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Kepler’s Laws
Johannes Kepler (1571–1630) discovered three laws of planetary motion in the early seventeenth century. These laws were discovered empirically, after studying for many years data
collected primarily by the Danish astronomer Tycho Brahe (1546–1601). The first mathematical derivation of Kepler’s laws appeared in 1687 in the book Principia Mathematica by Isaac
Newton (1642–1727). In his treatise, Newton invented the calculus, and applied his results to
solve several problems in physics. The ideas set forth by Newton in that book probably represent one of the greatest intellectual achievements of all time. Kepler’s first two laws were
announced in 1609:
Kepler’s First Law: The planets revolve around the sun in elliptical orbits, with the sun at
one focus.
Kepler’s Second Law: For each planet, the line joining it to the sun sweeps out equal areas
in equal times.
After studying Brahe’s data for 10 more years, Kepler announced his third law in 1619:
Kepler’s Third Law: The square of the period of revolution of a planet is proportional to the
cube of the semimajor axis of its elliptical orbit.
We can derive Kepler’s first and second laws here using the vector calculus studied in class.
Background from rotational dynamics
If a force F(t) acts on a particle with vector location r(t), the torque, τ(t), produced on the
particle by the force about the origin is defined by the equation τ(t) = r(t) × F(t). The angular
momentum, L(t), of the particle about the origin is defined by the equation L(t) = r(t)×mr (t).
Torque and angular momentum are the rotational analogues of force and momentum. Just as
force is the time rate of change of momentum, torque is the time rate of change of angular
momentum, i.e., L (t) = τ(t) (show this).
Particle motion under a central force
Suppose a particle P moving through 3-space is acted upon by a force in the direction of the
line from P to a fixed point, say the origin O of R3 (i.e., the force is a central force). Then we
can show that the motion of the particle occurs in a fixed plane containing O. First note that
central forces produce no torque about the origin: τ(t) = r(t) × F(t) = 0 since r(t) and F(t)
are always parallel for central forces. But if τ(t) = 0, then L(t) must be a constant vector, say
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L0 , since L (t) = τ(t). In other words, the angular momentum remains constant for a particle
under a central force. But r is orthogonal to L0 by definition of L. So r(t) is orthogonal to the
fixed vector L0 for all t. So, the particle remains in a plane containing the origin.
Inverse square central forces
Now suppose F is a central force which obeys an inverse square law. For example, if F is a
gravitational force exerted on an object of mass m at P by an object of mass M at 0, then
F = − GMm
u where G is the gravitation constant and u =
r2
r
r .
GM
GM
Then r = − r
2 · u = − r3 r. In
C
general, for an inverse square central force, r = − r
3 · r for some constant C.
Derivation of Kepler’s first law
Let h = L0 /m. Then h is a constant vector, and h = r × r . Thus
d C
(r × h) = r × h + r × h = r × h = −
u × (r × r )
dt
r2
C
d
= −
(ru)
u
×
ru
×
r2
dt
d
C
ru
+
ru
u
×
ru
×
= −
r2
dt
C
2
u
×
r
= −
u
×
u
= −Cu × u × u
2
r
= −C(u(u · u ) − u (u · u)) [since a × (b × c) = b(a · c) − c(a · b)]
= C u
[since u · u = 0 and u · u = 1]
Thus r ×h = Cu+q for some constant q so r·(r ×h) = ru·(GMu+q) = Cr+rq cos θ
where θ is the angle between r and q. Also, r · (r × h) = (r × r ) · h = h · h = h2 , so
h2 = Cr + rq cos θ. So
r = r =
h2 /C
h2
=
C + q cos θ
1 + (q/C) cos θ
Thus P moves along a conic having eccentricity e =
q
C
and O as a focus, which establishes
Kepler’s first law. Now suppose r describes the location of a body around the sun, taking the
sun as O. Then we use C = GM in the above formulas. When the body is at perihelion (i.e.,
when r is minimized), θ = 0. If we let r0 and v0 denote the vectors r and v at perihelion, then
h = h = r0 × v0 = r0 v0 = r0 v0 . But r0 =
h2 /C
1+q/C
=
r0 v02
C+q ,
r0 v02 −GM
r0 v02
q
2
=
=
GM
GM
GM − 1. So if GM = r0 v0 , then e
GM < r0 v02 < 2GM, then 0 < e < 1, and the orbit is elliptical.
the orbit is parabolic. And if r0 v02 > 2GM, then e > 1, and the
Thus e =
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so q = r0 v02 − C = r0 v02 − GM.
= 0 and the orbit is circular. If
If r0 v02 = 2GM, then e = 1, and
orbit is hyperbolic.
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Derivation of Kepler’s second law
Assume the motion of the planet is in the xy-plane so that the constant vector h is parallel to
the unit vector k. Introduce polar coordinates r and θ in the xy-plane so that r = (r cos θ)i +
(r sin θ)j. Thus
h = r × r = i
j
r cos θ
r cos θ
r sin θ
− r θ sin θ
r sin θ
+ r θ cos θ
k
0 = r 2θk
0
Thus, r 2 dθ
dt = h = constant. Recall the formula for the area inside a polar curve given by
r = r (θ) from θ = α to θ = β:
1
Area =
2
β
α
[r (θ)]2 dθ
Now let t0 be a fixed time and t a variable time later than t0 . From the area formula, changing
variable from θ to τ (τ is not torque here, just a dummy variable in place of t), we have
Area =
Therefore,
1
2
t
t0
r2
dθ
dτ
dτ
dA
1 dθ
1
= r2
= h = constant
dt
2
dt
2
which is Kepler’s second law.
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