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7. When should I observe my target? • Rough answer: estimate the date when it transits the meridian at midnight. • Sun is at RA = 0h on March 21 (spring equinox) • So objects are best observed around: – – – – Sept 21 for RA = 0h Dec 21 for RA = 6h Mar 21 for RA = 12h Jun 21 for RA = 18h • Add 2 hours per month. • Also make sure itʼs in the right hemisphere! How long can I observe my target? • This depends on the targetʼs declination. • Seeing and transparency degrade rapidly at sec z > 2. • Wait till star has risen to z = 60 degrees, i.e. altitude 30 degrees. • How much longer does it take to reach the meridian? • It will take the same time again from meridian crossing to setting below 30 degrees. 1 The Meridian • A great circle divides the sky exactly into 2 hemispheres. • The meridian is the great circle that runs overhead through the zenith Z and the celestial pole P. – Angle PZ = 90 - latitude φ Z 90 -φ P Right Ascension and Declination • The equivalent of longitude on the sky • Line of constant RA is a great circle running through star S and pole P. – Angle SP = 90 - declination, δ Z 90 δ 90- -φ P S 2 Hour Angle • The meridian intersects with a line of constant RA through the star, at the pole. • The angle H = ZPS is the hour angle of the star. – Negative when star is east of the meridian – Positive when star is west of the meridian – Increases with time. Z 90 -φ H δ 90- P S Altitude and zenith distance • Draw a third great circle through zenith Z and star S. • Angle ZS is the zenith distance z of the star. – Altitude = 90 - z – Airmass = sec z Z 90 z -φ H δ 90- P S 3 Spherical triangles: the essentials • All three sides are great circles. sin a sin b sin c • Sine rule: = = sin A sin B sin C • Cosine rule: cos c = cos a cos b + sin a sin bcos C B € a € c C b A When is star at a given zenith distance? • We know the angles ZP, PS and ZS – need to know the hour angle H. – So use cosine formula: cos z = cos(90 − φ )cos(90 − δ ) + sin(90 − φ )sin(90 − δ )cos H ⇒ cos z = sin φ sin δ + cos φ cos δ cos H. Z 90 z € -φ H δ 90- P S 4 How many hours above 2 airmasses? • Cosine formula gives H in degrees: cos H = cos z − sin φ sin δ cos φ cos δ • Divide by 15 to convert to hours. • Object rises above 2 airmasses when z=60 € deg. • H gives time until meridian crossing • Object is observable for twice this duration Question Latitude @ St Andrews=56.34 deg. 1) Above what declination will stars never set from St Andrews 2) Ra=12.0, dec=0.0 How long does field spend above Airmass=2 5 Answer 1 • Lowest point of star is when H=180, i.e., ZPS are points on the same great circle. • In this case we have the condition: Z 90=90-φ+90-δ, δ=90-φ=33.66 90 z -φ H δ 90- P S • Stars above 33.66 deg never set in St Andrews sky Answer 2 Use: cos H = Sub in values: cosH = cos z − sin φ sin δ cos φ cos δ cos(60) − sin(56)sin(0) cos(56)cos(0) € Evaluate: € H = 26.77deg H = 1h47m Total time above 60deg = 3h34m € 6 What time does the star rise/ set? • Youʼve already worked out the hour angle H when the star rises above z = 60 deg. – H is negative when star rises, positive when it sets. • H = Local Sidereal Time (LST) - RA of target – So LST = RA - |H| when star rises – And LST = RA + |H| when star sets. • Local Sidereal Time = GMST - (West longitude) – So GMST = LST when star rises + West Longitude • Look up the UT (GMT) at this Greenwich Mean Sidereal time on the date concerned. – http://www.roman-britain.org/astronomy/astro.htm – The Astronomical Almanac For 12h 0dec field • LST field rises = 10h13m • LST field sets =1h47m • On 1st Dec RA overhead at midnight is ~: 0+ 71 24 = 4h40m 365.25 • Hence my field rises at: midnight +(10h13m-4h40m)=5h34m am • And sets at: 5h34m+3h34m=9h8m am € • Will rise and set at same local time in Australia etc. • Note slight modification due to longitude offset. 7