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J = mq × q˙
and if we use the expression for ˙q obtained in (4), we have
q × q˙ = (r cos θˆı + r sin θˆ) × [( ˙r cos θ − r θ˙ sin θ)ˆı + ( ˙r sin θ + r θ˙ cos θ)ˆ] = [r^2 θ˙ cos^2 θ + r r˙ cos θ sin θ − r sin θ( ˙r cos θ − r θ˙ sin θ)]ˆk = r^2 θ˙k,ˆ
so that J = mr^2 θ˙ˆk.
- We use equation (2) to solve for θ˙ in terms of r:
θ^ ˙ = j mr^2
Combining this and equation (1) we express E in terms of r:
E =
m r˙^2 + Veff(r) (6)
where
Veff(r) = V (r) + j^2 2 mr^2
The only thing to note here is that θ˙^2 = j^2 /m^2 r^4.
- We solve (6) for ˙r to obtain
r˙ =
m
(E − Veff(r)). (7)
It should be noted that in our use of the symbol for the positive square root we are not asserting that r˙ is positive! It is entirely possible that the above root is negative! This, as we will discuss below (in # 5) will not effect the form of our solution for r in terms of θ.
- Using (5) and (7) show that
dθ dr
θ˙ √ 2 m (E^ −^ Veff(r))
j/mr^2 √ 2 m (E^ −^ Veff(r))
By the chain rule, we have that
θ˙ = dθ dr
r,˙
which when combined with (7) (and subsequently (5)) gives:
dθ dr
θ˙ √ 2 m (E^ −^ Veff(r))
j/mr^2 √ 2 m (E^ −^ Veff(r))
Upon integration we arrive at
θ = θ 0 +
(j/mr^2 )dr √ 2 m (E^ −^ Veff(r))
Now let’s specialize to the case of gravity, where f (r) = −k/r^2 and thus V (r) = −k/r for some constant k.
- 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.
Figure I shows a sketch of Veff in the case that |j| > m (this is the case where V (^) eff′(r) < 0 for r < j^2 / 2 mk) and II shows Veff where |j| < m (where V (^) eff′(r) > 0 for r < j^2 / 2 mk.) In both sketches, the zero is at r = j^2 / 2 mk and the r-axis is a horizontal asymptote as r → ∞. Let us briefly discuss the behavior of a particle with energy E < 0 with |j| > m. Such a particle is shown in III. As was discussed in the example in class, the particles radius r would oscillate within the classically allowed region (the r values lying between the intersection points of E and Veff(r)). The particle would be moving fastest at the minimum value of Veff and would change from moving away from the origin to moving towards it (or vice a versa) at the intersection points.
- Carry out the integration in (8). We must compute ∫ (j/mr^2 )dr √ 2 m (E^ +^ k/r^ −^ j
(^2) / 2 mr (^2) )
We should note that if the sign of the radical for ˙r
- Show that equation (10) describes an ellipse, parabola, or hyperbola in polar coordinates, depending on the value of the parameter e, which we call the eccentricity. Begin by making a shift (a rotation) of θ 0 in θ. We will call the new coordinates that result from this shift r′^ and θ′. We have that
r′^ =
p 1 + e cos θ′
by (10), or equivalently
r′^ + er′^ cos θ′^ = p.
Making the standard change to cartesian coordinates, the above reads
√ x^2 + y^2 + ex = p.
Now a little algebra yields
x^2 + y^2 = p^2 − 2 ex + e^2 x^2 ,
or put a little differently,
(1 − e^2 )x^2 + 2ex + y^2 = p^2 ; (11)
which we immediately recognize as the equation of a conic. The particular conic that (11) describes will be determined by the value of e. If e = 0, for instance, then (11) reduces to
x^2 + y^2 = p^2 ,
a circle centered at the origin with radius p. If e = 1, then (11) reduces to
2(x − p^2 /2) = y^2 ,
a parabola with vertex (in the original polar coordinates) (p^2 / 2 , θ 0 ) opening towards the origin. Let’s exhaust all of the cases. If e 6 = 0 or 1, then we may rewrite (11) as ( x + (^1) −epe 2
p2 1+ 1 −ee^22
y^2 p^2 (1 + e^2 )
We see that in this case (12) represents either a hyperbola ( e > 1) with vertices (in rotated cartesian coordinates) ( −ep 1 − e^2
, ±p(1 − e^2 )^1 /^2
opening in the y direction or an ellipse (0 < e < 1) with center (again in rotated cartesian coordi- nates)
( −ep 1 − e^2
- How are the three kinds of orbits related to the energy E? Recall that e is given by
e =
2 Ej^2 mk^2
so that
E =
mk^2 2 j^2
(e^2 − 1).
Using this, we compile the following chart:
Orbit(type) Energy Circular E = −mk^2 / 2 j^2 Parabolic E = 0 Hyperbolic E > 0 Elliptic −mk^2 / 2 j^2 < E < 0
