Show that the angular radius of the star is given by....

[FaraDazed]

Homework Statement

A star has an effective temperature $T_{eff}$ and is observed to have a flux $F$. Show that the angular radius in arcseconds of the star (as seen from Earth) is given by

$\theta = (\frac{2.06 \times 10^5}{T_{eff}^2}) \sqrt{\frac{F}{\sigma}}$

Homework Equations

$L = 4 \pi R^2 \sigma T_{eff}^4 \\ T_{eff} = (\frac{L}{4 \pi R^2 \sigma})^{\frac{1}{4}} \\ F = \frac{L}{4 \pi d^2}$

And probably..
[tex]
\omega = \frac{A}{R^2}
[/itex] ?

The Attempt at a Solution

[/B]
I am a bit lost here! I missed several of the lecture containing this material, but from looking at what I have go already I should be able to do it?

First I started by rearranging the eqn for Flux for L
$F = \frac{L}{4 \pi d^2} \\ L = 4 \pi d^2 F$
And then subsituted that into the eqn for T_eff
$T_{eff} = (\frac{L}{4 \pi R^2 \sigma})^{\frac{1}{4}} \\ T_{eff} = (\frac{4 \pi d^2 F}{4 \pi R^2 \sigma})^{\frac{1}{4}} \\ T_{eff}^4 = (\frac{ d^2 F}{R^2 \sigma}) = (\frac{d}{R})^2 (\frac{F}{\sigma}) \\ (\frac{R}{d})^2 = \frac{1}{T_{eff}^4}(\frac{F}{\sigma}) \\ (\frac{R}{d}) = \sqrt{\frac{1}{T_{eff}^4}(\frac{F}{\sigma})} \\ \frac{R}{d} = \frac{1}{T_{eff}^2} \sqrt{\frac{F}{\sigma}} \\$

That is as far as I have got. I assume in mine, R/d is the angle in radians, so I hope that it relates to the angle in arcsecond by that 2.06 * 10^5 factor? Or I am way off. I'd appreciate some help/advie with this please.

EDIT: yes It does and I can see how now. My bad.

 

[gneill]

Yep. Looks good.