Most stars are main-sequence stars, a group of stars for which size, mass, surface temperature, and radiated power are closely related. The sun, for instance, is a yellow main-sequence star with a surface temperature of 5800 K. For a main-sequence star whose mass M is more than twice that of the sun, the total radiated power, relative to the sun, is approximately P/P_sun=1.5(M/M_sun)^3.5. The star Regulus A is a bluish main-sequence star with mass 3.8M_sun and radius 3.1R_sun. What is the surface temperature of Regulus A?

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Accepted Answer

Hey, everyone. This problem is dealing with the Stefan Boltzmann theorem. Let's see what it's asking us. So we're told that the energy from a main sequence star is a function of its mass and radius. The power radiated by a main sequence star of a given mass M is given by the equation P equals 1.5 multiplied by M divided by MS to the 3.5 power, all multiplied by P where M sub S and P sub S is the mass and power of the sun. We are asked to consider a hypothetical star V that has a mass of 2.1 times the mass of the sun and a radius of 2.4 times the radius of the sun. And we are asked to determine the surface temperature of this star. They give us the sun's surface temperature as approximately 5780 Kelvin. And we're told that both of the, both the star V and our son have the same emissivity. Our multiple choice answers here are a 65 Kelvin B 7200 Kelvin C 7900 Kelvin or D 8300 Kelvin So the key to this problem is we're calling the Stefan Boltzmann equation which tells us that power is equal to area multiplied by the emissivity multiplied by the Stefen Boman constant sigma multiplied by the temperature to the fourth power. And so I made it look or the power of star V given using the equation for power given to us in the equation or sorry, give it to us. In the problem statement, we can write that the power of star V is equal to 1.5 times or multiplied by the mass of B divided by the mass of the sun to the 3.5 power multiplied by the power of the sun. And so the mass of star V was given as 2.1 multiplied by the mass of the sun. And we can use the Stefen Boldman theory or Stefen Boldman formula to solve her temperature. So what that looks like is you have the area of V then multiplied by the mi multiplied by the step in both constant multiplied by the temperature of V to the fourth power. And that equals 1.5 multiplied by 2.1, the mass multiplied by the mass of the sun divided by the mass of the sun to the 3.5 power multiplied by the area, the sun multiplied by the emissivity multiplied by the sun con bolt and constant multiply by the temperature of the sun to the fourth hour. And so as we look to simplify things, we emi are the same. So those cancel stuff involvement, constant cancels the mask, the sun cancel and then area is given by IRS square. And so the area of V is going to be pi multiplied by the radius of V which is 2.4 multiplied by the radius of the sun square. And so the area of the sun is simply going to be pi multiplied by the radius of the sun squared. And so we can plug that in. So we'll have pi multiplied by two, sorry, 2. multiplied by the radius of the sun. That quantity squared multiplied by TV to the fourth times or sorry equals 1.5 multiplied by 2.1 to the 3. multiplied by pi R S squared multiplied by A T S to the fourth. So now our, our S squares cancel P cancel and we are left with 5. multiplied by TV. Four equals 20.1 multiplied by T S four. And so solving for TV will take the fourth route of 20.1 times or sorry, 5780 Kelvin to the fourth divided by 5.76. We plug that into our calculator and we get 7900 K. So looking at our multiple choice answers that aligns with answer choice C so that's all we have for this one. We'll see you in the next video.

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Key Concepts

Main-Sequence Stars

Stellar Mass-Luminosity Relation

Surface Temperature and Color