# More solar constant stuff

1. Jan 30, 2006

### stunner5000pt

A large sunspot with an apparent diameter of 100,000 km forms on the photospheric disk of the Sun. Its effective temperature is 1700 K cooler than the surrounding photosphere which has an effective temperature of 5700 K.

(a) Calculate the percentage reduction in the solar constant, measured at the Earth, as a result of this sunspot. Use a solar photosphere radius of 7x10^5 km.

in this case the energy emanated by the sun is the
total area of the photosphere x $$\sigma T^4$$
to get the usual solar consant we simply divide by 4
the change in the solar constant would be to subtract the deficit?

2. Jan 30, 2006

### SpaceTiger

Staff Emeritus
Yes, that's the luminosity.

Why do you divide by four? You want the relationship that takes you from luminosity to flux.

The "percentage reduction" probably refers to:

$$100 \times \frac{\Delta f}{f}$$

where $\Delta f$ is difference between the usual solar constant (flux) and that with the sunspot.

3. Jan 30, 2006

### stunner5000pt

im not quite sure what the relationship between luminosity and flux is ... is it
energy radiated by teh sun/the total area of the sphere that encompasses the space between the sun and the earth?
$$\frac{\pi R_{photosphere}^2 \sigma T^4}{4 \pi (R_{sun} + 1AU)^3}$$

Last edited: Jan 30, 2006
4. Jan 30, 2006

### SpaceTiger

Staff Emeritus
That's right. Your formula isn't quite right, though...

In the numerator, what is the area of the surface of the sun? Is it a circle or a sphere? In the denominator, are you sure that's the proper area for a sphere? Also, how does Rsun compare to 1 AU?

5. Jan 30, 2006

### stunner5000pt

OOPS
i got mixed between volume and area (dont know why i was thiking volume)

also Rsun << 1 AU so
$$\frac{4\pi R_{photosphere}^2 \sigma T^4}{4 \pi (R_{sun} + 1AU)^2}$$
$$\frac{R_{photosphere}^2 \sigma T^4}{(1 AU)^2}$$
ok that gives the solar constant

the flux due to the sun is now the area of hte sun less the area oh the spot. But the area of the sun is 100 times that of the are of the spot...
flux due the circular spot
the attached diagram is what i have in mind
the length of the arc of the sun is diamater of the sunspot
the the length of the arc near earth is the diameter of the area where the solar energy is distributed.

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6. Jan 31, 2006

### stunner5000pt

is that correct now?? Also could you have a look at the other thread where the photn flux has to be determined? MY prof gave us a rough idea

7. Jan 31, 2006

### SpaceTiger

Staff Emeritus
I don't know, it says the figure hasn't been approved.

8. Feb 1, 2006

### stunner5000pt

it has now!
that figure would be in 3d however.. would be a cone ?
could we simply use similar triangles here? in taht the square of corresponding radii of the circles is propertional to the areas of the circles?

9. Feb 2, 2006

### SpaceTiger

Staff Emeritus
That sounds right, though you may be overcomplicating a bit. Why don't you write your work out in equation form, it will be easier to check that you're on the right track.

10. Feb 2, 2006

### stunner5000pt

ok so
$$\frac{7 x*10^5 km}{100 000 km} = \frac{1 AU}{r}$$
now that we have the area of the curlce of the big cone and that hte energy emanated is the same for both the circles
$$\rho_{sunspot} A_{sunspot} = \rho_{big circle} A_{big circle}$$

nowwe know the density of the energy reachng the earth. This is delta f? th susual flux ,f is 1380 W/m^2 ?

11. Feb 2, 2006

### SpaceTiger

Staff Emeritus
Let's approach this a different way. Find the percentage reduction in total light emitted from the sun. How does this relate to the solar constant?

12. Feb 3, 2006

### stunner5000pt

im afriad you've lost me on taht one ... what do you mean?

13. Feb 4, 2006

### SpaceTiger

Staff Emeritus
With the sunspot present, the sun will emit less light. This is what causes the reduction in the solar constant. Can you explain why the sunspot results in a reduction of luminosity?

14. Feb 4, 2006

### stunner5000pt

well the sun will emit less radiation on account of this low temperature spot.

so what is the difference delta f...
it would simply be the luminosity of the spot itself?

15. Feb 4, 2006

### SpaceTiger

Staff Emeritus
In both cases, what are the total luminosities (sun+spot)? You have the formulae, just plug the appropriate numbers into them.

16. Feb 5, 2006

### stunner5000pt

te question refers to disc rather than the sun's surface so i am using pi r^2 rather than 4 pi r^2
luminosity of the spot
$$\pi * (50000 * 10^3)^2 * \sigma * 4000^4 = 1.1 * 10^{23} W$$
luminosity of the sun
$$\pi * (7 * 10^8)^2 * \sigma * 5700^4 = 9.3 * 10^{25} W$$
dividing both of these by the the area of a hemisphere (2 pi AU^2) will yield the flux we want
delta f is calculated by the flux of the spot - flux of sun

the answer is supposed to be in th vicinity of less than 5%

Last edited: Feb 6, 2006
17. Feb 6, 2006

### stunner5000pt

is that correct now? Im simply taking the irradiance of the spot and the irradiance of the sun. Thus to find delta f we simply subtract the two and divide by the sun's irradiance and that solves our problem?

also i must thank you for your patience... sometimes i say silly things or i overcomplicate things (like here).

thanks a lot spacetiger!

18. Feb 6, 2006

### SpaceTiger

Staff Emeritus
The luminosity of the different regions of the sun will depend on the surface area at the sun, not the apparent surface area on the sky. Your calculation for the spot should be roughly correct, since the spot is much smaller than radius of curvature of the sun and would therefore be roughly circular. The sun, however, is a sphere and the surface of a sphere is not $\pi r^2$.

Remember that this is just the first step. You will eventually need to find the flux at the earth. Technically, the problem doesn't give enough information for you to do this because it doesn't tell you how the spot is oriented relative to us. However, you should probably assume that it's right in the middle of the disk.

19. Feb 6, 2006

### stunner5000pt

so the flux is to divide the irradiance by the quantity
4 pi AU^2 right?
i mean photon flux is given is W/m^2 isnt it ?

20. Feb 6, 2006

### SpaceTiger

Staff Emeritus
For the entire sun, yes. For the spot, you can use the cone approximation you were using before.