# Quantum Physics Intensity of Radaiation of Stars Help Needed 🌟

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#1
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
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. Calculate the intensity of the radiation from Proxima Centauri on Earth.

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)

The second closest star to the Sun is Alpha Centauri A, which is 4.37 light years away and has a power output of 5.813 x 10^26 W. Although Alpha Centauri A is part of a binary star system to the unaided eye Alpha Centauri A and B appear as one star.

3. Calculate the intensity of the radiation from Alpha Centauri A on Earth.
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. Despite Proxima Centauri being the closest star to our Sun it was not discovered until 1915. Use your results to explain this.

This is the question I am having the greatest difficulty with. I believe that despite Proxima Centauri being closer to the sun it was not discovered until 1915 as it has a low illumination, since its intensity being the amount of energy it carries per unit area per unit time is significantly lower than say other stars, e.g Alpha Centauri A which is further away.
Alpha Centauri A has a far greater intensity but is actually 0.17 light years further away.
I think that I need to perform some calculations to support my answer here or find why the intensity is lower for Proxima Centauri. This section is worth four marks but I cannot think of four points to make. I would be very grateful of any help 👍
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2 months ago
#2
Is this university Astrophysics?
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#3
(Original post by Despaxir)
Is this university Astrophysics?
No, A-Level Physics 😁
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2 months ago
#4
1.4 MW is unbelievably weak for a star... literally unbelievable - it's equivalent of 700 two bar household electric fires.

there's a minimum size for star that is about 1/10 the diameter of the sun... still far to big to glow visibly if it's radiating the same amount of power as 700 electric fires.

Given reasonable power output figures you could attempt to show by inverse square law or the distance modulus formula that the difference in distance is insignificant in comparison to the difference in luminosity.
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#5
(Original post by Joinedup)
1.4 MW is unbelievably weak for a star... literally unbelievable - it's equivalent of 700 two bar household electric fires.

there's a minimum size for star that is about 1/10 the diameter of the sun... still far to big to glow visibly if it's radiating the same amount of power as 700 electric fires.

Given reasonable power output figures you could attempt to show by inverse square law or the distance modulus formula that the difference in distance is insignificant in comparison to the difference in luminosity.
Thank you for your reply. Yes, that is exactly what I thought, 1.4MW is very weak indeed, it is a peculiar statement in the question. Maybe it is just a mistake? Are my other answers to questions 1-3 correct or would they be wrong because this measure of power is so low?

Regarding question 4 I am still not sure how to explain that despite Proxima Centauri being the closest star to our Sun it was not discovered until 1915. I know that you suggest using to inverse square law to show the difference in distance is insignificant in comparison to the difference in luminosity, but I am uncertain how to do so.
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.
I also know that the inverse square law of apparent brightness is represented by the equation; 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.

The distance from Proxima Centauri to Earth is 3.97 * 10^16 m, therefore, to find the apparent brightness:
b=L/4πd^2
b=1,400,000/4*π*(3.97 * 10^16)^2
b~7.07*10^-29 Wm^-2 to 3.s.f

The distance from Alpha Centauri A to Earth is 4.13*10^16m, therefore, to find the apparent brightness:
b=L/4πd^2
b= 5.813 x 10^26 W/4*π*(4.13*10^16)^2
b~ 2.71*10^-8 Wms^-2Wm^-2 to 3.s.f

But I not think calculating this has shown anything to substantiate an answer as it is just the intensity which I have previously calculated?

I do also know that the intensity of light is inversely proportional to the square of the distance.
I∝1/d^2
Would I use this to input the values for the distances to find the intensity of the stars to show the difference in distance is insignificant in comparison to the difference in luminosity?

Sorry I think I am overcomplicating matters and do realise 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!
Last edited by AN630078; 2 months ago
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2 months ago
#6
Bear in mind that in exam questions you are given a number of points for each section and it's not worth doing extensive calculations for a 1 point question, a brief answer is all that is required.

Alright, you can look at the formula b=L/4πd^2

and see a few things,

1. L has a linear effect on b

i.e. if you had stars at the same distance but of different luminosities the apparent brightness would be proportional to the luminosity...

if you had a star with Luminosity 1 and apparent brightness 1 then a star at the same distance with 2x that luminosity would have apparent brightness 2 and a star with 3x the luminosity would have apparent brightness 3

2. d has an effect on b given by the inverse square rule

i.e. if you had stars of the same luminosity at different distances, the apparent brightness would be proportional to 1/d2

if you had stars with the same luminosity at distance 1, 2 and 3 the apparent brightnesses would be in the ratio 1:1/4:1/9

if you had stars of the same luminosity at distances of 4.13 and 3.97 the ratio of apparent brightness would be

1/4.132 : 1/3.972

1/17.1 : 1/15.8

0.0585 : 0.0633

0.924

so the more distant star would have an apparent brightness of 0.924 (92.4%) of the nearer star

however the case of prioxima and alpha the luminosities differ by many orders of magnitude and the linear relationship between luminosity and apparent brightness is very much more significant than the small inverse square difference from their similar distancves.
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#7
Joinedup
Thank you for your reply. Yes I agree, this question specifically is worth four points but if the distance is unimportant here how should I comment with four points besides from the fact that Proxima Centauri emits far less radiation than Alpha Centauri and this would be harder to discern?

Thank you for evaluating the formula b=L/4πd^2.
Yes I can understand how L has a linear effect on b
and d has an effect on b given by the inverse square rule.

How if you had stars of the same luminosity at distances of 4.13 and 3.97 would ratio of apparent brightness would be 1/17.1 : 1/15.8 from 1/4.13: 1/3.97 ?

But I can see how the more distant star would have an apparent brightness of 0.924 (92.4%) of the nearer star.
Last edited by AN630078; 2 months ago
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2 months ago
#8
Using

(Original post by AN630078)
Joinedup
Thank you for your reply. Yes I agree, this question specifically is worth four points but if the distance is unimportant here how should I comment with four points besides from the fact that Proxima Centauri emits far less radiation than Alpha Centauri and this would be harder to discern?

Thank you for evaluating the formula b=L/4πd^2.
Yes I can understand how L has a linear effect on b
and d has an effect on b given by the inverse square rule.

How if you had stars of the same luminosity at distances of 4.13 and 3.97 would ratio of apparent brightness would be 1/17.1 : 1/15.8 from 1/4.13: 1/3.97 ?

But I can see how the more distant star would have an apparent brightness of 0.924 (92.4%) of the nearer star.
for stars of the same luminosity
The star at distance 4.13 would have an apparent brightness proportional to 1/4.132 and 4.132 is 17.1

The star at distance 3.97 would have an apparent brightness proportional to 1/3.972 and 3.972 is 15.8

you can sort by apparent magnitude and see that the stars with the lowest apparent magnitude (i.e. the brightest looking) have generally been known about for longest... they are naked eye objects that can be seen without a telescope, the limit for naked eye objects is about 5 or 6 apparent magnitude. alpha centauri is 0.01 magnitude and so was very clearly visible before telescopes of any sort had been invented.

proxima has apparent magnitude of 11.09 and requires a large telescope to detect, and we've only really had telescopes of the necessary size for a short time... mainly starting from the early 20th century.
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#9
(Original post by Joinedup)
Using

for stars of the same luminosity
The star at distance 4.13 would have an apparent brightness proportional to 1/4.132 and 4.132 is 17.1

The star at distance 3.97 would have an apparent brightness proportional to 1/3.972 and 3.972 is 15.8

you can sort by apparent magnitude and see that the stars with the lowest apparent magnitude (i.e. the brightest looking) have generally been known about for longest... they are naked eye objects that can be seen without a telescope, the limit for naked eye objects is about 5 or 6 apparent magnitude. alpha centauri is 0.01 magnitude and so was very clearly visible before telescopes of any sort had been invented.

proxima has apparent magnitude of 11.09 and requires a large telescope to detect, and we've only really had telescopes of the necessary size for a short time... mainly starting from the early 20th century.
Thank you for your reply. Oh yes I can understand your calculations thank you for explaining how their apparent brightness is proportional to the calculated figures. Thank you for providing a link to the Wikipedia page, it was very interesting and I enjoyed reading it. Moreover, I appreciate your evaluation of the discovery of Proxima Centauri and naked eye object detection. Just to summarise and attempt to answer this question with four points;

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. 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.

Would this be correct? 😄
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1 month ago
#10
(Original post by AN630078)
Thank you for your reply. Oh yes I can understand your calculations thank you for explaining how their apparent brightness is proportional to the calculated figures. Thank you for providing a link to the Wikipedia page, it was very interesting and I enjoyed reading it. Moreover, I appreciate your evaluation of the discovery of Proxima Centauri and naked eye object detection. Just to summarise and attempt to answer this question with four points;

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. 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.

Would this be correct? 😄
basically, but be a bit careful about 'lesser magnitude' though - lower magnitude means more brightness.
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#11
(Original post by Joinedup)
basically, but be a bit careful about 'lesser magnitude' though - lower magnitude means more brightness.
Right, yes of course, I knew something was wrong with my wording I could not pinpoint it! I think I had been reading for too long😂 Thank you very much again for your help, do you think I could add anything salient to my answer? 😀
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