How does distance from the sun affect the apparent brightness?

[Mariko]

I have a bit of a hard time with these concepts mathematically and if i can have one person explain it would be a bit lighter on me and i would appreciate the help I just need in steps the way you would go about working this out:

Q:the Earth is about 150 million km from the sun, the apparent brightness of the sun in our sky is about 1,300 watts per square meter. Determine the apparent brightness we would measure for the sun if we were located half the Earth's distance from the sun.

by the way i know its the inverse square law but its been so long since my last math class I'm really rusty so i don't mind remedial steps acctually i need them Thx!

 

[turbo]

I have a bit of a hard time with these concepts mathematically and if i can have one person explain it would be a bit lighter on me and i would appreciate the help I just need in steps the way you would go about working this out:

Q:the Earth is about 150 million km from the sun, the apparent brightness of the sun in our sky is about 1,300 watts per square meter. Determine the apparent brightness we would measure for the sun if we were located half the Earth's distance from the sun.

by the way i know its the inverse square law but its been so long since my last math class I'm really rusty so i don't mind remedial steps acctually i need them Thx!

If the Earth found itself suddenly located twice as far away from the Sun, how much would the Solar flux be attenuated? Does that help?

Additional hint: EM flux follows an inverse square law. You need to understand this law to make the relationship real to you. Getting "the answer" might be important to you today, but it is of zero value to you if you do not understand how the answer is derived.

 

[dextercioby]

If a radius of a sphere is halved,what happens to its surface...?

Daniel.

 

[Mariko]

You need to understand this law to make the relationship real to you. Getting "the answer" might be important to you today, but it , is of zero value to you if you do not understand how the answer is derived.

Im not asking you to do it for me I am trying to understand this - please don't feel like I am just trying to get the answer because in all reality i know the answer its (mv)1=L over d2

now how this answer came to be is why I am here asking you this and your hints i still don't get I am terrible at math! I look in my book it tells me this:
the increase in the area that the radiation must cover is propotional to the square of the distance the radiation has traveled. i don't even know how to write that in mathmatical form to figure out that i must sound stupid...

 

[dextercioby]

Isn't the answer [itex]5200 \frac{\mbox{W}}{\mbox{m}^{2}} [/itex] ...?

Daniel.

 

[SpaceTiger]

now how this answer came to be is why I am here asking you this and your hints i still don't get I am terrible at math! I look in my book it tells me this:
the increase in the area that the radiation must cover is propotional to the square of the distance the radiation has traveled. i don't even know how to write that in mathmatical form to figure out that i must sound stupid...

Alright, try imagining this situation:

There are a million grains of sand sitting inside of an explosive canister in the middle of space. Suddenly, the canister explodes and the grains are shot off evenly in all directions, all at once and at the same speed. Now, let's imagine taking a freeze-frame of the sand. What shape will it make? Well, if it was ejected evenly in all directions, then it should make a sphere around the canister. Why? Because if they're all moving at the same speed, then they should have all gone the same distance between the time of the explosion and the time of the freeze-frame. In other words, they're all at the same "radius" from their point of origin (the canister).

Alright, so we have a freeze-frame of the sand at one instant in time. What happens if we freeze at another instant (let's say a few seconds later)? Well, all the grains ought to move a little further in the same direction and your sphere should expand. In other words, the grains of sand will still make a sphere, but a bigger one. Since the sphere is bigger, however, the sand must be more spread out. After all, there should be the same number of sand grains as there were before, but in order to make a bigger sphere with the same number of grains, you'll need to spread the grains out (make them more diffuse).

So what's the connection? Well, imagine now that you're sitting a certain distance from the canister and you have a bowl in your hand. If you hold this bowl in front of you as the canister explodes, you ought to be able to collect some sand. How much sand will you collect? Well, that depends on how far away you're standing. If you're standing near the canister, the sand will be in a small sphere in which the grains are very close together, so you'll collect a lot of sand. If you stand back however, the grains will be spread out into a larger sphere and only a few grains will hit your bowl.

So that tells us conceptually how it works, but we need a formula for how "spread out" the grains of sand are so that we know how many hit the bowl. It turns out that this can be simply described by the "surface density" of sand grains. This is given by:

[tex]\kappa=\frac{N}{4\pi r^2}[/tex]

where [tex]\kappa[/tex] is the surface density, N is the number of sand grains initially in the canister, and r is their distance from their point of origin (the position of the canister). Let's make sure we understand what this formula means. First, we look at the denominator, you should recognize that as something:

[tex]A=4\pi r^2[/tex]

That's the area of a sphere. This makes sense because we said that the sand grains were spread out over a sphere. So why are we dividing the number of sand grains by the area of the sphere? Since the number of grains is staying constant (none are created or destroyed), then that formula is just saying that we're dividing the sand evenly over the spherical surface. Since N is constant, as time goes on, the grains travel further ("r" increases) and the grains become more spread out ([tex]\kappa[/tex] decreases).

Now, to complete the connection of our above situation to the one with luminosity and flux, just replace the sand grains with photons of light and the canister with the sun. There's one other important difference however, and that is that the photons are not emitted all at once, but steadily as time goes on. This means that you can split the light coming from the sun into many spheres throughout space, each consisting of photons emitted at a particular instant in time. Thus, whatever you're detecting the light with (like the bowl with the sand) will continue detecting it as time goes on, but the rate at which it picks up photons will follow the same dependence with distance as I described above. The equation is:

[tex]F=\frac{L}{4\pi d^2}[/tex]

where F is the flux, L is the luminosity, and d is the distance.

I'm sorry if the above explanation was too simple, but I wanted to make sure to capture all of the essential concepts in the situation.

 

[Mariko]

SpaceTiger I appreciate the time and thought you took to explain this to me and It is perfect i see now thank you very much I am very grateful to you. :!)

 

[SpaceTiger]

I'm glad it helped, cause it did take some time to put together.

This is a really important concept in astronomy (as well as other things, I'm sure), so I felt it was worth writing out in full.

 

[Chronos]

kudos, ST. I would have missed a meal trying to cobble that much information together - although it is highly probable I would not suffer any lasting adverse effects from said experience.