When viewed from the sun, the brightness of the planet with a given size and albedo changes according to the fourth power of inverse distance.
I found the statement in one encyclopedia but I can't find any mathematical proof of this.
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$\begingroup$ Look for "inverse square law" $\endgroup$– Greg MillerJul 27, 2022 at 15:42
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$\begingroup$ Physics doesn't prove. It measures ;) $\endgroup$– planetmakerAug 1, 2022 at 16:45
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1 Answer
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The brightness of reflected light of the planet depends on the brightness of the incident light and the distance (d) from the planet to the observer.
Light obeys an inverse square law (think of light as spreading out in a sphere) So the brightness of the planet is inversely proportional to the distance from the planet to the observer (B is proportional to $\frac{1}{d^2}\ $) (assuming things like the planet is fully illuminated etc)
But the brightness of the incident light also obeys an inverse square law with respect to the distance from the planet to the sun (r). The brightness of the incident light is proportional to $\frac{1}{r^2}\ $
So the brightness is proportional to $\frac{1}{d^2}\ $×$\frac{1}{r^2}\ $ But if your observer is near the sun, then d=r and so the brightness is proportional to $\frac{1}{r^4}\ $
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1$\begingroup$ To paraphrase: as a planet moves, $ \frac{1}{r^2}\ $ light from the Sun reaches the planet. Then $ \frac{1}{r^2}\ $ is of the reflected light reaches back to the Sun. $\endgroup$ Jul 27, 2022 at 18:35
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$\begingroup$ One doubt: Can't we just apply 1/(2r)^2 $\endgroup$ Jul 28, 2022 at 7:39
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$\begingroup$ @Particleking 1/(2r)^2 would be correct for a big (at least half sun sized) flat mirror at distance r. But (unlike a mirror) a planet disperses the light in all directions, so it has to be treated as a new light source. $\endgroup$ Jul 28, 2022 at 9:43