Mathematics for Physics 3: Dynamics and Differential Equations
21
93
Lecture 21: The orbit equation for inverse square-law forces
In the previous lectures we derived the equations of motion under the influence of central forces F (r) = Fr (r)r̂ in three
dimensions. The angular equation in plane polar coordinates amounts to the conservation of angular momentum,
r2 θ̇ ≡ h = |L|/m
(345)
while the radial equation depends on the details of the force as follows:
r̈ −
h2
Fr (r)
=
.
r3
m
(346)
We now wish to solve these equations, at least for the specific case of Newtonian gravity, the inverse square law force.
We shall also look at other power laws.
Exact linearsation of the orbit equation
Before specialising in a particular power law behaviour let us have another look at the radial equation of motion
(346). This equation presents a major difficulty in that it is in general, non-linear.
We now wish to show that in certain cases (i.e. for particular force laws, as discussed below) the equation can be
linearised by a simple change of variables: u(θ) ≡ 1/r. To rewrite our equations of motion in terms of u(θ) we need
an expression for ṙ and r̈. Applying the chain-rule we obtain:
ṙ =
d(1/u) dθ
u� (θ)
= − 2 h u2 = −hu� (θ)
dθ dt
u
(347)
where we used (345) to substitute θ̇ = h/r2 = h u2 . Applying the chain-rule, we obtain for the second derivative,
� �
�� dθ
d
�
r̈ =
− hu (θ)
= −hu�� (θ) hu2 = −h2 u2 u�� (θ) .
(348)
dθ
dt
Substituting this expression for r̈ into (346) we get:
−h2 u2 u�� (θ) − h2 u3 =
and after dividing by −h2 u2 ,
Fr (1/u)
.
m
Fr (1/u)
.
(349)
mh2 u2
The l.h.s. of this equation is linear in u. This, however, does not yet guarantee that the equation is linear, since the
r.h.s. may not be. This depends on the force law.
To have a linear orbit equation the force law must be of the following form:
u�� (θ) + u = −
Fr (1/u) = λu2 + µu3
=⇒
Fr (r) =
λ
µ
+ 3
r2
r
Luckily Newtonian gravity, where the force admits the inverse square law, Fr (r) = λ/r2 belongs to this class of forces
for which the orbit equation in terms of u(θ) linearises, and indeed, it can therefore be solved analytically.
Exact solution for the orbit equation for an inverse square law force
Consider now the orbit equation for u(θ) for the attractive inverse square law force, Fr (r) = −k/r2 , namely
u�� (θ) + u =
k
.
mh2
(350)
In Newtonian gravity k = GM m so the r.h.s becomes GM/h2 .
We recognise that (350) is just the equation of a simple harmonic oscillator with frequency ω = 1. The general
solution is:
k
u(θ) = A cos(θ) + B sin(θ) +
(351)
mh2
Let us now determine A and B for specific initial conditions (yet the conclusions we shall draw concerning the possible
orbit solutions remain general). Consider a case where a particle of mass m is projected at distance a from the centre
of the force with velocity v, in a direction perpendicular to the radius vector from the centre to the projection point.
Without loss of generality we define θ = 0 to correspond to time t = 0. We then have, at t = 0, tangent velocity
Mathematics for Physics 3: Dynamics and Differential Equations
94
vθ = aθ̇ = v while the radial velocity vanishes, vr = 0. The tangent motion simply fixes the angular momentum:
according to (347) h = r2 θ̇ = avθ = av. Using the initial condition for the radial velocity together with (347) we
have:
ṙ
vr
u� (θ = 0) = − = − = 0
h
h
This should be compared to the derivative of u according to our solution (351)
u� (θ) = −A sin(θ) + B cos(θ)
(352)
leading to the conclusion that B = 0. Finally the initial condition for the position yields
u(θ) = A cos(0) + B sin(0) +
so
A=
k
1
=
2
mh
a
(353)
1
k
−
a mh2
We therefore get the following orbit solution:
1
=
r
�
1
k
−
a mh2
or
r=
�
cos(θ) +
k
mh2
l
1 + � cos(θ)
(354)
(355)
where we defined
mh2
l
,
�≡ −1
(356)
k
a
We recognise that our solution for the orbit (355) describes a conic section where r measures the distance from the
focal point to the orbit (the centre of the force is at the focal point). Depending on the value of the eccentricity �
(a dimensionless parameter), it is a circle (� = 0), an ellipse (0 < � < 1), a parabola (� = 1), or a hyperbola (� > 1).
l≡
Figure 17: The three conic sections described by eq. (355), here shown for l = 3 and � = 0.61 (ellipse), � = 1
(parabola) and � = 1.5 (hyperbola) . Note that in all cases the origin (r = 0) corresponds to the centre of force: this
is a focal point of these conic sections.
The other parameter l has dimensions of distance; it is called the semi-latus rectum, and it is given by
�
�
1
1 1
1
=
+
l
2 r+
r−
l
l
where r+ = 1−�
(apoapsis) and r− = 1−�
(periapsis) are the maximum and minimum values of r along the orbit.
Given the initial conditions above, namely projection at distance a from the centre of force, perpendicular to the
radius vector r, with velocity vθ , we have: h = (aθ̇)a = vθ a , and therefore l = mvθ2 a2 /k.
Let us now see how the initial conditions determine the eccentricity �, and in particular whether the orbit is
mvθ2 a
bound or unbound. We defined � ≡ al − 1. So with the stated initial conditions we have � =
− 1 . Eq. (355)
k
then yields:
Mathematics for Physics 3: Dynamics and Differential Equations
• a circular orbit r = l for � = 0, that is for a =
• an elliptic orbit (0 < � < 1) for
95
k
GM
= 2 .
mvθ2
vθ
GM
2GM
<a<
.
2
vθ
vθ2
• a parabolic orbit (� = 1) for a =
2GM
.
vθ2
• a hyperbolic orbit (� > 1) for a >
2GM
vθ2
The various scenarios obtained above have been determined directly using the solution of the equation of motion.
It is interesting to see how these correspond to the energy considerations we have done in the previous lecture. To
this end recall that with the projection described above:
L = mavθ
E=
m 2 GM m
v −
2 θ
a
Indeed, as can be seen in figure 16, bounded motion corresponds to E < 0, while unbounded to E ≥ 0. The transition
point E = 0 is preciely the conditions for a parabolic orbit described above.
The other threshold corresponds to circular orbits, in which case the projected body has just enough energy to
GM
2
satisfy (337). One then has a = 2 and E = − m
2 vθ .
vθ