 #1
AN630078
 242
 25
 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 (EarthProxima 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 (EarthAlpha 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 everwidening 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!
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 (EarthProxima 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 (EarthAlpha 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 everwidening 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!