Question

Sirius A has a luminosity of 26 LSun and a surface temperature of about 9400 K.What is its radius?

Answer 1

Answer

Given,

Sirius A surface temperature,T = 9400 K

Sirius A luminosity,L = 26 L₀

L₀ is the luminosity of sun.

Radius of sun =  695700000 m

Temperature on sun surface = 5780 K

Luminous intensity is given by:-

$L=4 \pi R^{2} \sigma T^{4}$

Now

$\frac{L}{L_{0}}=\frac{4 \pi R^{2} \sigma T^{4}}{4 \pi R_{0}^{2} \sigma T_{0}^{4}}=26$

$\Rightarrow \frac{R^{2} T^{4}}{R_{0}^{2} T_{0}^{4}}=26$

$\Rightarrow R^{2}=26 \times \frac{R^{2} T_{0}^{4}}{T^{4}}=26 \times \frac{695700000^{2} \times 5780^{4}}{9400^{4}}$

$R=1341246640\ m=1.34 \times 10^{9}$