Page 579 - Advanced_Engineering_Mathematics o'neil

15.3 Bessel Functions 559

is C and line BA is the semimajor axis, whose length we will denote 2a. The focus is the sun
at S. P is the planet at some time t, which we set at zero when P is at A. A (dashed) circle is
drawn with center C and through A and B. Q is the point of intersection of this circle with the
line through P perpendicular to AB. P is imagined to move along the circle at a constant speed
∗
so that its location coincides with that of P at both A and B.

We are interested in the two angles
ψ = ∠AC P and ϕ = ∠AC Q.
∗
ψ is the mean anomaly of the planet, and ϕ the eccentric anomaly.The true anomaly is
θ = ∠ASP, but we will not be concerned with θ here. The part of Kepler’s problem we
will discuss is to express ϕ in terms of ψ.

Note that ψ is proportional to the time t it takes for the planet to move from A to P. Thus,
we might equivalently state the problem as expressing ϕ in terms of t.
As is customary, let e be the eccentricity of the ellipse. Also let T denote the period of the
planet (time required for one complete orbit).
Recall that the area of an ellipse with semimajor axis of length 2a and semiminor axis 2b is
πab. Now, since ψ and ∠ASP are both proportional to t, then
t ψ area of elliptical sector ASP
= =
T 2π πab
area of circular sector ASQ
= .
πa 2
But
area of ASQ = area of elliptical sector AC Q
− area of triangleSC Q

1 2 1
= a ϕ − ae(a sin(ϕ)).
2 2
Therefore
2
2
1
ψ 1 2 a ϕ − a e sin(ϕ)
2
= ,
2π πa 2
so
ψ = ϕ − e sin(ϕ).
We want ϕ in terms of ψ, so we must somehow solve this equation for ϕ. Observe that ψ − ϕ is
an odd periodic function of ϕ, having period 2π. We can therefore write a Fourier expansion
∞

ψ − ϕ = b n sin(nψ)
n=1
in which
2 π
b n = (ϕ − ψ)sin(nψ)dψ.
π 0
Integrate by parts, keeping in mind that
π
(ϕ − ψ)(0) = (ϕ − ψ)(π) = 0 and cos(ψ)dψ = 0.
0

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October 14, 2010 15:20 THM/NEIL Page-559 27410_15_ch15_p505-562