{Note: If your browser does not distinguish between
"a,b" and "α, β " (the Greek letters
"alpha, beta") then I am afraid you will not be able to
make much sense of the equations on this page.}
Any coordinates given in the horizontal or alt-az
system We need a system of celestial coordinates which is
fixed on the sky, The North Celestial Pole (NCP) and the South
Celestial Pole (SCP) Any great circle between the NCP and the SCP is a
meridian. A typical star comes up over the horizon (rises) somewhere in the
eastern sector;
depend on the place of
observation
(because the sky appears different from
different points on Earth)
and on the time of
observation
(because the Earth rotates, and each
star appears to trace out a circle centred on North Celestial Pole).
independent of the observer's time and place.
For this, we change the fundamental circle
from the
horizon to the celestial equator.
lie directly above North and South Poles of
Earth.
The NCP and SCP form the poles of a great circle on
celestial sphere, analogous to the equator on Earth.
It is called
the celestial equator and it lies directly above the Earth's
equator.
The one which also passes through the zenith and
the nadir
is "the" celestial meridian, or the
observer's meridian.
(It is identical to the principal vertical.)
This provides our new zero-point;
in this case, we use the
point where it crosses the southern half of the equator.
it moves round to the right, climbing higher in
the sky;
it reaches its highest point when it's due south, i.e.
on the meridian;
it continues moving right, and sinking lower;
and it disappears below the horizon (sets) somewhere in the
western sector.
(Note that this is only true in the northern
hemisphere;
in the southern hemisphere, the star will move to the
left,
and reach its highest point when it's due north.
In what
follows, I assume we are in the northern hemisphere.)
The
star's highest point, due south, is called (upper) transit or
culmination.
The star will also cross the meridian again,
in the opposite direction,
at the lowest point in its daily path.
This is called lower transit, and it occurs below the
horizon unless the star is circumpolar.
Stars close to North Celestial Pole never set;
if
a star's north polar distance is less than the altitude of the Pole,
then that star cannot reach the horizon.
These are defined as
north circumpolar stars.
Similarly, stars close to the
South Celestial Pole will never rise:
these are south
circumpolar stars.
All others are equatorial stars,
which rise and set.
The division between circumpolar and
equatorial stars
depends on the altitude of the North Celestial
Pole,
i.e. on the observer's latitude.
To fix the coordinates of an object X on the
celestial sphere,
draw the meridian through X.
The
declination δ of X is the
angular distance
from the celestial equator to X,
measured
from -90° at the SCP to +90° at the NCP.
Any point on
celestial equator has declination 0°.
Alternatively, the
North Polar Distance of X = 90° - δ.
Any two objects with the same declination lie on a parallel of
declination.
The Hour Angle or HA (H) of object X
is
the angular distance between the meridian of X and "the"
celestial meridian.
It is measured westwards in hours,
0h-24h, since the Earth rotates 360° in 24 hours.
time interval |
angle |
1 hour |
15° |
1 minute |
15' |
1 second |
15" |
An object on the meridian (culminating) has H =
0h.
Its HA then steadily increases as the object moves westwards.
At lower transit, when it is due north (and possibly below the
horizon), H = 12h.
At H = 23h, it is just one hour short of
culminating again.
This system is still dependent on the time of
observation,
but an object's declination generally doesn't change
rapidly,
and its Hour Angle can be determined quite simply,
given the time and the location.
A telescope can be built on
an equatorial mounting,
with its axis pointing at the NCP.
Once it is set on a star,
if the telescope rotates about its
polar axis at the correct speed (15° per hour),
the star will
stay in view.
Exercise:
The
most northerly star of the Southern Cross, γ
Crucis, has declination -57°.
At what latitude will it just be
visible?
At what latitude will it pass directly overhead?
At what latitudes will it never set?
Click here for the answer.
Previous section: Coordinate
systems: the horizontal or "alt-az" system
Next
section: Coordinate systems: the second
equatorial or "RA-dec" system
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