Compute Planet's Apparent Visual Magnitude

Question

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.

Answer 1

Planetary magnitudes vary not only according to the Sun’s luminosity, their own average albedo, and their distance from the Earth, but also from:

• Variations in their albedo across their surface.
• Their phase angle, for planets that we sometimes see as a crescent.
• Their inclination, for planets like Saturn and Uranus that have a different albedo at their equator than at their poles.
• Neptune keeps getting brighter. No one knows why.

All of these effects are detailed in an informative paper Computing Apparent Planetary Magnitudes for The Astronomical Almanac by Mallama and Hilton, revised in 2018:

https://arxiv.org/pdf/1808.01973.pdf

Their source code for computing planetary magnitudes given all of these effects can be found here:

https://sourceforge.net/projects/planetary-magnitudes/

Answer 2

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 parsecs 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 the albedo and the geometry of the system that are not taken into account here, but it should be a fair approximation.

Answer 3

If you want correct values, you have to take into account the effects mentioned in Brandon Rhodes' answer. Nevertheless, here's how to do a quick-and-dirty calculation.

The absolute magnitude of a planet is defined as the apparent magnitude if the Sun-planet and planet-observer distances are 1 au, at opposition.

Assuming a diffuse disc reflector model, the absolute magnitude $H$ of a planet of diameter $D_{\rm p}$ is given by

$$H = 5 \log_{10} \left( \frac{D_0}{D_{\rm p} \sqrt{a_p}} \right)$$

Where $a_p$ is the planet's geometric albedo, and $D_0$ is given by

$$D_0 = 2\,{\rm au} \times 10^{H_\ast / 5}$$

Where $H_\ast$ is the absolute magnitude defined at a reference distance of 1 au, which can be calculated from the usual 10 parsec absolute magnitude $M_\ast$ as follows:

$$H_\ast = M_\ast + 5 \log_{10} \left( \frac{1 \rm \ au}{10 \rm \ pc} \right) \approx M_\ast - 31.57$$

For the Sun, this results in $D_0 \approx 1329\ \rm km$.

To get the apparent magnitude of the planet, you can then use

$$m_{\rm p} = H + 5 \log_{10} \left( \frac{d_{\rm p\ast} d_{\rm po}}{1 \mathrm{\ au}^2} \right) - 2.5 \log_{10} q(\alpha)$$

Where $d_{\rm p\ast}$ is the distance between the planet and the star, $d_{\rm po}$ is the distance between the planet and the observer, and $q(\alpha)$ is the phase integral at the phase angle $\alpha$.

For the diffuse disc reflector, $q(\alpha) = \cos \alpha$. For a Lambertian sphere,

$$q(\alpha) = \frac{2}{3} \left( \left(1 - \frac{\alpha}{\pi} \right) \cos \alpha + \frac{1}{\pi} \sin \alpha \right)$$

At opposition, $\alpha = 0$, giving phase integrals of $q(0) = 1$ for the diffuse disc, and $q(0) = \tfrac{2}{3}$ for the Lambertian sphere.

Real planets have more complex phase functions which need to be determined empirically (see Brandon Rhodes' answer).

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As a quick check, let's plug in values for Jupiter, taken from the NASA Jupiter Fact Sheet. With a (volumetric mean) diameter of 139,822 km and a geometric albedo of 0.538, the computed absolute magnitude is -9.44, versus the actual value of -9.40.

Using the computed value of -9.44, a distance from Jupiter of 5.204 au to the Sun and 4.204 au to the Earth, and using the Lambertian sphere $q(0) = \tfrac{2}{3}$, the apparent magnitude works out at -2.30, while for a diffuse disc $q(0) = 1$ the apparent magnitude works out at -2.74. The brightest the real planet gets is around -2.94. Fortunately for me, the value comes out in the right ballpark, despite the gross approximations being made to the reflection behaviour of real planets.

My computed values using $q(0) = 1$ for Saturn, Uranus and Neptune are 0.54, 5.57 and 7.75 respectively, which aren't too far off the real values either.

Answer 4

This answer is from Fundamental Astronomy 6ed, Section 7.8

The first several steps are the same with @Brandon Rhodes

We got: ($A$ is the Bond albedo)

$$ F = AC\Phi(\alpha) \frac{1}{4\pi\Delta^2} \pi R^2 \frac{L_\odot}{4\pi r^2} $$

The difference is this normalisation constant $C$:

Since all the radiation reflected from the planet is found somewhere on the surface of the sphere, we must have: ($\mathrm{d}S = \Delta^2\mathrm{d}\Omega$)

$$ \frac{C}{4\pi\Delta^2} \int_S \Phi(\alpha)\,\mathrm{d}S = 1 $$

$$ C = \frac{2}{\int_0^\pi \Phi(\alpha)\sin\alpha\,\mathrm{d}\alpha} $$

We denote the phase integral:

$$ q = 2\int_0^\pi \Phi(\alpha)\sin\alpha\,\mathrm{d}\alpha = \frac{4}{C} $$

and geometric albedo:

$$ p = \frac{AC}{4} $$

For a Lambertian surface (defined as $A = 1$):

$$ \Phi(\alpha) = \begin{cases}& \cos\alpha,\quad 0\leq\alpha\leq\pi/2 \\ & 0,\quad \text{otherwise}\end{cases} $$

which leads to:

$$ C = 4,\quad A = 1,\quad p = q = 1 $$

If we replace the planet with a Lambertian surface of the same size, we get:

$$ F_L = 4 \Phi(\alpha) \frac{1}{4\pi\Delta^2} \pi R^2 \frac{L_\odot}{4\pi r^2} $$

the geometric albedo (which we see on wikipedia etc.) is the ratio of the flux densities at phase angle $\alpha = 0$ reflected by a planet and a Lambertian surface of the same cross section.

$$ p = \frac{F}{F_L} $$

and finally we have:

$$ F = p\Phi(\alpha) \frac{R^2}{\Delta^2} \frac{L_\odot}{4\pi r^2} $$

which is different of a $4$ factor from the first equation, that is because the normalisation constant $C = 4$.

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I ran into this problem as I am working on the albedo of the Moon. I saw the last equation, but I do not know why the $4$ is missing.

But if I calculate using the first equation, I came to the Moon's albedo is $\sim 0.45$, about four times the correct data ($\sim 0.13$).

I use the apparent magnitude of the full moon (-12.74mag) and the sun (-26.83mag), the solar luminosity (3.828e+26W), the moon's radius (1738km), the moon's distance to earth (384400km) and the sun (1.5e+8km).