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Stars
Distance
Getting distances to stars:
1. Geometry: ‘Stellar Parallax’
> Only direct method!
2. Inverse-Square Law
L
F 
2
4 d
* Measure F
* Know (or guess!) L
Find d
Stellar Parallax
Triangulation:
d
p
x
Baseline
q
x and q (or x and p)  d
For getting distances to stars, we want longest
possible baseline:
x
d
p
Earth
p  parallax angle
d  distance
x = 1 AU; measure p  d
As a practical matter, how do we get p?
B
d
p
A
Star appears to shift position against background
– the parallax effect.
A
B
Shift proportional to
2p
Clearly, as d increases, p decreases. Astronomers
find:
1
d
p
p: arcseconds (1 arcsec = 1/3600o)
If p = 1 arcsec,
1
1
d    1 parsec
p
1
1 parsec = 3.26 light year = 206,265 AU
Nearest star: Proxima Centauri
p = 0.772 arcsec
1
1
d 
p
0.772
d = 1.295 pc = 4.22 ly
Inverse-Square Law
Luminosity: total amount of energy radiated
per second (“wattage”)
Watt? Watt?
50 W
100 W
Twice the
luminosity
Star
Luminosity
Sun
1
Proxima Centauri
0.00082
Alpha Centauri
1.77
Sirius
26.1
Betelgeuse
15,000
Rigel
70,000
These stars would
appear to be about
equally “bright.”
Does this mean
they’re equally luminous?
(Apparent) Brightness  Luminosity
2 stars – differ in luminosity –
may appear equally bright!
On the other hand . . . two stars that differ
in brightness need not differ in luminosity.
Brighter
Dimmer
How much dimmer?
How much brighter?
Sphere, radius = d
d
L
1 m2
Flux (F)  amt. of
light energy flowing
per second through
1 m2
All of star’s light must pass through sphere . . .
So the energy is spread over the sphere’s surface.
Amt. of energy
per sec through
1 m2
=
Total energy per second flowing
Total number of sq meters
L
F 
2
4 d
Inverse-square
law of light!
For a given star (i.e., specified luminosity):
d
1
F 2
d
d (ly)
F (watt/m2)
1
100
2
25
5
4
10
1
100 100
F  2 
 4
5
25
Flux & distance  Luminosity

L
2
F 
 L  4 d F
2
4 d
Sun’s flux at Earth:
F = 1370 W/m2
d = 1 AU = 1.5 x 1011 m
L = 4(1.5 x 1011)2 x 1370
= 3.9 x 1026 Watt (!)