1. Feb 3, 2006

### stunner5000pt

A solar flare releases 10^25 Joules of energy in one hour over an area amounting to 1% of the area of the Sun’s disk (radius 6.96 x 10^5 km).

(a) Calculate the Solar irradiance in W m-2 associated with this flare.

(b) Compare this with the black body radiation emitted by the Sun assuming a temperature of 6000 K.

(c) If the flare's energy is mostly in 5 keV x-rays calculate the associated x-ray photon flux striking a spacecraft in low earth orbit

area of the sun = 1.54 x 10^18 m^2
two areas
area of the remaining of the sun = 1.52 x 10^18 m^2
area of the flare = 1. 54 x 10^16 m^2
irradiance =$$\sigma T^4 (\mbox{Area})$$

from th energy we can calculate the associated wavelength. Should the energy be per SECOND? Or leave it unconverted...
$$E = \frac{hc}{\lambda}$$
from the wavelength we can caluclate the temperature
$$\lambda_{max} T = 2897 \mu m K$$
from the temperature we can caluclate the irradiance of the flare.

good so far?
More to come!

Please help me out! This is the final problem in an assignment worth 10% of my mark!

2. Feb 3, 2006

### nrqed

Hi...
I am a bit confused by what you wrote. You did not say anything about part a). I am assuming you know how to answer that?

As for part b), you seem to be doing it the correct way.

I am a bit confused by this last part. I don't see the need to introduce the wavelength. If you have the energy of each photon (5 keV) and you know the total energy released in one hour, then it's a simple step to calculate the number of photons emitted per second. Then you can calculate the flux received on Earth (the only tricky thing is to whether your prof wants you to use the simplest assumption of a point source, which is clearly wrong but easy, or the "real" way of limiting the flux received to a fixed solid angle. )

Hope this helps

Pat

3. Feb 4, 2006

### stunner5000pt

to clear a misundertanding

actually whatever i wrote was my attempt at solving part a) alone.

which i have those areas... but what i have done was try to compute the enrgy of each photon (i think this involves ignoring part c for now?) and then and since we know the number of photons emited by the flare... we can caluclate the flux on earth? (perhaps there is a simpler approach to this... because i think this way is wayyyyyyyyyyyy toooo long!)

4. Feb 4, 2006

### nrqed

Oh! Then, if this is your attempt to part a) alone, it *is* too complicated. From the units of W/m^2, all you have to do is to take the energy emitted, divide by the area of the flare and by the time it lastes, in seconds (since a watt is a joule per second). Thats all!!

Pat

5. Feb 4, 2006

### stunner5000pt

oh ok i havdnt looked at it that way
so then
$$irradiance = \frac{10^{25}}{\pi (6.96*10^8)^2 (3600)} = 1.8 * 10^3 W/m^2$$ which agres wiuth the ranges for this answer, posted by my prof

for b) i simply use $$\sigma T^4$$
which comes to 7.4 x 10^7 J which does not agree with my profs ranges because he says the answer is around 1000 to 10000 times less than part a) s answer
for c)
we know the energy of each photon $$5keV = 5000 eV = 8.0*10^{-16} J$$
so then the totla number of photons is 10^25J/(8.0*10^-16)/(3600) = 3.5 x10^36 photons/sec
divide by the area of the big sphere of radius 1AU and that gives 4.9x10^13 photon/sec
but that is about twice as much as my profs estimated answer of
25 (UK) trillion photons m-2 s-1 whatever UK means...

6. Feb 4, 2006

### nrqed

I am confused. Shouldnt you have divided by the area of the *flare* (not of the Sun). (btw: even if you divided by the area of the Sun, the formula is 4 Pi r^2, not Pi r^2). So I would get a value 100/4 = 25 times larger than yours

..
Mmm.... I dont see how to get a factor of 1000 to 10000..let me think a few minutes

Well, I get one fourth of your answer because of the factor of 4 in the area equation. So I am 2 times less than your prof. I have to think about where this factor of 2 could come from....

Pat

7. Feb 4, 2006

### stunner5000pt

but doesnt the question state DISC and not sphere?
so then my answer would be correct...?

8. Feb 4, 2006

### nrqed

Ah ok. I see. Yes, it could be that the question refers to the 1% as being a projected area. Fair enough.

I still dont see how to get a factor of 1000 or more with sigma T^4:grumpy:

Pat

9. Feb 5, 2006

### stunner5000pt

for the falre shouldnt it be the area of teh FLARE divided out?
in that case (1% of the sun's area)
$$\mbox{irradiance} = \frac{10^{25}}{(\pi (6.96*10^8)^2)0.01 (3600)} = 1.8 * 10^5 W/m^2$$
the profs 'ranges' for the answer was between 10^3 and 10^5 W/m^2...

for part b... welll again we are assuming it is a sphere and not a cirlcei wth the sigma T^4 formula. So then if we didved by 4 then that would give us radiation for a flat disc... sine a cirlce is 4 times the area of a disc...

is c) fine then? where does that factor of 2 come from tho..

10. Feb 5, 2006

### nrqed

Yes, (see my post number 6 where I say that you forgot to take into account the factor of 1%)

You will get the correct answer if you assume that the photons are emitted over half a sphere instead of an entire sphere (makes sense, since the flare is on one side of the disk only). In that case, you will divide by 2 Pi r^2 instead of 4 Pi r^2. and you will get the answer of your prof (you had used Pi r^2 when you did it, which is not the equation of the area of a sphere, so you where getting a result two times too large. I had used 4 pi r^2 so I had got a result two times too small).

As for the sigma T^4, there could be a factor of 2 in going from the area of half a sphere to the area of a disk, but I don`t see how to get a factor of a 1000 or more.

Pat

11. Feb 5, 2006

### stunner5000pt

perhaps there is some other formula that I am unaware of using? What else can be used to calculate blackbody radiation??