For the sake of simplicity, let’s say that the solar flux at the moon’s distance from the sun is the same as the solar flux at Earth’s distance from the sun.
a) The moon has essentially no atmosphere, and “moonlight” is solar flux reflected by the moon. What is the average intensity of “moonlight” immediately above the illuminated hemisphere of the moon? For the sake of simplicity, let’s assume that “moonlight” is reflected with even intensity from the entire illuminated hemisphere of the moon. Write an expression, then solve. (hint: We know that the area of a disk, such as the moon’s disk, is Pi*r2, and that the surface area of a sphere is 4Pi*r2. So, what’s the surface area of a hemisphere relative to a disk?)
b) Starting with the value you calculated for (a), calculate the flux of “moonlight” at the Earth-moon distance from the moon. Continue, and calculate the flux of “moonlight” after it goes through the atmosphere. (Tip: Light that passes through the atmosphere is light that is not reflected by the atmosphere.)
an example of solubility is salt or sugar dissolving in water.
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