Question

The sun is 1.5 x 1011 m away from earth. About 1370 W/m2 of electromagnetic radiation is incident on the earth from the sun. Assuming that the sun emits radiation isotropically:

a. How much energy per unit time does it emit?
b. How much power is incident on a rectangular detector on the surface of the earth with dimensions 1.5 m x 3m?

Answer 1

The sun emits a total power output which can be calculated using the solar constant and the distance from the sun to the earth. The power incident on a given detector area is derived by multiplying the solar constant with the area of the detector.

Step-by-step explanation:

The energy per unit time that the sun emits, assuming it emits radiation isotropically, can be deduced from the solar constant and the distance from the sun to the earth. With the solar constant being approximately 1370 W/m² and the distance to the sun being 1.5 × 10±1 m, we can use the equation P = 4πr² × intensity to find the total power output.

Power received by the detector: The power incident on a rectangular detector with dimensions 1.5 m x 3 m can be found by multiplying the detector's area by the solar constant (1370 W/m²). With an area of 4.5 m², the detector would receive a power of 4.5 m² × 1370 W/m² which equals 6165 W.