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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,

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

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).

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.

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).

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.

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,

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

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).

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 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.

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,

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

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).

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.

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,

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

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).

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 -9.49, a distance 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.

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,

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

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).

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 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.