Exercise:

A minor planet passes very near the Earth,
at a distance of 200,000 km.
What will be its horizontal parallax?
(Take the Earth to be a sphere of radius 6378 km.)

Horizontal parallax P = a/r radians,
where a = 6378 km, r = 200,000 km.
So P = 0.0319 radians = 1.827°

At St.Andrews (latitude +56°20'),
the minor planet is observed to cross the meridian
at an apparent altitude of +35°.
What does its declination appear to be?

Meridian altitude a = (90°-φ) + δ
So the apparent declination is
δ = a - (90°-φ) = +1°20'

What is its true declination, after correcting for geocentric parallax?

Apparent altitude = 35°
so apparent zenith angle z'= 55°.

Angle of parallax p = P sin(z') = 1.497°

Parallax increases the zenith angle.
The true zenith angle z must be less than the observed zenith angle z'.
So the true altitude must be greater than the observed altitude.

So in this case the true declination must be
greater than the observed declination, by 1.497°,
making it +2°50'.

Back to "geocentric parallax".