Blue Supergiants. A typical blue supergiant star (the type...

Blue Supergiants. A typical blue supergiant star (the type that explodes and leaves behind a black hole) has a surface temperature of $30,000 \mathrm{K}$ and a visual luminosity $100,000$ times that of our sun. Our sun radiates at the rate of $3.86 \times 10^{26} \mathrm{W}$ .(Visual luminosity is the total power radiated at visible wavelengths.) (a) Assuming that this star behaves like an ideal blackbody, what is the principal wavelength it radiates? Is this light visible? Use your answer to explain why these stars are blue. (b) If we assume that the power radiated by the star is also $100,000$ times that of our sun, what is the radius of this star? Compare its size to that of our sun, which has a radius of $6.96 \times 10^{5} \mathrm{km}$ . (c) Is it really correct to say that the visual luminosity is proportional to the total power radiated? Explain.

Blue Supergiants. A typical blue supergiant star (the type that explodes and leaves behind a black hole) has a surface temperature of $30,000 \mathrm{K}$ and a visual luminosity $100,000$ times
that of our sun. Our sun radiates at the rate of $3.86 \times 10^{26} \mathrm{W}$ .(Visual luminosity is the total power radiated at visible wavelengths.) (a) Assuming that this star behaves like an ideal blackbody, what is the principal wavelength it radiates? Is this light visible? Use your answer to explain why these stars are blue. (b) If we assume that the power radiated by the star is also $100,000$ times that of our sun, what is the radius of this star? Compare its size to that of our sun, which has a radius of $6.96 \times 10^{5} \mathrm{km}$ . (c) Is it really correct to say that the visual luminosity is proportional to the total power radiated? Explain.

Key Concepts

Luminosity and Visual Luminosity

Luminosity refers to the total amount of energy a star emits per unit time across all wavelengths, while visual luminosity is the portion of that output in the visible spectrum. Understanding the distinction is important in astrophysics, as different observational techniques and instruments may capture only a portion of the total emitted energy, and spectral analysis is necessary to estimate the full radiative output.

Stefan-Boltzmann Law

The Stefan-Boltzmann Law states that the total energy radiated per unit surface area of a blackbody is directly proportional to the fourth power of its temperature. In astrophysical contexts, it allows for the calculation of a star's total luminosity when its temperature and radius are known, or vice versa.

Blackbody Radiation

This concept refers to an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence, and re-emits energy in a continuous spectrum characteristic of its temperature. It is fundamental in astrophysics for modeling the radiation emitted by stars, which closely resemble ideal blackbodies.

Wien's Displacement Law

Wien's Displacement Law establishes a relation between the temperature of a blackbody and the wavelength at which its radiation is most intense. This law is used to determine the peak emission wavelength for stars, and it explains how higher temperature stars shift their peak emission towards shorter, bluer wavelengths.

Transcript

00:01 In this question we have a blue super giant star which we want to compare with our own sun.

00:10 For this particular blue giant, its temperature is 30 ,000 at its surface.

00:18 We want to find what is its principal wavelength.

00:21 So basically the peak wavelength that it emits.

00:24 We're going to find that using the wayne's displacement law.

00:28 So wains displacement law states that the peak wavelength emitted by body is just equals to a constant divided by its temperature.

00:38 So the constant they're talking about 2 .9 times 10 power minus 3 in terms of kelvin's meter.

00:48 We divided divided by our temperature, our surface temperature is 30 ,000 kelvin's we get 9 .7 times 10 power minus 8 meters which is about 90 ,000.

01:04 7 nanometers.

01:08 Now you know that this is in the ultraviolet range because it's smaller than 400 nanometers which is the visible range, right? the smallest wavelength of visible range.

01:21 So it's actually not very not in the visible spectrum.

01:28 And therefore it would since its peak is more biased towards the ultraviolet, therefore from what we observe in the visible range will be mostly blue light because you will have its peak near the blue region and right this will maybe be the blue region 400 nanometers its actual pick is in the ultraviolet range but of course it will have a significant proportion of its energy or photons with wavelength there is nearly 400 kilometers so that's why we see it as blue now for the next part we want to find what is the radius of the star considering that it is about the hundred times right is power emitted compared to our sun so the actual power of the star we take a hundred times by the power of our sun now we know that by stephen bozeman law, that this power, in terms of watts, is equals to sigma times the surface area, times its temperature to the power of fault.

03:12 So we know its temperature, we know the sigma, the constant.

03:18 It's the first bozeman's constant...

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