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Considering that a star of certain declination is crossing the local meridian at the observer's zenith at an unknown location on the earth. Here, the declination of the star is equal to the observer's latitude. Knowing the Right Ascension of the star, how could the observer calculate his longitude?

Example: The star Miaplacidus (Dec: -69° 42' and RA: 9h 13m) crossing at the observer's zenith!

Latitude = 69° 42' S

[ Longitude= RA(Decimal Conversion) X 15° ... I'm not sure how to proceed from here ... ]

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You need a clock. Do you have one for this exercise? –  Envite Dec 12 '13 at 10:45
    
Nope :/ Perhaps proceed with a time or consider september 21 midnight... ST=LT –  Karthikeyan KC Dec 12 '13 at 16:28
    
Given the right ascension and declination of a star, how you will find its longitude and latitude? –  user1770 May 29 at 14:54
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2 Answers 2

Let's think it is september 21 midnight AT GREENWHICH (you didn't specify midnight where).

Let's go to http://www.csgnetwork.com/siderealjuliantimecalc.html

There input september 21, midnight at Greenwich. It says Sidereal time (ST) is 00:00:6.62 or, in hours, 0.001839h

This number is the difference between any star RA and its Azimuth at Greenwich.

Miaplacidus has a RA of 9h13min (9,216667h) so its Azimuth at Greenwich is 9,218506 at that moment.

Since the star is at Azimuth 0 for you, you are -9,218506h away from greenwich, that is (multiplying by 15) -138,27759º away from Greenwich. This is your Longitude.

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In other "words", the connection between the time of transit $t_\mathrm{tr}$ of an object, its right ascension $\alpha$ and the geographical longitude of an observer $l$ is the (apparent) siderial time at Greenwich $\theta_0$ (if you know your local siderial time $\theta$, you don't need $\theta_0$ or $l$): $$t_\mathrm{tr} = \alpha - \theta_0 - l = \alpha - \theta.$$

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