# Chapter 5

Chapter 5 – The Celestial Sphere. If there is anything you notice that looks unusual then let me know… but these calculation are ones I have done to assist me in my process of thinking and most of them make sense to me… almost! 1. Draw a diagram to show the celestial sphere, putting in the celestial poles, the celestial equator, and the poles and equator of the Earth.

2. a) Define the following terms: Ecliptic, Hour Angle, Great Circle, Culmination, First Point of Aries, Zenith Distance.
Ecliptic: The projection of the Earth’s orbit (around the sun) onto the celestial sphere. It can also be defined as the apparent path the Sun takes through the Zodiac.
Hour Angle: The time which has passed since a celestial object has crossed the meridian.
Great Circle: A circle, whose plane passes through the centre of a sphere; such as Earth.
Culmination: The highest altitude of a celestial object above the horizon.
First Point of Aries: The vernal equinox.
Zenith Distance: The angular distance of a celestial object from the zenith.

b)What is meant by the right ascension and declination of a star? How does this differ from altitude and azimuth?
The right ascension the equivalent of longitude on a celestial sphere. It is measured from the First Point of Aries. The declination is the equivalent of Latitude on a celestial sphere. That is measured from the celestial equator, either North or South, to the celestial body.
This differs from altitude and azimuth because, the first pair are more or less fixed, and are used based on the Earth’s and the Celestial Sphere. The second pair are more based on where the observer is on the surface of the Earth.

c)Why does a clock which keeps GMT run at a slower rate than an observatory clock which keeps sidereal time? What is the difference per hour?
A GMT clock runs at a slower rate because it keeps to a constant 24 hour day routine, whilst a clock running on sidereal time runs on a 23 hour, 56 minute and 4 second day, losing roughly 4 minutes of a GMT day. The difference per hour is roughly 10 seconds.

3. At 0 hours GMT on 6 April 1967, the Greenwich Sidereal Time was 12h 54m. Find:
a) The local hour angle of Betelgeuse, RA 5h 53m, dec. +7°24’, for an observer at Broadstairs, lat. 51°21’N, long. 1°26’E, at 18h GMT.

Longitude/360x24x60
1.43/360x24x60
5.72 minutes =5m 43s.

GST+18h = 30h 54m
30h 54m + 5m 43s = 30h 59m 43s
30h 59m 43s + increment for 18h (2m 57s) = 31h 2m 40s

Hour angle = LST -RA
HA = 31h 2m 40s – 5h 53m
HA = 25h 9m 40s

b) The altitude of Betelgeuse when it is on the meridian at Broadstairs.
90-Latitude+Star declination
=90-51°21’+7°24’
=38°39’+7°24’
=46°3’

Altitude=46°3’

c) Whether Vega, dec + 38°44’, will be circumpolar as seen from Broadstairs.
declination>90-latitude
=90-51°21’
=38°39’
38°44’> 38°39’

Yes, Vega will be circumpolar as seen from Broadstairs.

4. An observer at Bristol, lat. 51°27’N, long. 2°33’W, finds that the star Altair, RA 19h 48m, dec. +8°44’, is at transit at 22h 20m GMT. Find:
a) The altitude of Altair at upper transit.

90-Latitude+Star declination
=90-51°27’+8°44’
=38°33’+8°44’
=47°17’

Altitude=47°17’

b)The Greenwich Sidereal Time of the transit at Bristol.
RA=LST
Longitude/360x24x60

2.55°/360=0.00708333
=0.00708333×24
=0.17×60
=10.2
=10m 12s
10m 12s + LST= 10m 12s+19h 48m = 19h 58m 12s
GST= 19h 58m 12s

c)The GMT of the transit of Altair as seen from Bristol one week later.
Four minutes lost each day, and there is 7 days in one week. 7×4=28 minutes.
To see Altair at Bristol one week later, the time would have to be 21h 52m.

5. At 0 hours on 1 March 1968, the Greenwich Sidereal Time was 10h 35m 26s. Find:
a) The GMT on the same date when the star Mizar, RA 13h 21m 54s, dec. +55°11’, was at upper transit at the McGill Observatory, Montreal, lat. 45°30’20”N, long. 73°34’42”.

longitude/360×24
=73°34’42’’/360
=0.204384258×24
=4.905222211
=4h 54m
4h 54m+RA = 4h 54m+13h 21m 54s = 18h 15m 54s
18h 15m 54s – GST at 0 hours = 18h 15m 54s – 10h 35m 26s = 7h 40m 28s – 1m 16s for the difference between sun and star time.
=7h 39m 12s

b)The altitude of Mizar at this transit.

90-latitude+declination
=90-45.501°+55.183°
=44.499°+55.183°
=99.682°
90-9.683= 80.318°

c)The local hour angle of Mizar as seen from McGill Observatory at 20h 0m GMT on the same date.

longitude/360×24
73.57833/360×24
4.90522 h = 4 hours 54 minutes 19s

GMT-LST
20h – 4h 54m 19s = 15h 5m 41s
15h 5m 41s + increment for 20h (3m 17s) = 15h 8m 58s
15h 8m 58s + GST (10h 35m 26s)= 25h 44m 24s

HA = LST – RA
HA = 25h 44m 24s – 13h 21m 54s
HA = 12h 22m 30s