Central Force Problem: Nature of Orbits for F=-ar/(r^3) & U=-(a/r)

[neelakash]

Homework Statement

Central force F=-ar/(r^3) & Central potential energy,U=-(a/r)
(not U_eff)
Find the nature of orbits if (i)a>0 and (ii)a<0

Homework Equations

The Attempt at a Solution

If we remember the attractive central force E=E(r) diagram,i.e.the one showing the graph of U_eff,we only need to know E_total=K+U.
Where only PE is given.
We see,

U= -integration[F.dr]=integration[dW]=-integration[dK]=-K

Then K=a/r and U=-(a/r)

So,the E=K+U=0

Then in positive and negative both caes we get a parabolic orbit.

Please check if i am correct.

 

[neelakash]

Consider a>0.
Then, -(m(v^2)/r)=-(a/r^2)
or,m(v^2)=(a/r)
Hence total energy E=K+U= -(1/2)(a/r)

=>elliptic orbit.

Consider a<0.

Then,F=+b/r where b=-a>0
and U_p=b/r

K is itrinsically positive.So,total energy positive.

From e=sqrt[1+{(2L^2*E)/(m*b^2)}]
e>1.
hence hyperbolic trajectory.