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