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

Question

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?

Answer 1

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.