🚀 go-pugleaf

In article <1863@saber.UUCP> jc@saber.UUCP (John Cincotta) writes:
>How many photons per second per square meter are received from a star of 
>say 10th magnitude on the surface of the earth??

From Allen's "Astrophysical Quantities" (Chapter 10):

   log f      (V) = -0.4m  - 8.43
        lambda           V

                                         -2         -1    -1
where f      (V) is visual flux in erg cm   Angstrom   sec   outside the Earth's
       lambda

atmosphere near 5500 Angstroms.  "This relation is almost unchanged from B to
M stars."

Therefore (plugging in numbers), for m  = 10
                                      V
                   -13       -2   -1        -1
f	= 3.72 x 10    erg cm  sec  Angstrom
 lambda

Now a photon at a wavelength of 5500 Angstrom has an energy of

E = h(c/lambda)

               -27                    18
  = (6.626 x 10   erg-sec) (2.998 x 10   Angstrom/sec) / (5500 Angstrom)

                            -12
  =		  3.612 x 10   ergs

this yields
                                     -2         -1
Flux(5500 Angstrom) = 0.11 photons cm   Angstrom

                                       -2        -1
		    = 110 photons meter  Angstrom


For a 1000 Angstrom bandpass (roughly since I am too lazy now to do the
integral and the whole business is approximate anyway)  this is about
110,000 photons / square meter for that 10th magnitude star.
-- 
Bill Sebok			Princeton University, Astrophysics
{allegra,akgua,cbosgd,decvax,ihnp4,noao,philabs,princeton,vax135}!astrovax!wls