Question

in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's intensity for an outer planet located at a distance 6 r from the sun?

Answer 1

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

What is the Inverse Square Law formula?

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

Where

• I₁ = Intensity at distance 1 (W/m²)
• I₂ = Intensity at distance 2 (W/m²)
• d₁ = distance 1 from a light source (m)
• d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

$\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}$

$\frac{200 }{I_{2} } = \frac{36r^{2} }{r^{2}}$

$I_{2} = \frac{200} {36}$

I₂ = 5.55 W/m²

Hence, the sun's intensity for a planet at a distance 6r from the sun is 5.55 W/m².

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