# The surface temperature of Arcturus is about half as hot as the Sun’s, but Arcturus is about 100 times more luminous than the Sun. What is its radius, compared to the Sun’s?

May 26, 2015

Let,
$T =$Arcturus's surface temperature
${T}_{0} =$Sun's surface temperature
$L =$Arcturus's Luminosity
${L}_{0} =$Sun's Luminosity

We are given,
$\setminus \quad L = 100 {L}_{0}$

Now express luminosity in terms of temperature.

The power radiated per unit surface area of a star is $\setminus \sigma {T}^{4}$ (Stefan-Boltzmann law).

To get the total power radiated by the star (its luminosity) multiply the power per unit surface area by the surface area of the star$= 4 \setminus \pi {R}^{2}$, where $R$ is the radius of the star.

Luminosity of a star $= \left(\setminus \sigma {T}^{4}\right) 4 \pi {R}^{2}$

Using this, $L = 100 {L}_{0}$ can be written as

$\left(\setminus \sigma {T}^{4}\right) 4 \pi {R}^{2} = 100 \cdot \left(\sigma {T}_{0}^{4}\right) 4 \pi {R}_{0}^{2}$

where ${R}_{0}$ references the radius of the sun.

Rearranging the above equation gives

$\frac{R}{R} _ 0 = 10 \cdot {\left({T}_{0} / T\right)}^{2}$

We are given that $T = {T}_{0} / 2$. substituting into the above equation gives

$\frac{R}{R} _ 0 = 10 \cdot {\left(2\right)}^{2} = 40$