# What is the luminosity in solar units of a star having three times the radius of the Sun and a surface temperature of 10,000K? In W/m2?

*print*Print*list*Cite

### 1 Answer

The luminosity of a star is related to its surface temperature and radius as:

`L = 4piR^2 sigma T^4`

Where L is the luminosity in Watts, R is the radius in meters, `sigma` is the Stefan-Boltzmann constant (5.67 x 10^-8 Wm^-2K^-4), and T is the star's surface temperature in Kelvin. In solar units,

`L_(Star)/L_(Sun) = R^2/(R_S^2 ) ×T^4/(T_S^2 ) `

`=3^2*(10000/5800)^4` (assuming the surface temperature of Sun to be 5800K)

`=80`

Therefore, the star has a luminosity, 80 times that of the Sun.

In units of `Wm^-2` , the luminosity is obtained by putting the values of `sigma` and the solar radius,

`L_(Sun)=4 xx pi xx(1.5 xx 10^11m)^2xx1.4xx10^3 wm^-2=3.85xx10^26 W`

`L_(Star)=(80xx3.85xx10^26W)/(4xxpixx(3xx1.5xx10^11m)^2)`

`=12000 Wm^-2`

**Sources:**