**Question:** Your post about that Phobos eclipse was fascinating, thanks. Awesome video by Perseverance. But there was one part I didn’t quite get. When you were talking about how fast Phobos moves across the sky, you said that its elevation above the horizon affects its speed. I have a hard time visualizing why, and I couldn’t follow your explanation. Any chance you could create a diagram showing how that works? — ERB, Chicago, IL

**Answer:** Yes, my “explanation” in last week’s Phobos Annular Ellipse post was a bit clunky. Thanks for your question. I think this graphic should make it more clear. Take some time to study it. The key point is: *When observing of Phobos from the surface of Mars, its distance from the observer is a function of its elevation above the horizon. This affects both the angular size and apparent speed of the moon.
*

We can use the geometry in that graphic to learn more about the eclipse and elucidate some of the numbers I cited last week. Here’s the given orbital data for Phobos:

orbital period: T = 0.319 days = 27,562 s

orbital radius: R = 9376 km

orbital speed: V = circumference / period = 2πR / T = 2.138 km/s

angular speed: ω = circumference / period = 360° / 27,562 s = 0.013°/s

Note: The 0.013°/s angular speed is what would be seen from the *center* of Mars. Because the orbit of Phobos is circular its true speed is constant. From the surface its speed appears to vary. More on that later.

In contrast, the Sun’s angular speed through the sky is essentially constant. It’s apparent motion is a consequence of the 24.6 hour rotation period of Mars: ω = circumference / period = 360° = 360° / 88,560s = 0.0041°/s.

If we want to predict the duration of an eclipse, we need to account for their mutual motion. Since Phobos moves from west to east, and the Sun moves from east to west, their *relative speed* during an encounter would be: ω = 0.013°/s + 0.0037°/s = 0.017°/s.

The angular size of the interacting bodies also affects eclipse duration. On the date of this eclipse Mars was 2 months from perihelion, where it’s distance from the Sun is 2.07×10^{8} km. At that distance the Sun will have an angular diameter of θ = tan^{-1}(1.4×10^{6} km / 2.07×10^{8} km) = 0.39°

Because of its extreme distance from Mars, the Sun’s angular size barely changes as it moves through the sky. Not so for Phobos. Since it is *tidally locked* to Mars with its longest dimension of 27 km facing the planet, its angular diameter would be seen as:

θ = tan^{-1}(diameter / distance) = tan^{-1}(27 km / 8712 km) = 0.18° (when rising or setting)

θ = tan^{-1}(diameter / distance) = tan^{-1}(27 km / 5966 km) = 0.26° (when overhead)

The distances and angles (blue) were derived from the given geometry (yellow) in the graphic. I leave verification of those numbers as an exercise for students. Finding the angular speed of Phobos when overhead involves solving a right triangle. Getting the speed when rising or setting is a bit more complex and requires the law of cosines. We need to find values for the two θs opposite the green vectors representing the 2.138 km/s speed of Phobos. Here’s the numbers you should get:

ω = 0.013°/s (when rising or setting)

ω = 0.021°/s (when passing overhead)

So because eclipse duration depends on both angular speed and angular diameter, and because both speed and diameter depend on distance, and because distance depends on elevation above the horizon, the duration of a Phobos eclipse depends on where it happens in the sky. Eclipses are longer when the Sun is rising, and shorter when the Sun is overhead.

Want to learn more about Phobos? Watch an animation of its motion through the sky along with Mars’ smaller and more distant moon Deimos. Their orbits are distinctly unearthly:

https://sky-lights.org/2020/07/06/qa-the-unearthly-moons-of-mars/

Or explore an interactive 3D model of Phobos compiled from NASA images:

https://solarsystem.nasa.gov/moons/mars-moons/phobos/in-depth/

Next Week in Sky Lights ⇒ Conjunction of Jupiter and Venus