# Trappist-1 planet albedo

Is the albedo of any of the Trappist-1 planets know? Is it high enough so if you are on the surface of one of the planets, you can see the others?

I looked up the distances from the star and size of the planets and it seems a few of them will appear larger than the Moon in the night sky, unless the albedo is really low that is.

• I think that the consensus is that they are all close enough together and large enough in size that you'd be able to see them regardless of their albedos. – Phiteros Feb 25 '17 at 3:31

## 1 Answer

So far there have not been any published measurements of the albedos of the TRAPPIST-1 planets. While albedos have been measured for some exoplanets, the planets in question are irradiated far more severely than even the innermost TRAPPIST-1 planets.

Using the formulae in my answer to this question about computing planetary apparent magnitudes, we can make a back-of-the-envelope check. The definition of absolute magnitude becomes the apparent magnitude of a planet as the magnitude it would have at opposition if it were located 1 au from TRAPPIST-1, and the observer located 1 au from the planet.

Defining the absolute magnitude of the TRAPPIST-1 planets as the apparent magnitude if the planet were 1 AU from TRAPPIST-1, and the observer were 1 AU from the planet, the uniform disc model gives the following formula for the absolute magnitude $$H$$:

$$H = 5 \log_{10} \left( \frac{671\,000\ \rm km}{D_{\rm p} \sqrt{A_{\rm g}}} \right)$$

Where $$D_{\rm p}$$ is the diameter of the planet and $$A_{\rm g}$$ is the geometric albedo. The 834,000 km value is the equivalent of the 1,329 km value used for asteroid diameter estimation in our Solar System, adjusted for the fainter absolute magnitude of TRAPPIST-1 (using the V-magnitude 18.798 and parallax 80.4512 mas from SIMBAD, I calculate an absolute magnitude of 18.33).

The least favourable case for visibility of an outer planet from an inner one would be for planet h viewed from planet b: TRAPPIST-1h is both the smallest and outermost of the TRAPPIST-1 planets. To pick an unfavourable case for visibility, let's use a geometric albedo of 0.05 similar to the P-type asteroid 46 Hestia. Using a diameter of 0.752 times the Earth (NASA Exoplanet Archive), the absolute magnitude works out as roughly 12.5. The apparent magnitude is then given by

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

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

Again using the uniform disc model which gives $$q(0) = 1$$, the apparent magnitude of h from b, when b and h are located on the same side of TRAPPIST-1 is about -0.2 for the Hestia-like case, which is similar to Saturn viewed from Earth. On the other hand, the angular diameter in this configuration is about 3.7′, so unlike Saturn it would be visible as a disc rather than a point.