Power and intensity question!?

OP

Original post by Joinedup

The reason for the 1/r^2 law is that if you have a point source of sound (or light etc.) giving out a known power (say 60W) which is radiating equally in every direction and you have it located at the centre of a transparent sphere with a known surface area (say 1m^2) you can say that the intensity at the sphere surface must be 60/1 60W/m^2 all the power radiated from the source must cross the surface (by conservation of energy)

the surface area, A, of a sphere is related to the distance from the centre, r, by
A=4πr^2 e.g. if the radius doubles the area increases by factor 2^2 or 4 and if the radius triples the area increases by factor 2^3 or 8

in each case the 60W radiated at the centre must equal 60W crossing the surface of the sphere however far away it is from the source.
---

part C is giving you 5.6*10^-2 W m^-2

which is 5.6*10^-2 Watts per meter squared... power/cross sectional area... it's already an intensity, you don't need to do any more working out.

the power of a wave is proportional to Amplitude^2 if you hold frequency, speed and mass constant.

hope that helps.

for part c I did 5.6*10^-2 / 3.5 * 10^-3
it is 16 times larger than our original value so its amplitude bust be 4x larger, then did 4 * 0.45?

Now need help on 7 though. For b
I used intensity is proportional to 1/r^2
the distance is 1.5*10^11 m
so our 1.5kw of energy is a (1/(1.5*10^11))"th" of the actual intensity, so did
1.5kw * distance squared from sun to get 3.15 * 10 ^22
however this seems enormous.
as when i then go onto C and re arrange to do intensity * cross sectional area = power
I get 1.9396 * 10^41
So my mistake seems to be calculating the intensity of the sun, however cannot see my mistake.
The intensity of the sun is 1/distance^2 so 1/(1.5*10^11)^2
therefore if we get 1.5kw, surely if we times it by distance squared that is equal to the full intensity of the sun? It makes logical sense to me

Reply 3

Original post by Danny.L

for part c I did 5.6*10^-2 / 3.5 * 10^-3
it is 16 times larger than our original value so its amplitude bust be 4x larger, then did 4 * 0.45?

Now need help on 7 though. For b
I used intensity is proportional to 1/r^2
the distance is 1.5*10^11 m
so our 1.5kw of energy is a (1/(1.5*10^11))"th" of the actual intensity, so did
1.5kw * distance squared from sun to get 3.15 * 10 ^22
however this seems enormous.
as when i then go onto C and re arrange to do intensity * cross sectional area = power
I get 1.9396 * 10^41
So my mistake seems to be calculating the intensity of the sun, however cannot see my mistake.
The intensity of the sun is 1/distance^2 so 1/(1.5*10^11)^2
therefore if we get 1.5kw, surely if we times it by distance squared that is equal to the full intensity of the sun? It makes logical sense to me

Well you would expect the sun to have a very large power output. It's the sun.

Reply 4

OP

Original post by Texx

Well you would expect the sun to have a very large power output. It's the sun.

yes, very smart of you.
However, when I have looked in my book which uses similar values for distance and for intensity of light when it reaches earth, its power of the sun is 3.7*10^26
to the power of 41 is considerably larger and thus making me think i'm incorrect.

Reply 5

Original post by Danny.L

yes, very smart of you.
However, when I have looked in my book which uses similar values for distance and for intensity of light when it reaches earth, its power of the sun is 3.7*10^26
to the power of 41 is considerably larger and thus making me think i'm incorrect.

Because you need to root the distance. And covert kW to SI units.

Also, calm down, it's amazing you get touchy because I offered the suggestion that you weren't in fact incorrect.

(edited 9 years ago)

Reply 6

OP

Original post by Texx

Because you need to root the distance. And covert kW to SI units.

Also, calm down, it's amazing you get touchy because I offered the suggestion that you weren't in fact incorrect.

Edit: it's the surface area of the sun you'll want to include in the total power calculation, also.

Not touchy, just looked sarcastic in your original response, so i responded accordingly.
Anyway,
I understand the SI units, that's stupid of me.
However, the root of distance I'm trying to get my head around. care to explain more why? Not sure I fully understand

Reply 7

OP

Original post by Texx

Because you need to root the distance. And covert kW to SI units.

Also, calm down, it's amazing you get touchy because I offered the suggestion that you weren't in fact incorrect.

we have been told that we're 1.5 * 10^11 m from the sun, and thus the intensity of light here on earth is
(1.5*10^11)^2 less bright than on the surface of it, correct?
So why would we root it?
if we move 10 times further away the light is 100 times dimmer.
So if we are told the intensity is 10w at 10 m away the intensity is 1/10^2 of the actual magnitude.
so 10w is 1/100 th of the actual magnitude. So 10w * 100 (distance squared) surely equals original intensity. Which would be 1000w
Do you see where I'm coming from. I'm not sure I understand why a root is coming into place

Reply 8

OP

Original post by Texx

Because you need to root the distance. And covert kW to SI units.

Also, calm down, it's amazing you get touchy because I offered the suggestion that you weren't in fact incorrect.

After working it out.
I took the energy earth recieves, 1400w * 2.25 * 10^22 (distance squared) and got 3.15*10^25, which looks suspiciously close to what I was expecting the power of the sun to be.
It's almost like I have unintentionally done intensity * cross sectional area opposed to scaling up the intensity.
although if I were to do intensity * cross sectional area, wouldn't I actually do 4pi(1.5*10^11)^2
the 4pi gives the circle created at the distance squared.
I am so confused, I have no idea why my original idea isn't working but is now seem to have accidently worked out my power of the sun with it.
I need someone to step-by-step walk though this w/ me

Reply 9

Original post by Danny.L

Not touchy, just looked sarcastic in your original response, so i responded accordingly.
Anyway,
I understand the SI units, that's stupid of me.
However, the root of distance I'm trying to get my head around. care to explain more why? Not sure I fully understand

Yes, you're right, sorry I had a dumb moment. Hang on, I'll try to explain it properly.

Reply 10

OP

Original post by Texx

Yes, you're right, sorry I had a dumb moment. Hang on, I'll try to explain it properly.

Thanks, appreciate it

Reply 11

Original post by Danny.L

After working it out.
I took the energy earth recieves, 1400w * 2.25 * 10^22 (distance squared) and got 3.15*10^25, which looks suspiciously close to what I was expecting the power of the sun to be.
It's almost like I have unintentionally done intensity * cross sectional area opposed to scaling up the intensity.
although if I were to do intensity * cross sectional area, wouldn't I actually do 4pi(1.5*10^11)^2
the 4pi gives the circle created at the distance squared.
I am so confused, I have no idea why my original idea isn't working but is now seem to have accidently worked out my power of the sun with it.
I need someone to step-by-step walk though this w/ me

Alright:

Imagine the earth is a point on a sphere with radius 1.5x10^11m, every point of this sphere is receiving light from the sun with the same intensity of 1.4x10^3 W per m. The surface area of this sphere is 4 x pi x r^2. The total power of the sun is the surface area multiplied by the intensity of the light at this radius. Does that make sense?

Sorry for the stupid moment with mentioning rooting it... I think I'll go have a nap.

Reply 12

OP

Original post by Texx

Alright:

Imagine the earth is a point on a sphere with radius 1.5x10^11m, every point of this sphere is receiving light from the sun with the same intensity of 1.4x10^3 W per m. The surface area of this sphere is 4 x pi x r^2. The total power of the sun is the surface area multiplied by the intensity of the light at this radius. Does that make sense?

Sorry for the stupid moment with mentioning rooting it... I think I'll go have a nap.

So whatever I was doing originally was complete nonsense?
Seemed logical to me though.
I understand the principle, I guess. What I am unsure about is why the distance is the radius, I am now imagining looking side-on with the sun at the centre and earth to the far right, you're saying we treat the distance from the 2 as the radius, I guess I can understand that.
Okay so the light at this radius is 1400w.
the surface area is 4 Pi r^2 although I cant see how that works now, as surely as you get towards the point the intensity of light increases.
I am imagining it as a 2d circle looking at it with the earth on the right, sun middle and the surface area is the whole circle, which I do not understand as all of those bits arent being supplied with 1400w, but will be receiving more energy as they are closer. But I guess I'll just accept it.

My question is.
Is there any possible way to work out intensity directly?
It wants intensity first, but the only way I see is to calculate power and then we can get intensity from that?
Thanks so far, most helpful person of tonight. This has to be in tomorrow as well :mad:

Reply 13

Original post by Danny.L

So whatever I was doing originally was complete nonsense?
Seemed logical to me though.
I understand the principle, I guess. What I am unsure about is why the distance is the radius, I am now imagining looking side-on with the sun at the centre and earth to the far right, you're saying we treat the distance from the 2 as the radius, I guess I can understand that.
Okay so the light at this radius is 1400w.
the surface area is 4 Pi r^2 although I cant see how that works now, as surely as you get towards the point the intensity of light increases.
I am imagining it as a 2d circle looking at it with the earth on the right, sun middle and the surface area is the whole circle, which I do not understand as all of those bits arent being supplied with 1400w, but will be receiving more energy as they are closer. But I guess I'll just accept it.

My question is.
Is there any possible way to work out intensity directly?
It wants intensity first, but the only way I see is to calculate power and then we can get intensity from that?
Thanks so far, most helpful person of tonight. This has to be in tomorrow as well :mad:

Well, the intensity of the light at the surface should be correct, just not the total power output.
(edit: that wasn't very clear, I mean your initial method for working out the intensity at the surface should yield the correct answer. :smile:

but the total power output is found by picturing the sphere, and can't be done the way you originally tried.)
I actually only realised it because I had a vague memory of doing a similar question at A-level. It's one of those things that makes sense when you are explained it but wouldn't necessarily occur to you otherwise. Yeah it's quite hard to visualise, the distance to the earth is the radius and not the diameter because the sun is emitting light in every direction, so it's emitting light of the same intensity in a direction away from the earth too. The tough thing is you're working out the surface area of a kind of imaginary sphere, made up of points that are all the same distance from the sun as the earth is. Try thinking about it in 3D.
You're right, a point closer to the sun receives light with greater intensity, but it's not the surface area of a circle you're working out. It's hard to explain without a drawing, I've tried to attach one. Imagine the yellow patch is a square metre on the surface of the sphere. It's receiving 1.4 kW of power. Every square metre on that sphere is receiving the same.

(edited 9 years ago)

OP

Original post by Texx

Well, the intensity of the light at the surface should be correct, just not the total power output.
(edit: that wasn't very clear, I mean your initial method for working out the intensity at the surface should yield the correct answer. :smile:

but the total power output is found by picturing the sphere, and can't be done the way you originally tried.)
I actually only realised it because I had a vague memory of doing a similar question at A-level. It's one of those things that makes sense when you are explained it but wouldn't necessarily occur to you otherwise. Yeah it's quite hard to visualise, the distance to the earth is the radius and not the diameter because the sun is emitting light in every direction, so it's emitting light of the same intensity in a direction away from the earth too. The tough thing is you're working out the surface area of a kind of imaginary sphere, made up of points that are all the same distance from the sun as the earth is. Try thinking about it in 3D.
You're right, a point closer to the sun receives light with greater intensity, but it's not the surface area of a circle you're working out. It's hard to explain without a drawing, I've tried to attach one. Imagine the yellow patch is a square metre on the surface of the sphere. It's receiving 1.4 kW of power. Every square metre on that sphere is receiving the same.

So I have done it correct for 7)b)i)?
I have looked at in another way,
using:
Power = intensity * surface area
Power = 1400 * 4Pi (1.5 *10^11)^2
power = 3.96 * 10^26
so intensity = (3.26 * 10^26)/(4Pi(7*10^8)^2)
Intensity = 64.3 Gwatts per m^-2
However doing it my original way, the intensity of light on the sun must be 2.25 *10^22 larger than that on earth so
1400 * (2.25 * 10^22) = 3.15 *10^22
Which one is correct?
It seems to me the 64.3 Gw is more reasonable as if I use the 3.15 *10^22 that means the power becomes something * 10^41, which I am certain is wrong!
What do you think?

(edited 9 years ago)

Reply 15

OP

Original post by Texx

Well, the intensity of the light at the surface should be correct, just not the total power output.
(edit: that wasn't very clear, I mean your initial method for working out the intensity at the surface should yield the correct answer. :smile:

but the total power output is found by picturing the sphere, and can't be done the way you originally tried.)
I actually only realised it because I had a vague memory of doing a similar question at A-level. It's one of those things that makes sense when you are explained it but wouldn't necessarily occur to you otherwise. Yeah it's quite hard to visualise, the distance to the earth is the radius and not the diameter because the sun is emitting light in every direction, so it's emitting light of the same intensity in a direction away from the earth too. The tough thing is you're working out the surface area of a kind of imaginary sphere, made up of points that are all the same distance from the sun as the earth is. Try thinking about it in 3D.
You're right, a point closer to the sun receives light with greater intensity, but it's not the surface area of a circle you're working out. It's hard to explain without a drawing, I've tried to attach one. Imagine the yellow patch is a square metre on the surface of the sphere. It's receiving 1.4 kW of power. Every square metre on that sphere is receiving the same.

Sorry, yeah I understand you, I got surface area and volume mixed up. I understand it. Almost like we're collecting all of the energy at the "distance" away from the sum and adding it all back up again to get the power. Still unsure on what is correct for the intensity. Both methods make sense, 1 uses inverse square law, another the formula! But The formula gives a large, but much more reasonable calculation.
When i used it to calculate intensity on Neptune I got 1.556 W/m^-2 which is exactly the same as the value if you were to divide neptune distance from sun by earths distance = 30
30^2 =900
1400 (earths light) / 900 = 1.555(rec)
So I believe the formula is correct, just curious as to why the inverse square law doesnt work!
For some reason I feel very agitated when I don't understand something, this being one of those times!

Reply 16

Joinedup

20

Original post by Danny.L

So whatever I was doing originally was complete nonsense?
Seemed logical to me though.
I understand the principle, I guess. What I am unsure about is why the distance is the radius, I am now imagining looking side-on with the sun at the centre and earth to the far right, you're saying we treat the distance from the 2 as the radius, I guess I can understand that.
Okay so the light at this radius is 1400w.
the surface area is 4 Pi r^2 although I cant see how that works now, as surely as you get towards the point the intensity of light increases.
I am imagining it as a 2d circle looking at it with the earth on the right, sun middle and the surface area is the whole circle, which I do not understand as all of those bits arent being supplied with 1400w, but will be receiving more energy as they are closer. But I guess I'll just accept it.

My question is.
Is there any possible way to work out intensity directly?
It wants intensity first, but the only way I see is to calculate power and then we can get intensity from that?
Thanks so far, most helpful person of tonight. This has to be in tomorrow as well :mad:

Power first would be a good way of getting to the intensity at the surface of the sun

or you could also do it by ratios which would look like

I₁ R₁^2 = I₂ R₂^2

I₁ is intensity at earth
R₁ is earth's orbital radius

I₂ is intensity at sun's surface
R₂ is the radius of the sun

it's *as if* you're treating the sun as if it was a sphere with a point source inside it

Reply 17

OP

Original post by Joinedup

Power first would be a good way of getting to the intensity at the surface of the sun

or you could also do it by ratios which would look like

I₁ R₁^2 = I₂ R₂^2

I₁ is intensity at earth
R₁ is earth's orbital radius

I₂ is intensity at sun's surface
R₂ is the radius of the sun

it's *as if* you're treating the sun as if it was a sphere with a point source inside it

Yeah, I showed my teacher my working today, he said it wa great. However I could get the intensity directly via the method you used. He went a step further and showed I1R1^2 = K I2R2^2 = k
therefore they equal each other.
He also spotted what I had done wrong for my working by multiplying 1400 Wm-2 by (1.5 *10^11)^2. That was finding the intensity 1m away from the centre of the sun, but we wanted it at the surface, so needed to then divide by (7 * 10*)^2 (which is radius of sun squared). I think that's correct! How does that sound to you?

Reply 18

OP

Original post by Texx

Well, the intensity of the light at the surface should be correct, just not the total power output.
(edit: that wasn't very clear, I mean your initial method for working out the intensity at the surface should yield the correct answer. :smile:

but the total power output is found by picturing the sphere, and can't be done the way you originally tried.)
I actually only realised it because I had a vague memory of doing a similar question at A-level. It's one of those things that makes sense when you are explained it but wouldn't necessarily occur to you otherwise. Yeah it's quite hard to visualise, the distance to the earth is the radius and not the diameter because the sun is emitting light in every direction, so it's emitting light of the same intensity in a direction away from the earth too. The tough thing is you're working out the surface area of a kind of imaginary sphere, made up of points that are all the same distance from the sun as the earth is. Try thinking about it in 3D.
You're right, a point closer to the sun receives light with greater intensity, but it's not the surface area of a circle you're working out. It's hard to explain without a drawing, I've tried to attach one. Imagine the yellow patch is a square metre on the surface of the sphere. It's receiving 1.4 kW of power. Every square metre on that sphere is receiving the same.

Thanks for your help btw, my teacher spotted why 1400 * (1.5*10^11)^2 was wrong straight away, was quite impressed, actually! :biggrin:

It was because doing that finds the intensity of light at the centre of it, 1m away, but we wanted the surface which is 7*10^8 m away, so the intensity at that point is (1400 * (1.5*10^11)^2) all / by (7*10^8)^2
Like we work out the intensity at the core, then work our way back out. Hope that's correct, I figured that'd be right.

Reply 19