Download Week 6

Announcements
•Homework: Chapter 13 # 45, 48, 49 & 50
•Exam 2 is next time. Will cover telescope
and lens stuff from last week and
distance-luminosity-brightness stuff
from today.
Stars!
The fundamental question:
How far away is that star
The most direct method of determining stellar
distances is parallax.
The parallax angle is half the total displacement
a star makes over the course of a year
The distance in parsecs is
simply one over the parallax
angle in arcseconds
1
d (parsecs) =
θ (arcseconds)
1parsec = 3.261ly
= 3.0857 ×1016 m
Examples
The star Tau Ceti has a parallax of 273.96 milliarcsecs.
How far away is Tau Ceti in parsecs and lightyears?
The star Epsilon Eridanus is the third closest star to Earth
with a parallax of 310.94 milliarcseconds. How far away is
it in parsecs and lightyears
Example Solution: Tau Ceti
1
1
d=
(parsec) =
3.6502parsec
=
−3
θ (arcsec) 273.96 ×10 arcsec
Converting to lightyears using 1 parsec = 3.26156 lightyears
3.6502 pc × 3.26156
ly
pc
=
11.905ly
Example Solution: Epsilon
Eridanus
1
1
d=
(parsec) =
=
3.2161parsec
−3
θ (arcsec) 310.94 ×10 arcsec
Converting to lightyears using 1 parsec = 3.26156 lightyears
3.2161 pc × 3.26156
ly
pc
10.489ly
=
The distance-luminosity
relationship can also be used
to find distance
Luminosity
Brightness =
2
4π r
Luminosity depends on the
size of the star
L ∝ 4π R
2
Luminosity also depends on
Temperature
L ∝ σT
4
Putting it all together
L = 4π R σ T
2
4
R is the radius of the star in meters
T is the temperature of the star in Kelvin
σ is the Stephan-Boltzmann constant
=
σ 5.67 ×10−8
W
m2 ⋅ K 4
Example
The bright star in the top left corner of Orion, Betelgeuse,
has a radius 936 times that of the Sun and a surface
temperature of 3500 K. What is the luminosity of this star?
If Betelgeuse is 640 ly from Earth, what is the brightness
of the light from Betelgeuse that reaches Earth?
Example Solution
L = 4π R 2σ T 4
(
= 4π 936 RSun × 6.9599 ×10 m
8
)(
2
5.67 ×10−8
W
m2 K 4
) ( 3500K )
= 4.54 ×1031W
The Sun has a luminosity of 3.9 x 1026 W so this is
about 116,000 times more luminous than the Sun
4
Example Solution 2
Now that we know the luminosity we can
use the distance-brightness-luminosity
relationship to find the brightness
We want the answer to be in Watts per
square meter so we need to convert
lightyears to meters
L
4.54 ×1031W
−8 W
=
B
=
=
9.86
×
10
2
m2
2
15
4π r
4π 640ly × 9.46 ×10 m ly
(
)