chapter 3 stars: distances & magnitudes nick devereux 2006 revised 8/2012
TRANSCRIPT
Chapter 3
Stars: Distances & Magnitudes
Nick Devereux 2006 Revised 8/2012
Astrophysical Units & Constants
In addition to the usual list of physical constants – listed inAppendix A (pg A-2), there is another list of astronomicalconstants that we must be familiar with and these are listed in on page A-1.
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The Sun as the “yardstick”
Since the distances and masses are so large in astronomy,the basic units of measurement are expressed in terms of theSun.
The Astronomical Unit (AU) is the distance between the Earth andthe Sun,
1 AU = 1.496 x 1011 m
Which is perhaps more familiar to you as 93 million miles.
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for larger distances, there is the parsec (pc)
1 pc = 3.086 x 1016 m
The parsec
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Angular unitsAstronomers can measure the angular extent on the sky forcelestial objects, even if they don’t know how far away theyare, and therefore unable to attribute a linear size.
Radians
360o = 2 radians
So, 1o = 2 radians
or, 1o = radians
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Angular Units (continued)Arc seconds (‘’) and Arc Minutes (‘)
1‘ = 60 ‘’
1o = 60 ‘ = 3600 ‘’
From previous page… 1o = radians
So, radians = 3600 ‘’
xradians = 1‘’
Or, 1‘’= 4.8 x 10-6 radians and 206265 ‘’ = 1 radian Nick Devereux 2006
Getting back to the pc…..
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The pc
(radian) = arc /radius
1‘’/ ‘’/ radian) = 1 AU / 1pc
So, 1 pc = 1 AU x 206265
1 pc = 1.496 x 1011 x 206265 m
Or, 1 pc = 3.086 x 1016 m
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In words, a parsec is the distance at which the separation between the Earth and the Sun could be resolved with a medium sized telescope…
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What does resolved mean?
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The Resolving Power of a Telescope
Depends on both the size of the telescope mirror, D,and the wavelength, ,of the light under observation.
(radian) = 1.22 /D
with and D in the same units.
For the Hubble Space Telescope D = 2.4mand = 0.05‘’ @ = 5500 Å .
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Question: How far away would you have to hold a dime (2cm in diameter) for it to subtend an angle of 1‘’, 0.l‘’ ?
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Mass
Usually, the masses of stars, galaxies, clusters of galaxiesare given in terms of the mass of the sun,
1 M = 1.99 x 1030 kg
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Measuring Brightness
Brightness is measured in a variety of ways;
eg. Magnitude, Flux, and Luminosity
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Luminosity of the Sun
The luminosity of the Sun,
1 L = 3.9 x 1026 W
You may recall that Watts = Joules/sec, so the luminosity of the Sun is a measure of the rate of flow of energy through the surface of a star.
Concept: Think of luminosity as the rate at which a star emits packets (photons) of energy…
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FluxThe further you move away from the star, the flux of photons,(measured in units of W/m2) passing through a 1m x 1m area goes down as the reciprocal of the distance squared;
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Flux (continued)
Quantitatively,
the flux f = L/4D2
Units: W/m2
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MagnitudesThe magnitude scale dates back to the Greek astronomer Hipparchus (200 BC).
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Definition of Magnitude
The human eye perceives, as linear, what are actually logarithmic differences in brightness.
m = -2.5log f + c
m is the apparent magnitudef is the fluxc is a constant related to the flux of a zero magnitude star
Note the –2.5, the brighter the star (f increases), the more negative the magnitude (m decreases).
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Differences in magnitudes are equivalent to ratio’s of fluxes
This obviates the need to know the constant c, or, the zero point,of the magnitude scale because;
m1 = -2.5log f1 + cm2 = -2.5log f2 + c
m1 – m2 = 2.5 log f2 / f1
Note the c’s cancelled (c – c = 0)
Also, beware the subscripts are reversed on either side f the equals sign.
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Question: A binary star system has one star that is 8 times brighter than the other. What is the magnitude difference between the two stars?
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Absolute Magnitude
We are unable to tell just by looking at the night sky if one staris fainter than another because it is intrinsically fainter (ie. lowerluminosity) or just further away.
To realistically compare stars on an equal basis we introduce the concept of Absolute magnitude (M) which is the magnitude starshave if they are all placed at the same reference distance of 10 pc.
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--------------------------------------------------------------------- * (2) d(pc)
----------------------------* (1) 10 pc
f1 = L/410)2 and f2 = L/4 d2
M – absolute magnitude = -2.5log f1 +cm – apparent magnitude = -2.5log f2 +c
Then,
M-m = 2.5log f2 / f1
M-m = 2.5log L 4 10)2 /4 d2 .L
M-m = 2.5log 100/d2 Nick Devereux 2006
Distance ModulusM-m = 5 – 2.5log d2
M-m = 5 – 5log d
So, the absolute magnitude,
M = m + 5 – 5log d
(remember, the distance d must be in pc)
On rearranging,
m - M= 5log d – 5
Where the quantity m – M is known as the distance modulus
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Trigonometric ParallaxOf course, to calculate the absolute magnitude of a star, we mustknow it’s distance. Distances to nearby stars can be found usingTrigonometric parallax.
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Parallax Angle Question:
Given the definitions for angular units provided earliershow that the parallax angle, measured in arc seconds is equal to the reciprocal of the distance to the star in pc.
So that,
= 1/d
The nearest star Centauri at a distance of 1.3 pc has a parallax angle = 1/1.3 = 0.77’’
. All other stars have even smaller parallaxes. Nick Devereux 2006