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".