How can i calculate the brightness of moon's light at night on the earth?

  • Thread starter rony01
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  • #1
rony01
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hello Forum

i am a novice physics learner. actually i want to calculate or construct an equation through which i can measure the maximum brightness of moon's light on our planet.

can anyone give me any idea from where to start?
 

Answers and Replies

  • #2
cragar
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well there is an experiment to measure the energy output of the sun , And this might work on a full moon . Set out a glass of water and see if the light from the moon heats up the glass of water , and measure the temperature increase , and measure the time it took and then you can get the energy per second it was putting out over that cross-sectional area . Now i feel stupid for writing this because i don't know if it will work, I don't know if there is enough light coming off the sun . Maybe you could do something with a solar cell.
 
  • #3
tiny-tim
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welcome to pf!

hello rony01! welcome to pf! :smile:

for an equation, you can assume that the full moon is a light bulb which radiates a certain percentage (the "albedo" of Moon-rock) of the light from the sun falling on it …

and assume it radiates equally over a 2π solid angle (ie only on one side!) …

then use the angular diameter of the Moon as seen from the Earth :wink:
 
  • #4
Andy Resnick
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Just to be clear, you are asking about the *brightness*, as opposed to the irradiance?
 
  • #5
arydberg
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First you need a photo receptor. For dim sources i suggest a photo resistor from a night light removed and connected to a resistance meter. The resistance will be a function of the light on it. If you are able to get measurable changes by directing the photo resistor on the moon then you need to calibrate it. I suggest a candle placed 1 foot from the photo resistor. ( one candle power ) in a dark room. Add more candles to get a curve. This is all a little sloppy but it should work and get you started. If you are in a city it may be hard to find moonlight without other light sources interfering.
 
  • #6
rony01
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i think it would be easier to calculate in that to measure the unit LUMEN in which i must require the periphery and the radiation angle. for the radiation angle it will be easily calculated with stick perpendicular to Earth and then using F = Iv × 2π × (1 - cos(A/2)), this law might be easier.


but thanks for the comments!
 
  • #7
arydberg
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Great. What did you get?
 

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