# Central Force problem

by Ed Quanta
Tags: central, force
 P: 297 A particle of mass m is subjected to a repulsive force k/r^5, where k is a constant. Initially it is at a very large distance from the force center O and has a velocity of magnitude such that, if it were directed along a path through the center of the force, the closest distance of approach would be a. Actually, it is projected, with the same velocity, along a path which would pass the force center at a distance b if it were not deflected. Find the minimum linear velocity (in terms of k,a,b,m) that the particle experiences as it follows its trajectory. I have no idea how to go about doing this. Any suggestions would be well appreciated.
 PF Gold P: 2,921 Ed, looks as a trivial problem to practice with the idea of potential energy etc.
 P: 297 Thank you, and that will probably make calculations a lot messier. I apologize for being an idiot, but I don't see how this problem is so trvial. I don't suppose you could be more specific?
 Sci Advisor HW Helper P: 1,772 Central Force problem IT starts as a balanced "KE to PE" equation. At a "great distance" the potential energy is zero, so all energy is kinetic. At the closest distance, all the energy is potential. Use integration to calculate the work done to bring the object from "infinity" to distance "a." (work=force "dot" distance but the force increases as distance,r, decreases.) THis amount of work is equal to maximum PE at "a" and maximum KE at a great distance; from this you find v.
 P: 297 I'm sorry, but I am still not sure how the distance b fits into this problem. What is meant by the mass would travel a distance b through the force center if not deflected? Is b greater than the distance a?
 P: 618 Ed Quanta asks: "What is meant by the mass would travel a distance b through the force center if not deflected?" But that's not what the problem you originally posted says. What it does say is: "Actually, it is projected, with the same velocity, along a path which would pass the force center at a distance b if it were not deflected." So the starting velocity of the particle is directed along a straight line whose closest approach to the force center is the distance b. That is if the particle continued along this path, it would miss the force center by the distance b. Of course this means that the particle isn't aimed directly at the center which in turn means thatit will be pushed even further off course and will miss the force center by some distance greater than b. Does that help?
 P: 297 So I am calculating the work done by an integral from a to infinity which is just really using the fact that potential energy is 0 when the particle is an infinite distance from the force center. And thus its potential energy is greatest when its at its closest approach to the force center a distance a. Then we calculation an integral from b to infinity and understand the work done by the force to push away the particle. And then we just examine where v is lowest during this trajectory wouldn't it be where kinetic energy is the lowest at point a? I guess the whole point is expressing this in terms of b as well though. Thanks jdavel, I think I understand.
 P: 618 arivero posted: "looks as a trivial problem to practice with the idea of potential energy etc." I don't think so. This is a two part problem. And unless I'm missing something here (which is certainly possible!) only the first part can be solved using just PE, KE and total energy conservation. But that first part only gives the initial conditions (speed and direction) of a particle in an unusual central potential. The second part requires that the actual trajectory of the particle be found. Doesn't it? Also, to Ed Quanta: It looks to me as though this is a problem from an introductory QM or modern physics class where you've learned about the Rutherford scattering experiment. I can't remember whether that analysis requires that you find the distance of nearest approach of the alpha particles to the nuclei. If so, you might find help in that analysis.
 P: 297 I think what you said is pretty much on the money jdavel. Thanks for the help.
 P: 1,322 The potential energy function is simply the integral of the force, so $U(x) = -\frac{k}{4r^4}$ The kinetic energy is simply $T(v) = \frac{1}{2}mv^2$ The force is conservative, so no work is done by any non-conservative forces. Therefore the energy of the object is conserved. At point (a) the object has no kinetic energy, so the total energy is simply given by $U(a)$ At its initial point the total energy is given by $U(r_o) + T(v)$ Set the initial and final energies equal to each other and solver for v. Or did I misread the problem?
 P: 618 JohnDubYa asked: "Or did I misread the problem?" You didn't misread; you just didn't quite finish reading. In the first half of the problem you were given the force and the fact that v(a)=0, from which you correctly showed how to find v(oo). In the second part of the problem, you're given that the speed of the particle at infinity is the same as before, but now its aimed along a path whose closest approach to the force center (FC) is the distance b. If the particle actually followed this straight line path, you could figure out how fast it would be going at b (just like you did before), and since b is the closest approach, it would be the point of greatest PE, so v(b) would be the minimum speed. But since the particle isn't aimed directly at the the FC, it will get pushed off this straight line path and never go through b. So it's maximum PE will be at some other place, say b', and v(b') is now the minimum speed. Finding b' doesn't look easy to me. What do you think?
 P: 1,322 You are right, I quit reading too early. This looks like the classic scattering problem, with b as the impact parameter. The solution should be found in many nuclear physics books, or by doing a Google search on "classical scattering cross-section impact parameter."

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