6
$\begingroup$

I have heard that gas giants are primarily huge solid bodies like regular rocky planets that exponentially gained more and more gas in their atmosphere through their increase in mass which they use to pick up even more mass ,gaining even more gas, and practically any planet we know today above a certain size threshold is a gas giant

That being said, assuming an area in space with very little or no gas but a bountiful supply of rocks, how big can a solid planet consisting of these coalesced rocks get?

Improve this question
$\endgroup$
3
  • $\begingroup$ Kind of related: astronomy.stackexchange.com/q/39921/16685 $\endgroup$
    – PM 2Ring
    Commented Mar 26, 2021 at 17:50
  • $\begingroup$ If there are soft ice rocks, there will be a snowball effect, otherwise it will all bounce around. $\endgroup$
    – bandybabboon
    Commented Mar 28, 2021 at 6:05
  • $\begingroup$ Sorry, @bandyb, I don't think that's a helpful comment. $\endgroup$
    – PM 2Ring
    Commented Sep 28 at 4:22

2 Answers 2

Reset to default
6
$\begingroup$

That very much depends on the structure of the solids, i.e. whether they exist in the form of small dust, or ready-to-smash planetesimals.

In any case, the available median dust mass for planet formation is about 158 $\rm m_{earth}$ (see Tychoniec et al., (2020)). If you leave all this mass as dust, without any gas interactions, the dust will not coalesce into planetesimals, as a hydrodynamic instability is required to jump over the meter-sized barrier (Johansen et al., (2014)).

However, if you somehow allow all this dust to be converted into planetesimals, then the size of the planet you can form will be given by how narrow you can pack the planetesimals. The absolute upper mass of the formed planet will be given by the 158 $\rm m_{earth}$, but realistically , that is going to be lower, as planetesimals and collision ejecta are lost during the smashing phase of planet formation.

As you were asking about the size of this hypothetical planet, if we assume no compressional effects and hence the same mean density as earth, you would get a solid ball of the size of $\rm 158^{1/3} \; r_{earth} \approx 5.4\; r_{earth}$.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ Why is it only limited to 5.4 times earth radius though? What exactly is the limiting fator and what happens if we go beyond that point? Is it because that's how tight we can compress rocks? $\endgroup$
    – Hash
    Commented Mar 27, 2021 at 18:24
  • $\begingroup$ There is no more solid mass around a typical star, even during planet formation (give our take factor of 3 our so) . He even assumed a very low density, not taking Denser form under pressure in the interior into account $\endgroup$
    – planetmaker
    Commented Mar 28, 2021 at 1:03
  • $\begingroup$ I'm intrigued by the final equation of 158^(1/3)rearth≈5.4rearth. Why is the cubed root of mass only used to approximate the radius? If the equation for the volume of a sphere is used & the actual average density of Earth, the radius is ≈ 2; [r^3 = 3(158)/4π(5.514)]. Am I wrong? $\endgroup$
    – Fred
    Commented Mar 28, 2021 at 3:24
  • 2
    $\begingroup$ @Fred: It is $4/3 \pi r^3 = m/\rho$, which means $(r/r_{\oplus})^3 = m/m_{\oplus}$ at same density. $\endgroup$
    – AtmosphericPrisonEscape
    Commented Mar 28, 2021 at 17:29
  • $\begingroup$ @Hash: This is an upper limit for the radius, at same density as Earth's. If you allow the mass to compress, i.e. density to increase, the radius would be smaller. For the simple radius estimate, see my comment above. $\endgroup$
    – AtmosphericPrisonEscape
    Commented Mar 28, 2021 at 17:31
0
$\begingroup$

Good answer, but the question is, "What's the largest possible rock?" s The largest would be a rock that'between one and two solar diameters.

Since the sun has the same average density as water, I imagine a large star would have about the same average density as rock, so the diameter of the biggest rock would be about the same.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Why would it be between one and two solar diameters? $\endgroup$
    – PM 2Ring
    Commented Sep 27 at 3:53
  • $\begingroup$ Because one diameter is a solar mass. Two diameters is eight solar masses. 2.4 is the supernova limit. $\endgroup$
    – Miss Understands
    Commented Sep 27 at 19:49
  • $\begingroup$ Ok. The limit is probably close to the lower side, due to the high metallicity of the rock. ProfRob has some info on the influence of molecular weight on the Chandrasekhar limit at astronomy.stackexchange.com/q/28123/16685 & astronomy.stackexchange.com/a/46712/16685 I don't know if the rock could get hot enough to undergo fusion. $\endgroup$
    – PM 2Ring
    Commented Sep 27 at 20:13
  • $\begingroup$ @PM2Ring That guy put the limit at 5.7M⊙. No way, raye. Mass has weight. Weight crushes stars into black holes. That doesn't go away when the nuclei are less dense. $\endgroup$
    – Miss Understands
    Commented Sep 28 at 0:01

You must log in to answer this question.

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