Henry Braun
Footnote 18 Project
MAT 294 3:40
Kepler's Laws
Johannes Kepler created these three
laws of planetary motion in order to explain the astronomical data
gathered by Tycho Brahe.
The textbook establishes the following
notation:
First law:
The first law was proven in section
13.4 of Calculus:
Early Transcendentals:
This is the polar equation for an
ellipse with one focus at the origin, in this case the sun. e is the eccentricity
of the planet's orbit and ed is its semi-latus
rectum.
Second Law:
h =
(u
u')
u = r / r =
u' =
(since
is a function of t, the chain
rule must be used.)
(u x u') =
h=
h =
A(t) is the area swept out by r(t) in the time
interval [t0,t]. dA is the
area swept out by r(t) over an infinitesimal angle
.
We can see that the area dA swept out
of by r(t) over an
infinitesimal angle
is dA
=
Therefore:
The fundamental theorem of Calculus
then shows that:
Since h is a constant, this proves
Kepler's second Law.
Third Law:
T is the planet's period, that is, the
time it takes for the planet to complete one elliptical orbit. We
also know A to be an ellipse with major and minor axes 2a and 2b. From
this it follows that:
From the definitions of e and d, we know that:
Since ed is also the semi-latus rectum l of
the orbit, we also know that:
(definition of the
semi-latus rectum)
equating the two definitiions for ed, we get:
Substituting
The square of the period is
proportional to the cube of the length of the planet's major axis.
Kepler's 3rd law is proved.
Major axis of the Earth's Orbit
The length of Earth's orbit's major
axis can easily be found using the proof of Kepler's 3rd law and Maple:
T must be converted into seconds for
the units to work out correctly:
The Earth's Major Axis is
m long.
Radius of a Geosynchrous orbit
A satellite in a geosynchronous orbit
around the earth will stay in exactly the same position above the
planet; that is, it will have a period of exactly 24 hours. This
also implies that it will have a circular orbit, that is, a = b = r.
T must be converted into seconds for
the units to work out correctly:
A geostationary satellite must orbit
3.59*10
m above the surface of the Earth.