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;
<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>
where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²
<em>P </em>sun = 2.863 10²³ × 1360
<em>P </em>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;
<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars
where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²
<em>I </em><em>sun-mars = </em>3.85×10²⁶W / 6.53 × 10²³m²
<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²
<em>I </em><em>sun-mars = </em>590 W/m²
