Remaining Radiative Transfer Solutions

We now look at the remaining solutions to the last two problems:

1) For many stars, the solar constant S can be computed if its angular diameter is known. If the angular radius of a star is: a= (R/r) with r the distance to Earth and R the star’s linear radius then:

π F= S (r/R)^2

(Note: a is measured in radians where 1 rad = 57.3 deg)

If the Sun’s angular radius is 959.63 arcsec then find the solar constant S.


Solution:

We re-arrange to find:

S = π F/ (r/R)^2

Where we already know that the flux:

π F = 6.3 x 10^ 7 Jm^-2 s^-1 (See problem (a) previous instalment on radiative transfer)

a = 959.63 " but this must be in radians before using the equation.

One radian = 57.3 degrees = 57.3 deg/rad x (3600"/ deg)= 206 280 "

Then: a = (R/r) = 959.63"/ 206 280"/rad = 0.00465 rad

So: (r/R) = 1/0.00465 rad = 215 rad^-1

Therefore:

S = [6.3 x 10^ 7 Jm^-2 s^-1]/ [215 rad^-1]^2 = 1362 W/m^2


2) Find the solar constant S for alpha Lyrae (Vega) if we know (from Hanbury and Brown’s measurements) that its angular diameter is 0.0032 arcsec, and it has a (B- V) color index of 0.00. Show all working and state any assumptions.

Solution:

We know: π F = σ (T_eff)^4

So need to obtain T_eff from the color index data at the link:

http://brane-space.blogspot.com/2011/09/tackling-intermediate-astronomy_22.html

There, we find: for a (B - V) index = 0.00 then log T_eff = 4.03

Then antilog (4.03) = 10 700, so T_eff = 10 700 K

π F = σ (T_eff)^4 = (5.67 x 10^-8 W m^-2 K^-4) (10 700 K)^4 = 7.4 x 10^8 Jm^-2 s^-1

The solar constant for Vega is obtained from:

S = π F/ (r/R)^2

and we know: a = (R/r) = 0.0032 arcsec

Or, a = 0.0032 "/ 206 280"/rad = 1.55 x 10^-8 rad

(r/R) = 1/1.55 x 10^-8 rad = 6.4 x 10^7 rad^-1

Therefore the solar constant for Vega is:

S = [7.4 x 10^8 Jm^-2 s^-1/ [6.4 x 10^7 rad^-1]^2 = 1.7 x 10^-7 W/m^2