# Intensity of Stars and the Inverse Square Law Problem

• AN630078
In summary: Alpha Centauri A has a luminosity of 2.71*10^-8 Wms^-2 while Proxima Centauri has a luminosity of 7.07*10^-29 Wm^-2. This means that the power emitted by Alpha Centauri A is 4.3*10^-29 Wm^-2 greater than the power emitted by Proxima Centauri.
AN630078
Homework Statement
Hello, I have a question concerning the intensity of radiation of different stars and performing some associated calculations. I have answered all sections fully but I would nonetheless be very grateful for any advice or guidance upon improving my answers or perhaps adopting a more suitable approach. I am having rather a bit of difficulty with question 4 and cannot formulate a sound conclusion, therefore, I would be tremendously grateful of any help specifically related to this question. Thank you very much in advance

Proxima Centauri is a red dwarf star located 4.2 light years away that is the nearest star to the sun. The power emitted by Proxima Centauri is 1.4 MW.

1. Prove that 4.2 light years is approximately 3.97 * 10^16 m
2. Calculate the intensity of the radiation from Proxima Centauri on Earth.
3. Calculate the intensity of the radiation from Alpha Centauri A on Earth.
4. Despite Proxima Centauri being the closest star to our Sun it was not discovered until 1915. Use your results to explain this.

(Also I realise that 1.4MW is a very weak power emitted by a star, I think that this may be an error in the question or perhaps this value has been chosen for ease of calculation. Anyhow, I have tried to proceed as if it were realistic).
Relevant Equations
Intensity (Wm^-2)=Power(W)/Area(m^-2)
I∝1/d^2
1. speed of light = 3*10^8 ms^-1
length of an Earth year=365 days*24 hours*60minutes*60seconds=3.15*1 0^7 seconds
distance=speed*time
1 light year= 3*10^8 *3.15*10^7=9.46*10^15m
4.2 light years = 4.2*9.46*10^15m=3.9732*10^16m ~ 3.97*10^16m (to 3sf)

2. Intensity=power/area
Power=1.4MW=1,400,000 W
The intensity at Earth is total power spread over a sphere of radius 3.97*10^16m (Earth-Proxima Centauri distance)
Area of sphere=4πr^2
Ares of sphere=4*π*(3.97*10^16)^2
Area~1.98*10^34 m^2 (to 3sf)

Intensity=1,400,000 W/1.98*10^34
Intensity=7.0686610...*10^-29 ~7.07*10^-29 Wm^-2 (to 3sf)

3. Intensity = power/area
4.37 light years~4.13*10^16m (to 3sf)
The intensity at Earth is total power spread over a sphere of radius 4.13*10^16m (Earth-Alpha Centauri distance)
Area=4*π*(4.13*10^16)^2
Area~2.14*10^34 m^2 (to 3sf)
Intensity = 5.813 x 10^26 / 2.14*10^34
Intensity= 2.712004..*10^-8 ~ 2.71*10^-8 Wms^-2 (to 3sf)

4. This is where I am having the most difficulty, I think I could use the inverse square law to explain why Proxima Centauri was not discovered until 1915 and substantiate this with calculations but I am not sure where to begin?
I know that the intensity of light is inversely proportional to the square of the distance.
I∝1/d^2
How could I show the difference in distance is insignificant in comparison to the difference in luminosity of the stars? Would that be an appropriate approach here?

Or should I try to use the inverse square law for apparent brightness:
b=L/4πd^2
- apparent brightness, b in Wm^-2
- luminosity L in W
- distance d in meters
Since we know the distance of both Proxima Centauri and Alpha Centauri to Earth we can calculate the apparent brightness (a measure of the amount of light received by Earth from a star or other object—that is, how bright an object appears in the sky, as contrasted with its luminosity) of the stars.
Moreover, I understand that light fades with increasing distance and the energy received is inversely proportional to the square of the distance. For example, if two stars had the same luminosity but one is twice as far away it would appear four times dimmer than the nearer star.
As light waves travel away from a star they expand in all directions covering an ever-widening space, but the total amount of light available does not change. As the same expanding shell of light covers a larger area, there must be less light in any given place, thus light and other EM radiation becomes weaker as it gets farther from its source. The increase in the area that the light must cover is proportional to the square of the distance that the light has traveled.

Sorry I think I am overcomplicating matters and do realize that this question is simpler than I am making it, I just do not know how to conduct the necessary calculations to support my answer. My apologies, just utter brain freeze it would seem!

This is the problem with memorizing formulas. Answer (2) is the apparent brightness of Proxima Centauri (as seen from earth). I should qualify this by saying people sometimes quote the brightness in logarithmic units on a relative scale (with Vega being the original reference star m=0) and call it magnitude. So you are already finished! Just compare (2) and (3)...
Does look like brain freeze to me... too much studying.

hutchphd said:
This is the problem with memorizing formulas. Answer (2) is the apparent brightness of Proxima Centauri (as seen from earth). I should qualify this by saying people sometimes quote the brightness in logarithmic units on a relative scale (with Vega being the original reference star m=0) and call it magnitude. So you are already finished! Just compare (2) and (3)...
Does look like brain freeze to me... too much studying.
Thank you very much for your reply . From my earlier calculations the apparent brightness of the Proxima Centauri, found to be 7.07*10^-29 Wm^-2, is far lower than than the apparent brightness of Alpha Centauri A (2.71 × 10^-8 Wms^-2 specifically). However, this contrasts with the fact that despite the apparent brightness of Alpha Centauri A being 2.71 × 10^-8 Wms^-2 greater than Proxima Centauri, it is 1.6*10^15 m further away. It is therefore probable that Proxima Centauri was not discovered until 1915 since although being closer to Earth than other stars, it has exhibits significantly less brightness and is harder to discern.

Sorry, but I think that my answer is far too simplistic to adequately answer the question here, I think I need more supporting evidence to substantiate my answer as it is worth four marks. I know this is just a revision exercise but I want to ensure I arrive at a sound conclusion in case I am faced with a similar problem in the future.

1. Where did you get the power emitted by alpha centauri A (you don't mention it in the problem statement??
2. The only thing of interest here is ratios of numbers: by what ratio is the power less? Statement of absolute differences mean very little in this context
3. The difference in distance to Earth is negligible here in fact ...they are thought to be part of the same star system.
Also stars emit radiation (power) at wavelengths invisible to the eye. The photometric (what your eye sees) emission of an object and radiometric (total power) emission can be very different depending on the temperature of the star. Is that important here?

So I don't know what you are expected to say other than Proxima puts out a lot less radiation.

: said:
1. Where did you get the power emitted by alpha centauri A (you don't mention it in the problem statement??
2. The only thing of interest here is ratios of numbers: by what ratio is the power less? Statement of absolute differences mean very little in this context
3. The difference in distance to Earth is negligible here in fact ...they are thought to be part of the same star system.
Also stars emit radiation (power) at wavelengths invisible to the eye. The photometric (what your eye sees) emission of an object and radiometric (total power) emission can be very different depending on the temperature of the star. Is that important here?

So I don't know what you are expected to say other than Proxima puts out a lot less radiation.
1. The power emitted by Alpha Centauri A was given in the question
2. Ok, how would I find the ratio of the powers? Would this be 7.07*10^-29:2.71*10^-8 ?
3. I did not realize the difference in distance was negligible here, thank you for stating that.

Yes, I would say that the in this instance the photometric emission is very different between the stars, since although it may be difficult to see Alpha Centauri A with the naked eye it would be easier to discern than Proxima Centauri. Similarly, the radiometric emission of Alpha Centauri A far surpasses Proxima Centauri.
Sorry I am still a bit confused here. "So I don't know what you are expected to say other than Proxima puts out a lot less radiation." maybe I need to incorporate the inverse square law to substantiate that Proxima Centauri emits less radiation to fufil all four points?

According to your numbers Iproxima/Ialpha ~10-20
They are approximately the same distance away.The inverse square law will make less than a 10% difference in the amount that gets to Earth from the each. This does not matter in this context..

hutchphd said:
According to your numbers Iproxima/Ialpha ~10-20
They are approximately the same distance away.The inverse square law will make less than a 10% difference in the amount that gets to Earth from the each. This does not matter in this context..
Yes I found that the intensity of Proxima Centauri and Alpha Centauri A differ by many orders of magnitude, however the linear relationship between luminosity and apparent brightness ( in the formula b=L/4πd^2) is more significant than the small inverse square difference from their similar distances.

So if the distance is unimportant here how should I comment on question 4 with four points besides from the fact that Proxima Centauri emits far less radiation?

Hello, I was rereading this thread and realized that I do not think I answered question 4 fully, and admittedly i am still a little uncertain. Anyhow;

Proxima Centauri was not discovered until 1915 because it has a lesser apparent magnitude than Alpha Centauri A, as it emits less radiation. This is evident in the difference of previous calculations of the intensity of the two stars.
I Proxima Centauri/ I Alpha Centauri A = 2.71*10^-8/7.07*10^-29=3.87*10^20
Therefore, the intensity of Proxima Centauri is ~ 10^20 orders of magnitude lesser then Alpha Centauri A despite it being nearer to Earth. Since intensity of radiant energy is the power transferred per unit area this means that the magnitude of EM radiation received discernible as visible light is far lesser than Alpha Centauri A, resulting in greater difficulty in discovering it.
Moreover, the naked eye can see stars with an apparent magnitude up to +6m, the apparent magnitude of Alpha Centauri A is 0.01 whereas Proxima Centauri has an apparent magnitude of 11.09. Alpha Centauri A was discovered with the naked eye far prior to the invention of telescopes for this reason, however, Proxima Centauri is too faint to be seen with the naked eye. Consequently, it was not discovered until 1915 following the development of telescopes and instrumentation of the necessary size and capacity to detect stars of such apparent magnitudes at the onset of the 20th century.

Could I add anything to my answer, or use my results further in my explanation?

Again the issue of apparent magnitude has to do with distance and that is not a big deal as I mentioned. It is the absolute magnitude that is important here.
Also in terms of naked eye brightness your eye sees only a narrow band of emission and proxima emits an even smaller potion of its smaller emission in the visible because it is very red so is is visually really really darker.
If you can't see it, you can't see it. QED.

AN630078
hutchphd said:
Again the issue of apparent magnitude has to do with distance and that is not a big deal as I mentioned. It is the absolute magnitude that is important here.
Also in terms of naked eye brightness your eye sees only a narrow band of emission and proxima emits an even smaller potion of its smaller emission in the visible because it is very red so is is visually really really darker.
If you can't see it, you can't see it. QED.
Thank you for your reply. I found that the absolute magnitude of Proxima Centauri is +15.6 and of Alpha Centauri A is 4.38, which is a measure of the luminosity of the stars. But, I thought from my calculations that Alpha Centauri A had a greater intensity, and would therefore have a greater absolute magnitude?

How can I correct my answer or expand upon it with information I have previously calculated?

AN630078 said:
The power emitted by Proxima Centauri is 1.4 MW.
.
(Also I realize that 1.4MW is a very weak power emitted by a star, I think that this may be an error in the question or perhaps this value has been chosen for ease of calculation. Anyhow, I have tried to proceed as if it were realistic).

1.4MW is wildly wrong and physically nonsensical. If this is work to be handed-in I would check the value with your teacher first to avoid embarressment! Then correct your working if needed.

The actual power output (luminosity, L) is about 0.17% that of the sun (https://www.solarsystemquick.com/universe/proxima-centauri)

##L = 3.8\times 10^{26} \times \frac {0.17}{100} = 6.5\times10^{23} W##
This is 17 orders of magntiude bigger than 1.4MW!

## 1. What is the inverse square law?

The inverse square law is a principle in physics that states that the intensity of a physical quantity, such as light or sound, is inversely proportional to the square of the distance from the source of that quantity.

## 2. How does the inverse square law apply to stars?

In the context of stars, the inverse square law explains why the brightness, or intensity, of a star decreases as the distance from the star increases. This is because the light from a star spreads out in all directions, so the further away you are from the star, the less light you receive.

## 3. How does the intensity of a star affect its appearance?

The intensity of a star is directly related to its apparent magnitude, which is a measure of how bright a star appears to an observer on Earth. The higher the intensity of a star, the lower its apparent magnitude will be, making it appear brighter in the night sky.

## 4. Can the inverse square law be applied to all types of stars?

Yes, the inverse square law applies to all types of stars, regardless of their size, temperature, or distance from Earth. This is because it is a fundamental principle of physics that governs the behavior of light and other physical quantities.

## 5. How is the inverse square law used in astronomy?

In astronomy, the inverse square law is used to calculate the distance to stars and other celestial objects. By measuring the intensity of a star's light, astronomers can determine its distance from Earth using the inverse square law and other mathematical equations.

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