6
$\begingroup$

The surface gravities of the Sun’s planets are all close to 1g, 0.38-2.53 (about a factor of e, by chance). The cloud-top gravity of the gas giants is not too interesting here, but I suppose that the gravity on their solid surfaces (if any) does not tend to be higher. So called super-Earth exoplanets seem likely to generally accumulate light gas atmospheres, which decreases their expected solid surface gravity relative to their size as measured during for example transits (I suppose so anyway).

Are there good reasons to expect that in the zoo of planets in the universe, our Earth scores high in the surface gravity top-list? That it is unusual with solid real estate to stand on which has substantially (as in a factor of a few tenths) greater gravity than we have here?

Surface gravities from Wikipedia:

  • Mercury 0.38
  • Venus 0.90
  • Earth 1.00
  • Mars 0.38
  • Jupiter 2.53
  • Saturn 1.07
  • Uranus 0.89
  • Neptune 1.14
$\endgroup$

3 Answers 3

9
$\begingroup$

According to this paper, fig. 4, solid planets of arbitrary mass (up to 3,000 times Earth's mass) don't grow much about three- or four-times the Earth in diameter. That's because interior parts of the planet become compressed by the high pressure. Planets heavier than 3,000 times Earth's mass are in the transition zone to stars. In the case of rocky planets they would become substellar objects, similar in their physical properties (electron degeneracy in the core) to white or black dwarfs, not capable to start nuclear fusion due to lack of light elements in their core.

Surface gravity is proportioal to the mean density $\rho$, and to the radius $r$ of the planet: $g=\frac{4\pi}{3}G\rho r$, with $G$ the gravitational constant.

A planet with roughly the composition of Earth, and 3,000 times the mass of Earth would have about 3-times the Earth's radius, hence the 27-fold volume, and the 3000/27 = 111-fold density. Its surface gravity would hence be 333 g. That's close to the upper bound of surface gravity of what would be called a planet. With a pure iron planet we could go a little beyond. Any surface gravity below this upper bound is possible for a solid planet, at least in theory.

According to page 1284 of the paper:

Massive solid exoplanets of hundreds to thousands of Earth masses may be able to form around massive stars (B and O stars; 5-120 solar masses) where the protoplanetary disk would contain enough heavy elements.

B and O type stars are rare (0.13% of main sequence stars) and short-lived (less than 100 million years; more accurately less than $10^{10}\cdot 18^{-2.5}$ years, for mass > 18 solar masses). Hence large solid planets of the described type will also be rare, although a priori not impossible.

The planet with the highest estimated surface gravity discovered thus far (March 8, 2014) is CoRoT-Exo-3b (add "surface gravity" column to the exoplanets table, and sort by that column):

CoRoT-Exo-3b has a radius of 1.01 ± 0.07 R_Jup and transits around its F3-type primary every 4.26 days in a synchronous orbit. Its mass of 21.66 ± 1.0 M_Jup, density of 26.4 ± 5.6 g cm-3, and surface gravity of logg = 4.72 clearly distinguish it from the regular close-in planet population, making it the most intriguing transiting substellar object discovered so far.

A surface gravity of logg = 4.72 means surface gravity equals $10^{4.72}\mbox{ cm}/\mbox{s}^2=52,480 \mbox{ cm}/\mbox{s}^2 = 53.5~g.$ This research has made use of the Exoplanet Orbit Database and the Exoplanet Data Explorer at exoplanets.org.

CoRoT-3b on Wikipedia.

$\endgroup$
3
  • $\begingroup$ The formation of planets and stars is different, and the "transition zone" you mean, does not exist (brown dwarf desert: there are no detected objects that are in that range of masses). Another problem with you answer is that for those rocky planets you talk about, they would be uncapable of doing nuclear fusion, but still become white dwarfs or black dwarfs. White dwarfs are stellar remnants (meaning they were stars before, so they have done fusion before), and black dwarfs are theoretical objects for which the universe is too young to have them. $\endgroup$ Commented Mar 7, 2014 at 8:23
  • $\begingroup$ @usethedeathstar crunchthenumbers.net/crunch-the-numbers/2012/11/4/…; the brown dwarf desert refers to close-up brown dwarfs, meaning orbital periods shorter than the central star's rotation period. See also spectral types T and Y for stars. (en.wikipedia.org/wiki/Stellar_classification) $\endgroup$
    – Gerald
    Commented Mar 7, 2014 at 11:31
  • $\begingroup$ still, they cant be white or black dwarfs, they have to be brown dwarfs or subbrown dwarfs. The 13 Jupiter masses is when they start fusing deuterium, not hydrogen. $\endgroup$ Commented Mar 7, 2014 at 12:00
4
$\begingroup$

Surface gravity is a function of two things:

  1. Mass
  2. Radius

Mass is proportional to the cube of the radius times the mean density. Gravity is proportional to the square of the radius. As a result, is you have two planets with the same density, the larger planet will have the higher gravity.

But mean density can be a big factor.

Earth:

  • Mean density 5.515 g/cm3
  • Mean radius 6371.0 km

Venus:

  • Mean density 5.243 g/cm3
  • Mean radius 6051.8 km

Jupiter:

  • Mean density 1.326 g/cm3
  • Mean radius 69911±6 km

(source: Wikipedia pages for Earth, Venus, and Jupiter)

So Venus is both slightly smaller and slightly less dense than Earth, giving it 90% of Earth gravity. Jupiter is much larger (more than 10 times) but much less dense (about one fourth) giving it 10/4 or about 250% of Earth gravity.

My understanding is that the Earth is fairly small, but unusually dense. Of planets that we can detect with current technology, I would expect that more of them would be larger and/or heavier, as those properties would make them easier for us to detect.

Of all the planets in the universe, I expect that we will find a distribution not too unlike our own Solar System and that Earth will fall just on the high side of average.

$\endgroup$
6
  • $\begingroup$ Mercury's density is 5.4 so it fits well with its neighbours. Still, it has the same surface gravity as the larger but less dense Mars. So I thought that maybe there are more general physical processes which tend to top out planetary solid surface gravity at about 1g. $\endgroup$
    – LocalFluff
    Commented Mar 6, 2014 at 18:34
  • $\begingroup$ Am i reading this wrong, or did you just call 8 datapoints a distribution? Basing the theory of planet formation on what we know of our solar system is doing small number statistics. $\endgroup$ Commented Mar 7, 2014 at 17:26
  • $\begingroup$ True, but it's the only complete data we have so far. The alternative would be to close the question as "primarily opinion based". $\endgroup$ Commented Mar 7, 2014 at 17:45
  • $\begingroup$ @usethedeathstar, $\endgroup$
    – LocalFluff
    Commented Mar 7, 2014 at 18:09
  • 1
    $\begingroup$ @LocalFluff Between 10 and 30g (log g between 4 and 4.5 cm/s^2) is rather common among gas giants (about Jupiter-size), see table 2 in this paper: iopscience.iop.org/0004-637X/646/1/505/pdf/64046.web.pdf $\endgroup$
    – Gerald
    Commented Mar 8, 2014 at 14:41
0
$\begingroup$

Ultimately we don't know enough about exoplanets to be sure; for now all our data is skewed toward more massive planets which are easier to detect using Doppler wobble, or large diameter planets (almost certainly gas giants) which are easy to detect by their host star dimming when they eclipse it relative to us. More data is coming in every day, and as fantastic as Keplar has been, I think we need to at least hold out for the James-Webb to be online before we draw any really hard conclusions form the data.

Without the data, all we can rely on is our theories of planet formation, which we're fairly good at.
Earth is probably more dense than an average planet of it's size, as a result of it colliding with a roughly mars-sized object (nicknamed Theia) early in it's development. Theia's core would have been absorbed into earth's core, but the outer layers of both were stripped away, creating a ring which would coalesce into our moon. This would leave earth with a higher mass core than a planet forming at its distance would have.

We can see this in the densities of the terrestrial planets;

--Object-------Density (g cm−3)-----Semi-major axis (AU)-

-------------Mean----Uncompressedm-----------------------

-Mercury-----5.4---------5.3-------------0.39------------

-Venus-------5.2---------4.4-------------0.72------------

-Earth-------5.5----------4.4--------------1.0-------------

-Mars--------3.9----------3.8--------------1.5-------------

Credit, Wikipedia

Planets closer to their star are naturally going to have higher densities as a result of mass differentiation; denser material settling to the core of a planet or the center of a solar accretion disk.

Looking at it from the perspective of habitability,
We know that density is positively correlated to surface gravity, so we can expect that earth would have a slightly higher than average surface gravity for a planet in the habitable zone around a star in the category of one solar mass.

That being said, most stars do not have one solar mass, most of the stars in the universe are red dwarfs, which are much dimmer and lighter than our sun, and would have a closer, narrower habitable zone. A habitable planet around a red dwarf would probably be smaller and lighter, but denser, due to its lower mass accretion cloud and closer proximity to it's star respectively.

I think we could expect the majority of exoplanets to be planets similar to mercury orbiting red dwarfs.
If this is the case we can expect that earth would have a high surface gravity relative to terrestrial planets (although FAR more massive terrestrial planets of similar diameter exist) and about average gravity when taking all planets into account.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .