bluesky Posted May 4, 2007 Report Posted May 4, 2007 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. Quote
Qfwfq Posted May 9, 2007 Report Posted May 9, 2007 bluesky said: ...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? bluesky said: 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] Quote
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