Assume a brown dwarf has a surface temperature of \(1000 \mathrm{K}\) and approximately the same radius as Jupiter. What is its luminosity compared to that of the Sun? How many brown dwarfs like this one would be needed to produce the luminosity of a star like the Sun?

The Stefan-Boltzmann Law is fundamental in astrophysics for understanding stellar luminosity. It states that the total energy radiated per unit surface area of a black body in unit time (luminosity, L) is directly proportional to the fourth power of the black body's surface temperature (T). The mathematical representation of this law is given by: \[ L = 4 \pi R^2 \sigma T^4 \]Here,

• \(R\) is the radius of the star,
• \(T\) is the surface temperature,
• \(\sigma\) is the Stefan-Boltzmann constant, approximately equal to \(5.67 \times 10^{-8} \mathrm{W} \mathrm{m}^{-2} \mathrm{K}^{-4}\).

This equation helps us understand how changes in a star’s radius and temperature affect its luminosity. For instance, if the temperature is doubled, the luminosity increases by a factor of 16 (\(2^4\)).

A brown dwarf is a celestial object that is too large to be called a planet and too small to sustain hydrogen-1 fusion reactions in its core, the characteristic that defines true stars. Typically, they have masses between about 13 to 80 times that of Jupiter. Because they cannot sustain fusion, brown dwarfs do not shine as brightly as stars and are much cooler in temperature.In our exercise, we consider a brown dwarf with a surface temperature of \(1000\,\mathrm{K}\), significantly cooler than the Sun's surface temperature of \(5778\,\mathrm{K}\). Despite having characteristics more like gas giants than stars, brown dwarfs are valuable for studying the range of objects in the universe.

Comparing the radii of different celestial objects helps us contextualize their sizes. For instance, in the original exercise, the brown dwarf's radius is considered to be approximately equal to Jupiter's radius (\(R_J = 7.14 \times 10^7 \ \text{m}\)). The Sun, on the other hand, has a much larger radius of \(6.96 \times 10^8 \ \text{m}\).Using these values, we can see that the Sun's radius is roughly ten times that of Jupiter's. This comparison shows the vast difference in sizes between a brown dwarf and a star like the Sun, significantly affecting their respective luminosities.

Stellar temperature is crucial in determining a star's luminosity and color. The higher the temperature, the more luminous the star. For instance, the Sun has a surface temperature of \(5778\,\mathrm{K}\). In contrast, a brown dwarf like the one described has a much cooler surface temperature of \(1000\,\mathrm{K}\).Temperature influences the luminosity through the Stefan-Boltzmann Law. Since luminosity is proportional to the fourth power of the temperature, small changes in temperature can lead to significant changes in luminosity. For instance, even though the brown dwarf has a similar radius to Jupiter, its cooler temperature means its luminosity will be enormously less than that of the Sun.

To find the luminosity of the brown dwarf compared to the Sun, we use the simplified form of the Stefan-Boltzmann Law: \[ \frac{ L_{BD} }{ L_{\odot} } = \frac{ R_{BD}^2 \ T_{BD}^4 }{ R_{\odot}^2 \ T_{\odot}^4 } \]Given:

• \( R_{BD} = R_J \approx 7.14 \times 10^7 \ \mathrm{m}\)
• \( R_{\odot} = 6.96 \times 10^8 \ \mathrm{m}\)
• \( T_{BD} = 1000 \ \mathrm{K}\)
• \( T_{\odot} = 5778 \ \mathrm{K}\)

By substituting these values in, we get: \[ \frac{ (7.14 \times 10^7)^2 \times (1000)^4 }{ (6.96 \times 10^8)^2 \times (5778)^4 } \approx 3.05 \times 10^{-10} \]Thus, the brown dwarf's luminosity is approximately \(3.05 \times 10^{-10}\) times that of the Sun. It means we would need about \(3.28 \times 10^9\) brown dwarfs like this to produce the same luminosity as the Sun.