# The Naked Scientists Forum

### Author Topic: Distant astronomy objects  (Read 1201 times)

#### allan marsh

• Full Member
• Posts: 91
##### Distant astronomy objects
« on: 04/08/2014 17:56:10 »
A stellar object say 1000 light years away emits visible light which like a lamp diverges until after 1000 light years something reaches earth.

How, assuming divergence do we get a multi photo image, ie many photons and not by then just one lone photon after all that divergence?

#### UltimateTheory

• Sr. Member
• Posts: 107
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##### Re: Distant astronomy objects
« Reply #1 on: 04/08/2014 20:04:08 »
Do you know inverse square law?

f.e.
Energy/Area = Initial Energy / 4*PI*r^2
or
Power/Area = Initial Power / 4*PI*r^2

In the case of Sun, each 1 m^2 of Earth (including what is absorbed by atmosphere) is receiving 1360 W/m^2
Reverse equation:
1360 W/m^2*4*PI*150 mln km^2 = 3.8453*10^26 Watts (power of Sun)
(knowing it, you can apply inverse square law again with any other distance, and calculate f.e. energy that Jupiter, Saturn or anything further is receiving)

If we assume average photon has wavelength = 532 nm (green in middle of visible spectrum). Such photon has h*c/wavelength = 3.734*10^-19 J.

1360 W/m^2 is equivalent to 3.64*10^21 photons with energy 3.734*10^-19 J each per m^2 per second.

If you will do similar calcs for 1000 light years distant object that is Sun-alike (same power), you will get something like 915,000 photons per m^2. It's far far from 1 photon per m^2.
« Last Edit: 04/08/2014 20:20:34 by UltimateTheory »

#### chiralSPO

• Global Moderator
• Neilep Level Member
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##### Re: Distant astronomy objects
« Reply #2 on: 04/08/2014 20:58:24 »
Do you know inverse square law?

f.e.
Energy/Area = Initial Energy / 4*PI*r^2
or
Power/Area = Initial Power / 4*PI*r^2

In the case of Sun, each 1 m^2 of Earth (including what is absorbed by atmosphere) is receiving 1360 W/m^2
Reverse equation:
1360 W/m^2*4*PI*150 mln km^2 = 3.8453*10^26 Watts (power of Sun)
(knowing it, you can apply inverse square law again with any other distance, and calculate f.e. energy that Jupiter, Saturn or anything further is receiving)

If we assume average photon has wavelength = 532 nm (green in middle of visible spectrum). Such photon has h*c/wavelength = 3.734*10^-19 J.

1360 W/m^2 is equivalent to 3.64*10^21 photons with energy 3.734*10^-19 J each per m^2 per second.

If you will do similar calcs for 1000 light years distant object that is Sun-alike (same power), you will get something like 915,000 photons per m^2. It's far far from 1 photon per m^2.

Is that 915,000 photons m–2s–1?

#### UltimateTheory

• Sr. Member
• Posts: 107
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##### Re: Distant astronomy objects
« Reply #3 on: 04/08/2014 21:30:48 »
Is that 915,000 photons m–2s–1?

Yes, 915,000 photons per m^2 per second at 1000 light years (distance 1000*365.25*24*3600*c) from Sun-alike star (that has initial 3.8453*10^26 Watts).

Power [W/m^2] = Initial Power / (4*PI*(1000*365.25*24*3600*c)^2) = 3.8453*10^26 / 1.12476*10^39 = 3.41878E-013 Watts/m^2

I am making simplification that photon has average wavelength = 532 nm (E=3.734*10^-19 J).

3.41878E-013 / 3.734*10^-19 = ~915000

#### The Naked Scientists Forum

##### Re: Distant astronomy objects
« Reply #3 on: 04/08/2014 21:30:48 »