3
$\begingroup$

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 ... ]

$\endgroup$
  • $\begingroup$ You need a clock. Do you have one for this exercise? $\endgroup$ – Envite Dec 12 '13 at 10:45
  • $\begingroup$ Nope :/ Perhaps proceed with a time or consider september 21 midnight... ST=LT $\endgroup$ – Ken Dec 12 '13 at 16:28
3
$\begingroup$

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 $l$): $$t_\mathrm{tr} = \alpha - \theta_0 - l = \alpha - \theta.$$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.