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---------------------- 2606 - BETELGEUSE - use temperature to learn size?
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- This review uses the star Betelgeuse in the Constellation Orion the Hunter. Betelgeuse is the star at the left shoulder of the Orion Hunter. It is a Magnitude 0.41 which is a little dimmer than the Magnitude of the Reference star Vega which is set at zero Magnitude, m = 0.
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- All positive numbers greater than zero are dimmer stars. Each increase in Magnitude corresponds to a decrease in brightness by a factor of 2.5, that is 100 ^1/5. Therefore an increase in 5 magnitudes corresponds to a decrease in brightness of 100.
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- Betelgeuse is 427 lightyears away. It is 38,000 times the luminosity of the Sun. It is Spectral Type M with a surface temperature of 3500 Kelvin It has a life expectancy of 200 billion years compared to our Sun that has a life expectancy of 10 billion years.
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- Its light spectrum curve peaks at 830 nanometers wavelength in the infrared. It is a Red Supergiant star. But, how do we know all this stuff? Our goal in this review is to learn how big it is? What is its diameter?
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- We can show that its radius is a function of its luminosity. And, its luminosity is function of its temperature.
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- Luminosity is the total power output measured in watts. For example: Our Sun has a luminosity, Lsun = 3.8*10^26 watts. That translates to us on Earth receiving 1,300 watts per square meter of solar power, except at night.
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- It makes sense when you think about it a star’s brightness is directly proportional to its size. The bigger the star the brighter the star. And, its temperature. The hotter the star the brighter it is:
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----------------------- The Luminosity (L) = Constant * radius^2 * Temperature^4
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- Any time a function is a proportionality it can be turned into an equality with the proper constant of proportionality. L is proportional to the temperature raised to the forth power and the radius squared. In this case the constant is Ks , the Stefan-Boltzman Constant
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------------------------ The Luminosity (L) = Ks * ( 4*pi*r^2 )^2 * T^4
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------------------------- Ks = L / (T^4 * 4* pi* r^2)
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------------------------ (4 * pi * r^2) is the surface area of a sphere
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------------------------- (radius)^2 = L / ( 4*pi* Ks ) * T^4
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- Our goal is to find the radius of Betelgeuse but we do not know Luminosity , or Ks, or temperature.
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- Let’s first find the temperature, T. We measure the light from Betelgeuse and we take a spectrograph that identifies the peak wavelength amplitude. This is the same as frequency where the color is the brightest amplitude. Remember Betelgeuse is a Red Supergiant star. The light curve amplitude versus wavelength is the shape of a Blackbody curve peaking at 830 nanometers.
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- Now we use Wien’s law that is the peak wavelength is indirectly proportional to the temperature.
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--------------------------- Peak Wavelength = Kw / T
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- As the temperature gets higher the wavelength gets shorter. In this case the constant of proportionality is Kw = 2,900,000 nanometers / Kelvin, determined by empirical measurements.
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------------------------- T = 2,900,000 nanometers / 830 nanometers
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------------------------ T = 3,500 Kelvin
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------------- THE SURFACE TEMPERATURE OF BETELGEUSE IS 3,500 KELVIN.
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- The Sun is a yellow star and its surface temperature is 5,800 Kelvin. Betelgeuse star is cooler and redder at 3,500 Kelvin. Next we need to determine the constant Ks which is called the Stefan-Boltzman’s constant.
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- One of the mysteries of science is that all of nature’s constants can not be derived mathematically. They have to be measured empirically, with experiments and observations. For example, light had to be measured to determine that the speed of light, “c” = 299,800 kilometers per second. Ks is a constant so it would be the same for every star. Let’s find it for our Sun than it will be the same for Betelgeuse.
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-------------------------------- Ks = L / ( T^4 * 4* pi* r^2 )
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----------------------- Luminosity of the Sun = Lsun = 3.8*10^26 watts.
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----------------------- Apparent Brightness = Lsun / 4*pi*d^2
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- This is the inverse square law for light. The Apparent Brightness can be measured on the surface of the Earth to be 1,300 watts per square meter. The distance is that from the Earth to the Sun 149.6*10^9 meters. Imagine a giant sphere centered on the Sun. The surface area of the sphere is 4*pi *d^2.
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----------------- 1.3*10^3 watts / m^2 = Lsun / 4 * pi * (1.496*10^11)^2 m^2
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-------------------- Lsun = 3.7*10^26 watts. ---------- Google says 3.8*10^26 watts
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--------------------------- THE SUN’S LUMINOSITY IS 3.8 * 10^26 WATTS.
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--------------------------------- Ks = Lsun / Tsun^4 * 4* pi* r^2
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-------------------------------- Tsun = 5,800 Kelvin
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--------------------------------- Tsun^4 = 1.132*10^15 k^4
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-------------------------------- r = 6.96*10^8 m
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------------------------------- 4*pi*r^2 = 6.087*10^18 m^2
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------------------- Ks = (3.8*10^26 watts) / ((1.132*10^15 K^4) * (6.087*10^18 m^2))
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-------------------- Ks = 5.52 * 10^-8 watts/m^2/K^4
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- ---------- Google says: THE STEFAN-BOLTZMAN CONSTANT IS 5.67*10^-8 watts/m^2/K^4
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- We have the surface temperature of the star Betelgeuse, we have the Stefan-Boltzman constant of proportionality, next we need the luminosity, “Lbet“, in order for our equation to be able to calculate the radius.
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- We can measure the Apparent Brightness of Betelgeuse to be Magnitude, mbet = 0.41 as it appears from Earth. By mathematically setting all brightness magnitudes to the same distance away we can compare their Absolute Magnitudes.
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- The reference distance is set for 10 parsecs, or 36.2 lightyears for all Absolute Magnitudes. The Sun is an Absolute Magnitude 4.8. We need to calculate the Absolute Magnitude of Betelgeuse and compare that to the Sun’s in order to learn the Luminosity of Betelgeuse.
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----------------------------------- mbet = Mbet + 5* log ( d / 10)
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- In this case the distance “d” is in parsecs. 1 parsec = 3.26 lightyears. Betelgeuse is 427 lightyears away. This distance can be calculated measuring the “redshifts” of the star’s light spectrum. See Review 1501 to learn about measure distance with redshifts.
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----------------- d = 427 LY / 3.26 LY / parsec = 131 parsecs
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------------------ 0.41 = Mbet + 5*log ( 131 / 10 )
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------------------ 0.41 = Mbet + 5 ( 1.12 )
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------------------- Mbet = - 5.19
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------------------- Msun = 4.8
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------------ THE ABSOLUTE MAGNITUDE OF THE BRIGHTNESS IS -5.19.
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- Distance of Betelgeuse calculated using the brightness Magnitudes:
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-------------------- d bet = 10 ^((m-M+5)/5)
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--------------------- d bet = 10 ^(0.41 +5.19+5)/5) = 10 ^(2.12)
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---------------------- d bet = 132 parsecs * 3.27 lightyears / parsec
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--------------------- d bet = 431 lightyears
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---------------- Google says: THE DISTANCE TO BETELGEUSE IS 427 lightyears
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- Our Sun is the brightest star in the sky
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-------------- Apparent Brightness , msun = Msun + 5*log (d/10)
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----------------------- d sun = 1.496*10^11 meters ( 93 million miles )
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------------------------ d sun = 1.496*10^11 / 3.0857*10^16 = 0.458 *10^-5 parsecs
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----------------------- 1 parsec = 3.0857 *10^16 meters
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----------------------- Msun = 4.8
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- THE ABSOLUTE MAGNITUDE OF OUR SUN’S BRIGHTNESS AT A DISTANCE OF 10 PARSECS IS 4.8.
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---------------------- msun = 4.8 + 5*log (0.485*10^-6)
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---------------------- msun = 4.8 + 5*log ( -5.314)
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----------------------- msun = 4.8 - 26.57
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----------------------- msun = 22, the brightest star in the sky
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---------------- THE APPARENT BRIGHTNESS OF THE SUN IS THE SKY IS +22.
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- We can see that at a standard distance of 10 parsecs Betelgeuse is much, much brighter than our Sun. The equation for calculating Luminosities from Absolute Magnitudes is:
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----------------- Lbet / Lsun = 100 ^(Msun - Mbet) / 5
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----------------- Lbet / Lsun = 100 ^( 4.8 + 5.19) / 5
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----------------- Lbet / Lsun = 100 ^2
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----------------- Lbet / Lsun = 10,000
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- This calculation puts Betelgeuse brightness at 10,000 brighter than our Sun. Google Wikipedia says it is 38,000 times brighter. I do not know where my error is in this calculation.
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- If the Sun and Betelgeuse were sitting next to each other at 10 parsecs distance. How much brighter would Betelgeuse be than our Sun?
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-------------------- Mbet / Msun = 100 ^(Msun - Mbet) / 5
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-------------------- Mbet / Msun = 100 ^(4.8 +5.19) / 5 = 100^(10/5) = 100^2
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-------------------- 10,000 times brighter magnitude
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- GOOGLE SAYS: THE LUMINOSITY OF BETELGEUSE IS 38,000 the LUMINOSITY OF THE SUN.
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- We need the luminosity of the Sun next:
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-------------------- Lsun = ( 4*pi*Rs^2 ) * Ks * Ts^4
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-------------------- Rs = 6.9599*10^8 meters
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-------------------- Rs^2 = 48.4 * 10^16
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-------------------- ( 4*pi*Rs^2 ) = 608.7 * 10^16
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-------------------- Ks = 5.67*10^-8 watts / m^2 / K^4
------------------- Ts = 5800 K
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-------------------- Ts^4 = 1131*10^12 K^4
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---------------------- Lsun = ( 6.087*10^18 ) ) * ( 5.67 *10^-8 ) * ( 1.131*10^15 ) watts
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----------------------- Lsun = 3.9* 10^26 watts
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------------ Google says: THE LUMNINOSITY OF THE SUN IS 3.8268*10^26 watts
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- THE SUN PUTS OUT 38,000,000,000,000,000,000,000,000 WATTS OF POWER.
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----------------------- Lbet = 38,000 * Lsun
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----------------------- Lbet = ( 3.8*10^4 ) * ( 3.8268*10^26 )
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----------------------- Lbet = 14.55*10^30 watts
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- BETELGEUSE PUTS OUT 14,550,000,000,000,000,000,000,000,000,000 WATTS OF POWER.
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- Now, we have enough information t solve for the radius of Betelgeuse, Rb
-------------------- Lbet = ( 4*pi*Rb^2 ) * ( 5.67 *10^-8 ) * Tb^4
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-------------------- Rb = ?
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-------------------- Ks = 5.67*10^-8 watts / m^2 / K^4
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------------------ Tb = 3500 K
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------------------- Tb^4 = 150 *10^12 K^4
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-------------------- 14.55*10^30 watts = ( 4*pi*Rb^2 ) * ( 5.67 *10^-8 ) * ( 150 *10^12 )
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-------------------- 14.55*10^30 watts = ( 4*pi*Rb^2 ) * ( 8.5 *10^6 )
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------------------- 1.71*10^24 watts = ( 12.567Rb^2 )
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------------------ Rb^2 = .136 *10^24
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------------------- Rb = 369 * 10^9
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- BETELGEUSE RADIUS IS 369 BILLION METERS. THE RADIUS OF MARS ORBIT IS 227.9 BILLION METERS.
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- Therefore the Solar System inner planets cold all fit inside the star Betelgeuse. Think about it next time you look at the southern horizon and see the Constellation Orion the Hunter .
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- What is Betelgeuse’s rate of change in luminosity with a change in surface temperature?
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-------------------------- L = 4*pi*Ks*T^4
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- Take the differential to get the rate of change.
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---------------------- dL / dT = 16 pi*r^2*Ks*T^3
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---------------------- dL / dT = 16 * pi* (369*10^9)^2 * ( 5.67*10^-8 )*( 3500 )^3
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---------------------- dL / dT = 50.27* ( 13.62*10^22) * ( 5.67*10^-8 )*( 42.875*10^9 )
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---------------------- dL / dT = 166 * 10^26 watts per Kelvin degree.
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------------------------ Sun Luminosity is 3.83 *10^26 watts, so this is equivalent to a change of 43 Sun luminosities per degree change from the 3,500 Kelvin surface temperature of Betelgeuse.
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- February 5, 2020 1522 2606
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--- Some reviews are at: -------------- http://jdetrick.blogspot.com -----
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--- to: ------ jamesdetrick@comcast.net ------ “Jim Detrick” -----------
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--------------------- Wednesday, February 5, 2020 --------------------
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