Exercise:
A stars true position is
Right Ascension 6h 0m 0s, declination 0° 0' 0",
and
it lies at a distance of 25 parsecs.
On the date of the Spring
Equinox,
how far will it appear to be shifted by annual parallax,
and in what direction?
First convert from RA and dec.
into ecliptic coordinates.
sin(δ) = sin(β) cos(ε) + cos(β) sin(ε) sin(λ)
sin(β) = sin(δ) cos(ε) - cos(δ) sin(ε) sin(α)
cos(λ) cos(β) = cos(α) cos(δ)
where α = 6h0m = 90°, δ = 0°
So this star has λ = 90°, β = -ε = -23°26'.
Its distance = 25 pc, so annual parallax Π = (1/25)" = 0.040"
At the spring equinox, the Sun has ecliptic longitude
λS = 0°
so λS-λ = -90°.
Δλcosβ = Π
sin(λS-λ) = 0.040"
cosβ = 0.918
so Δλ = 0.044".
Δβ= -Π cos(λS-λ) sinβ = 0
So star is shifted 0.044" eastwards, by parallax.
Back to "annual parallax".