# Solar Constant (1 Viewer)

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#### 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

Homework Helper
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

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stunner5000pt said:
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

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

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stunner5000pt said:
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

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stunner5000pt said:
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

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stunner5000pt said:
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

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stunner5000pt said:
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
no seconds anymore
but hte photon fux is supposed to be m-2 s-1 isnt it ?

#### SpaceTiger

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stunner5000pt said:
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

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stunner5000pt said:
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

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stunner5000pt said:
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...

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

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