Measuring stellar distances.
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1 Measuring stellar distances
2 This method can be used to measure distances up to 100pc Some new technology allows measuring distances up to 200pc using this method p= 1/d Stellar Parallax.htm
3 This method works for distances between 100pc < d < 10Mpc The relative intensities and wavelengths from distant and nearby stars in their emission spectra are used to estimate temperature and spectral type Using an HR diagram, the star s luminosity can be estimated and its distance can be calculated
4 Matter b/w star and observer can affect light received Some light might be absorbed by dust and affect brightness Dust can scatter different frequencies in different ways When the brightness is found, the distance can be calculated As the distance increases, the uncertainty in L increases and therefore the uncertainty in b increases as well Beyond 10Mpc the error is too large to make reliable calculations
5 The maximum wavelength of a distant star is measured 600nm and its apparent brightness is 1.0 x W m 2. What is its distance from Earth in ly?
6 The brightness of a star as seen from Earth Historically: m =1 à brightest m = 6 à dim, just visible A bright star of +1 is about 100 times brighter than a dim star +6
7 The absolute magnitude M of a star is the apparent magnitude that it would have if it would be observed at a distance of 10pc If the real distance of the Star i> 10pc, M is more negative than m Dimmest stars have M = + 15 Brightest stars have M = - 10 The closer a star, the brighter it appears Apparent magnitude depends on distance To compare star s brightness, we use absolute magnitude M at 10 pc
8 Absolute Magnitude M If a star is exactly 10pc away: m = M If a star is closer to us than 10pc, it will appear brighter than at 10pc: m < M If a star is more than 10pc away, it will appear fainter than at 10pc: m > M Apparent magnitude m
9 R = ratio The difference b/w apparent magnitudes m=6 1=5 Since a star of +1 is about 100 times brighter than a star +6, m dim / m bright = R 5 = 100 R = 100 1/5 = A star m = 0 is times brighter than a star m = 1 This means each step of magnitude is equivalent to a change in brightness by a factor of 2.512
10 If the difference is 2, the ratio of apparent brightness would be (2.512) 2 = 6.31 If the difference is 3, the ratio of apparent brightness would be (2.512) 3 =15.85 What does this mean mathematically? (2.512) m 2 m 1 = b 1 / b 2 This is variation in brightness Remember log 10 x = y 10 y = x (2.512) m 2 m 1 = b 1 / b 2 log b 1 / b 2 = m 2 m 1 Change of base for logs: log a b = log b / log a log b 1 / b 2 / log = 1/ log x log b 1 / b 2 = 2.5 log b 1 / b 2 m 2 m 1 = 2.5 log b 1 / b 2
11 m 2 m 1 = 2.5 log b 1 / b 2 Switch fraction using - m 2 m 1 = 2.5 log b 2 / b 1 m 2 m 1 = log b 2 - (- 2.5 log b 1 ) log a/b =log a log b m log b 2 = m log b 1 = constant This means using ratios, apparent magnitude can be used to find an expression for brightness
12 Polaris has apparent magnitude of and Betelgeuse How much brighter is Betelgeuse than Polaris?
13 Suppose A star has m and apparent brightness b at a distance d in pc This means b 1/ d 2 Take ratios b/w the Sat at 10pc and Earth: b E / b S = 1/ d 2 / 1/ 10 2 = 10 2 / d 2 = 100/ d 2 But (2.512) M m = b 1 / b 2 Hence 100/ d 2 = (2.512) M m
14 M m = 5 5logd m M = 5 log d 5 M m = 5 log 10/d M = absolute magnitude m = apparent magnitude D is distance b/w earth and star in pc
15 To calculate d m M = 5 log d 5 5 log d = m M +5 log d = m M+5/5 d= 10 m M+5/5 To calculate M M m = 5 5logd = -2.5 log 100/ d 2 To compare star s brightness m 2 m 1 = 2.5 log b 2 / b 1 To compare star s luminosity M 2 M 1 = 2.5 log L 2 / L 1 = 5 log 10/d M=m 5 log 10/d
16 Star m Distance /pc M Sirius Canopus Alpha Centauri Procyon Betelgeuse Sun x 10 6?
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