<p>Good answer, but the question is, &quot;What's the largest possible rock?&quot;
s
The largest would be a rock that'between one and two solar diameters.</p>
<p>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.</p>

CC BY-SA 4.0
2024-09-27T03:53:09.530Z
Why would it be between one and two solar diameters?
PM 2Ring
CC BY-SA 4.0
2024-09-27T19:49:35.220Z
Because one diameter is a solar mass. Two diameters is eight solar masses. 2.4 is the supernova limit.
Miss Understands
CC BY-SA 4.0
2024-09-27T20:13:59.810Z
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 https://astronomy.stackexchange.com/q/28123/16685 & https://astronomy.stackexchange.com/a/46712/16685 I don't know if the rock could get hot enough to undergo fusion.
PM 2Ring
CC BY-SA 4.0
2024-09-28T00:01:42.470Z
@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.
Miss Understands
CC BY-SA 4.0
2024-09-26T16:26:49.673Z
2024-09-27T23:53:58.603Z
2024-09-27T23:53:58.603Z
Miss Understands
<p>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.</p>
<p>In any case, the available median dust mass for planet formation is about 158 <span class="math-container">$\rm m_{earth}$</span> (see <a href="https://ui.adsabs.harvard.edu/abs/2020A%26A...640A..19T/abstract" rel="nofollow noreferrer">Tychoniec et al., (2020)</a>). 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 (<a href="https://arxiv.org/abs/1402.1344" rel="nofollow noreferrer">Johansen et al., (2014)</a>).</p>
<p>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 <span class="math-container">$\rm m_{earth}$</span>, but realistically , that is going to be lower, as planetesimals and collision ejecta are lost during the smashing phase of planet formation.</p>
<p>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 <span class="math-container">$\rm 158^{1/3} \; r_{earth} \approx 5.4\; r_{earth}$</span>.</p>

CC BY-SA 4.0
2021-03-27T18:24:35.937Z
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?
Hash
CC BY-SA 4.0
2021-03-28T01:03:12.630Z
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
planetmaker
CC BY-SA 4.0
2021-03-28T03:24:58.373Z
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?
Fred
CC BY-SA 4.0
2021-03-28T17:29:57.583Z
@Fred: It is $4/3 \pi r^3 = m/\rho$, which means $(r/r_{\oplus})^3 = m/m_{\oplus}$ at same density.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-03-28T17:31:50.430Z
@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.
AtmosphericPrisonEscape
CC BY-SA 4.0
2024-10-07T20:32:18.853Z
@PM2Ring: The absolute upper limit is the mass of solids in the protoplanetary disc (plus late infall, minus T-Tauri outflows). Also 160 earth masses is plenty of rocks, don't you think?
AtmosphericPrisonEscape
CC BY-SA 4.0
2024-10-07T21:52:27.373Z
@PM2Ring: Can you please be consistent in what you are critizicing. First you say "i don't like that you only allow for finite mass" now you say "I think there is only finite mass". Independent of that, I answered OP's question.
AtmosphericPrisonEscape
CC BY-SA 4.0
2024-10-07T22:36:37.110Z
@PM2Ring: Claiming you're consistent while you're not, doesn't change the fact that you're not. Anyway, feel free to provide a lower-upper mass limit to the possible formed planet.
AtmosphericPrisonEscape
CC BY-SA 4.0
2021-03-27T16:43:38.050Z
2024-09-27T03:49:11.090Z
2024-09-27T03:49:11.090Z
AtmosphericPrisonEscape
<p>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</p>
<p>That being said, assuming an area in space with very little or no gas but a bountiful supply of rocks, how big can a <strong>solid planet</strong> consisting of these coalesced rocks get?</p>

CC BY-SA 4.0
2021-03-26T17:50:31.757Z
Kind of related: https://astronomy.stackexchange.com/q/39921/16685
PM 2Ring
CC BY-SA 4.0
2021-03-28T06:05:34.430Z
If there are soft ice rocks, there will be a snowball effect, otherwise it will all bounce around.
bandybabboon
CC BY-SA 4.0
2024-09-28T04:22:31.507Z
Sorry, @bandyb, I don't think that's a helpful comment.
PM 2Ring
<planet><planetary-formation><gas-giants>
Assuming a hypothetical system without gas but only solid rocks, how big of a planet can form through the coalescing of these rocks?
Hash
# Assuming a hypothetical system without gas but only solid rocks, how big of a planet can form through the coalescing of these rocks?

(Asked by: username_0 on 2021-03-26)

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?

**Tags:** planet, planetary-formation, gas-giants

> **Comments:**
> **username_1**: Kind of related: https://astronomy.stackexchange.com/q/39921/16685  
> **username_2**: If there are soft ice rocks, there will be a snowball effect, otherwise it will all bounce around.  
> **username_1**: Sorry, @bandyb, I don't think that's a helpful comment.  

---

## 2 answers

### Answer by username_4 on 2021-03-27. Score: 6 | Accepted answer

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)](https://ui.adsabs.harvard.edu/abs/2020A%26A...640A..19T/abstract)). 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)](https://arxiv.org/abs/1402.1344)).

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 $\rm 158^{1/3} \; r\_{earth} \approx 5.4\; r\_{earth}$.

> **Comments:**
> **username_0**: 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?  
> **username_5**: 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  
> **username_6**: 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?  
> **username_4**: @username_6: It is $4/3 \pi r^3 = m/\rho$, which means $(r/r_{\oplus})^3 = m/m_{\oplus}$ at same density.  
> **username_4**: @username_0: 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.  
> **username_4**: @PM2Ring: The absolute upper limit is the mass of solids in the protoplanetary disc (plus late infall, minus T-Tauri outflows). Also 160 earth masses is plenty of rocks, don't you think?  
> **username_4**: @PM2Ring: Can you please be consistent in what you are critizicing. First you say "i don't like that you only allow for finite mass" now you say "I think there is only finite mass". Independent of that, I answered OP's question.  
> **username_4**: @PM2Ring: Claiming you're consistent while you're not, doesn't change the fact that you're not. Anyway, feel free to provide a lower-upper mass limit to the possible formed planet.  

---

### Answer by username_3 on 2024-09-26. Score: 0

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.

> **Comments:**
> **username_1**: Why would it be between one and two solar diameters?  
> **username_3**: Because one diameter is a solar mass. Two diameters is eight solar masses. 2.4 is the supernova limit.  
> **username_1**: 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 https://astronomy.stackexchange.com/q/28123/16685 & https://astronomy.stackexchange.com/a/46712/16685 I don't know if the rock could get hot enough to undergo fusion.  
> **username_3**: @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.  

---

