Question

The intensity of the sunlight that reaches Earth's upper atmosphere is approximately \(1400 \mathrm{W} / \mathrm{m}^{2} .\) (a) What is the total average power output of the Sun, assuming it to be an isotropic source? (b) What is the intensity of sunlight incident on Mercury, which is $5.8 \times 10^{10} \mathrm{m}$ from the Sun?

Short answer

Question: Calculate (a) the total average power output of the Sun and (b) the intensity of sunlight incident on Mercury. Answer: (a) The total average power output of the Sun is \(3.8 \times 10^{26} \mathrm{W}\). (b) The intensity of sunlight incident on Mercury is \(9.1 \times 10^{3} \mathrm{W} / \mathrm{m}^{2}\).

Step-by-step solution

(a) Calculate the total average power output of the Sun

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: \(P_{Sun} = I_{Earth} \times 4 \pi r_{Earth}^2 = 1400 \times 4 \pi (1.5 \times 10^{11})^2\) \(P_{Sun} = 3.8 \times 10^{26} \mathrm{W}\)

(b) Calculate the intensity of sunlight incident on Mercury

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 \(9.1 \times 10^{3} \mathrm{W} / \mathrm{m}^{2}\).