<p>You really need a full-blown stellar evolution model to answer this precisely and I'm not sure anyone would ever have done this with an oxygen-dominated star.</p>
<p>To <em>zeroth</em> order the answer will be the similar to a metal-rich star - i.e. about 0.075 times the mass of the Sun. Any less than this and the brown dwarf (for that is what we call a star that never gets hot enough at its centre to initiate significant fusion) can be supported by electron degeneracy pressure.</p>
<p>A star/brown dwarf with the composition you suggest would be a bit different.  The composition would be thoroughly and homogeneously mixed by convection. Note that other than a thin layer near the surface, the water would be completely dissociated and the hydrogen and oxygen atoms completely ionised. Hence the density of protons in the core would be lower for the same mass density than in a &quot;normal star&quot;.  However, the temperature dependence is so steep I think this would be a minor factor and nuclear fusion would be significant at a similar temperature.</p>
<p>Of much greater importance is that there would be fewer electrons and fewer particles at the same density. This decreases both the electron degeneracy pressure and normal gas pressure at a given mass density.
The star is therefore able to contract to much smaller radii before degeneracy pressure becomes important and can thus reach higher temperatures for the same mass as a result.</p>
<p>For that reason I think that the minimum mass for hydrogen fusion of a &quot;water star&quot; would be <em>smaller</em> than for a star made mainly of hydrogen.</p>
<p>But how much smaller? Back-of-the-envelope time!</p>
<p>Use the virial theorem to get a relationship between perfect gas pressure and the temperature, mass and radius of a star. Let gravitational potential energy be <span class="math-container">$\Omega$</span>, then the virial theorem says</p>
<p><span class="math-container">$$ \Omega = -3 \int P \ dV$$</span></p>
<p>If we only have a perfect gas then <span class="math-container">$P = \rho kT/\mu m_u$</span>, where <span class="math-container">$T$</span> is the temperature, <span class="math-container">$\rho$</span> the mass density, <span class="math-container">$m_u$</span> an atomic mass unit and <span class="math-container">$\mu$</span> the average number of mass units per particle in the gas.</p>
<p>Assuming a constant density star (back of the envelope) then <span class="math-container">$dV = dM/\rho$</span>, where <span class="math-container">$dM$</span> is a mass shell and <span class="math-container">$\Omega = -3GM^2/5R$</span>, where <span class="math-container">$R$</span> is the &quot;stellar&quot; radius. Thus
<span class="math-container">$$\frac{GM^2}{5R} = \frac{kT}{\mu m_u} \int dM$$</span>
<span class="math-container">$$ T = \frac{GM \mu m_u}{5k R}$$</span>
and so the central temperature <span class="math-container">$T \propto \mu MR^{-1}$</span>.</p>
<p>Now what we do is say that the star contracts until at this temperature, the phase space occupied by its electrons is <span class="math-container">$\sim h^3$</span> and electron degeneracy becomes important.</p>
<p>A standard treatment of this is to say that the physical volume occupied by an electron is <span class="math-container">$1/n_e$</span>, where <span class="math-container">$n_e$</span> is the electron number density and that the momentum volume occupied is <span class="math-container">$\sim (6m_e kT)^{3/2}$</span>. The electron number density is related to the mass density by <span class="math-container">$n_e = \rho /\mu_e m_u$</span>, where <span class="math-container">$\mu_e$</span> is the number of mass units per electron. For ionised hydrogen <span class="math-container">$\mu_e=1$</span>, but for oxygen <span class="math-container">$\mu_e=2$</span> (all the gas would be ionised near the temperatures for nuclear fusion). The average density <span class="math-container">$\rho = 3M/4\pi R^3$</span>.</p>
<p>Putting these things together we get
<span class="math-container">$$h^3 = \frac{ (6m_e kT)^{3/2}}{n_e} = \frac{4\pi \mu_e}{3}\left(\frac{6 \mu}{5}\right)^{3/2} (Gm_e R)^{3/2} m_u^{5/2} M^{1/2}$$</span><br />
Thus the radius to which the star contracts in order for degeneracy pressure to be important is
<span class="math-container">$$ R \propto \mu_e^{-2/3} \mu^{-1} M^{-1/3}$$</span></p>
<p>If we now substitute this into the expression for central temperature, we find
<span class="math-container">$$ T \propto \mu M \mu_e^{2/3} \mu M^{1/3} \propto \mu^2 \mu_e^{2/3} M^{4/3}$$</span></p>
<p>Finally, if we argue that the temperature for fusion is the same in a &quot;normal&quot; star and our &quot;water star&quot;, then the mass at which fusion will occur is given by the proportionality
<span class="math-container">$$ M \propto \mu^{-3/2} \mu_e^{-1/2}$$</span> .</p>
<p>For a normal star with a hydrogen/helium mass ratio of 75:25, then <span class="math-container">$\mu \simeq 16/27$</span> and <span class="math-container">$\mu_e \simeq 8/7$</span>.  For a &quot;water star&quot;,  <span class="math-container">$\mu = 18/13$</span> (18 mass units divided by an oxygen nucleus, 2 protons and 10 electrons) and <span class="math-container">$\mu_e= 9/5$</span> (18 mass units divided by 10 electrons). Thus if the former set of parameters leads to a minimum mass for fusion of <span class="math-container">$0.075 M_{\odot}$</span>, then by increasing <span class="math-container">$\mu$</span> and <span class="math-container">$\mu_e$</span> this becomes smaller by the appropriate factor <span class="math-container">$(18\times 27/13\times 16)^{-3/2} (9\times 7/5\times 8)^{-1/2} = 0.223$</span>.</p>
<p>Thus a water star would undergo H fusion at <span class="math-container">$0.017 M_{\odot}$</span> or about 17.5 times the mass of Jupiter!</p>
<p>NB This only deals with hydrogen fusion. The small amount of deuterium would fuse at lower temperatures. A similar analysis would give a minimum mass for this to occur of about 3 Jupiter masses.</p>

CC BY-SA 3.0
2017-03-18T20:47:05.657Z
A splendid analysis of a water star, much of which was beyond my expertise. But 13 M♃ is sufficiently small that its radius would be about thrice that of Saturn, far too small for Saturn even to try floating — ignoring the minor practical issues. So the comment on my son’s poster, which I recall being used in my long-lost youth, is really stupid. Thank you.
jdaw1
CC BY-SA 3.0
2017-03-18T22:20:58.367Z
@jdaw1 I think the radius would be about half that of a 0.075 solar mass object, so about half that ofJupiter.
ProfRob
CC BY-SA 3.0
2017-03-18T22:24:59.660Z
Truly, you are the expert, so I must accept your calculation. But water is not very compressible, so something that is 13 M♃, with water being less dense than Jupiter, would surely be bigger than Jupiter?
jdaw1
CC BY-SA 3.0
2017-03-18T22:26:58.450Z
@jdaw1 Water is not present at several million degrees...
ProfRob
CC BY-SA 3.0
2017-03-18T22:35:10.457Z
Ahh, yes, at a just-sub-fusion mass the centre would be ultra-compressed steam. Thank you. Does that mean that the water planet would be larger if it were much lighter than 13 M♃? Might a 7 M♃ planet be larger than a 13 M♃?
jdaw1
CC BY-SA 3.0
2017-03-18T23:57:06.640Z
Would this star still be water, that is, H₂O, or would the hydrogen and oxygen separate from one another, one forming the core and the other outside that? If so, which goes where?
KRyan
CC BY-SA 3.0
2017-03-19T00:32:49.237Z
@KRyan edit made so that is now crystal clear. There is H and O - completely ionised and thoroughly mixed.
ProfRob
CC BY-SA 3.0
2017-03-19T01:12:31.740Z
@jdaw1  Water is very much compressible at the pressure inside a planet or gas giant.    I just want to add, the chemistry would make a "water world" kind of impossible long before 12 or 13 Jupiter masses.  The chemistry inside the planet would likely split the water molecules and you'd have a hydrogen atmosphere gas giant that looks nothing like a water world at 1 Jupiter mass, probably even less.   The practical limit to a water world that looks like a water world is probably lighter and smaller than Saturn.
userLTK
CC BY-SA 3.0
2017-03-19T15:44:28.163Z
So a sufficiently large ocean is indeed impossible, not only for the reason I had intuited, but for other reasons that come into effect much earlier. Thank you again.
jdaw1
CC BY-SA 3.0
2017-03-20T01:08:13.290Z
I'm sure this answer is wrong, if only it precludes the possibility of SDSS J1240+6710. We just don't know...
Aron
CC BY-SA 3.0
2017-03-20T07:56:13.033Z
@Aron perhaps you could explain what you mean? There is no "experimental evidence" on the issue. The star you mention is a white dwarf., supported by electron degeneracy pressure and containing almost no hydrogen. The temperature for oxygen fusion is much higher than H fusion by a factor of >500. Factoring that in, my **back-of-envelope** calculation suggests a minimum mass for O fusion of about 0.7 solar masses. A correct stellar evolution calculation would show that a C/O core needs to grow to just over 1 solar mass to begin fusion. I'll accept that level of accuracy.
ProfRob
CC BY-SA 3.0
2017-03-20T08:01:25.603Z
@RobJeffries My point was tongue in cheek. You are using physics which we know could not have produced Dox, given that fusion upto Oxygen is impossible at the confinement found on something Dox sized. Some process must exist for lower temperature burn, which we do not know of. But it does put a good upper bound to the problem ;)
Aron
CC BY-SA 3.0
2017-03-20T10:41:28.773Z
@Aron I still don;t understand your point. And what is Dox? The question is of course hypothetical, but a star does not have to produce what it is made of. The Sun contains iron.
ProfRob
CC BY-SA 3.0
2017-03-20T21:51:44.940Z
@Aron if you are sure that the answer is wrong then please submit an answer that you believe better.
jdaw1
CC BY-SA 4.0
2020-09-01T08:33:55.123Z
Rob, wouldn't $\mu_\mathrm{water}$ be 18/13, rather than 18/11? The total number in the denominator being 2 protons, 2 H-electrons, 1 oxygen nucleus, and 8 O-electrons? Or am I misunderstanding the definition of $\mu$?
pela
CC BY-SA 4.0
2023-07-17T08:29:50.500Z
@pela you are correct. A slip on my part.
ProfRob
CC BY-SA 4.0
2017-03-18T20:00:32.187Z
2023-07-17T08:35:21.920Z
2023-07-17T08:35:21.920Z
ProfRob
<p>How large can a ball of water be without fusion starting?</p>

<p>Peculiar question: some explanation might be necessary. My young son is into ‘space’ and astronomy. One of his posters says that Saturn could float, if a sufficiently large ocean could be found. Obviously that wouldn’t work: Saturn’s atmosphere would peel off and join or become the atmosphere of the larger body, and then Saturn’s dense core would sink.  </p>

<p>But could such an ocean even exist without fusion starting?</p>

CC BY-SA 3.0
2017-03-20T03:11:15.703Z
Why is there an assumption that this ocean is a big ball of water?   Surely it is pictured being a vast puddle on an even vaster plain on a vast hollow planet?   Then there would be no fusion.   IE I don't think that just because a ball of water can't be created that is big enough means that the proposition itself is fundamentally unsound.
GreenAsJade
CC BY-SA 3.0
2017-03-20T12:58:49.303Z
Why the need to ask "why", @GreenAsJade? The OP has painted the scenario of Saturn swimming in the ocean of a vastly larger "planet", so let's roll with that. It's not about Saturn, but about the planet (a.k.a. a sun-sized body/blob of water).
AnoE
CC BY-SA 3.0
2017-03-20T15:06:17.577Z
Marginally related: https://what-if.xkcd.com/4/ , "a mole of moles"
Carl Witthoft
CC BY-SA 3.0
2017-03-21T01:33:28.633Z
@AnoE The reason I asked why is because the answers conclude that Saturn could not float in an ocean of water, based on the assumption that the ocean we speak of is a large spherical blob of water that would fuse.   However, the "childrens' story" that "Saturn would float" is not based on such an assumption.     If you are going to get all sciency-pedantic about a story for children who's purpose is simply to make them think about what density means, then you need to be sciency-pedantic about the assumptions.   The OP assumed the ocean is a blob of water, but no real ocean is a blob.
GreenAsJade
CC BY-SA 3.0
2017-03-21T21:30:25.320Z
@GreenAsJade That’s a fair response. The water needs to be almost as deep as Saturn’s diameter. If it were on a very large hollow planet (engineering details TBD), could that work? Would there be problems with the ‘horizontal’ quantity of water, stretching to the horizon for multiple Saturn diameters? This would imply multiple Saturn volumes of water in close proximity: are we back to the consequences of gravity?
jdaw1
<gravity><saturn><brown-dwarf>
How large can a ball of water be without fusion starting?
jdaw1
# How large can a ball of water be without fusion starting?

(Asked by: username_0 on 2017-03-18)

How large can a ball of water be without fusion starting?

Peculiar question: some explanation might be necessary. My young son is into ‘space’ and astronomy. One of his posters says that Saturn could float, if a sufficiently large ocean could be found. Obviously that wouldn’t work: Saturn’s atmosphere would peel off and join or become the atmosphere of the larger body, and then Saturn’s dense core would sink.

But could such an ocean even exist without fusion starting?

**Tags:** gravity, saturn, brown-dwarf
> **Comments:**
> **username_1**: Why is there an assumption that this ocean is a big ball of water?   Surely it is pictured being a vast puddle on an even vaster plain on a vast hollow planet?   Then there would be no fusion.   IE I don't think that just because a ball of water can't be created that is big enough means that the proposition itself is fundamentally unsound.  
> **username_2**: Why the need to ask "why", @GreenAsJade? The OP has painted the scenario of Saturn swimming in the ocean of a vastly larger "planet", so let's roll with that. It's not about Saturn, but about the planet (a.k.a. a sun-sized body/blob of water).  
> **username_3**: Marginally related: https://what-if.xkcd.com/4/ , "a mole of moles"  
> **username_1**: @AnoE The reason I asked why is because the answers conclude that Saturn could not float in an ocean of water, based on the assumption that the ocean we speak of is a large spherical blob of water that would fuse.   However, the "childrens' story" that "Saturn would float" is not based on such an assumption.     If you are going to get all sciency-pedantic about a story for children who's purpose is simply to make them think about what density means, then you need to be sciency-pedantic about the assumptions.   The OP assumed the ocean is a blob of water, but no real ocean is a blob.  
> **username_0**: @GreenAsJade That’s a fair response. The water needs to be almost as deep as Saturn’s diameter. If it were on a very large hollow planet (engineering details TBD), could that work? Would there be problems with the ‘horizontal’ quantity of water, stretching to the horizon for multiple Saturn diameters? This would imply multiple Saturn volumes of water in close proximity: are we back to the consequences of gravity?  

---

## 1 answers

### Answer by username_4 on 2017-03-18. Score: 49 | Accepted answer

You really need a full-blown stellar evolution model to answer this precisely and I'm not sure anyone would ever have done this with an oxygen-dominated star.

To *zeroth* order the answer will be the similar to a metal-rich star - i.e. about 0.075 times the mass of the Sun. Any less than this and the brown dwarf (for that is what we call a star that never gets hot enough at its centre to initiate significant fusion) can be supported by electron degeneracy pressure.

A star/brown dwarf with the composition you suggest would be a bit different. The composition would be thoroughly and homogeneously mixed by convection. Note that other than a thin layer near the surface, the water would be completely dissociated and the hydrogen and oxygen atoms completely ionised. Hence the density of protons in the core would be lower for the same mass density than in a "normal star". However, the temperature dependence is so steep I think this would be a minor factor and nuclear fusion would be significant at a similar temperature.

Of much greater importance is that there would be fewer electrons and fewer particles at the same density. This decreases both the electron degeneracy pressure and normal gas pressure at a given mass density.
The star is therefore able to contract to much smaller radii before degeneracy pressure becomes important and can thus reach higher temperatures for the same mass as a result.

For that reason I think that the minimum mass for hydrogen fusion of a "water star" would be *smaller* than for a star made mainly of hydrogen.

But how much smaller? Back-of-the-envelope time!

Use the virial theorem to get a relationship between perfect gas pressure and the temperature, mass and radius of a star. Let gravitational potential energy be $\Omega$, then the virial theorem says

$$ \Omega = -3 \int P \ dV$$

If we only have a perfect gas then $P = \rho kT/\mu m\_u$, where $T$ is the temperature, $\rho$ the mass density, $m\_u$ an atomic mass unit and $\mu$ the average number of mass units per particle in the gas.

Assuming a constant density star (back of the envelope) then $dV = dM/\rho$, where $dM$ is a mass shell and $\Omega = -3GM^2/5R$, where $R$ is the "stellar" radius. Thus
$$\frac{GM^2}{5R} = \frac{kT}{\mu m\_u} \int dM$$
$$ T = \frac{GM \mu m\_u}{5k R}$$
and so the central temperature $T \propto \mu MR^{-1}$.

Now what we do is say that the star contracts until at this temperature, the phase space occupied by its electrons is $\sim h^3$ and electron degeneracy becomes important.

A standard treatment of this is to say that the physical volume occupied by an electron is $1/n\_e$, where $n\_e$ is the electron number density and that the momentum volume occupied is $\sim (6m\_e kT)^{3/2}$. The electron number density is related to the mass density by $n\_e = \rho /\mu\_e m\_u$, where $\mu\_e$ is the number of mass units per electron. For ionised hydrogen $\mu\_e=1$, but for oxygen $\mu\_e=2$ (all the gas would be ionised near the temperatures for nuclear fusion). The average density $\rho = 3M/4\pi R^3$.

Putting these things together we get
$$h^3 = \frac{ (6m\_e kT)^{3/2}}{n\_e} = \frac{4\pi \mu\_e}{3}\left(\frac{6 \mu}{5}\right)^{3/2} (Gm\_e R)^{3/2} m\_u^{5/2} M^{1/2}$$  
Thus the radius to which the star contracts in order for degeneracy pressure to be important is
$$ R \propto \mu\_e^{-2/3} \mu^{-1} M^{-1/3}$$

If we now substitute this into the expression for central temperature, we find
$$ T \propto \mu M \mu\_e^{2/3} \mu M^{1/3} \propto \mu^2 \mu\_e^{2/3} M^{4/3}$$

Finally, if we argue that the temperature for fusion is the same in a "normal" star and our "water star", then the mass at which fusion will occur is given by the proportionality
$$ M \propto \mu^{-3/2} \mu\_e^{-1/2}$$ .

For a normal star with a hydrogen/helium mass ratio of 75:25, then $\mu \simeq 16/27$ and $\mu\_e \simeq 8/7$. For a "water star", $\mu = 18/13$ (18 mass units divided by an oxygen nucleus, 2 protons and 10 electrons) and $\mu\_e= 9/5$ (18 mass units divided by 10 electrons). Thus if the former set of parameters leads to a minimum mass for fusion of $0.075 M\_{\odot}$, then by increasing $\mu$ and $\mu\_e$ this becomes smaller by the appropriate factor $(18\times 27/13\times 16)^{-3/2} (9\times 7/5\times 8)^{-1/2} = 0.223$.

Thus a water star would undergo H fusion at $0.017 M\_{\odot}$ or about 17.5 times the mass of Jupiter!

NB This only deals with hydrogen fusion. The small amount of deuterium would fuse at lower temperatures. A similar analysis would give a minimum mass for this to occur of about 3 Jupiter masses.

> **Comments:**
> **username_0**: A splendid analysis of a water star, much of which was beyond my expertise. But 13 M♃ is sufficiently small that its radius would be about thrice that of Saturn, far too small for Saturn even to try floating — ignoring the minor practical issues. So the comment on my son’s poster, which I recall being used in my long-lost youth, is really stupid. Thank you.  
> **username_4**: @jdaw1 I think the radius would be about half that of a 0.075 solar mass object, so about half that ofJupiter.  
> **username_0**: Truly, you are the expert, so I must accept your calculation. But water is not very compressible, so something that is 13 M♃, with water being less dense than Jupiter, would surely be bigger than Jupiter?  
> **username_4**: @jdaw1 Water is not present at several million degrees...  
> **username_0**: Ahh, yes, at a just-sub-fusion mass the centre would be ultra-compressed steam. Thank you. Does that mean that the water planet would be larger if it were much lighter than 13 M♃? Might a 7 M♃ planet be larger than a 13 M♃?  
> **username_5**: Would this star still be water, that is, H₂O, or would the hydrogen and oxygen separate from one another, one forming the core and the other outside that? If so, which goes where?  
> **username_4**: @KRyan edit made so that is now crystal clear. There is H and O - completely ionised and thoroughly mixed.  
> **username_6**: @jdaw1  Water is very much compressible at the pressure inside a planet or gas giant.    I just want to add, the chemistry would make a "water world" kind of impossible long before 12 or 13 Jupiter masses.  The chemistry inside the planet would likely split the water molecules and you'd have a hydrogen atmosphere gas giant that looks nothing like a water world at 1 Jupiter mass, probably even less.   The practical limit to a water world that looks like a water world is probably lighter and smaller than Saturn.  
> **username_0**: So a sufficiently large ocean is indeed impossible, not only for the reason I had intuited, but for other reasons that come into effect much earlier. Thank you again.  
> **username_7**: I'm sure this answer is wrong, if only it precludes the possibility of SDSS J1240+6710. We just don't know...  
> **username_4**: @Aron perhaps you could explain what you mean? There is no "experimental evidence" on the issue. The star you mention is a white dwarf., supported by electron degeneracy pressure and containing almost no hydrogen. The temperature for oxygen fusion is much higher than H fusion by a factor of >500. Factoring that in, my **back-of-envelope** calculation suggests a minimum mass for O fusion of about 0.7 solar masses. A correct stellar evolution calculation would show that a C/O core needs to grow to just over 1 solar mass to begin fusion. I'll accept that level of accuracy.  
> **username_7**: @RobJeffries My point was tongue in cheek. You are using physics which we know could not have produced Dox, given that fusion upto Oxygen is impossible at the confinement found on something Dox sized. Some process must exist for lower temperature burn, which we do not know of. But it does put a good upper bound to the problem ;)  
> **username_4**: @Aron I still don;t understand your point. And what is Dox? The question is of course hypothetical, but a star does not have to produce what it is made of. The Sun contains iron.  
> **username_0**: @Aron if you are sure that the answer is wrong then please submit an answer that you believe better.  
> **username_8**: Rob, wouldn't $\mu_\mathrm{water}$ be 18/13, rather than 18/11? The total number in the denominator being 2 protons, 2 H-electrons, 1 oxygen nucleus, and 8 O-electrons? Or am I misunderstanding the definition of $\mu$?  
> **username_4**: @pela you are correct. A slip on my part.  

---

