Fundamental Astronomy

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6. Celestial Mechanics
The solution of (6.5) now gives the relative orbit
of the planet. The equation involves the radius vector
and its second time derivative. In principle, the solution
should yield the radius vector as a function of time,
r = r(t). Unfortunately things are not this simple in
practice; in fact, there is no way to express the radius
vector as a function of time in a closed form (i. e. as
a finite expression involving familiar elementary functions).
Although there are several ways to solve the
equation of motion, we must resort to mathematical
manipulation in one form or another to figure out the
essential properties of the orbit. Next we shall study one
possible method.
6.2 Solution of the Equation of Motion
The equation of motion (6.5) is a second-order (i. e.
contains second derivatives) vector valued differential
equation. Therefore we need six integration constants
or integrals for the complete solution. The solution is
an infinite family of orbits with different sizes, shapes
and orientations. A particular solution (e. g. the orbit
of Jupiter) is selected by fixing the values of the six
integrals. The fate of a planet is unambiguously determined
by its position and velocity at any given moment;
thus we could take the position and velocity vectors at
some moment as our integrals. Although they do not
tell us anything about the geometry of the orbit, they
can be used as initial values when integrating the orbit
numerically with a computer. Another set of integrals,
the orbital elements, contains geometric quantities describing
the orbit in a very clear and concrete way. We
shall return to these later. A third possible set involves
certain physical quantities, which we shall derive next.
We begin by showing that the angular momentum
remains constant. The angular momentum of the planet
in the heliocentric frame is
L = m2r × ˙r . (6.7)
Celestial mechanicians usually prefer to use the angular
momentum divided by the planet’s mass
k = r × ˙r . (6.8)
Let us find the time derivative of this:
˙k = r × ¨r + ˙r × ˙r .
The latter term vanishes as a vector product of two
parallel vectors. The former term contains ¨r, which is
given by the equation of motion:
˙k = r × (−µr/r 3 ) =−(µ/r 3 )r × r = 0 .
Thus k is a constant vector independent of time (as is L,
of course).
Since the angular momentum vector is always perpendicular
to the motion (this follows from (6.8)), the
motion is at all times restricted to the invariable plane
perpendicular to k (Fig. 6.2).
To find another constant vector, we compute the
vector product k× ¨r:
k× ¨r = (r × ˙r) × (−µr/r 3 )
=− µ
[(r · r)˙r − (r · ˙r)r] .
r3 The time derivative of the distance r is equal to the
projection of ˙r in the direction of r (Fig. 6.3); thus, using
the properties of the scalar product, we get ˙r = r · ˙r/r,
which gives
r · ˙r = r ˙r . (6.9)
Hence,
k× ¨r =−µ(˙r/r − r˙r/r 2 ) = d
dt (−µr/r).
The vector product can also be expressed as
k× ¨r = d
(k× ˙r),
dt
Fig. 6.2. The angular momentum vector k is perpendicular
to the radius and velocity vectors of the planet. Since k is
a constant vector, the motion of the planet is restricted to the
plane perpendicular to k