What is the relationship between solar radiation intensity and lunar phases?

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Homework Help Overview

The discussion revolves around the relationship between solar radiation intensity and lunar phases, specifically focusing on the scenario of a full moon. Participants explore the implications of the distances involved, including the Earth's distance from the Sun and the Moon's distance from the Earth.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of solar radiation intensity at the distance of the full moon, questioning the setup and the assumptions about distances. Some express uncertainty about the problem statement and seek clarification on the relationship between the distances involved.

Discussion Status

There is an ongoing exploration of the calculations involved, with some participants offering guidance on the relationships between intensities and distances. Multiple interpretations of the problem are being considered, and participants are engaging with each other's reasoning.

Contextual Notes

Participants mention the need for specific values such as the luminosity of the Sun and the irradiance at the Earth's surface, indicating that they are working within the constraints of provided data and physical tables.

charlie05
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Homework Statement


What is the intensity of solar radiation in a place where is the full moon ?

I know luminosity of the Sun and Earth's distance from the Sun RE and Moon´s distance from the Earth RM. Trajectory of Earth and Moon are circular.

Homework Equations


distance of the site from the sun is d = RE + RM

The Attempt at a Solution


Radiation intensity decreases with the square of the distance.
Intensity of radiation is amount of radiant power per unit area
 
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charlie05 said:

Homework Statement


What is the intensity of solar radiation in a place where is the full moon ?

I know luminosity of the Sun and Earth's distance from the Sun RE and Moon´s distance from the Earth RM. Trajectory of Earth and Moon are circular.

Homework Equations


distance of the site from the sun is d = RE + RM

The Attempt at a Solution


Radiation intensity decreases with the square of the distance.
Intensity of radiation is amount of radiant power per unit area
Sorry, your problem statement didn't translate very well. Is there a diagram associated with this question? Are they asking what is the intensity of sunlight at the combined distance Re + Rm? As if you were on the Moon and the Earth was nearly between the Moon and the Sun?
 
sorry for my bad English :-(
I think - full Moon - position is Sun - Earth - Moon
in full Moon - Earth is between the Sun and Moon ...and distance between the Sun and the Moon is the sum of Earth's distance from the Sun RE and Moon´s distance from the Earth RM. At this distance I have to calculate solar radiation intensity.
 
charlie05 said:
sorry for my bad English :-(
I think - full Moon - position is Sun - Earth - Moon
in full Moon - Earth is between the Sun and Moon ...and distance between the Sun and the Moon is the sum of Earth's distance from the Sun RE and Moon´s distance from the Earth RM. At this distance I have to calculate solar radiation intensity.
Okay, makes sense. :smile:

So it sounds like you have all the information you need to do the calculation. You said you have the radiant intensity at the Earth's surface, right? And you know the intensity decreases with the square of the distance. So can you show us how you would calculate the lower intensity when you add in the extra distance to the Moon?
 
I know only luminosity of Sun...3,827.10 26 W...but maybe I do not to calculate from this information...
but I think, I can find the radiant intensity at the Earth's surface in some tables... IE = 1360 W/m2

IE/IM = RE2 / ( RE + RM )2
 
charlie05 said:
IE = 1360 W/m2
That looks about right -- I always remember that it's about 1kW/m^2 at the Earth.
charlie05 said:
IE/IM = RE2 / ( RE + RM )2
That doesn't look quite right to me. Remember that the intensity is proportional to the inverse of the square of the distance. Intensity gets lower the farther away you get... :smile:
 
conversely... :-)
IM/IE = RE2 / ( RE + RM )2
 
charlie05 said:
conversely... :-)
IM/IE = RE2 / ( RE + RM )2
Yeah, that looks better. What answer do you get? Can you show your work?
 
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RE = 150.106 km
RM = 384.103 km
IE = 1360 W/m2

IM = 1353 W/m2 ?
 
  • #10
charlie05 said:
RE = 150.106 km
RM = 384.103 km
IE = 1360 W/m2

IM = 1353 W/m2 ?
Looks good to me. Do you have a way to check the answer? Is the value of the intensity at the Earth a standard value that you got from your class or textbook?
 
  • #11
value on Earth I found my physical tables..solar irradiance
 
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  • #12
thank you very much for your help :-)
 
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