Question

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 $5800\mathrm{K}$. 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 $P/P_{\text{sun}}=1.5{\left(M/M_{\text{sun}}\right)}^{3.5}$. The star Regulus $\mathrm{A}$ is a bluish main-sequence star with mass $3.8M_{\text{sun}}$ and radius $3.1R_{\text{sun}}$. What is the surface temperature of Regulus $\mathrm{A}$?

Short answer

The surface temperature of Regulus $A$ is$12000K.$

Step-by-step solution

Given Information

Star with a surface temperature $=5800K$

Star's radiating power relative to the sun is $P/P_{\text{sun}}=1.5{\left(M/M_{\text{sun}}\right)}^{3.5}.$

Regulus $A\text{'}s$mass$=3.8M_{sun}$

Regulus $A\text{'}s$radius$=3.1R_{sun}$

Explanation

We know that the ratio of the power $P$produced by a star of mass $M$follows the following relation

$\frac{P}{P_s}=1.5{\left(\frac{M}{M_s}\right)}^{3.5}$

We can therefore express the power as

$P=1.5{\left(\frac{M}{M_s}\right)}^{3.5}{P}_s$

where $P_s$and $M_s$are the Sun's power and mass, respectively.

The power $P$radiated by a star at surface temperature will be

$P=Aϵ\sigma T^4$

The temperature therefore can be found as

$T=\sqrt[4]{\frac{P}{Aϵ\sigma }}$

We should remember that since stars are of spherical shapes, their surface areas will be $A=4\pi R^2$, where $R$is their radius. Let us now substitute the expression for the power $P$of a star relative to the power of the sun:

$T=\sqrt[4]{1.5{\left(\frac{M}{M_s}\right)}^{3.5}\frac{P_s}{Aϵ\sigma }}$

Now, let us substitute for the power radiated by the Sun, as a star of surface area $A_s$and surface temperature $T_s$:

$T=\sqrt[4]{1.5{\left(\frac{M}{M_s}\right)}^{3.5}\frac{A_sϵ\sigma T_s^4}{Aϵ\sigma }}$

Explanation

As we see, we can simplify the emissivities of the stars (assuming they are the same) and the Stefan-Boltzmann constants ($\sigma$). Also, let us substitute for the areas in the next step, and also take the sun's temperature out of the fourth root, thus reaching

$T=T_s\sqrt[4]{1.5{\left(\frac{M}{M_s}\right)}^{3.5}\frac{4\pi R_s^2}{4\pi R^2}}$

Simplifying by $4\pi$we finally reach to our parametric answer of

$T=T_s\sqrt[4]{1.5{\left(\frac{M}{M_s}\right)}^{3.5}{\left(\frac{R_s}{R}\right)}^2}$

As a last step, we only need substitute $M=3.8M_s$and $R=3.1R_s$, and the Sun's temperature $T_s=5800\mathrm{K}$, thus finding

$T=5800·\sqrt[4]{1.5·3.8^{3.5}{\left(\frac{1}{3.1}\right)}^2}=11724\mathrm{K}$

Rounding to a meaningful significance, we get

$T=12000\mathrm{K}$

Final Answer

Hence, the surface temperature of Regulus $A$islocalid="1648120167773" $12000K.$