Astronomy Principles and Practice Fourth Edition.pdf

Stationary points 159
Figure 12.8. The velocities of the Earth and a superior planet produce retrograde motion of the planet (P 1 )at
opposition and direct motion of the planet (P 2 ) at quadrature.
At the following quadrature, the positions of planet and Earth are P 2 and E 2 ,whereSE 2 P 2 =
90 ◦ . The Earth’s orbital velocity V ⊕ is now along the line P 2 E 2 but the planet’s velocity V P has a
component V P sin α, perpendicular to E 2 P 2 . The other component V P cos α lies along the line P 2 E 2
and, like the Earth’s velocity V ⊕, does not contribute to the observed angular velocity of the planet.
This geocentric angular velocity at quadrature is, therefore,
V P sin α
E 2 P 2
and is seen to be in the same direction as the orbital movement. It is thus direct at quadrature.
12.9 Stationary points
Some time between opposition and quadrature, the planet’s geocentric angular velocity must change
from being retrograde to being direct. The point where it is neither retrograde nor direct is said to be a
stationary point. We wish to obtain an expression for the elongation E at a stationary point in terms
of the distances of planet and Earth from the Sun. From such an expression and the measured value of
the elongation, the planet’s heliocentric distance can be calculated.
It is also possible to obtain an expression giving the value of the angle θ between the heliocentric
radius vectors of the planet and Earth when the stationary point is reached, in terms of the two planets’
distances b and a respectively. This expression, with a knowledge of the synodic period of the planet,
enables a prediction to be made of the time of the next stationary point after an opposition of the planet.
In figure 12.9, let the positions of Earth and planet at a stationary point be E and P. The velocity
of P relative to E must lie along the geocentric radius vector EP if the planet appears stationary. This
velocity is represented by the line PB; the parallelogram PABC is the parallelogram of velocities
where PA = V P , PC =−V ⊕ and PB is the resultant of these velocities.
Since the orbits are circular, the angle θ between the heliocentric radius vectors must be the angle
PCB between the velocity vectors. Also it is readily seen that APB = 90 + φ.
Extend CP to meet the line SE produced at D. ThenDPA = θ and EPD = 90 − (θ + φ).
Hence, in △PCB, using the sine formula, we have
sin[90 − (θ + φ)]
V P
=
sin(90 + φ)
V ⊕