Astronomy Principles and Practice Fourth Edition.pdf

Sidereal time 91
Figure 9.3. An equatorial telescope mounting.
The right ascensions and declinations of celestial objects are catalogued. This information,
together with the knowledge of the LST, enables an equatorially-mounted telescope to be pointed at an
object of interest, even if it is invisible to the unaided eye. Anticipating section 22.2, it may be said that
such a telescope is free to rotate about an axis parallel to the Earth’s rotational axis (P ′ Q ′ in figure 9.3)
and also about an axis perpendicular to it (AB in figure 9.3).
A graduated circle at A on the declination axis enables the telescope to be swung about this axis
and then clamped so that it sweeps along the correct parallel of declination for the object of interest
when the telescope is rotated about the other axis P ′ Q ′ , known as the polar axis. The object will be
in the telescope’s field of view when the other graduated circle at Q ′ is set to the value of the object’s
hour angle at that time.
The correct angle is obtained from a knowledge of the local sidereal time and the object’s right
ascension. Thus, using equation (9.1),
HA⋆ = LST − RA⋆.
The telescope drive (a motor that turns the telescope about the polar axis) is then locked in to
compensate for the apparent rotation of the heavens due to the Earth’s rotation about its axis pq.
In this way, the object, once caught in the field of view of the telescope, is kept there, for hours if
necessary.
Example 9.1. A star has right ascension 9 h 46 m 12·s7. When it transits on two consecutive nights the
times on the clock are 9 h 46 m 21·s0and9 h 46 m 18·s6. Calculate the clock error and the rate.
The problem is best attempted by setting out the data in tabular form.
Let RA⋆ denote the star’s right ascension. Now by equation (9.1),
LST = HA⋆ + RA⋆.
But at transit, HA⋆ = 0, hence LST = RA⋆. Hence, by defining the error as ‘clock time − LST’, we
have: