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The intensity of the sunlight that reaches Earth's upper atmosphere is approximately 1400 W/m2.

What is the intensity of sunlight incident on Mercury, which is 5.7e10 m from the Sun?

2006-11-17 06:05:01 · 2 answers · asked by p_rob22 1 in Science & Mathematics Mathematics

2 answers

The light intensity is inversely proportional to the SQUARE of the distance.

Use this formula:

Earth sunlite * Earth dist. ^ 2 = Mercury sunlite * Mercury dist. ^ 2

1400 * Earth dist. ^ 2 = Mercury sunlite * 5.7e10 ^ 2

You use multiplication, not a proportion, because the relationship is inverse.

Radagast is wrong in three ways. First, the proportion is INVERSE, not direct, so we use multiplication, not division. Second, the relationship depends on the square of the distance, not the distance itself. And third, he means similar triangles, not congruent ones.

Radagast's formula will give you lower light intensity on Mercury than on Earth, clearly nonsense.

2006-11-17 06:18:57 · answer #1 · answered by ? 6 · 1 0

Think of this as two congruent triangles, where one side is the distance from the sun, the other, the intensity of sunlight.

So, LightIntensity(earth) / distance from sun(earth) = LightIntensity(mercury) / distance from sun(mercury)

2006-11-17 06:21:19 · answer #2 · answered by Radagast97 6 · 0 1

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