Kepler's laws
Conservation of angular momentum
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Conservation of angular momentum.
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Torque tau is defined as the
instantaneous time rate of change of angular momentum l:
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tau == dl/dt.
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equation 1
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Angular momentum is a quantity which plays the same part in rotational
mechanics as linear momentum does in linear mechanics.
We know from the previous section that tau = o -- the Sun never
applies a torque to a planet. Therefore dl/dt must also be the zero
vector:
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dl/dt = o.
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equation 2
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If the time derivative of something is zero, that means that thing
does not change as time passes; in other words, it remains constant.
This is usually only applied to scalars, however. In vectors, if the
time derivative of a vector is the zero vector, then that vector does
not change magnitude or direction. In other words, the angular
momentum vector of a planet is a constant vector:
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l = constant.
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equation 3
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Because the Sun does not apply a torque to a planet from its
gravitational influence, the angular momentum of the planet remains
constant; it is conserved. This is the main concept behind Kepler's
second law.
What is the mathematical expression for angular momentum, though?
We can find an expression for angular momentum from our expression
for torque, substituting in dl/dt for tau:
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dl/dt = r cross F.
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equation 4
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We can use Newton's law of motion, F = m a, and substitute this into
our equation:
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dl/dt = r cross (m a).
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equation 5
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The acceleration of a body is equal to its instantaneous rate of
change of velocity; that is,
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a == dv/dt.
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equation 6
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Making this substitution
(and also exploiting the fact that the cross product is associative
with respect to scalar factors), we find that
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dl/dt = r cross (m dv/dt)
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equation 7
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dl/dt = m (r cross dv/dt).
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equation 8
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If we solve this differential equation, we find that
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l = m (r cross v).
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equation 9
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The magnitude of the angular momentum is
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l = |l|
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equation 10
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l = |m (r cross v)|
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equation 11
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l = m |r cross v|.
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equation 12
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This relates the angular momentum of a planet to its mass, position,
and velocity.
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