4Orbits
IA Dynamics and Relativity
4.4 The Kepler problem
The Kepler problem is the study of the orbits of two objects interacting via
a central force that obeys the inverse square law. The goal is to classify the
possible orbits and study their properties. One of the most important examples
of the Kepler problem is the orbit of celestial objects, as studied by Kepler
himself.
Shapes of orbits
For a planet orbiting the sun, the potential and force are given by
V (r) =
mk
r
, F (r) = −
mk
r
2
with
k
=
GM
(for the Coulomb attraction of opposite charges, we have the same
equation with k = −
Qq
4πε
0
m
).
Binet’s equation then becomes linear, and
d
2
u
dθ
2
+ u =
k
h
2
.
We write the general solution as
u =
k
h
2
+ A cos(θ − θ
0
),
where A ≥ 0 and θ
0
are arbitrary constants.
If
A
= 0, then
u
is constant, and the orbit is circular. Otherwise,
u
reaches a
maximum at
θ
=
θ
0
. This is the periapsis. We now redefine polar coordinates
such that the periapsis is at θ = 0. Then
Proposition. The orbit of a planet around the sun is given by
r =
`
1 + e cos θ
, (∗)
with
`
=
h
2
/k
and
e
=
Ah
2
/k
. This is the polar equation of a conic, with a
focus (the sun) at the origin.
Definition
(Eccentricity)
.
The dimensionless parameter
e ≥
0 in the equation
of orbit is the eccentricity and determines the shape of the orbit.
We can rewrite (
∗
) in Cartesian coordinates with
x
=
r cos θ
and
y
=
r sin θ
.
Then we obtain
(1 − e
2
)x
2
+ 2e`x + y
2
= `
2
. (†)
There are three different possibilities:
– Ellipse: (0 ≤ e < 1). r is bounded by
`
1 + e
≤ r ≤
`
1 − e
.
(†) can be put into the equation of an ellipse centered on (−ea, 0),
(x + ea)
2
a
2
+
y
2
b
2
= 1,
where a =
`
1 − e
2
and b =
`
√
1 − e
2
≤ a.
ae
b
a
`
O
a
and
b
are the semimajor and semiminor axis.
`
is the semilatus rectum.
One focus of the ellipse is at the origin. If
e
= 0, then
a
=
b
=
`
and the
ellipse is a circle.
–
Hyperbola: (
e >
1). For
e >
1,
r → ∞
as
θ → ±α
, where
α
=
cos
−1
(1
/e
)
∈
(
π/
2
, π
). Then (
†
) can be put into the equation of a hyperbola centered on
(ea, 0),
(x − ea)
2
a
2
−
y
2
b
2
= 1,
with a =
`
e
2
− 1
, b =
`
√
e
2
− 1
.
a
O
`
b
This corresponds to an unbound orbit that is deflected (scattered) by an
attractive force.
b
is both the semiminor axis and the impact parameter. It is the distance
by which the planet would miss the object if there were no attractive force.
The asymptote is y =
b
a
(x − ea), or
x
p
e
2
− 1 − y = eb.
Alternatively, we can write the equation of the asymptote as
(x, y) ·
√
e
2
− 1
e
, −
1
e
!
= b
or r · n = b, the equation of a line at a distance b from the origin.
– Parabola: (e = 1). Then (∗) becomes
r =
`
1 + cos θ
.
We see that
r → ∞
as
θ → ±π
. (
†
) becomes the equation of a parabola,
y
2
= `(` − 2x). The trajectory is similar to that of a hyperbola.
Energy and eccentricity
We can figure out which path a planet follows by considering its energy.
E =
1
2
m( ˙r
2
+ r
2
˙
θ
2
) −
mk
r
=
1
2
mh
2
du
dθ
2
+ u
2
!
− mku
Substitute u =
1
`
(1 + e cos θ) and ` =
h
2
k
, and it simplifies to
E =
mk
2`
(e
2
− 1),
which is independent of θ, as it must be.
Orbits are bounded for
e <
1. This corresponds to the case
E <
0. Un
bounded orbits have
e >
1 and thus
E >
0. A parabolic orbit has
e
= 1,
E
= 0,
and is “marginally bound”.
Note that the condition
E >
0 is equivalent to

˙
r >
q
2GM
r
=
v
esc
, which
means you have enough kinetic energy to escape orbit.
Kepler’s laws of planetary motion
When Kepler first studied the laws of planetary motion, he took a telescope,
observed actual planets, and came up with his famous three laws of motion. We
are now going to derive the laws with pen and paper instead.
Law
(Kepler’s first law)
.
The orbit of each planet is an ellipse with the Sun at
one focus.
Law
(Kepler’s second law)
.
The line between the planet and the sun sweeps
out equal areas in equal times.
Law
(Kepler’s third law)
.
The square of the orbital period is proportional to
the cube of the semimajor axis, or
P
2
∝ a
3
.
We have already shown that Law 1 follows from Newtonian dynamics and
the inversesquare law of gravity. In the solar system, planets generally have very
low eccentricity (i.e. very close to circular motion), but asteroids and comets
can have very eccentric orbits. In other solar systems, even planets have have
highly eccentric orbits. As we’ve previously shown, it is also possible for the
object to have a parabolic or hyperbolic orbit. However, we tend not to call
these “planets”.
Law 2 follows simply from the conservation of angular momentum: The area
swept out by moving dθ is dA =
1
2
r
2
dθ (area of sector of circle). So
dA
dt
=
1
2
r
2
˙
θ =
h
2
= const.
and is true for any central force.
Law 3 follows from this: the total area of the ellipse is
A
=
πab
=
h
2
P
(by
the second law). But b
2
= a
2
(1 − e
2
) and h
2
= k` = ka(1 − e
2
). So
P
2
=
(2π)
2
a
4
(1 − e
2
)
ka(1 − e
2
)
=
(2π)
2
a
3
k
.
Note that the third law is very easy to prove directly for circular orbits. Since
the radius is constant, ¨r = 0. So the equations of motion give
−r
˙
θ
2
= −
k
r
2
So
r
3
˙
θ
2
= k
Since
˙
θ ∝ P
−1
, the result follows.