- 3035 - STARS - How many stars are in the sky? How many stars can you count on a clear night. I sure you would estimate several thousand. Astronomers have been fascinated with counting the stars for centuries. Over recent decades they have even developed a mathematical formula for calculating the number of stars you can see.
--------------- 3035 - STARS - How many stars are in the sky?
- Of course how dim a star is depends on how far away it is. How many you see depends on how sharp your naked eyes are. Many backward gazers can see stars that are a Brightness Magnitude of “6”, M = 6.
-
- With a small telescope you can see to a Magnitude “10”, M = 10. The larger the number the dimmer the star. The Hubble Space Telescope can see to a Magnitude of “ 25, M = 25.
-
- What are these Magnitude numbers? To learn we need to explain “ Apparent Brightness” and “Absolute Brightness” in stars. Remember brighter stars further away can “look” dimmer. Then we need to explain “ parsecs” for measuring astronomical distances. These will be saved for the footnotes.
-
- Let’s first go for the number of stars ,”N” , we can count given the Magnitude of Apparent Star Brightness, “M”:
-
---------------- Log10(N) = -0.0003M^3 + 0.0019M^2 + 0.484 *M -3.82
-
- This is a third order polynomial equation that is valid for star Brightness Magnitudes ranging from 4 to 25. The polynomial is Log10 (y) = x^3 + x^2 + x + constant which is an exponential function. It gives an answer of the number of stars per square degree in the sky. The Full Moon is ¼ square degree.
-
- Let’s assume you do not have the sharpest eyes but on a clear night how many stars could you count up to a Magnitude of M = 5?
-
----------------- Log10(N) = -0.0003(5)^3 + 0.0019(5)^2 + 0.484 *(5) -3.82
-
----------------- Log10(N) = -0.0003(125) + 0.0019(25) + 0.484 *(5) -3.82
-
----------------- Log10(N) = -0.0375 + 0.0475 + 2.42 -3.82
-
----------------- Log10(N) = -1.39
-
----------------- N = 10^-1.39 = 0.0407 stars per square degree.
-
- There are 41,253 square degrees in the night sky.
-
- Number of stars you can count brighter that Magnitude 5 = 0.0407 * 41,253 = 1,679 stars. If you had sharp eyes and could see all the stars up to a Magnitude 6 you could count 5,077 stars
-
- How many stars if you use a backyard telescope that can see up to a Magnitude 10?
-
----------------- Log10(N) = -0.0003(10)^3 + 0.0019(10)^2 + 0.484 *(10) -3.82
-
----------------- Log10(N) = -0.0003(1000) + 0.0019(100) + 0.484 *(10) -3.82
-
----------------- Log10(N) = - 0.3 + 0.19 + 4.84 - 3.82 = 5.03 - 4.12 =
-
----------------- Log10(N) = 0.91
-
----------------- N = 10^0.91 = 8.13 stars per square degree.
-
----------------- 8.13 per square degree * 41,253 square degrees - 335,316 stars
-
- You can see over 60 times as many stars with a small telescope. What about the Hubble Space telescope that can see up to a Magnitude , M = 25?
-
----------------- Log10(N) = -0.0003(25)^3 + 0.0019(25)^2 + 0.484 *(25) -3.82
-
----------------- Log10(N) = -0.0003(15,625) + 0.0019(625) + 12.1 -3.82
-
----------------- Log10(N) = - 4.69 + 1.19 + 12.1 - 3.82 = 13.29 - 8.51
-
----------------- Log10(N) = 4.78
-
----------------- N = 10^4.8 = 60,255 stars per square degree.
-
- Hubble’s view is ¼ square degree, about the size of the Full Moon. * 60,255 stars per square degree = 15,054 stars. So the telescope needs to point in many directions to count more stars.
-
- If it had the time to cover the sky we see it would see 60,255 * 41,253 = 2,487,100,000 stars. 2.5 billion stars. And, that is just the night sky we see. Hubble can see the entire cosmic sphere
-
- “Apparent Brightness of a star is how bright it is to an observer on Earth. The actual or “ Absolute” or “Intrinsic” is the brightness, or “ Luminosity” at the power source. How bright we see it depends on how far away it is.
-
- Astronomers invented a Magnitude scale. Giving the bright star Vega a zero and giving fainter or dimmer stars a higher positive number. In the beginning the number assignments were eyeball subjective. Eventually a mathematical formula was developed for this scale. The formula developed assumed that an increase of 5 Magnitudes corresponded to a decrease in brightness by 100.
-
------------------------------- 100^ 1/5 = 2.5
-
- To compare the brightness of 2 stars, b1 and b2 having brightness Magnitudes of M1 and M2:
-
------------------------ b1/b2 = 2.5 ^(M2-M1) = 100 ^ ((M2-M1) / 5)
-
- To compare al the stars as to their “ absolute brightness” astronomers arbitrarily put them at the same distance away from us, a distance of 10 parsecs.
-
---------------------- 1 parsec = 3.26 lightyears distance.
-
----------------------10 parsecs = 191,757,000 million miles away.
-
----------------- Apparent Brightness Magnitude, m = Absolute Magnitude, M + 5 10g ^ d/10. “d” is the distance in parsecs.
-
- So if you know the apparent brightness and the absolute brightness you can calculate the distance for the star using:
-
------------------------- distance = d = 10 ^ ((m-M+5)/5)
-
------------------------- One parsec = 3.26 lightyears
-
- One parsec = 206,265 astronomical units. And AU is the Sun-Earth distance of 93 million miles.
-
- As the Earth orbit’s the Sun nearby stars appears to change their location in the sky relative to the most distant stars. This is called “ parallax”. The angle of the parallax shift is the parallax angle which is usually measured in arc seconds.
-
--------------------- distance in parsecs = d = 1 Astronomical Unit / parallax angle in arc seconds.
-
--------------------- Angle in arc seconds = 206,265 * Distance in AU / distance to the star.
-
- Our closest star, Proxima Centauri, has a parallax angle of 0.76 arc seconds. Some 7,000 stars have had their distances measured this way to an accuracy of better than 5%.
-
------------------------Apparent Magnitude -------- Absolute Magnitude
-
- Sirius -------------------- -1.46 ------------------------- 26
-
- Arcturus ------------------- -.06 -------------------------- 170
-
- Vega ---------------------- 0.04 -------------------------- 60
-
- Capella ------------------- 0.85 -------------------------- 77
-
- Rigel ---------------------- 0.14 ----------------------- 70,000
-
- Procyon ------------------- 0.37 ------------------------- 7.4
-
- Betelgeuse ---------------- 0.41 ----------------------- 38,000
-
- Spica ---------------------- 0.91 ------------------------ 23,000
-
- Deneb --------------------- 1.26 ----------------------- 170,000
-
February 7, 2021 STARS - How many stars? 3035
----------------------------------------------------------------------------------------
----- Comments appreciated and Pass it on to whomever is interested. ----
--- Some reviews are at: -------------- http://jdetrick.blogspot.com -----
-- email feedback, corrections, request for copies or Index of all reviews
--- to: ------ jamesdetrick@comcast.net ------ “Jim Detrick” -----------
--------------------- --- Monday, February 8, 2021 ---------------------------
No comments:
Post a Comment