1. Stars

2. Distance

3. Getting distances to stars:
Geometry: ‘Stellar Parallax’
> Only direct method!
2. Inverse-Square Law
* Measure F
* Know (or guess!) L
Find d

4. Stellar Parallax
Triangulation: d p x q
Baseline
x and q (or x and p)  d

5. For getting distances to stars, we want longest
possible baseline:
Earth p d x
p  parallax angle
d  distance
x = 1 AU; measure p  d

6. As a practical matter, how do we get p?B d p A
Star appears to shift position against background
– the parallax effect.
Shift proportional to
2  p A B

7. Clearly, as d increases, p decreases. Astronomers
find:
p: arcseconds (1 arcsec = 1/3600o)
If p = 1 arcsec,
1 parsec = 3.26 light year = 206,265 AU

8. Nearest star: Proxima Centauri
p = arcsec
d = pc = 4.22 ly

9. Inverse-Square Law
Luminosity: total amount of energy radiated
per second (“wattage”)
Watt? Watt?
100 W
50 W
Twice the
luminosity

10. Star
Luminosity
Sun 1
Proxima Centauri
Alpha Centauri
1.77
Sirius
26.1
Betelgeuse
15,000
Rigel
70,000

11. they’re equally luminous?
These stars would
appear to be about
equally “bright.”
Does this mean
they’re equally luminous?

12. (Apparent) Brightness  Luminosity
2 stars – differ in luminosity –
may appear equally bright!

13. On the other hand . . . two stars that differ
in brightness need not differ in luminosity.

14. Brighter
Dimmer
How much dimmer?
How much brighter?

15. Sphere, radius = d d L
Flux (F)  amt. of
light energy flowing
per second through
1 m2
1 m2

16. = 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
Inverse-square
law of light!

17. For a given star (i.e., specified luminosity):
d (ly) F (watt/m2) 1 2 5 10
100 25 4 1

18. Flux & distance  Luminosity 
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 (!)