Central force problem

[bluesky]

A particle moves in a circular orbit under the action of a force

f®=-(k/r^2).If k is suddenly reduced to half its value, what would be the nature of the orbit?

 

I used e=sqrt[1+(2*L^2*E)/(mk^2)]

 

My attempt:

Clearly,the particle moves under attractive central force.Now,for the circular orbit,eccentricity e=0 and as the motion is bound,the energy is negative.

If k is reduced to k/2, eccentricity changes to

e=1+(8*L^2*E')/(mk^2)=6*L^2*E'/(mk^2)

 

I assumed that E has changed to E'.Unless,I do not know E',I cannot infer anything.Please help.

[Qfwfq]

...and as the motion is bound,the energy is negative.

How sure are you that the motion will still be bound, after k has diminished by half?

 

I assumed that E has changed to E'.Unless,I do not know E',I cannot infer anything.Please help.

Well the potential energy certainly has changed, while the sudden change of k wouldn't affect kinetic energy...

 

BTW it's a good idea to use our LaTeX feature to make math more easily readable:

 

[math]e=\sqrt{1+\frac{2L^2E}{mk^2}}[/math]

 

[math]e=1+\frac{8L^2E'}{mk^2}=\frac{6L^2E'}{mk^2}[/math]