Thus the energy looks just like the energy of a particle of mass min a potential Veff on the half-line
{0< r < ∞}. We have reduced the problem to a 1-dimensional problem! Veff is called the effective
potential. Note that the second term creates the effect of a repulsive force equal to j2/mr3, called
the centrifugal force.
3. Show that
˙r=r2
m(E−Veff (r)).(4)
We could solve this differential equation to find ras a function of t, but it’s nicer to find ras a
function of θ, since this allows us to see the shape of the particles’ orbits. In fact it turns out to be
easier to first find θas a function of rand then solve for rin terms of θ— so that’s what we’ll do.
4. Using equations (3) and (4) show that
dθ
dr =j/mr2
q2
m(E−Veff (r))
.
Conclude that
θ=θ0+Z(j/mr2)dr
q2
m(E−Veff (r))
(5)
Now let’s specialize to the case of gravity, where f(r) = −k/r2and thus V(r) = −k /r
for some constant k.
5. Sketch a graph of the effective potential Veff (r) in this case, and say what a particle moving
in this potential would do, depending on its energy E.
6. Show using equation (5) that
θ=θ0+ arccos
j
mr −k
j
q2E
m+k2
j2
.
This is the only part of this homework where you really need to sweat. However, some ways to do
it are easier than others, so think a bit before you plunge into an enormous masochistic calculation
— if you do it intelligently, you will only need a medium-sized masochistic calculation! For example,
you may want to derive a general formula for
Zdx
√ax2+bx +c
and then use it to do the integral in equation (5).
7. Reduce the clutter a bit more by defining
p=j2/km, e =r1 + 2Ej2
mk2.
Show that in terms of these variables we have
θ=θ0+ arccos p/r −1
e
2
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