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so that by (1)
E =
m 2 ( ˙r^2 +
j mr^2
r^2 ) + V (r).
m 2
r˙^2 +
j^2 2 mr^2
=
m 2
r˙^2 + Veff (r)
where
Veff =
j^2 2 mr^2
Thus the energy looks just like the energy of a particle of mass m in 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 first term creates the effect of a repulsive force equal to j^2 /mr^3 , called the centrifugal force.
- From this new description of energy, we have
r˙ = ±
m
(E − Veff (r)) (4)
so that from (3) dθ dr
θ˙ r ˙
j^2 /mr^2 √ 2 m (E^ −^ Veff^ (r))
We now have θ as a function of r since
θ(r) = θ(r 0 ) +
∫ (^) r
r 0
dθ dr (s)ds
= θ(r 0 ) ±
∫ (^) r
r 0
j^2 /ms^2 √ 2 m (E^ −^ Veff^ (s))
ds (5)
for any r 0 in (0, ∞).
Now let us specialize to the case of gravity, where f (r) = −k/r^2 and thus V (r) = −k/r for some constant k.
- First let’s examine the graph of the effective potential:
Veff (r) =
j^2 2 mr^2
k m
For a given total energy E of q, we have from equation (4) the classically allowed values for r (i.e., {r | Veff (r) ≤ E}). In this case, we see from the following figure, a set of values for E which will yield a unique behavior for r as a function of t. In addition, equation (4) gives us the analogy between ˙r with respect to r and that of the velocity of a skateboarder with respect to her position on the ramp (r, Veff (r)) under a roof of height E. Suppose the total energy of q is E 1 and is equal to the minimum value of the effective potential. In this case there is only one classically allowed value:
r =
j^2 mk
E 3
j^2 mk
Veff
E 2
−mk^2 2 j^2 E^1
r
Figure 1: Gravitational effective potential
and the only interesting thing here is to ask about the mood of the skateboarder. Suppose now that the total energy is E 2 such that −mk^2 2 j^2
< E 2 < 0.
We see the radius accelerate down the ramp away from the attracting body, then decelerate up the ramp, touch the roof and return with the same speed pattern to do it again. If the total energy E 3 is such that E 3 ≥ 0 we have enough energy to escape the pull of the attracting body and the distance from this object becomes unbounded. It should be noted, when E 3 = 0 we have the minimum value of energy to escape this pull
- Now, to find θ as a function of r, by equation (5) we must find the following antiderivative: ∫ j/r^2 √ −j^2 ( (^) r^12 ) + 2km( (^1) r ) + 2mE
dr.
Use the substitution x = j/r to transform the above into ∫ −dx √ ax^2 + bx + c
where a = −1, b = 2km/j and c = 2mE.
To solve this problem, first set α to satisfy
cos α =
x + 2 ba √ b^2 − 4 ac 4 a^2
thus w(t), and hence q(t) is on the circle of radius p.
If 0 < e < 1, then 1 − e^2 > 0 so that (8) can be written as
( x + (^1) −pee 2
p 1 −e^2
) 2 +^
y^2 ( √p 1 −e^2
thus w(t), and hence q(t) is on an ellipse with foci at (0, 0) and ( 1 −−pee 2 , 0).
If e = 1, solving for x in (8) gives
x = −
y^2 2 p
p 2
thus q(t) is on a parabola.
If e > 1, then e^2 − 1 > 0 so that (8) can be written as
( x + (^1) −pee 2
p 1 −e^2
) 2 −^
y^2 ( √p e^2 − 1
and hence q(t) is on an hyperbola with foci at (0, 0) and ( 1 −−pee 2 , 0).
- So when e = 0 we have an orbit of a circle, solving for the total energy we see
E =
−mk^2 2 j^2
which is the minimum value of the effective potential as expected. When 0 < e < 1 we’re on an ellipse, and this occurs when −mk^2 / 2 j^2 < E < 0. When E = 0, e = 1 and we’re on a parabola. And when E > 0, e > 1 so we’re on a hyperpola.
