I have been fascinated by astronomy since I was a child, but could never understand the coordinate system. Today at the age of 33 after finally watching this video, it all made sense. Seriously, thank you!
trickybilly Thanks for the feedback - yes, that would have been a good thing to include. When I get a chance I'll drop a note on the video mentioning that.
Eugene Shubert - For some reason I am unable to reply directly to your comment (check your settings to see if you have disabled replies?). But to answer your question, there isn't a formula for directly finding an angle between two arbitrary points using standard RA and DEC coordinates because they are different units. One would need to convert RA into an angle and use the spherical law of cosines or 3D vectors. Or we fire up Stellarium, turn on the angle measure plugin, and let the computer do the hard work for us :-)
At 9:50 you might confuse some people by drawing a link between RA's minutes and seconds and Declination's minutes and seconds. Same name but completely different. ....RA is actually measuring time but Declination, angles on a sphere hence when we use minutes and seconds in astronomy as a finer angular increment of degrees, we actually ought to be careful and more properly say "ARC MINUTES" or "ARC SECONDS"
I assume that right ascension and declination are the standard coordinates for the celestial sphere. What is the formula for the angle between two arbitrary directions as a function of these standard coordinates?
How are you supposed to plot coordinates and input them into a telescope when the sky is rotating around polaris, meaning yes the stars are in fixed points on the celestial sphere but the earth is rotating, which makes the stars appear to be in different locations in the sky... So how is it we're able to input RA & DEC into a telescope without having a set time! So confused by this... Please help.
Actually, you hit the nail on the head. You DO need to input time, that way the scope knows the sidereal time (ie, the RA value of the sky due south), and can figure it out from there. Many scopes now have GPS built in, so they can figure out both location AND time by themselves.
Ed Hitchcock Oh, I see. Thanks! I appreciate it. Would I be correct in saying that the GPS is mostly for tracking celestial bodies within our solar system? (Seeing as the stars are in fixed points and only rely on the value of time, whereas the planets and other celestial bodies are moving in their orbits and change location depending on when and where you view them?)
knight654654 The GPS just tells the scope's computer your physical location on earth, and the time. The computer then uses that information to figure out where everything is. It has the coordinates of thousands of stars and objects, and the orbital data for the moon, planets and asteroids etc, from which it can calculate the precise position at any given time.
I dont understand right ascension completely. When talking about declination you said that 90 degrees is at the celestial north pole and -90 degrees is at the celestial south pole. But what is the point where right ascension is 0 hours and 24 hours? Like on the earth longitude is 0 in greenwitch observatory and goes to +180 degrees to the east and -180 degrees to the west. What about right ascension?
Since the Earth keeps rotating, right ascension is like a clock. Just as the day starts at 0:00 (midnight) and goes to 23:59 before clicking back to 0:00, RA starts at 0:00 (in Pisces) all the way around to 23:59, and then back to 0:00.
I have been fascinated by astronomy since I was a child, but could never understand the coordinate system. Today at the age of 33 after finally watching this video, it all made sense. Seriously, thank you!
Wow, I am happy I could help :-)
This video was VERY useful, as I'm a biology teacher and need to teach this to my Y11s tomorrow afternoon. Thanks!
Glad it helped :-)
Ed Hitchcock Well, the students seemed to get it just fine. Installing and using Stellarium on my projector was a great help. Thanks again.
This is a great explanation! Well done.
I'm a student of computer science but I love astronomy and stuff like these...
Great video. I have learned much information which was omitted elsewhere. One suggestion would be defining 0h RA in a video like this. Thank you
trickybilly Thanks for the feedback - yes, that would have been a good thing to include. When I get a chance I'll drop a note on the video mentioning that.
Eugene Shubert - For some reason I am unable to reply directly to your comment (check your settings to see if you have disabled replies?). But to answer your question, there isn't a formula for directly finding an angle between two arbitrary points using standard RA and DEC coordinates because they are different units. One would need to convert RA into an angle and use the spherical law of cosines or 3D vectors. Or we fire up Stellarium, turn on the angle measure plugin, and let the computer do the hard work for us :-)
Good job. Thanks.
At 9:50 you might confuse some people by drawing a link between RA's minutes and seconds and Declination's minutes and seconds. Same name but completely different. ....RA is actually measuring time but Declination, angles on a sphere hence when we use minutes and seconds in astronomy as a finer angular increment of degrees, we actually ought to be careful and more properly say "ARC MINUTES" or "ARC SECONDS"
An excellent point, thanks.
I assume that right ascension and declination are the standard coordinates for the celestial sphere. What is the formula for the angle between two arbitrary directions as a function of these standard coordinates?
Thanks a lot. It helped me.
How are you supposed to plot coordinates and input them into a telescope when the sky is rotating around polaris, meaning yes the stars are in fixed points on the celestial sphere but the earth is rotating, which makes the stars appear to be in different locations in the sky... So how is it we're able to input RA & DEC into a telescope without having a set time! So confused by this... Please help.
Actually, you hit the nail on the head. You DO need to input time, that way the scope knows the sidereal time (ie, the RA value of the sky due south), and can figure it out from there. Many scopes now have GPS built in, so they can figure out both location AND time by themselves.
Ed Hitchcock Oh, I see. Thanks! I appreciate it. Would I be correct in saying that the GPS is mostly for tracking celestial bodies within our solar system? (Seeing as the stars are in fixed points and only rely on the value of time, whereas the planets and other celestial bodies are moving in their orbits and change location depending on when and where you view them?)
knight654654 The GPS just tells the scope's computer your physical location on earth, and the time. The computer then uses that information to figure out where everything is. It has the coordinates of thousands of stars and objects, and the orbital data for the moon, planets and asteroids etc, from which it can calculate the precise position at any given time.
Ed Hitchcock Oh :D... I just saw Venus, Mars and Uranus c: Thanks btw.
knight654654 Nice. Jupiter is in the eastern sky - have a look at the king of planets :-)
Which software is that ?
Stellarium. It's free.
I dont understand right ascension completely. When talking about declination you said that 90 degrees is at the celestial north pole and -90 degrees is at the celestial south pole. But what is the point where right ascension is 0 hours and 24 hours? Like on the earth longitude is 0 in greenwitch observatory and goes to +180 degrees to the east and -180 degrees to the west. What about right ascension?
Since the Earth keeps rotating, right ascension is like a clock. Just as the day starts at 0:00 (midnight) and goes to 23:59 before clicking back to 0:00, RA starts at 0:00 (in Pisces) all the way around to 23:59, and then back to 0:00.
Ed Hitchcock I see. Thanks, that helped. And the video was very informative.