Question

Let’s consider several models of the Earth’s radiation properties. In each part, assume that the Earth's radiative output balances the input radiation from the Sun, and that there are no other energy or heat sources. 1) Consider a 'far Earth' (FE) that has an orbital distance 1.6 times farther from the sun, but the same diameter as the real Earth (RE). Assume that both are blackbodies with a uniform temperature. What is the ratio of temperatures between those two planets? (Note: you won't need to look up the Earth-Sun distance or the diameter of the Earth) TFE/TRE=

Answer 1

0.391

Step-by-step explanation:

Since they both have the same diameter, the temperature of both earth's surfaces will be proportional to the solar intensity incident on each surface.

The solar intensity is inversely proportional to the square of their distance.

For true earth Te, distance from the sun is r.

For far Earth Fe, distance from the sun is 1.6r

Solar intensity = power p ÷ area of propagation.

Area of propagation = 4¶r^2 (r is their distances from the sun)

Te = P/(4 x 3.142 x r^2) = p/12.568r^2

TTE is proportional to p/12.568r^2

Fe = p/4¶(1.6r)^2 = p/(4 x 3.124 x 2.56 x r^2) = p/32.174r^2

TFE is proportional to p/32.174r^2

Ratio of TFE/TTE =

p/32.174r^2 ÷ p/12.568r^2

= 32.174 ÷ 12.568 = 0.3906

Approximately 0.391