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Chapter 22: Problem 37

A \(1.0-\mathrm{m}^{2}\) solar panel on a satellite that keeps the panel oriented perpendicular to radiation arriving from the Sun absorbs $1.4 \mathrm{kJ}\( of energy every second. The satellite is located at \)1.00 \mathrm{AU}\( from the Sun. (The Earth Sun distance is defined to be \)1.00 \mathrm{AU} .$. How long would it take an identical panel that is also oriented perpendicular to the incoming radiation to absorb the same amount of energy, if it were on an interplanetary exploration vehicle 1.55 AU from the Sun? ( W) tutorial: sunlight on Io).

Short Answer

Expert verified
Answer: It takes approximately 2.405 seconds for the solar panel at a distance of 1.55 AU to absorb the same amount of energy as the one at a distance of 1.00 AU.

Step by step solution

01

Calculate the initial intensity

First, we will find the intensity of the sunlight per square meter when the solar panel is at a distance of 1.00 AU from the sun. Intensity (I) is given by the power (P) divided by the area (A). The absorbed energy is given as 1.4 kJ per second. Therefore, the power absorbed is P_absorbed = 1.4 kW. The area of the solar panel is given as A = 1.0 m². Now, we can calculate the intensity I_initial, I_initial = P_absorbed / A = (1.4 kW) / (1.0 m²) = 1.4 kW/m².
02

Calculate the ratio of intensities

Next, we need to find the ratio of the intensities when the panel is at 1.00 AU and 1.55 AU. The inverse square law dictates that the intensity is inversely proportional to the square of the distance. Thus, Intensity ratio = (I_initial / I_final) = (distance_final² / distance_initial²) = (1.55 AU / 1.00 AU)² = (1.55)² = 2.4025
03

Calculate the final intensity

Now, we can find the intensity of the sunlight when the solar panel is at a distance of 1.55 AU from the sun. I_final = I_initial / Intensity ratio = 1.4 kW/m² / 2.4025 = 0.5824 kW/m²
04

Calculate the power absorbed by the solar panel at the new distance

Next, we will calculate the power (P_final) absorbed by the solar panel when it is at a distance of 1.55 AU from the sun. P_final = I_final * A = (0.5824 kW/m²) * (1.0 m²) = 0.5824 kW
05

Calculate the time taken to absorb the same amount of energy at the new distance

Finally, we will determine the time (t) it would take for the solar panel to absorb the same amount of energy when it is at a distance of 1.55 AU from the sun. Energy_absorbed = P_initial * t_initial = P_final * t_final Using the given values of P_initial = 1.4 kW and P_final = 0.5824 kW, t_final = (P_initial * t_initial) / P_final = (1.4 kW * 1 s) / 0.5824 kW ≈ 2.405 s It would take about 2.405 seconds for the solar panel at a distance of 1.55 AU from the sun to absorb the same amount of energy as the one at a distance of 1.00 AU.

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Most popular questions from this chapter

Polarized light of intensity \(I_{0}\) is incident on a pair of polarizing sheets. Let \(\theta_{1}\) and \(\theta_{2}\) be the angles between the direction of polarization of the incident light and the transmission axes of the first and second sheets, respectively. Show that the intensity of the transmitted light is $I=I_{0} \cos ^{2} \theta_{1} \cos ^{2}\left(\theta_{1}-\theta_{2}\right)$.
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