# 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 × 1026 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:

Short Full Number of zeroes
103 1000 3
104 10,000 4
105 100,000 5
106 1,000,000 6

Examples:

• 2 million = 2,000,000 = 2×106
• R = 696,000 km = 6.96×105 km = 6.96×108 m

# HR diagram - classifications Source: Spacepotato CC-BY SA 3.0

# Classification table

Class Temperature Colour Luminosity % of MS stars
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 T4

• In physics, "black body" is an ideal emitter (and absorber) of light
• T4 = 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 T4

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
= 6.42×107 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 R2
• where π = 3.141592... but we'll use 3.14
• From earlier: R = 6.96×108 m

A = 4π R2
= 4 × 3.14 × 6.96×108 × 6.96×108
= 4 × 3.14 × 4.84 × 1017
= 6.08×1018 m2

# Theoretical solar luminosity

• Plugging our values for F and A into L = F A, we find

L = 6.42×107 × 6.08×1018
= 3.90×1026 W
• This is very close to measured values (see earlier).
• This confirms that stars are indeed good approximations to black bodies.

• L is proportional to R2 and T4
• 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?

# Sizes and colours Source: LucasVB CC-BY SA 3.0

# 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.