979473 • Letter: C
Question
Consider the following potential energy in a two-body system:
U(r) = -A/r^n
Where nA>0 and n is an integer.
a) what does the condition on nA tell you about this fore?
b) Write down the effective potential, Ueff(r).
c) Sketch Ueff(r) for n= -2, 1 and 3 each on separate axes (assume that the angular momentum l is a constant and non-zero).
d) Find the radius ro for which the particles orbit at fixed separation (circular orbits).
e) For realistic orbits of this type, do you see any restrictions on n? Do your sketches confirm this conclusion?
Explanation / Answer
a) That it is an attractive force. F= - dU/ dr = -nA / r^(n+1)
b) Effective potential is a sum of the original potential energy plus the angular part of the kinetic energy.
Ueff(r) = U(r) + 1/2 m r^2^2= U(r) + 1/2L^2 /(mr^2), where L is the angular momentum
c) Seems like drawing a diagram is quite challenging in this website, so let me explain in words.
For n=-2, U(r) is a potential energy of the harmonic oscillator. Overall the graph looks somewhat parabolic. As r goes to 0 or infinity, U(r) diverges to + infinity.
For n=1, as r goes to 0, it diverges to + infinity. As r goes to infinity, it diverges to 0, but in between there is a small lump that corresponds to the local maxima. Between r=0 and this local maxima, there is a local minima, where the orbit is stable.
For n=3, as r goes to 0, it diverges to (IMPORTANT) - infinity. As r goes to infinity, it diverges to 0. There is no local minima anywhere, so any orbit in unstable.
d) dUeff / dr = 0 is the stable(fixed value of r) orbit.This corresponds to
nA/r^(n+1) -L^2 / (mr^3)=0
r^(n-2)= nAm / L^2
e)As one can see in part c, for n=3, the orbit is unstable. In fact, any value of n larger than 2 will result in Ueff diverging to - infinity as r goes to 0, which means there will not be a stable circular orbit. In other words, n must be smaller than 2.