How does the diameter of Polaris compare to the diameter of the sun?

[jai6638]

Hey.. need some help in solving this problem:


Q) The luminosity of Polaris is 10,000 times the luminosity of the sun. The surface temperature of Polaris is about 5800 kelvins. Using k=33,640,000 find how the diameter of polaris compares with the diameter of the sun. ( Equation: D= (K)(sqrt of L ) / (T)^2

Dp/Ds = ( k(sqrt of Lp ) / (Tp)^2 ) / ( k(sqrt of Ls) / (Ts)^2 )
= ( sqrt of 10,000Ls ) / (5800)^2 / ( sqrt of Ls ) / (Ts)^2)

Now I'm stuck.. Not sure what this problem is leading to.. i don't know the value of Ts and hence am not going to get an answer.. Any help is much appreciated..

 

[jai6638]

anyone...?

 

[UrbanXrisis]

d_{polaris}=\frac{k\sqrt{L}}{T^2}=\frac{33,640,000 \sqrt{10,000}}{5800 ^2}
d_{sun}=\frac{k\sqrt{L}}{T^2}=\frac{33,640,000 \sqrt{1}}{T^2}

ratio=\frac{d_{polaris}}{d_{sun}}

 

[jai6638]

The answer I'm getting is 100 ( i.e. the diameter of POlaris is 100 times that of the sun ) . Is that the correct answer?

thanks