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Saturday, January 2, 2016

Calculating Orbits: Kepler's Second Law

Kepler’s Second Law

Our aim is to derive Kepler’s Second Law:
The radius from the sun to the planet sweeps out equal areas in equal times.
To put this in language better suited to mathematical manipulation, let A denote the area swept out from time zero to time t.  Then Kepler’s Second Law saws DA is a constant.  We shall show that not only is it a constant but a constant with meaning to physics.
In this derivation, we shall have occasion to use a formula derived from Newton’s Law of Gravity:
The acceleration of a planet is inversely proportional to the square of the distance from the sun and is directed toward the sun.
Mathematically:
or
So
as the angle from \vec{r} to itself is zero and so \sin \theta =0.  Thus,
Let us also observe that
So that \vec{r}\times\vec{v} is constant.  We will denote this constant by \vec{H} and note that
where \vec{L} is angular momentum.  We can think of \vec{H} as being angular momentum per unit mass.
Now consider the following diagram of our two-body system as time \Delta t has elapsed.
For small \Delta t
so that the altitude of this triangle is v\sin \theta where \theta is the angle from \vec{r} to \vec{v},  If \Delta A is the area of this triangle, then
so that
where H is the magnitude of \vec{H}.
Therefore,

which is Kepler’s Second Law.

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