If you threw something over a certain cliff on a moon of Uranus, it would take over ten minutes to hit bottom.

By this measure, local fall time rather than absolute height, what are the near contenders to the prize for tallest cliff on a body in hydrostatic equilibrium in the solar system? Throwing the object horizontally at a few meters per second is ok. Not ok is throwing it beyond a tiny body's Hill sphere. Cliffs, not spheres.

Simplifying assumptions about atmospheres and exospheres are ok.

  • $\begingroup$ I'm not sure what "By this measure..." means. Are you asking for the "tallest" cliffs where "height" is measured in local fall time rather than in distance? $\endgroup$
    – uhoh
    Sep 2, 2019 at 3:44
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    $\begingroup$ Exactly. I've edited my question and your answer, to reflect that. $\endgroup$ Sep 2, 2019 at 4:57
  • $\begingroup$ Can we throw it fast enough so it lands at the bottom of the mountain? $\endgroup$ Sep 2, 2019 at 4:58
  • $\begingroup$ Only if the mountain is steep enough. Olympus Mons is too gentle a slope. That's why I said a few m/s. What's possible for a human in a not too fancy spacesuit. $\endgroup$ Sep 2, 2019 at 5:01
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    $\begingroup$ Perhaps you could limit this to bodies in hydrostatic equilibrium. For "potatoes", "needles" or "dust", I'm not sure that the idea of a cliff makes much sense. This would also avoid the need to worry about "throwing out the Hill sphere" and so on. If so, then the cliffs on Miranda probably take the prize. $\endgroup$
    – James K
    Sep 2, 2019 at 6:53

2 Answers 2


Let's apply @uhoh's answer's equation to a few bodies. Numbers come from the obvious Wikipedia pages unless otherwise stated. G = 6.6741e-11, in kg-m-s.

Verona Rupes on Miranda.
f = 20 km / 236 km, ρ = 1200 kg/m^3. So t = 711 s.

Enceladus has rifts 1 km deep.
f = 1 km / 252 km, ρ = 1600. So t = 133 s.

Those are the two bodies with smallest mean radius, of all solar system bodies in hydrostatic equilibrium.

Let's try a much smaller eccentric body, just to check.
The nucleus of the comet Wild 2 has cliffs "hundreds of meters tall" (Ulivi's "Robotic Exploration of the Solar System," part 3, p. 228).
f = 200 m / 6400 m, ρ = 600. So t = 610 s.

So Miranda's Verona Rupes beats even a comet. Compared to Miranda, any other round body is much bigger (f is much smaller) and significantly denser (ρ is bigger), so to beat this it would need truly brobdingnagian cliffs, which flybys would have seen by now at least with radar if not with visible light.

I could grind through rest of that table of bodies in h.eq., but it's a safe bet that Verona Rupes wins by a long shot, Enceladus is a distant second, and everything else is a giant hardly differentiated peloton.

  • $\begingroup$ didn't see your answer until today, +1 $\endgroup$
    – uhoh
    Dec 9, 2020 at 4:51

This answer shows that the close orbital period around a spherical body of uniform density is

$$T = \sqrt{\frac{3 \pi}{G\rho}}$$

and so the orbit's period $T$ is defined only by the body's density $\rho$, not its size. It takes about 90 minutes to orbit the Earth in a low orbit, and it would likewise take 90 minutes for an atom to orbit a spherical speck of dust with a similar average density to that of Earth. (uncharged, unpolarizable dust, ignoring other forces e.g. Coulomb, Van der Waals, Casimir, etc.)

Let's compare cliffs on a spherical planet, small asteroid, and piece of dust. Assume the height of each cliff divided by the radius of the object is the same, and call the fraction

$$f = \frac{h}{R}.$$

So if $f=$ 1%, then that's 64 km on Earth, 64 m on a 1 km radius asteroid, and 10 microns on a 1 mm radius piece of dust.

The fall time is

$$t=\sqrt{\frac{2h}{a}} = \sqrt{\frac{2fR}{a}}$$

where $a$ is the gravitational acceleration, keeping it simple by assuming it's constant though it changes slightly as we fall from such a great height.

$$a = \frac{GM}{R^2}$$

$$M = \frac{4}{3} \pi \rho R^3$$

$$a = \frac{4}{3} \pi G \rho R$$


$$t = \sqrt{\frac{3f}{2\pi G \rho}}$$

For spherical bodies of the same density with cliffs of heights as a fixed fraction of radius, the time it takes to fall down the cliff is independent of the size of the body, and varies as the inverse square root of the density of the body.

For a $0.01 R$ cliff on a body of density 5.51 g/cm^3 (Earth's average density) that works out to 114 seconds, or about 2 minutes, and compares to 5063 seconds or 84 minutes for an orbit that skims the surface.

  • $\begingroup$ So we should look for icy rubble piles without rocky cores, that have been whacked hard yet survived more or less intact. Something really prolate, in Saturn's rings? $\endgroup$ Sep 2, 2019 at 4:55
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    $\begingroup$ There are estimates of how high mountains can be. It is not "height of each cliff divided by the radius of the object is the same". $\endgroup$ Sep 2, 2019 at 4:57
  • $\begingroup$ Good principle, but crusts other than SiO2 will have a different latent heat of fusion aka melting. In particular, Miranda's SiO2 is mostly in its core, not its icy crust. $\endgroup$ Sep 2, 2019 at 5:07
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    $\begingroup$ @KeithMcClary you make a good point. The paper cited in How did early estimates of a “potato radius” set 1 eV ~ GMμ/R and get 200 to 300 km? uses some simple principles to estimate mechanical stability of bodies based on very simple assumptions, ignoring the differences between specific materials (e.g. ice vs rock) and I think there could really be simple expression for maximum cliff height versus body radius, and that would indeed be better to use than my fraction $f$. Consider writing another answer based on better science? $\endgroup$
    – uhoh
    Sep 2, 2019 at 5:24
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    $\begingroup$ I still hope for measurements. But if theory says Verona Rupes is several standard deviations beyond the mean, and it's years until new measurements, then I'll accept a theory answer! $\endgroup$ Sep 3, 2019 at 2:17

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