Ask Question
20 December, 16:11

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the Earth and the Sun is 1.5 x 1011 m. a) Assuming it radiates uniformly in all directions what is the total power output of the Sun? b) If the frequency increases by 1 MHz what would be the relative (percentage) change in the power output? c) For frequency in b) what is the intensity of the radiation from the Sun measured on Mars? Note that Mars is 60% farther from the Sun than the Earth is.

+5
Answers (1)
  1. 20 December, 16:25
    0
    Answer: (a) power output = 3.85*10²⁶W

    (b). There is no relative change in power as it is independent from frequency

    (c). 590 W/m²

    Explanation:

    given Radius between earth and sun to be = 1.50 * 10¹¹m

    Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

    Frequency = 60 MHz

    (a). surface area A of the sun on earth is = 4πR²

    substituting value of R;

    A = 4π (.50 * 10¹¹) ² = 2.863 10²³*m²

    A = 2.863 10²³*m²

    now to get the power output of the sun we have;

    P sun = I sun-earth A sun-earth

    where A = 2.863 10²³*m², and I is 1360 W/m²

    P sun = 2.863 10²³ * 1360

    P sun = 3.85*10²⁶W

    (c). surface area A of the sun on mars is = 4πR²

    now we substitute value of 2.28 * 10¹¹ for R sun-mars, we have

    A sun-mars = 4π (2.28 * 10¹¹) ²

    A sun-mars = 6.53 * 10²³m²

    now to calculate the intensity of the sun;

    I sun-mars = P sun / A sun-mars

    where P sun = 3.85*10²⁶W and A sun-mars = 6.53 * 10²³m²

    I sun-mars = 3.85*10²⁶W / 6.53 * 10²³m²

    I sun-mars = 589.6 ≈ 590 W/m²

    I sun-mars = 590 W/m²
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the Earth and ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers