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Sect. 13.3: Kepler’s Laws & Planetary Motion
Johannes Kepler
• German astronomer (1571 – 1630)
• Spent most of his career tediously
analyzing huge amounts of observational
data (most compiled by Tycho Brahe) on
planetary motion (orbit periods, orbit radii, etc.)
• Used his analysis to develop “Laws” of
planetary motion.
“Laws” in the sense that they agree
with observation, but not true
theoretical laws, such as Newton’s
Laws of Motion & Newton’s
Universal Law of Gravitation.
Kepler’s “Laws”
Kepler’s “Laws” are consistent with & are obtainable from
Newton’s Laws
• Kepler’s First Law
– All planets move in elliptical orbits with the Sun at
one focus
• Kepler’s Second Law
– The radius vector drawn from the Sun to a planet
sweeps out equal areas in equal time intervals
• Kepler’s Third Law
– The square of the orbital period of any planet is
proportional to the cube of the semimajor axis of the
elliptical orbit
Math Review: Ellipses
• The points F1 & F2 are each a
focus of the ellipse
– Located a distance c from the center
– Sum of r1 and r2 is constant
• Longest distance through center
is the major axis, 2a
a is called the semimajor axis
• Shortest distance through center
is the minor axis, 2b

Typical Ellipse
b is called the semiminor axis
• The eccentricity is defined as e = (c/a)
– For a circle, e = 0
– The range of values of the eccentricity for ellipses is 0 < e < 1
– The higher the value of e, the longer and thinner the ellipse
Ellipses & Planet Orbits
• The Sun is at one focus
– Nothing is located at the other focus
• Aphelion is the point farthest away from the Sun
– The distance for aphelion is a + c
• For an orbit around the Earth, this point is called the
apogee
• Perihelion is the point nearest the Sun
– The distance for perihelion is a – c
• For an orbit around the Earth, this point is called the
perigee
Kepler’s 1st Law
All planets move in elliptical orbits
with the Sun at one focus
• A circular orbit is a special case of an elliptical orbit
– The eccentricity of a circle is e = 0.
• Kepler’s 1st Law can be shown (& was by Newton) to be a direct
result of the inverse square nature of the gravitational force.
Comes out of N’s 2nd Law + N’s Gravitation Law + Calculus
• Elliptic (and circular) orbits are allowed for bound objects
– A bound object repeatedly orbits the center
– An unbound object would pass by and not return
• These objects could have paths that are parabolas
(e = 1) and hyperbolas (e > 1)
Orbit Examples
• Fig. (a): Mercury’s orbit has the largest eccentricity of the planets.
eMercury = 0.21
Note: Pluto’s eccentricity is ePluto = 0.25, but, as of 2006, it is
officially no longer classified as a planet!
• Fig. (b): Halley’s Comet’s orbit has high eccentricity
eHalley’s comet = 0.97
• Remember that nothing physical is located at the second focus
– The small dot
Kepler’s 2nd Law
The radius vector drawn from the Sun
to a planet sweeps out equal areas in
equal time intervals

• Kepler’s 2nd Law can be shown
(& was by Newton) to be a direct result
of the fact that N’s Gravitation Law
gives Conservation of Angular
Momentum for each planet.
• The Gravitational force produces no
torque (it is  to the motion) so that
Angular Momentum is Conserved:
L = r x p = MPr x v = constant
Kepler’s 2nd Law
• Geometrically, in a time dt,
the radius vector r sweeps
out the area dA = half the
area of the parallelogram
• The displacement is dr = v dt
L
• Mathematically, this means dA

 constant
dt 2Mp
• That is: the radius vector from the Sun to any
planet sweeps out equal areas in equal times
Kepler’s 3rd Law
The square of the orbital period T of any
planet is proportional to the cube of the
semimajor axis a of the elliptical orbit
• If the orbit is circular & of
GMSunMPlanet MPlanetv 2

radius r, this follows from
2
r
r
Newton’s Universal
2 r
Gravitation.
v
T
• This gravitational force
2
supplies a centripetal force

 3
4

2
3
T

r

K
r


S
for user in
GMSun 

Newton’s 2nd Law   
Ks is a constant, which
• Ks is a constant
is the same for all planets.
Kepler’s 3rd Law
• Can be shown that this also applies to an elliptical orbit
with replacement of r with a, where a is the semimajor
axis.
2

 3
4

2
3
T 
a

K
a

S
 GMSun 
• Ks is independent of the planet mass, & is valid for any
planet
• Note: If an object is orbiting another object, the value of
K will depend on the mass of the object being orbited.
For example, for the Moon’s orbit around the Earth,
KSun is replaced with KEarth, where KEarth is obtained by
replacing MSun by MEarth in the above equation.
Solar System Data
Table 13-2, p. 370
“Weighing” the Sun!
• We’ve “weighed” the Earth, now lets “weigh” the Sun!!
Assume: Earth & Sun are perfect uniform spheres. & Earth orbit
is a perfect circle.
• Note: For Earth, Mass ME = 5.99  1024kg
Orbit period is T = 1 yr  3 107 s
Orbit radius r = 1.5  1011 m
So, orbit velocity is v = (2πr/T), v  3 104 m/s
• Gravitational Force between Earth & Sun: Fg = G[(MSME)/r2]
Circular orbit is circular  centripetal acceleration
Newton’s 2nd Law gives: ∑F = Fg = MEa = MEac = ME(v2)/r
OR: G[(MSME)/r2] = ME(v2)/r. If the Sun mass is unknown,
solve for it: MS = (v2r)/G  2  1030 kg  3.3  105 ME