Question #69698

Explain how we estimate the effective surface temperature of the Sun. A main
sequence star has mass 2 × 10^31 kg and radius 3 × 10^9 m. Obtain an estimate of the
average temperature throughout the star.
1

Expert's answer

2017-08-17T04:41:07-0400

Answer on Question #69698, Physics / Astronomy | Astrophysics

Explain how we estimate the effective surface temperature of the Sun.

Answer:

According to Wien's Law,


λmax=2900000T\lambda_ {m a x} = \frac {2 9 0 0 0 0 0}{T}

λmax=\lambda_{max} = wavelength of maximum intensity, b=b = Wien's constant =2,900,000= 2,900,000 nm.K, T=T = temperature in Kelvin

The color at a wavelength of 500nm500\mathrm{nm} that corresponds to in the visible-light spectrum is green. The human eye is most sensitive for peaks at this wavelength. This wavelength is the peak length of thermal emission from the Sun.


T=2900000λmaxT = \frac {2 9 0 0 0 0 0}{\lambda_ {m a x}}


Then,


T=2900000500=5800KT = \frac {2 9 0 0 0 0 0}{5 0 0} = 5 8 0 0 K


A main sequence star has mass 2×10312 \times 10^{\wedge}31 kg and radius 3×1093 \times 10^{\wedge}9 m. Obtain an estimate of the average temperature throughout the star.

Answer:

The average temperature <T>< T > will lie between TcT_{c} and TsT_{s} .

We can use the virial theorem to find <T>< T>

2KEtot+PEtot=02 K E _ {t o t} + P E _ {t o t} = 02(32Nk<T>)35GmHM2R=02 \left(\frac {3}{2} N k < T >\right) - \frac {3}{5} G \frac {m _ {H} M ^ {2}}{R} = 0


Where, N=M/μmH\mathrm{N} = \mathrm{M} / \mu \mathrm{m}_{\mathrm{H}} , let μ=1\mu = 1

<T>=15GmHMkR< T > = \frac {1}{5} G \frac {m _ {H} M}{k R}

G=6.71011Nm2/kg2G = 6.7 \cdot 10^{-11} \, \text{N} \, \text{m}^2 / \text{kg}^2 (constant of gravitation), mH=1.71027kgm_H = 1.7 \cdot 10^{-27} \, \text{kg} (mass of hydrogen atom), M=21031kgM = 2 \cdot 10^{31} \, \text{kg} (mass of the Star), k=1.41023J/Kk = 1.4 \cdot 10^{-23} \, \text{J} / \text{K} (Boltzmann constant), R=3109mR = 3 \cdot 10^9 \, \text{m} (radius of the Star)

Then,


<T>=15×21011×1.71027×210311.41023×3109=3.24106K< T > = \frac {1}{5} \times 2 \cdot 1 0 ^ {- 1 1} \times \frac {1 . 7 \cdot 1 0 ^ {- 2 7} \times 2 \cdot 1 0 ^ {3 1}}{1 . 4 \cdot 1 0 ^ {- 2 3} \times 3 \cdot 1 0 ^ {9}} = 3. 2 4 \cdot 1 0 ^ {6} K

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