Astronomy - Ch. 2: Understanding the Night Sky (8 of 23) Understanding Celestial Coordinates
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In this video I will explain the coordinates of the celestial sphere.
This genius saves me so many hours and days of my life, that would be otherwise wasted reading through books i don't understand. and i am doing degree level astrophysics
Sir is it possible to do astronomy through having degree in mathematics
@@arjyadebsengupta8159 With polar coordinate systems it should be easily within your capability to learn astronomy . A telescope with the scales showing the angles will soon let you make observations . Remember that most telescopes only go to about a degree or two on the scales of the protractors so measurements are a bit limited . There is a great app called "Stellarium Web online " which gives star maps for your computer or mobile phone which will show you where to look for lots of stars , planets and nebulae .
I always wanted to learn how to find stars. And finally someone is teaching it beautifully. I picked stargazing recently and amazed by it.
+Yashraj Atodaria Have fun with it!
+Michel van Biezen: Some sources are saying that from equator to zenith is 90 degrees, whereas you said that from zenith to equator is 90 degrees. So which is the 90 degree angle? Zenith or the equator?
For eg: When i observe arcturus, it has 14 hrs RA and +19 Dec. I am observing it on 21st may. So i should be looking at 10 hrs and i live at +21 lat. So 19-21, gives me -2 which means it should be right over my head. But when i checked on stellarium it's showing me below the ground. What am i doing wrong?
Same here.
Very True!
Michel, teachers like you are a precious rarity
This course series is just great. Exactly one of the things missing on the tube, so a great niche hit! Thanks a lot for that!
teacher, it is great to see your tutorials about the astronomy, they are the ones i was looking for. i am thankful
Just perfect. You are what we called a natural Professor.!! Thank You for shared you creativity.
Regurgitating lies means absolutely sfa.
We appreciate the efforts you put laying down the profound timeless knowledge for a wide range of sciences. I continue to follow your lead since 6 years now.
Much appreciated
Six years? You are a dedicated follower. Glad you find our videos engaging and informative. 🙂
Great to see an astronomy teacher who eschews obfuscation.
2 SAT words in a single sentence. Impressive! Nice to see someone who has such control of the English language. (And thank you for the comment)
He eschews it and spits it out in understandable terms.
After watching this subscribing to your channel was a no brainer. Thank you Michel.
Welcome aboard! Enjoy the videos.
your series on the night sky is incredible. I love it. a must watch for kids and adults alike.
Thank you. Very infomative. I've had Celestial Navigation as a course but so many years ago I forgot...
Wonderful! This concept of declination I first understood as an amateur astronomer since I couldn't understand Right Ascension yet, so I knew about it but the way you explain it here (especially with the little equation) made even clearer sense...brilliantly explained, Thank you!
Once you find a few key stars this way, you'll get the hang of it. (Such as Vega, Anares, Spica, etc.)
This dude is awesome. I have been watching his videos for over 4 years, and he never fails to amaze me.
Glad you are enjoying our videos! 🙂
This course is amazing. you do such a great job at explaining and demonstrating!
This is so well explained! It has never been explained to me so clearly.
fantastic tutorial and I learned a lot from it,thanks
This was a great video! Thank you so much sir. I was struggling to understand the concept before and now it's so much clearer.
Yes, this is a tricky topic. Glad it is clear to you now.
Thank you, Mr. van Biezen, for this clear explanation!
Thank you. Excellet job, very clear and in short time.
Thank you Mr. van Biezen. You are a really great teacher. I didn't believe I would ever understand RA / Dec but thanks to your videos now I do.
We are glad you figured it out. It is a confusing topic.
Very good video, I have to keep watching, there's a lot of info here, thank you!
Enjoy!
Subbed, thank you for the wonderful work ! I needed some help understanding this!
Thank you so much for the neat explanation sir!
Dank je, Michel van Biezen, voor je duidelijke uitleg👍
I'm very appreciative for both this video and the one on Right Ascension. You explain both concepts so clearly and concisely with excellent examples, and I am elated to finally have this knowledge firm in my brain lol. Thank you Sir.
You are so welcome!
i am joining a space competition and this guy made the topic easy to understand thanks
Most knowledgable guy to ever live
After some searching I came across these, which are great!
Glad you found them.
This one was very helpful....i've been trying to get these co ordinates. Thanks Professor
Glad it was helpful!
Legend! Saves me so many hours
Great. Glad you found our videos.
Such an awesome video... 5 years.... Hardly 800 Likes, even 6 dislikes too. What to say?
The >44k views is something hopeful.
Your way of teaching is AWESOME sir.
always a fan of your classes!
Glad you like them! 🙂
Searching youtube for info on RA and DEC, ended up watching all the chapter 2 videos. Thanks!
Awesome! Thank you!
Very good videos. Excellent!
Glad you like them! Pass the word. 🙂
You are a wonderful teacher.
Thank you! 😃 Glad you like our videos.
Absolutely amazing explanation.... ☺️
Thank you 🙏🏻🌼
Glad you are enjoying our astronomy videos
You have so much great content
We have over 9000 videos on this channel. Glad you are enjoying them.
Thanks Professor. Superb video.
Greate explanation wonderful lessons
Right Ascension and Declination nicely explained
Thank you.
Hello sir, greetings from India. I learn a lot from your sweet and short lecture. You are a really good educator. Many TH-cam channels on basics of astronomy are there, but most of them are too much talkative. I like your way of teaching with tredition chalk and board method. Thank you sir...
You are most welcome
Excellent explanation .... see #9 for RA
Does this "WHERE TO LOOK" have any astronomical term? If so what is it?
how to measure my position with respect to position of celestrial sphere?
simply awesome.
Thanks for sharing this with us . Your explanation is very impressive
My pleasure
@@MichelvanBiezen Make more interesting videos so that common people also get interested
Hello, it is I again. Now for a more difficult Star. This time I am talking Polaris. I did find it but again I am trying to use the Right Ascension and declination way to do it. Right Ascension is 2h 31min and Dec is +89° 15'. The RA by my calculation for October 21 is -12 hrs and November 21 that would be -14 hrs. If I understand right that is 14 hours AHEAD of the sun rising, right? So would that mean I should see it at 1730 hours or 5:30pm looking North at about 49° the previous evening. Oh I am on the Eastern side of US and I location is at 40° as mentioned in an earlier post. If anyone else would like to answer please be my guest. As always thank you for any information. Take care and be safe.
you're great :)
Please help. Tonight I'm looking at Jupitar and when I calculate its angle of declination with me being at 46 degrees I get -39, but that's not where I"m seeing it in the sky. It's right ascension is 11 hours. So is the -39 degree then the highest it will get in the sky during it's transit through the eliptical? I guess I'm confused on how to use these 2 things together to find an exact position because it seems angle of declination will only tell you exactly where the star is at one specific time.
hello adamEater
People must support people like you sir😭 thank you sir 😊🙂👍
hello sir...from the age of five I've been into this field...I'm living for it and I also have a huge dream now I'm 14 and all I wanna say is the way you describe is really amazing and I really really do love your videos A LOTT!!! thank you for such good videos
Yes, I remember that I loved astronomy at a very young age as well. Have fun with it, there is an amazing universe to explore out there. 🙂
@@MichelvanBiezen Ohh!
I'm on my way...thank you sir
Mighty glad you stayed :)@@layanaclement2154
The angle of declination that you took into account is in the day of vernal equinox, right? Because as we move past it....that would also change its Ra and declination. Your next video explain how to take into accounts the day while calculating stars position as nom of hours they are behind the sun.
My question is that does the declination angle also changes from our point of reference along the year
Correct for your first question. Since the Earth's axis virtually doesn't change during the span of 1 year, the angle of declination doesn't change throughout the year.
Excellent lecture Sir. Regards
Thanks and welcome
I've hit gold!
Hi Michel, great series. I have a question ( well several actually, but this one is keeping me awake at night) As you have explained, the Right Ascension of an object in the sky appears to shift by approx 2 hours every month, starting on the 21st of March, until it is back where started the following year, due to Earth's path around the Sun, but my question is this, does the Angle of declination also shift, due to the Earth's angle of tilt, and if so, by how much?Oh wait, another question, where could I find the coordinates of an object on any given day of the year?
The angle of declination essentially stays the same, since the Earth's tilt (and the direction of the tilt) remains "stationary" over the time frame of interest. Because of the precession of Earth's tilt, the angle of declination will slowly change over many years.
Thanks once again Michel.
Can I suggest you download stellariam
i am utterly confused. I don't see how to use DEC and RA together. Since a particular constellation will move through the night (rise and fall) wont its declination change through the night as well? Is there a particular time of night declination is accurate?
The angle of declination is the same (constant) throughout the night. It doesn't look like that because we live on a sphere, but it is that way.
great video, very informative! I am confused by your explanation for finding Polaris, if you are at a lattitude of 34 deg, wouldn't you look up 34 from the horizon to find polaris, that is where i find it, exactly at my lattitude...
Polaris (the North Star) is located at about 90 degrees north (angle of declination). If you are located at 34 degrees north, the to see Polaris you have to look at (90 - 34) = 56 degrees north of the zenith, which is the point directly above your head.
I have been trying to learn this but a little confused. My position is 40°. I am doing Vega as an example. My calculation is -2 which if I understand is more or less straight up, right? As far as the Right Ascension now and if I did it right for October 21 is 4 hours or after the sun goes down. In turn that is about 60° from horizon up when looking West. Sunset for my area is 1800 (6pm). So how do I find Vega? Do I look straight up at -2° or do I look west at 60° and what time 1800(6pm) or 2200(10pm)? Any info would be great. Thank you.
Vega will be 60 degrees up from the western horizon at 6 PM , Vega will be 30 degrees up from the western horizon at 8 PM and will set over the horizon at 10 PM.
Thank you for replying. I love learning this stuff but just takes awhile to sink in. Thanks again for the online lesson. Take care.
Sir, I have a question. When the earth revolve around the sun the ecliptic plane will change continuously so the plane of earth's equator will not be in the plane of celestial equator but will be parallel to it,so how can we spot stars if planes are not concurrent?
We use the coordinate system: Angle of declination and right ascension. (We have videos on that).
Basically hour or even second there are charts that have been worked out
Stelarium is a wonderful tool thats free
I do have very small doubt..
One small question ...
The coordinates of a single Star X are Dec 08 ° 50 'and RA 17 hours 25 minutes.
The coordinates of another star Y are Dec 9 ° 57 'and RA 05 hour 36 minutes.
If I want to locate one of these two stars with the help of a polar aligned German equatorial mount, how do I know if this star is on the east side of the mount or on the west side of the mount?
Right now the idea comes from the mobile app but what if I only have a table with RA and DEC?
Dec's + and - sign, we can say the Northern Hemisphere and the Southern Hemisphere, but is there any idea for the East-West direction?
Thank you sir in advance....
🙏
My suggestion is to use a phone APP if you don't have a computer on your telescope that will locate the star for you automatically. (The modern technology that exist today is amazing)
Sir, what that hrs and minutes represents?
Does it changes according to countries?
The amount of time between the Sun rising over the horizon and the the time the object you are observing rising over the horizon.
@@MichelvanBiezen Thanks sir
Hi Professor, how do I Know if I'm located at +34º or 36? or maybe 40?
Measure the angle of declination when a known star passes through the line that runs from north to south and cuts through the zenith. Then measure the angle between the zenith and the star.
@@MichelvanBiezen Thank you Professor :) I'm watching all videos in your Astronomic Serie, they are awesome and easy to understand for those who don't know about Physics and Astronomy just like me. I came here by chance trying to understand why there are so many types of zodiacs and houses systems (it doesn't matter whether they work or not, nor if some system fits just for mere coincidence), I just want to know the exact position of a given point at a certain time on earth related to the fixed stars, to picture the celestial sphere to see if a given celestial body (planet, Moon, asteroid, etc.) matches any constellation on the ecliptic to be able to tell if it is true that Jupiter is in Cancer for instance or to tell if the Moon's ecliptic "touches" different constellations when it's located away from the nodes.
You speak of the Zenith and Meridian interchangeably (4 min mark). I was under the impression that they are 2 different points in the sky. The Zenith is 90 degrees from the equator while the Meridian is running though the poles, which could be to one side or the other of your local Zenith. I'd appreciate any clarification you can offer.
The zenith is the point directly above the observer, regardless of the location of the observer. A meridian is a line that runs from the north pole to the south pole. The prime meridian runs through Greenwich England
@@MichelvanBiezen thank you. I'm really enjoying your videos :)
The mass of an object is easily determined if another object orbits around it. We know the mass of the Earth by observing the orbit of the Moon (or by a satellite orbiting the Earth). Measuring the distance is much more difficult to do.
How do you find what direction is north without a compass, a watch, or the sun?
At night you can find north by finding Polaris (the north star).
You could learn to spot the north star (Polaris) at night, if you are on the north hemisphere
A rough method is to find a tree with moss on it. The moss will tend to be more concentrated to the N only in the Northern hemisphere tho
whats the difference between the altititude of a star and where to look ?. i read somewhere "alt=90 - (lat + dec) whereas "where to look = dec - position"
If you were at the equator, those would be the same. But if you are north of south of the equator, you have to keep in mind you position and adjust for it to locate the star.
@@MichelvanBiezen after almost losing my mind i figured out, (where to look = dec - position),is vision at the degree from zenith, while (alt = 90 - lat + dec) applies to vision at the found degree from the horizon. Both of them give different answers and if you add those answers you get 90 as the sun ( a quadrant of the sphere ie earth )
Great lesson. And not just this one, all of them are great! I've learned a lot from you. But wouldn't Polaris be at 34 degrees north of the observer. I am confused because the position of polaris should be the latitude of the observer. Am i wrong?
Polaris is "almost" directly above the North Pole, and if we are at 34 degrees north of the equation and Polaris is 90 degrees north of the equator, the difference is 90 - 34 = 56. Which means we need to look 56 degrees down from directly above our head (the zenith) to find Polaris.
@@MichelvanBiezen Say we are at the North pole (90 degrees north of the equator), then the Polaris should be right above our head, i.e. at 90 degrees. However, by this method, we would need to look at 90 - 90, which is 0 degrees or the horizon. Please explain....
@@adityathakur9215 it's a relative angle from the zenith (when you look straight up). So 0° means you look straight up and don't change
I have confusion like my position is on 37 degree and vega star is 38 that means it 1 degree so i can see above my head right? But it is in north direction below the horizon
That means that once per day or night, Vega will pass above you 1 degree to the North from the zenith, (the point directly above your head). Vega is visible during the evening during summer and early fall.
That means degrees are only useful for seasonal stars?
They are useful for all stars, but in order to know if stars are visible at a certain time, you also need to know the right ascension. (We have videos on that as well in this playlist).
I always want to know this thing even though I haven't scored good in my 12 th 😃😍☀️
Wouldn’t Alpha Centauri be visible (just above the horizon) about 5 degrees south of LA, for example in San Francisco?
Since the angle of declination = - 61 degrees (61 degrees below the equator) , it is not visible from San Francisco
San Fran is north of Los Angeles. Maybe you meant San Diego?
If I'm at the equator, Polaris will be on the horizon, right? If I'm at the North Pole, a star with 0* declination will be on the horizon, and if I'm at 60* N, Polaris will be North, at 60* above the horizon, right?
You are correct.
thnk uuuuuu
What about if u r at pole???
If you are the pole, Polaris would make a tiny circle directly above your head and all the stars would revolve around that point every 24 hours.
sweet bow tie, bow ties are cool.
latidude?
Sir how to calculate our position
Do you mean our position on the surface of the Earth?
@@MichelvanBiezenyes sir for example I live in India how can I measure my position for where I can see stars on my position here on earth
I thought latitude lines were east to west parallel to the equator and longitude lines were north to south passing through the poles
Kevin,
That is correct.
"Once every 24 hours" a star passes the meridian. Technically, it's once every 23:56:04.
That is correct.
~2:52 : a star will be back to the meridian 23h56m04s later, not 24 hours...This is called the sidereal day (86164 seconds).
If one wants to find a celestial object easily with an equatorial mount WITHOUT A GOTO SYSTEM, he/she needs:
1) to have a mount with the so called (and wrongly) "Right ascension graduation circle" on the main axis : these ARE NOT right ascensions, these are HOUR ANGLE ! And that circle MUST be FIXED : it shoudn't turn when you rotate the main axis ! If you can't fix that graduation circle, you're doomed or you must target a known object first which is a very uneffective way of finding a celestial object, it works but it's a pain in the ass...
2) to know what is the hour angle of an object and what YOUR local sidereal time is !
HERE's the formula EVERY ASTRONOMER SHOULD KNOW : HA (object) = LST (Local Sidereal Time) - RA (object).
=> YOUR LST is also equal to the hour angle of the Vernal Point :
HA (vernal point) = LST - RA (vernal point) HA (v.p.) = LST - OO:OO:OO HA (v.p.) = LST !!! => YOUR LST is ALWAYS equal to the Vernal Point hour angle (for everybody, anytime on earth)
Hour angle starts on YOUR meridian (due south for those in the northern hemisphere)=> hour angle of any object on YOUR meridian is equal to 00:00:00, and at that precise moment, YOUR LST is equal to the RA of that object :
HA (object on merdidian) = LST - RA (object) 00:00:00 = LST - RA (object) LST = RA (object)
3) to put and fix the graduation to "06H" when your tube is in Home Postiion (tube points to the celestial pole, counterweights down) for nord hemisphere ("18H", for south hemisphere). Once this is done, leave it there, don't move it anymore, you can use it later to roughly polar align your mount WITH ANY STAR !
4) To use a software (like Stellarium for example) which will give you the hour angle af any object anytime (in Settings | Information) or calculate it yourself :
a) www.iwc.com/ch/fr/sidereal.html => choose YOUR location to get YOUR sidereal time and use the formula
b) if you can program :"Astronomical Algorithms, 2d Edition" (Jean MEEUS), teenagers can do that if htey went to school up to 16...
5) To know that if an object has an hour angle between 00H and 12H, it is necessarily located in the WEST part of your sky, and in the EAST part of your sky if its hour angle is between 12H and 24H.
When you move the tube from the WEST part to the EAST part without rotating the main axe (and obvioulsy passing the celestial pole), you must mentally add 12H to the hour angle of your mount, and conversly.
6) If you master the hour angle, you'll finally find out that a polar scope is not needed ! You can use your telescope as a "polar scope" and any star can be used to polar align your mount (the drift method should be use afterwards to precisely align your mount)
I use it regularly to point the sun (with a filter of course...) or Venus (during the day, when its elongation is BIG ENOUGH TO AVOID THE SUN, like now in february 2022 !) and it always works like a charm...
Give it a try at home in your bedroom, pretend your mount is polar aligned, and compare the postion of your tube with what you see in a software (like Stellarium), no need to go outside to realise that it works !
Know the hour angle and all will be crystal clear ! And BEWARE manufacturers's manuals what's concerning RA graduation circle, THEY GOT IT WRONG, CONFUSING ALL ROOKIES AND THAT'S A SERIOUS MATTER ! It took me years to understand !
Clear skies !
Excellent summary of how to use a telescope without the computer guided mechanism!!
I use both on my EQ mount but even easier is a BIG scope on a dobsonian mount
If you have a wide enough field of view and lots of photons it makes just looking such fun
44 + 46 = 90
And the poster graphic says "Angel of declination" - what a celestial typo ;)
It does have a nice ring to it.
Proxima centauri is the closest star besides Sun
10:04 you forgot the sun is the closest and brightest star 😅😅😅😅
Didn't forget. It is understood that the Sun is not included in the "star" list. 🙂
@@MichelvanBiezen Oh!! But in our childhood we learnt that sun is the nearest star to the earth
@@MichelvanBiezen By the way thank you for clarifying ☺️☺️🙏🏻🙏🏻
You are correct.
Should one pronounce Betelgeuse as Beetlejuice? ;)
That is what many people do.
He couldn't find the brightest star in the night sky.😂😂
1 day = 1round about earth.............1year =1round about sun ....... 1mounth = ???? what
1 month is 1/12 of the revolution around the sun
Check your spelling.
Thank you. Can you tell us which word is misspelled?
@@MichelvanBiezen... latitude
Thank you for pointing that out.
Such a bad explanation...