Kepler's laws
Kepler's second law
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Kepler's second law.
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We now proceed to directly address Kepler's
second law, the one which states that a ray from the Sun to a planet
sweeps out equal areas in equal times. This ray is simply the vector
r that we've been using. (And we shall continue to use it; r,
remember, is defined as the vector from the Sun to the planet.)
What we're looking for is the area that this vector sweeps out.
Imagine the planet at some time t = 0, and then imagine at a short
time afterward t = deltat. In that time, the vector has moved by a
short displacement
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deltar = r|(t = deltat) - r|(t = 0).
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equation 1
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The three
vectors r|(t = 0), deltar, and r|(t = deltat) form a triangle. The
area of this triangle closely approximates the area swept out by the
vector r during that short time deltat.
We can write this small area represented by this triangle, deltaA,
as one-half of the parallelogram defined by the vectors r and
deltar, or
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deltaA = (1/2) |r cross deltar|.
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equation 2
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We'll divide both sides of this equation by deltat, the short time
involved. Because of this, and the associative properties of the
cross product, we find:
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deltaA/deltat = (1/2) (1/deltat) |r cross deltar|
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equation 3
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deltaA/deltat = (1/2) |(r cross deltar)/deltat|
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equation 4
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deltaA/deltat = (1/2) |r cross (deltar/deltat)|.
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equation 5
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As we choose smaller and smaller values of deltat, we get better and
better approximations of the area swept out by the ray. If we let
deltat approach zero by taking the limit of both sides, the
approximation approaches the real value and we find that
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dA/dt = (1/2) |r cross dr/dt|
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equation 6
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or
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dA/dt = (1/2) |r cross v|.
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equation 7
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Knowing
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l = m |r cross v|
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equation 8
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and dividing both sides by m, we find
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l/m = |r cross v|.
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equation 9
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We can substitute this into our expression for dA/dt and find that
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dA/dt = l/(2 m).
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equation 10
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That is, the instantaneous time rate of change of area is the magnitude
of the angular momentum divided by twice the mass of the planet. But
we know that the mass of the planet is constant, and we also know from
our work earlier that the angular momentum vector is constant (and
thus its magnitude certainly is). Therefore, the time derivative of
area swept out by this ray is constant. In other words, no matter
where on the orbit the planet is, its ray still sweeps out the same
amount of area. This is Kepler's second law.
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