# What is wrong with my calculations of Jupiter's apparent magnitude?

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?

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.