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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.

Particle king
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1 Answers1

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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}\ $

Particle king
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James K
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    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. – Greg Miller Jul 27 '22 at 18:35
  • One doubt: Can't we just apply 1/(2r)^2 – Particle king Jul 28 '22 at 07:39
  • no. That's not how fractions work – James K Jul 28 '22 at 08:05
  • @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. – Ralf Kleberhoff Jul 28 '22 at 09:43