Question

name:calculating solar energypractice activitypurpose• to determine the actual amounts of solar energy at various locations in the solar system.part 1: solar energy vs distance in space:the sun gets dimmer the farther you move from it, so less solar energy is available as you move farther from the sun. we will use mercury as an example.we will use earth as our standard, and we know that the earth receives 1388 w/m2 (watts per square meter) of incoming sunlight. we will use the size of the iss solar panels area of 3,360 m2 as a standard for comparison.| planet | distance from sun in au | fraction of sunlight received (compared to earth) | fraction of sunlight received x earth standard (1388 w/m2) | sunlight received x area of solar panels = solar power available to space station | solar cells are only about 15 efficient so we must multiple the solar power times 0.15 ||---------|--------------------------|---------------------------------------------------|-----------------------------------------------------------|----------------------------------------------------------------------------------|---------------------------------------------------------------------------------------|| mercury | 0.4 | $1/(0.4)^2 = 6.25$ | $6.25 \times 1388 = 8,675$ w/m2 | $8,675$ w/m2 $\times 3,360$m2 = 29,148,000 w or 29,148 kw | $29,148$ kw $\times 0.15 = 4,372$ kw || venus | 0.72 | | | | || mars | 1.5 | | | | |

Explanation:

Step1: Calculate Venus's sunlight fraction

$\frac{1}{(0.72)^2} \approx 1.93$

Step2: Find Venus's sunlight intensity

$1.93 \times 1388 = 2678.84 \, \text{W/m}^2$

Step3: Compute Venus's total solar power

$2678.84 \times 3360 = 8,900,902.4 \, \text{W} = 8900.90 \, \text{kW}$

Step4: Find Venus's usable solar power

$8900.90 \times 0.15 \approx 1335.14 \, \text{kW}$

Step5: Calculate Mars's sunlight fraction

$\frac{1}{(1.5)^2} \approx 0.44$

Step6: Find Mars's sunlight intensity

$0.44 \times 1388 = 610.72 \, \text{W/m}^2$

Step7: Compute Mars's total solar power

$610.72 \times 3360 = 2,052,019.2 \, \text{W} = 2052.02 \, \text{kW}$

Step8: Find Mars's usable solar power

$2052.02 \times 0.15 \approx 307.80 \, \text{kW}$

Answer:

Planet	Distance from Sun in AU	Fraction of sunlight received (compared to Earth)	Fraction of Sunlight received x Earth standard (1388 W/m2)	Sunlight received x area of solar panels = solar power available to space station	Solar cells are only about 15% efficient so we must multiple the solar power times 0.15
Mars	1.5	$\approx 0.44$	$\approx 610.72 \, \text{W/m}^2$	$\approx 2052.02 \, \text{kW}$	$\approx 307.80 \, \text{kW}$