If I observe a planet because of reflected sun-light, how does the radiant flux vary with distance?

Question

I wish to figure out how the flux of the sun's reflected light from a planet varies with distance, so assume an observer is on the sun, looking at a planet, and the light originating from the planet is just the sun's light being reflected back.

The solar flux a planet receives is inverse squarely proportional to it's distance from the sun, but I get stuck considering how to deal with the reflection.

I originally thought I could consider the planet as a new light source, with luminosity equal to the incident flux times some function of the phase angle, i.e

$L_p = k(\theta)L_{\odot}/d^2$.

In this case, to calculate the flux as seen from the sun we can apply the inverse square law again and obtain

$F = L_p/4\pi d^2 = k(0)L_{\odot}/4\pi d^4.$

But then I considered a thought experiment: if the planets were perfect mirrors, then I'd just be seeing the sun's light but at a distance of $2d$, so the flux I'd see would be inverse squarely proportional to the distance, a different result to the $d^{-4}$ proportionality above.

Where's the flaw(s) in my reasoning?

Answer 1

Nothing wrong with your reasoning. Planets aren't plane mirrors, they are more like the first case that you discuss. The reflected light from the Sun does not come straight back to the observer, it is reflected over a range of angles.

The luminosity of the planet will actually be $$L \simeq a \pi R_p^{2} \frac{L_{\odot}}{4\pi d^2}$$ where $R_p$ is the planet radius and $a$ some mean albedo.

To then work out the flux in some direction at distance $d$, you divide the luminosity by $4\pi d^2$ and multiply by a normalised, phase angle dependent function. Complications would include whether you saw the full illuminated surface - not a problem in the case you present.

The phase angle dependent formula is not straightforward. You can go with a diffuse Lambertian reflector, which goes as the cosine of the phase angle, but most real objects exhibit an "opposition surge" phenomenon, where the reflectance is much higher than Lambertian at small phase angles (less than a few degrees).

Bottom line, the light received at the Sun would decrease as $d^4$.

Just as a check, consider Uranus and Neptune as viewed from the Earth. They have similar albedos and at their closest approach to Earth are at distances of about 19 au and 29 au respectively (and just a little further away from the Sun). At their brightest, thes planets have apparent visual magnitudes of 5.32 and 7.78 respectively. This corresponds to a flux ratio (Uranus/Neptune) of a factor of 9.6. But the ratio of their distances squared would only be 2.3, whereas distance to the fourth power would be a factor 5.4. Still doesn't quite explain the ratio, but the albedo of Uranus is in fact a little higher than that of Neptune.