I was searching for a formula to calculate the apparent magnitude of a planet knowing its physical characteristics and stumbled upon this answer.

The final formula given is:

$$m_{planet} = V_{planet} + 5 \log_{10}\left( d_{e-p} \right) - 5$$

In which,

$$ V_{planet}=-2.5 \log_{10}\left(a_p \frac{ r_p^2 }{ 4 d_s^2 } \right) - V_{sun}$$


  • $m_{planet}$ is the planet in question's apparent magnitude,

  • $V_{planet}$ is the planet's absolute magnitude,

  • $a_{p}$ is the bond albedo of the planet,

  • $r_{p}$ is the radius of the planet,

  • $d_{s}$ is the distance of the planet from its star,

  • $d_{e-p}$ is the distance between the planet and the observer (here on Earth) in parsecs, and,

  • $V_{sun}$ is the star's absolute magnitude.

To test it out I tried to calculate the apparent magnitude of Jupiter as seen from Earth at an average opposition. The values then are:

  • $a_{p} = 0.343$,

  • $r_{p} = 69,911,000$ m,

  • $d_{s} = 778.57 \times 10^9$ m (5.204 AU),

  • $d_{e-p} = 628.97 \times 10^9$ m $= 0.0000203835294968$ parsecs (4.204 AU)

  • $V_{sun} = 4.83$

All the above values were taken from NASA's Jupiter fact sheet.

Plugging in the numbers and I get $m_{planet} = -10.38$, which is much brighter than the actual apparent magnitude of Jupiter at average opposition of $-2.x$. Where did I go wrong?


I think that should be $+ V_{Sun}$ in your second equation, not $- V_{Sun}$. Otherwise, a fainter Sun (larger magnitude) would give you a brighter Jupiter (smaller magnitude). Those need to go in the same direction.

Applying that change (assuming the rest of the calculation is right) would increase your answer by $2 \times 4.83 = 9.66$, which gets you much closer to the right magnitude.

Your estimate will now be a bit on the faint side - not sure why - but at least a lot closer than before.

  • $\begingroup$ Yes, I think so too. But a >2 magnitude difference is a tiny bit too much. I wonder if there's anything else wrong. $\endgroup$ – ChristieToWin Aug 8 '20 at 12:48
  • $\begingroup$ If you look at the answer after the one with that formula, it mentions other factors that determine the apparent magnitude of a planet. $\endgroup$ – Eric Jensen Aug 8 '20 at 12:50

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