Question

A star has twice the diameter and twice the surface temperature of our sun.how will it's luminosity compare with that of our sun

Answer 1

The luminosity of the star is

$L = 4\pi R^2 \cdot \sigma T^4$

$D=2D_{Sun} \Rightarrow R=2R_{Sun}$ , $T=2T_{Sun}$

Then $L = 4\pi (2R_{Sun})^2 \cdot \sigma (2T_{Sun})^4 = 4\cdot 4\pi R^2_{Sun} \cdot \sigma \cdot 16 T_{Sun}^4 = 4 \cdot 16 \cdot( 4\pi R^2_{Sun} \cdot \sigma \cdot T_{Sun}^4) = 64 L_{Sun}$

Answer: The luminosity will be 64 times higher.