- #1

FaraDazed

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## Homework Statement

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

[itex]

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

[/itex]

## Homework Equations

[itex]

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}

[/itex]

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

[itex]

F = \frac{L}{4 \pi d^2} \\

L = 4 \pi d^2 F

[/itex]

And then subsituted that into the eqn for T_eff

[itex]

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}} \\

[/itex]

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.

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