# A level help

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#1
Hi can someone help with this physics question. I’m struggling to get to an answer

3C48 is a quasar in the constellation Triangulum. It is believed to have a power output 4x10^11 times great than the Sun. At the Earth, the Suns intensity is 1.4x10^17 times greater than that of the quasar.
Calculate using the inverse square law the distance from Earth to this quasar in AU

Thanks a bunch
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3 months ago
#2
(Original post by Zoeva123)
Hi can someone help with this physics question. I’m struggling to get to an answer

3C48 is a quasar in the constellation Triangulum. It is believed to have a power output 4x10^11 times great than the Sun. At the Earth, the Suns intensity is 1.4x10^17 times greater than that of the quasar.
Calculate using the inverse square law the distance from Earth to this quasar in AU

Thanks a bunch
Hi Zoeva123
You're right! This one is pretty hard, are you sure its on an a level physics paper? Anyway I tried it for a bit and have managed to get a ratio of their distances (I may be wrong) which is AU (right?).
Anyway:
(Intensity = Work done/Area) and (the inverse of intensity^2 is proportional to the distance "r")
As we are given in the question P(power output) of Sun = P
P(power output) of Quasar = (4x10^11)P
And also I(intensity) of Sun = (1.4x10^17)I
I(intensity) of Quasar = I
If we do (Power Output)/(Intensity) to find the respective (m^2) and then root then we perhaps should have a correct ratio?
That would mean Sun(m^2) : Quasar (m^2) = (7.1428... x 10^-18) : (4x10^11)
Root both = Sun(m) : Quasar (m) = (2.6726... x 10^-9) : (6.324...x10^5)

Now lets divide both by the distance of the sun and then I suppose you have it in Arbitrary Units in terms of the Sun's distance from earth.

So obviously Sun = 1 Sun Unit
And the Quasar = 2.3664...x10^14 Sun Units

This could be quite wrong but hopefully someone can verify this working out!
1
#3
(Original post by The Trut)
Hi Zoeva123
You're right! This one is pretty hard, are you sure its on an a level physics paper? Anyway I tried it for a bit and have managed to get a ratio of their distances (I may be wrong) which is AU (right?).
Anyway:
(Intensity = Work done/Area) and (the inverse of intensity^2 is proportional to the distance "r")
As we are given in the question P(power output) of Sun = P
P(power output) of Quasar = (4x10^11)P
And also I(intensity) of Sun = (1.4x10^17)I
I(intensity) of Quasar = I
If we do (Power Output)/(Intensity) to find the respective (m^2) and then root then we perhaps should have a correct ratio?
That would mean Sun(m^2) : Quasar (m^2) = (7.1428... x 10^-18) : (4x10^11)
Root both = Sun(m) : Quasar (m) = (2.6726... x 10^-9) : (6.324...x10^5)

Now lets divide both by the distance of the sun and then I suppose you have it in Arbitrary Units in terms of the Sun's distance from earth.

So obviously Sun = 1 Sun Unit
And the Quasar = 2.3664...x10^14 Sun Units

This could be quite wrong but hopefully someone can verify this working out!
Sorry for the late reply. Yes it was on an aqa astrophysics past paper. I did end up figuring it out and your answer is correct. Thanks for the step by step method it’s helped clarify how to answer it as my workings were very random.
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