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Is it possible that a moon has a higher surface gravity than its planet? I guess it would mean that the moon has a higher mass, but then it would be the planet gravitating around the moon and the roles would be exchanged...

Still, is there a way?

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    $\begingroup$ So far as I know there's no hard and fast definition of a moon. Though a proposed criteria is a barycenter beneath the surface. If that's the case I suppose it's possible for a more massive and very dense "moon" to orbit about the barycenter beneath the surface of a low density "planet". $\endgroup$ – HopDavid Dec 2 '19 at 16:26
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    $\begingroup$ Imagine two bodies of exactly equal mass orbiting eachother about the midpoint in between them (theoretically possible though astronomically unlikely). Which one is the "moon" and which is the "planet" in this case? It becomes a question of how you define those words. We use "moon" to refer to the less massive of the two, but there's essentially no difference. The moon does not orbit the Earth, they both orbit a common point at the center of their combined mass (which happens to be near - but not quite at - the center of the Earth due to it being much more massive). $\endgroup$ – Darrel Hoffman Dec 3 '19 at 16:04
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    $\begingroup$ You have conflicting answers because the word "gravity" is not specific enough. Do you mean surface gravity (that which people would experience on the object) or gravitational force for any given distance from the planets centre? $\endgroup$ – UKMonkey Dec 3 '19 at 18:19
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    $\begingroup$ What if the moon us a mini black hole (whatever we count as "surface gravity" then $\endgroup$ – Hagen von Eitzen Dec 4 '19 at 19:24
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    $\begingroup$ Tangentially related (pun intended): if the "planet" is spinning fast then surface acceleration at its equator can be greatly reduced, making apparent gravity much lower than on the "moon". $\endgroup$ – Brian B Dec 4 '19 at 19:48
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Given a pair of objects that are gravitationally bound to each other, they will orbit around their common barycenter (center of mass of the system). The object to be most logically deemed the moon will be the one of lesser mass because it will be further from the barycenter than its companion.

For example, Pluto has a gravitationally bound companion named Charon. Because of the distance between them and their relative masses, both bodies orbit a point between them. Because Pluto is the more massive of the pair, that point is closer to Pluto than to Charon so it makes sense that Charon should be deemed to be a moon of Pluto.

Earth similarly orbits a barycenter it shares with the Moon, but the Earth-Moon barycenter falls within the body of the Earth (about 3/4 the distance from its center to its surface).

So, the "moon" will be the object of lower mass.

Whether a moon can have a higher surface gravity will depend upon the densities of the two objects. To be the secondary ("moon"), it would need to be a least a little less massive than its primary, but to have higher surface gravity, it would need to be more dense. One possible case might be a primary made of water ice and a secondary made of rock. At higher density, the object would have a smaller radius for its mass, placing objects at its surface closer to its center, which increases gravitational attraction.

Some simple math:

Mass will be proportional to density x radius3; surface gravity will be proportional to mass / radius2.

Let's consider two homogeneous spherical objects of equal mass, but one is half the radius of the other. The smaller one would have (1/(1/2))2 = 4 times the surface gravity of the larger one. Now, let's make the larger one ice (density = 1) and the smaller one silicate rock (density about 3). That will make the smaller one about 3 x (1/2)3 = 3/8 the mass of the larger one, but at half the radius, its surface gravity will be 4 x 3/8 = 1.5 times that of the larger one.

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    $\begingroup$ -1. This is an interesting explanation, but it doesn't answer the question. $\endgroup$ – Dubu Dec 3 '19 at 9:20
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    $\begingroup$ I think it does imply the answer without stating "the moon is the one further from the barycenter" so if there was a small superdense moon and a large but light planet, then we'd define it to be a small superdense planet and a large fluffy moon. $\endgroup$ – Criggie Dec 3 '19 at 9:41
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    $\begingroup$ +1 for (apparently) forcing people to read and understand the text to get the answer. $\endgroup$ – Jarrod Christman Dec 3 '19 at 12:38
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    $\begingroup$ As the other answers show, the moon, the object with lower mass, can have a higher surface gravity. $\endgroup$ – prosfilaes Dec 4 '19 at 1:03
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    $\begingroup$ The question has been edited to specifically ask about surface gravity, which means this answer is no longer correct. $\endgroup$ – David Dec 4 '19 at 3:48
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Gravity isn't just about mass, but about distance, too.

Our moon has a surface gravity of about 1/6th of Earth, because it is small and less dense than the Earth is. Surface gravity of a body is inversely proportional to the square of its radius, holding mass constant. That means that if you compressed the moon such that it was $\frac{1}{\sqrt{6}}$th of its current radius, it would have the same surface gravity as the Earth even though its mass woudn't have changed at all.

It would have to have a density of about 50 tonnes per cubic metre though, and that's heavier than any normal material so this situation couldn't arise around Earth. You'd need to arrange for a very dense metallic moon to orbit a small or very low density planet... perhaps one mostly made of ice or water. It would be a little surprising to have this arrangement, but not actually impossible. Just unlikely.

As an example, you could imagine a planet a bit like Callisto, which has a surface gravity of about 1/8th of Earth despite its size due to being made largely of ice and rock. A spherical moon with a radius of 200km made of iridium would have a slightly higher surface gravity, but still weigh less than 1/150th of its parent planet. The barycenter of the system will still be comfortably within the radius of Callisto for a plausible orbital distance of the metal moon... for a 130000km orbit, the barycenter will be about 854km from the centre of Callisto, leaving the pair with less "wobble" than the Earth-Moon system. Seems fairly convincingly a planet-moon relationship, rather than a binary planet. At least to me, anyway.

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    $\begingroup$ @CJDennis The Earth also has 83 times more mass. The moon has 1.2% the mass, yet manages 16.7% the surface gravity. $\endgroup$ – Draco18s no longer trusts SE Dec 3 '19 at 4:19
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    $\begingroup$ @CJDennis this is why I said proportional not equal. $\endgroup$ – Starfish Prime Dec 3 '19 at 8:24
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    $\begingroup$ @CJDennis g=mG/r² : So, gravity is directly proportional to both Mass and the Gravitational Constant, but inversely proportional to the square of the radius (or distance from the centre of mass - gravity is lower at the top of Mt Everest, but by less than 0.5%) The question you forgot to ask yourself is "If it is proportional to the square of the radius, then what is that a proportion of?" $\endgroup$ – Chronocidal Dec 3 '19 at 13:18
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    $\begingroup$ Fun fact: Surface gravity of a sphere is linearly proportional to radius (and density). Gravity is based on mass divided by distance squared. Mass equals density * volume. Volume of a sphere is based on radius cubed. Therefore surface gravity is proportional to radius cubed divided by radius squared = radius^1. $\endgroup$ – Foo Bar Dec 3 '19 at 16:03
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    $\begingroup$ @FooBar if you hold density constant, yes, if you hold mass constant, then density is inversely proportional to the cube of radius, and so gravity becomes inversely proportional to the square; the two are the same. Though I'd agree that it's might be more practical physically to hold density more-or-less constant when constructing your moon. $\endgroup$ – Pete Kirkham Dec 3 '19 at 16:47
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Yes, it is.

Given two spherical, uniform, bodies one with mass $m_1$ and radius $r_1$ and the other with mass $m_2$ and radius $r_2$, then the surface acceleration due to gravity will be equal when

$$r_2 = \sqrt{\frac{m_2}{m_1}} r_1$$

For the Moon to have the same surface gravity as the Earth, we can plug in suitable numbers, and you end up with a radius for the Moon of $707\,\mathrm{km}$. The actual radius of the Moon is $1737\,\mathrm{km}$.

So if you got some kind of huge crushing machine and squashed the moon down to about 6% of its current volume, then it would have surface gravity equal to the Earth's.

Sadly, I can't find an element dense enough to make the Moon out of for this to be true. The current density of the moon is about $3344\,\mathrm{kg\,m^{-3}}$: to get radius small enough its density needs to be about $49710\,\mathrm{kg\,m^{-3}}$. The densest element I can find is osmium, which is $22590\,\mathrm{kg\,m^{-3}}$, so that's disappointing.

On the other hand, there would be some compression due to gravitation itself.

And of course, this points to the correct mad-scientist approach to this problem. Simply take the Moon and keep compressing it. Eventually you will construct a small black hole, with a Schwarzschild radius of about a tenth of a millimetre. This thing has a surface gravity as high as you like.

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    $\begingroup$ Surface gravity of Uranus is less than surface gravity of Earth, so an Earth sized moon orbiting Uranus would work, without resorting to osmium or black holes. $\endgroup$ – James K Dec 2 '19 at 22:18
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    $\begingroup$ @JamesK - This comment (or an expanded version) ought to be the accepted answer. It's valuable to have an actual answer with a believable concrete example. $\endgroup$ – Mark Foskey Dec 3 '19 at 2:54
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    $\begingroup$ On the way to a black hole, there is white dwarf material (10^4 to 10^7 g/cm3) and neutron star material (4×10^17 kg/m3). Whether it would stable when only the Moon's mass is another matter (pun intended). $\endgroup$ – Peter Mortensen Dec 3 '19 at 3:51
  • $\begingroup$ @JamesK yes, I agree with Mark Foskey, you should give that answer. I was just trying to give the most absurd answer I could think of, really! $\endgroup$ – tfb Dec 3 '19 at 10:05
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    $\begingroup$ @JamesK was thinking of a similar example. A Neptune-sized moon orbiting a Saturn would also work. Curiously, Uranus and Venus have the same surface gravity. Mercury orbiting Mars is also an example (extremely close in gravity) $\endgroup$ – Chieron Dec 3 '19 at 11:33
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Yes, it is possible. As James K observed in a comment, the surface gravity of Uranus is slightly less than that of Earth, but its mass is 14 times larger. If Earth were orbiting Uranus, it would be a very large moon, but it would still be considered a moon, and thus a moon with a higher surface gravity than its planet.

The reason this is possible is that the "surface" is much farther from the center of Uranus than Earth's surface is from its center.

If you insist that both bodies be solid, so you aren't calling the cloudtops the "surface" as we do with Uranus, then it's still possible, but the masses couldn't be as different. Chieron mentions Mercury and Mars. If Mercury were slightly more dense, or Mars slightly less so, then Mercury would have a larger surface gravity than Mars, even though Mars would still be more massive. They are close enough in mass so that the common center of gravity they orbited about would be somewhere in between them, as happens with Pluto and Charon. But we always consider the largest body in a system the primary, and the smaller ones moons.

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    $\begingroup$ best candidate should then be Saturn - the only planet with a density of lower than water (0.95 g / cm^3 or 950 kg / m^3 ) $\endgroup$ – eagle275 Dec 5 '19 at 14:07
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Beyond the Roche limit orbiting material coalesces forming an object (a planet or moon, depending on whether said material is orbiting a star or a planet, respectively), within the Roche limit orbiting material disperses and forms rings.

Much as objects have a Schwarzschild radius which traps light (the escape velocity is equal to the speed of light) they also have a Hill sphere or Roche sphere (not to be confused with the Roche limit or Roche Lobe) of which the outer limit constitutes a zero-velocity surface.

Put more simply: A moon (the smallest of three discussed objects, in this example) can orbit another moon or a planet, but if it is not within the Hill sphere of the object it is orbiting the star of the system will perturb the orbit causing said object to either be ejected from the system or forced to orbit the central body.

There are star systems where stars orbit each other, and neither is a "moon" because they fall outside the definition (see above links) of moon (or planet for that matter).

For an orbit of sufficient stability (our moon will no longer orbit Earth in 50 billion years, were it not for our sun in 2.3 billion years) the formula for the Hill sphere is approximately:

$$r_{\mathrm {H} }\approx a(1-e){\sqrt[{3}]{\frac {m}{3M}}}.$$

It is an untested hypothesis that rather than becoming a tidally detached exomoon (or ploonet, not to be confused with a pluot) that said moon could gain enough mass (much as Europa was formed) while it's planet simultaneously lost enough mass that they could switch places; swap whom orbits whom.

In the paper: "Binary planet formation by gas-assisted encounters of planetary embryos" (30 Oct 2018), by Ondřej Chrenko, Miroslav Brož, David Nesvorný they". write in their abstract:

"We present radiation hydrodynamic simulations in which binary planets form by close encounters in a system of several super-Earth embryos. ... close encounters of two embryos assisted by the disk gravity can form transient binary planets which quickly dissolve. Binary planets with a longer lifetime ∼10$^4$ yr form in 3-body interactions of a transient pair with one of the remaining embryos. The separation of binary components generally decreases in subsequent encounters and due to pebble accretion until the binary merges, forming a giant planet core. We provide an order-of-magnitude estimate of the expected occurrence rate of binary planets, yielding one binary planet per ≃2–5 $\!\times 10^4$ planetary systems. Therefore, although rare, the binary planets may exist in exoplanetary systems and they should be systematically searched for.

Note that this is a simulation to support a theory that binary planets exist for a short period of time, neither planet is the "moon" of the other.

Another instance of binary objects can be found in the Jupiter trojans. A Jupiter trojan is not a moon of Jupiter because it does not orbit the planet, instead trojans share the orbit of the larger object remaining in a stable orbit around the sun approximately 60° ahead or behind Jupiter near one of its Lagrangian points L$_4$ and L$_5$.

Jupiter has a number of dynamical families and binaries. One pair that has been studied is the Patroclus-Menoetius binary Jupiter Trojan. For more information refer to the paper: "Evidence for Very Early Migration of the Solar System Planets from the Patroclus-Menoetius binary Jupiter Trojan" (11 Sep 2018), by Nesvorny, David Vokrouhlicky, William F. Bottke, Harold F. Levison.

The size of Patroclus and Menoetius is calculated in the paper: "Size and Shape from Stellar Occultation Observations of the Double Jupiter Trojan Patroclus and Menoetius" (26 Feb 2015), by Buie, Marc W.; Olkin, Catherine B.; Merline, William J.; Walsh, Kevin J.; Levison, Harold F.; Timerson, Brad; Herald, Dave; Owen, William M., Jr.; Abramson, Harry B.; Abramson, Katherine J.; Breit, Derek C.; Caton, D. B.; Conard, Steve J.; Croom, Mark A.; Dunford, R. W.; Dunford, J. A.; Dunham, David W.; Ellington, Chad K.; Liu, Yanzhe; Maley, Paul D. Olsen, Aart M.; Preston, Steve; Royer, Ronald; Scheck, Andrew E.; Sherrod, Clay; Sherrod, Lowell; Swift, Theodore J.; Taylor, Lawrence W., III; Venable, Roger

"This shape model has mean-ellipsoidal axes of 127 × 117 × 98 km for Patroclus and 117 × 108 × 90 km for Menoetius. The total volume of both bodies is 1.366 km$^3$. Combining this volume with the mass of 1.20 × 10$^{18}$ kg (Mueller et al. 2010) provides a system density of 0.88 g cm$^{−3}$. A volume-equivalent spherical size for Patroclus is D$_1$ = 113 km and Menoetius is D$_2$ = 104 km. Combining these sizes into a effective mean projected area gives D$_A$ = 154 km. These numbers can be compared with those from Mueller et al. (2010) of D$_A$ = 145 ± 15 km, D$_1$ = 106 ± 11 km, D$_2$ = 98 ± 10 km and both sets are consistent as well as the ratio of the equivalent diameters.".

As you can see, given the margin of error, these objects could be the same size, and while they do orbit each other they are neither moons nor planets.

There are also a number of trans-Neptunian objects (TNO), none are moons, planets, nor similarly sized co-orbital objects.

Is it possible that a moon has a higher surface gravity than its planet?

No.

... then it would be the planet gravitating around the moon and the roles would be exchanged.

Yes.

Trivia: The smallest object orbiting our sun is 66391 Moshup, with a diameter of 1.317 ± 0.040 km and mass of (2.49 ± 0.054) $\!\times 10^{12}$ kg. It's moon (Squannit) is approximately 360 metres in diameter.

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    $\begingroup$ exomoons and pluots; I learn something new every day here! $\endgroup$ – uhoh Dec 23 '19 at 1:05

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