Stars  Sizes and types
Andrew Conway
Spectra of stars
Source: NASA Public Domain
Harvard spectral classification
 Spectral classification:
 From hottest to coolest: O B A F G K M
 Mnemonic: Oh, Be A Fine Guy (or Girl) Kiss Me
 Further subdivided: e.g. G0 is hottest in G and G9 is coolest
 Luminosity class uses a Roman numeral:
 0 to IV are giants
 V is main sequence
 VI and VII are dwarfs
Solar values
The symbol ☉ is used to denote solar values.
 The Sun is a G2V star.
 The surface temperature is 5800 K.
 Radius: R_{☉} = 696,000 km
 Luminosity: L_{☉} = 3.85 × 10^{26} W
[Optional task: If you were to cover the surface of the Sun with enough 100 W light bulbs to match its luminosity, what size would each bulb need to be to all fit on the surface?]
Large numbers
To avoid writing out long numbers, we can use a short hand:
10^{3} 
1000 
3 
10^{4} 
10,000 
4 
10^{5} 
100,000 
5 
10^{6} 
1,000,000 
6 
Examples:
 2 million = 2,000,000 = 2×10^{6}
 R_{☉} = 696,000 km = 6.96×10^{5} km = 6.96×10^{8} m
Classification table
O 
≥ 33,000 K 
blue 
≥ 30,000 L_{☉} 
~0.00003% 
B 
10,000–33,000 K 
blue white 
25–30,000 L_{☉} 
0.13% 
A 
7,500–10,000 K 
white 
5–25 L_{☉} 
0.6% 
F 
6,000–7,500 K 
yellow white 
1.5–5 L_{☉} 
3% 
G 
5,200–6,000 K 
yellow 
0.6–1.5 L_{☉} 
7.6% 
K 
3,700–5,200 K 
orange 
0.08–0.6 L_{☉} 
12.1% 
M 
2,000–3,700 K 
red 
≤ 0.08 L_{☉} 
76.45% 
Data source
Luminosity and temperature
Stefan's law: the power radiated by a "black body" at temperature T per unit surface area is proportional to T^{4}
 In physics, "black body" is an ideal emitter (and absorber) of light
 T^{4} = T×T×T×T
Flux and Stefan's law
 Power per unit area is often called a flux.
 If T is in kelvin, then the flux F is
F = 5.67×10^{8} T^{4}
where F will be in W m^{2} (watts per square metre)
Flux at the solar surface
The solar surface is at temperature 5800 K
F 
= 5.67×10^{8} × 5800 × 5800 × 5800 × 5800 

= 64,164,532 W m^{2} 

= 64,200,000 W m^{2} 

= 6.42×10^{7} W m^{2} 
Since we only started with three significant figures (e.g. 5.67), we round to 3 significant figures in the last step.
Flux and luminosity
 The flux is power per unit area.
 Luminosity is total power over the entire surface.
 So flux times surface area will give luminosity.
 L = F × A
 where A is the area of the Sun in square metres.
Surface area of the Sun
 For a sphere A = 4π R^{2}
 where π = 3.141592... but we'll use 3.14
 From earlier: R = 6.96×10^{8} m
A 
= 4π R^{2} 

= 4 × 3.14 × 6.96×10^{8} × 6.96×10^{8} 

= 4 × 3.14 × 4.84 × 10^{17} 

= 6.08×10^{18} m^{2} 
Theoretical solar luminosity
 Plugging our values for F and A into L = F A, we find
L 
= 6.42×10^{7} × 6.08×10^{18} 

= 3.90×10^{26} W 
 This is very close to measured values (see earlier).
 This confirms that stars are indeed good approximations to black bodies.
Deducing star radius
 L is proportional to R^{2} and T^{4}
 Double R and L increases by a factor of 2×2=4
 Double T and L increases by a factor of 2×2×2×2=16
 So is, say, an O type star more luminous than a B type star because R or T is greater?
 Answer: both!
Red giants
 Betelegeuse:
 L = 100,000 L_{☉}
 R = 1000 R_{☉}
 If the Sun was 1000 times larger at the same surface temperature, then its luminosity would be 1000×1000=1,000,000 times larger.
 But Betelgeuse is only 100,000 times more luminous than the Sun.
 So we can deduce that Betelgeuse must have a lower surface temperature than the Sun.
 It is about 3400 K, which corresponds to a red colour.
 Hence red and giant.