Solar Constant

[stunner5000pt]

If the ‘Solar Constant’ as integrated between wavelengths of 100 nm and 1000 nm has a value of 1000 watt m-2 and the Sun emits like a perfect black body with a temperature of 5900 K, evaluate the 100 nm to 1000 nm solar photon flux in photons m-2 s-1 striking the surface of a satellite in low Earth orbit. If the satellite was only 0.5 AU from the Sun what would the photon flux be?

so would i treat the entire area between the sun's cneter and the earth as one giant sphere?
from teh surface area of this sphere we know the photon flux of this sphere per unit area
then can we do the ratio like this
r1 is the distance between the sun and earth 1 AU
r2 the disatnce between the sun and the satellite 0.5AU
\frac{1000Wm^{-2}}{x} = \frac{r_{1}^2}{r_{2}^2}
where x is the photon flux at the satellite's position.

[stunner5000pt]

is this the correct way of doing this problem??? ANyone??? Please??

[Tide]

The solar constant is provided in terms of power per unit area (energy flux). The question is asking for photon flux, i.e. photons per unit area so you will need to relate energy to the number of photons.

You should start with Planck's radiation law which gives energy flux and, since E = h \nu, you can relate energy to photons. Integrating over the interval from 100 nm to 1,000 nm will require numerical approximation but I think you'll discover that the variation of flux over the interval relatively small. This will allow you to use an average energy for photons so the energy flux divided by h \bar \nu should give a good approximation of the number of photons.

[stunner5000pt]

the asnswer is not supposedto be an approximate one... according to my prof

but the answer we're looking for hte photon flux... not the spectral intensity... which Plank's radiation law gives?
Aren't we trying to find something of the same units... whereby a simple ratio would do?

[SpaceTiger]

but the answer we're looking for hte photon flux... not the spectral intensity... which Plank's radiation law gives?
Aren't we trying to find something of the same units... whereby a simple ratio would do?

I don't really see how you're going to get an exact answer on this one. The units you're given are

watts m-2 = ergs s-1 m-2

The units of your answer are:

photons s-1 m-2 = s-1 m-2

As Tide said, you'll need Planck's Law:

E_{photon}=h\nu

to convert the energy flux into a photon flux. Your photons span a wide range in frequency and the blackbody curve changes quite a lot within that range, so the only way to get an "exact" answer is to numerically integrate the blackbody distribution. The next closest approximation, which can be done analytically, is to use the Wein distribution, since that range samples more of the high-energy tail.

Also, you will need to do the r^2 ratio calculation that you did in the beginning once you have the photon flux on earth. That will just increase the result by a factor of four.

[stunner5000pt]

My prof suggested a similar method to waht you said

he did say to integrate the radiation function (using mathematica/maple)
once i do this i have to find a ratio between the integral and the planck function. This ratio would give the photon flux? He wasnt very clear on what he said... but is this what should be done here?

[SpaceTiger]

My prof suggested a similar method to waht you said

he did say to integrate the radiation function (using mathematica/maple)

What units does the radiation function have? What are you integrating over? What units would the integral have? How do these compare to the units we're looking for?

[stunner5000pt]

so iam doin this integration
\int_{\lambda=100nm}^{\lambda=1000nm} \frac{2 h c^2}{\lambda^5} \left(e^{hc/\lambda kT} -1 right)


the units of the integrand are Joules arent they?

[SpaceTiger]

so iam doin this integration
\int_{\lambda=100nm}^{\lambda=1000nm} \frac{2 h c^2}{\lambda^5} \left(e^{hc/\lambda kT} -1 right)


the units of the integrand are Joules arent they?

If you mean:

\frac{2 h c^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT} -1}

then the units are Joules s-1 m-2 m-1 ster-1. It's a measure of intensity. I separated the units of length into the area (meters squared) and the wavelength (meters). What are you integrating over?

[stunner5000pt]

integrating over the wavelength
so then the units will become J s-1 m-2 sr-1 m-2

what i also told was to integrate the same function over the frequencies (corresponding to 100nm and 1000nm) and integrate the frequencies
Multiply this answer by the solar constnatn and divde the resulting answer from the integrand in post 8

[SpaceTiger]

integrating over the wavelength
so then the units will become J s-1 m-2 sr-1 m-2

Try again. Is the d\lambda on the top or bottom?



what i also told was to integrate the same function over the frequencies (corresponding to 100nm and 1000nm) and integrate the frequencies

That should work just as well.

[stunner5000pt]

ok so we integrate and that gives us an answer taht is J m-2 sr-1 correct/
the integral of planck's function from 100nm to 1000nm is 1.7x10^17J m-2 sr -1
for hte frequencies it gives 2.14 x 10^28 J s^2 m-3
so we know the eneryg per square metre comin from the sun
if we muliply out the area of the sun taht gives J sr-1
units dontseem to cancel out properly...

[SpaceTiger]

ok so we integrate and that gives us an answer taht is J m-2 sr-1 correct/

You're missing the s-1. You didn't integrate over it, so it won't disappear.


the integral of planck's function from 100nm to 1000nm is 1.7x10^17J m-2 sr -1

I'll take your word for it on that. Don't forget to add "per second".



for hte frequencies it gives 2.14 x 10^28 J s^2 m-3

The frequency version of Planck's function starts with different units:

J s-1 m-2 st-1 Hz-1

Integrating over frequency eliminates the the Hz-1.



so we know the eneryg per square metre comin from the sun
if we muliply out the area of the sun taht gives J sr-1

Don't forget the seconds, but otherwise that's right.



units dontseem to cancel out properly...

Where do you see a contradiction?

[stunner5000pt]

well ok dividing out those
J m-2 sr-1 s-1 m^2 / J m-2 sr -1 s-1
ad that gives m-2
no seconds anymore
but hte photon fux is supposed to be m-2 s-1 isnt it ?

[SpaceTiger]

but hte photon fux is supposed to be m-2 s-1 isnt it ?

What you're solving for here is the luminosity of the sun per unit steradian. To turn that into a flux at a particular distance, you need to divide by the geometrical dilution factor: 4\pi d^2.

[stunner5000pt]

luminosity is given in watts only right?
or J s-1
so what we want in W sr -1
J s-1 sr-1
which si s the integral for hte wavelength
and d here is AU?
the answer is a very small number
6.0 x 10^-7 W/m^2
the answer si supposed to be within range of 10^20 and 10^22 W/m^2

[SpaceTiger]

so what we want in W sr -1
J s-1 sr-1

I forgot to mention that you need to integrate over solid angle. Do you know how to do that? At a given location on the sun, over what range of angles does the surface emit?



and d here is AU?

Is that the SI unit for distance?

[stunner5000pt]

not familiar with the solid angle concept

d should 1.5 x 10^11 m not AU (i was pplugging in the right number for htecaulations, incidentally)

[SpaceTiger]

not familiar with the solid angle concept

What text are you using?

[stunner5000pt]

there isnt a text we are simply given notes by the prof
he hasnt covered something like tis, however
here are his notes...
http://www.yorku.ca/mcdade/physeats3280/

nothing of this sort has been covered in extensive detail
this assignment is mostly basd on notes 1 -3

[stunner5000pt]

for the solid angle wouldnt it be simply
\int_{0}^{\pi} \int_{0}^{\pi} \sin \theta d \theta d \phi

[SpaceTiger]

Ok, we won't get into this then. Just multiply by \pi to cancel out the "per steradian".

[SpaceTiger]

the answer si supposed to be within range of 10^20 and 10^22 W/m^2

I think the solar constant is more like 1380 W/m2

[stunner5000pt]

oops i meant 10^20 and 10^22 photons/ m-2 s-1

[SpaceTiger]

oops i meant 10^20 and 10^22 photons/ m-2 s-1

Well, what you've solved for is energy flux, not photon flux. If I have a frequency interval d\nu with an intensity given by the Planck function, what is the equivalent "photon intensity" in that interval? The photon intensity has units:

photons m-2 s-1 Hz-1 st-1

[stunner5000pt]

that loksl ike the units for the frequency integral who was J s-1 m-2 sr-1
multiply by pi to eliminate teh steradian
by how would we rid outrselves of the Joules

[SpaceTiger]

by how would we rid outrselves of the Joules

What is the energy of a photon?

[stunner5000pt]

E = h \nu = hc/ \lambda
but we have two values of the frequency
we cant just average them out ...

[SpaceTiger]

E = h \nu = hc/ \lambda
but we have two values of the frequency
we cant just average them out ...

Actually, you have an infinite number of frequency values because you're integrating over a continuous distribution. How might you alter the integral to count photons rather than Joules?

[stunner5000pt]

we could replace h with c/lambda or c/frequency... couldnt we?

so then the resulting integral is counting hte number of photons... and we dont even need the other integrals?

[SpaceTiger]

we could replace h with c/lambda or c/frequency... couldnt we?

Could you justify this for me?



so then the resulting integral is counting hte number of photons... and we dont even need the other integrals?

You won't need the other integrals.

[stunner5000pt]

couldnt we just write joules as kg m^2 s-2??
what i wrote earlier wasnt correct i was thinkin about the relation betwen the sped of light lambda and frequency
anyway even if we did like taht then we have a kg factor and kg isnt relevant to a photon