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Kepler’s Third Law and the Scale of the Solar System
By Dr. Bob Brazzle
One of the longstanding problems for ancient astronomers was that of determining the size of our Solar
System. Three astronomers in ancient Greece (Hipparchus, Aristarchus and Eratosthenes) developed
clever geometric methods for determining distances from Earth to the Moon and Sun and relating them
to the size of Earth [see http://curious.astro.cornell.edu/question.php?number=400]. However, the
poor precision of some measurements resulted in calculated distances with low accuracy.
In 1619, Johannes Kepler published a creative, indirect method for calculating a scale for the Solar
System using Earth’s orbital radius as the basic distance unit. His success came after several tedious
years of analyzing the meticulous data taken by Tycho Brahe. In fact, Kepler discovered a simple
mathematical relationship between a planet’s orbital radius and its orbital period (period is much
easier to measure than distance!). This mathematical relationship is now called Kepler’s Third Law of
Planetary Motion [see http://www-istp.gsfc.nasa.gov/stargaze/Skeplaws.htm].
Like the vast majority of physical “laws” of nature, Kepler’s 3rd Law is merely a description of how
two physical quantities relate to each other. In modern scientific inquiry, we are not satisfied with
descriptions alone: we also require explanations of how Mother Nature operates. A scientific
explanation is called a “theory”, and scientists demand that all theories make testable predictions. In
1686, Isaac Newton published such an explanation – that the force of gravity is responsible for the
relationship described in Kepler’s 3rd Law [see
http://www.mtholyoke.edu/courses/mdyar/ast223/orbits/orb_lect.html]. In this paper, I will show that
modern observations of planetary motions are well-described by Kepler’s 3rd Law, and well-explained
by Newton’s equations of gravity and motion.
Observations
Today, NASA accurately measures planetary distances to
incredible levels of precision. The main measurement
tool is the radio used to communicate with spacecraft in
orbit about a distant planet. Engineers send a signal to a
given spacecraft, which then relays the signal back to
Earth. By measuring the round-trip travel time and using
the precisely-known speed of light, they can calculate the
distance to the spacecraft [for a technical report, see
http://ipnpr.jpl.nasa.gov/progress_report2/XIX/XIXZ.PDF].
The table at right shows the current values for the orbital
distances and periods of the planets. These data are from
http://solarsystem.nasa.gov/planets/index.cfm.
Table 1: Planetary Data
Planet
Orbital
Orbital
radius*
period
9
(10 m)
(106 s)
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
57.9092
108.209
149.598
227.944
778.341
1426.66
2870.66
4498.40
7.60061
19.4141
31.5585
59.3551
374.356
929.292
2651.37
5200.42
* Technically, this is the semi-major axis of the
elliptical orbit.
Kepler’s Third Law of Planetary Motion
The data from Table 1 were entered into
an Excel Spreadsheet, and the resulting
graph appears at right, along with the
T = (5.47•10-10)•r1.50
equation for the curve that best fits the
data. Therefore, this equation is a
mathematical model describing the
relationship between two measurable
physical quantities of each planet’s
orbit. Although these data are much
more precise than what was available to
Kepler, he was able to find the same
fundamental relationship: namely that
an orbit’s period squared is proportional to the cube of the length of its semi-major axis. This can be
written in the form: T = ka3/2 {where the coefficient in the equation, “k”, depends on the measurement
units used}. According to the data, the value of k is about 5.47 • 10-10 in SI units, but we don’t have an
explanation why it would take this value. For that, we need a theory.
Theory
Isaac Newton realized that any object traveling in a circular path requires a “centripetal” force pulling
the object (centripetal means “directed toward the center”). The eight planets in our Solar system orbit
the Sun in nearly circular orbits, so we will use circles to simplify the situation. In this case, the
centripetal force is gravity, and Newton showed that the gravitational force between two objects is
Fg = GMm/r2 {for this example, G is the Universal Gravitational Constant, M is the mass of the Sun,
m is the mass of a given planet and r is the distance between their centers}. If there is a centripetal
force, there must also be a centripetal acceleration, and the magnitude of this is ac = v2/r {where v is
the straight-line speed and r is the radius of the circle}.
S
Fnet = ma =mv2/r = GMm/r2 {Newton’s 2nd Law, and gravity}
So v2 = GM/r {multiply both sides by r/m}
But v = 2r/T {from d=rt and c=2r for a circle, and T is period}
So
Free-body Diagram of
planet orbiting the Sun
4 2 r 2 GM
{after substitution}

r
T2
 4 2 

Then T  
GM


1/ 2
 r 3 / 2 {after algebraic manipulation}
Using known values for G and M in the final equation (G = 6.672 • 10-11 Nm2/kg2) and
(Msun = 1.989 • 1030 kg) [see http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6], we have
(42/GM)1/2 = 5.45 • 10-10 in close agreement with the empirical findings above.
Kepler’s 3rd Law: Questions and Discussion Points
Name____________________________
1) What is meant by the term, “empirical”? {Used in the last sentence}
2) What does the term “theory” mean, as it is used by scientists?
3) What do I mean by the term, “model” in the first paragraph on page 2?
4) When do I reference researched information?
5) What can I do to check the credibility of my sources?
6) What are the physical quantities being investigated?
7) What is the mathematical relationship between (or among) the relevant physical quantities?
Problems
 4 2  3
  r
Kepler’s 3rd Law (alternate form): T 2  
 GM 
Solving this equation for M, we have: M 
4 2 r 3
G T 2
1) The NEAR spaceship orbited the asteroid 433 Eros at a radius of 35,000 m above its center of mass.
The orbital period was 61,580 seconds. Calculate the mass of 433 Eros.
2) If the NEAR spaceship’s orbital radius became 4 times larger, how many times larger would its
orbital period become?
3) The orbital periods and radii of Saturn’s many moons have been precisely measured by NASA. An
analysis of these orbital quantities (same approach we used in the Solar System lab) reveals a best-fit
line to the data with the equation: T2 = 1.04 • 10-15 r3. Use the alternate form of Kepler’s 3rd Law to
calculate the mass of the planet Saturn.
Setting Newton’s Law of Gravitation equal to the centripetal force equation, yields the following:
GM
GMm mv 2
or v 2 
{v is the “orbital speed” of an object}

2
r
r
r
4) The Compton Gamma Ray telescope (named after Washington University physicist and Nobel Prize
winner Arthur Holly Compton) orbited at an altitude of 4.50 • 105 m above Earth’s surface. Use the
orbital speed equation (v2 = GM/r) and Earth’s radius (6.38 • 106 m) to determine its orbital speed.
5) Communication spacecraft are often placed in a “geosynchronous” orbit, in which the satellite
always remains above a fixed spot on Earth’s equator. This is achieved by the orbital speed matching
the ground speed of 464 m/s. Calculate the radius of a geosynchronous orbit.