Astronomy Principles and Practice Fourth Edition.pdf

Stationary points 161
Using equations (12.6) and (12.5), we obtain
Hence, rearranging, we find that
b − a cos θ
b cos θ − a = V ⊕
V P
= b1/2
a 1/2 .
cos θ = a1/2 b 1/2 (a 1/2 + b 1/2 )
a 3/2 + b 3/2 . (12.10)
If b, expressed in units of the Earth’s distance from the Sun, has values β, with the value of a
being unity, equation (12.10) may be written as
cos θ = β1/2 (1 + β 1/2 )
1 + β 3/2 . (12.11)
By symmetry, it is obvious that when the Earth was at the point E ′ (figure 12.9), where
ESP = θ, there was another stationary point. The total time during which the planet will be seen to
move in the retrograde direction is hence the time it takes the Earth’s radius vector to advance through
an angle 2θ with respect to the planet’s radius vector. It will be given by t R where
t R = 2θ
360 × S = θ S
(12.12)
180
S being the planet’s synodic period.
The time that elapses between opposition and the next stationary point is obviously t R /2.
The time during a synodic period that the planet’s motion is direct is t D ,where
( 360 − 2θ
t D =
360
)
× S =
(
1 − θ
180
)
S. (12.13)
Going back to △SEP, we may write
a
. sin η = sin φ (12.14)
b
where η is the elongation at a stationary point.
Now η = 180 − (θ + φ), so that from equation (12.6), we have
or
− cos η = V P
V ⊕
cos φ
V ⊕
cos η =−cos φ. (12.15)
V P
Squaring and adding equations (12.14) and (12.15), and using relation (12.5), we obtain
a 2
b 2 sin2 η + b a cos2 η = 1.
Hence, substituting (1 − cos 2 η) for sin 2 η, there results
( )
b
a − a2
b 2 cos 2 η = 1 − a2
b 2