5
$\begingroup$

In a hypothetical solar system there exist:

  • a sun of radius $r_s$ and absolute visual magnitude $V$
  • the Earth, with radius $r_e$ and distance from the sun of $1\,AU$
  • another planet, wit radius $r_p$, albedo $a$, distance from Earth $d_e$ (in units of $AU$), and distance from the sun $d_s$ ($AU$)

Assume that the second planet is magically lit up entirely by the sun (so it does not have phases when viewed from Earth).

Now how can I compute the apparent visual magnitude of the second planet with respect to Earth?

If there are other parameters required, please tell me.

| improve this question | | | | |
$\endgroup$
  • 1
    $\begingroup$ Do you also assume that the second planet is $d$ + 1 $\rm{AU}$ from the star? $\endgroup$ – Aaron Jul 23 '14 at 2:31
  • $\begingroup$ @Aaron no, sorry. I'll edit the question. $\endgroup$ – feralin Jul 23 '14 at 14:07
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Florin Andrei Jul 24 '14 at 20:42
  • $\begingroup$ @FlorinAndrei looking at that link, how can I compute the absolute magnitude $H$ of a planet? $\endgroup$ – feralin Jul 24 '14 at 21:03
5
$\begingroup$

Lets assume we're dealing with a superior planet, which is to say, the planet is orbiting at a greater distance than the Earth. This effectively ensures that there is no planetary phase to deal with.

Now, the sun has a luminosity of $L_{sun}$ such that the Solar flux as seen by the planet is $$ f_{planet} = \frac{ L_{sun} }{ 4 \pi d_s^2 }. $$ The cross section of the planet is about $A = \pi r_p^2$, so for a given albedo $a_p$ the reflective luminosity of the planet will be \begin{aligned} L_{planet} &= a_p f_{planet} A \\ &= a_p \pi r_p^2 \frac{ L_{sun} }{ 4 \pi d_s^2 }. \end{aligned}

Given that we know the absolute magnitude of the sun, the absolute magnitude of the planet follows as \begin{aligned} V_{planet} &= -2.5 \log_{10}\left[ \frac{ L_{planet} }{ L_{sun} } \right] - V_{sun} \\ &= -2.5 \log_{10}\left[ a_p \frac{ r_p^2 }{ 4 d_s^2 } \right] - V_{sun} \end{aligned}

The apparent magnitude of the planet as seen from Earth can then be calculated as $$ m_{planet} = V_{planet} + 5 \log_{10}\left[ d_{e-p} \right] - 5 $$ where the distance between Earth and the planet, $d_{e-p}$, should be in parsec and of course depends on the orbital phase of the two respective planets.

So the one additional parameter required was the Solar luminosity. There are probably a bunch of subtle effects with he albedo and the geometry of the system that are not taken into account here, but it should be a fair approximation.

| improve this answer | | | | |
$\endgroup$
  • $\begingroup$ Thanks for answering! One question, though: can solar luminosity not be computed from the sun's absolute visual magnitude (or the other way around) because the total luminosity does not have to all be in visual light? Or is there some other reason? $\endgroup$ – feralin Jul 29 '14 at 13:29
  • $\begingroup$ @feralin I guess you can, but you'd still need a second parameter to normalize the conversion, be it either the relative magnitude of the sun or an absolute magnitude of a calibration star. Using the relative magnitude of the sun would make sense here and probably gives a more direct final expression, but I think using luminosity gives a better sense of what's physically going on. $\endgroup$ – Michael B. Jul 29 '14 at 19:16
  • $\begingroup$ @feralin As a second note; if you want to model an inferior planet you can simply make the reflective surface - A - depend on (synodic) orbital phase. $\endgroup$ – Michael B. Jul 29 '14 at 19:23
  • $\begingroup$ sorry, but can you explain what "relative magnitude" is? I'm not getting good results from google... $\endgroup$ – feralin Jul 29 '14 at 20:11
  • $\begingroup$ @feralin sorry, I meant to say apparent magnitude $\endgroup$ – Michael B. Jul 29 '14 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.