91Ó°ÊÓ

In order to find the total average power output of the Sun, we will use the formula for intensity: \(I_{Earth} = \frac{P_{Sun}}{4 \pi r_{Earth}^2}\) Where \(I_{Earth}\) is the intensity at Earth's upper atmosphere, \(P_{Sun}\) is the total power output of the Sun, and \(r_{Earth}\) is Earth's distance from the Sun. From the exercise, we have: \(I_{Earth} = 1400 \mathrm{W} / \mathrm{m}^{2}\) \(r_{Earth} = 1.5 \times 10^{11} \mathrm{m}\) (average distance between the Earth and the Sun) Now we can solve for the total average power output of the Sun: \(P_{Sun} = I_{Earth} \times 4 \pi r_{Earth}^2 = 1400 \times 4 \pi (1.5 \times 10^{11})^2\) \(P_{Sun} = 3.8 \times 10^{26} \mathrm{W}\)

Now we will use the calculated power output to find the sunlight's intensity incident on Mercury using its distance from the Sun: \(I_{Mercury} = \frac{P_{Sun}}{4 \pi r_{Mercury}^2}\) Where \(r_{Mercury}\) is Mercury's distance from the Sun. From the exercise, we have: \(r_{Mercury} = 5.8 \times 10^{10} \mathrm{m}\) Now we can solve for the intensity of sunlight incident on Mercury: \(I_{Mercury} = \frac{3.8 \times 10^{26}}{4 \pi (5.8 \times 10^{10})^2}\) \(I_{Mercury} = 9.1 \times 10^{3} \mathrm{W} / \mathrm{m}^{2}\) In summary, (a) the total average power output of the Sun is \(3.8 \times 10^{26} \mathrm{W}\) and (b) the intensity of sunlight incident on Mercury is \(9.1 \times 10^{3} \mathrm{W} / \mathrm{m}^{2}\).