Right, I think I have my first approximation for the perfectly spherical, homogeneous, incompressible sphere of water at the triple point - I've derived a function that equates mass into surface acceleration (g). It's not pretty. I'll use the caret (^) to signify raising to a power. First the preamble:
1. Density (should be symbol
rho, but I can't be bothered to find it):
rho = m/v
2. Volume of a sphere:
v = 4/3 . pi . r^3
3. General equation for (magnitude) of force:
F = m.a
4. (Magnitude) of force between two masses at a distance r from their centres of mass:
F = G . m1 . m2 /r^2
5. Rearranging (1) for
m and substituting (2) for
v:
m = 4/3 . pi . rho . r^3
6. Rearranging (5) for
r:
r = ( 3 . m / 4 . pi . rho)^1/3
7. Rearranging (3) for
a and substituting (4) for
F and letting
m1 be the mass of the body of water (
m) and
m2 the mass upon which
F is being calculated (note that
m2 is cancelled out in the substitution):
a = G . m / r^2
8. Substitute (6) for r:
a = G . m / ( 3 . m / 4 . pi . rho) ^ 2/3
So, from (6) and (8) we can calculate the radius and surface acceleration due to gravity of our hypothetical planetary blob of water, given that
G and
pi are constant and
rho is too when our water is magically incompressible and not given to changing phase under inordinate pressure.
Ideally, rearrange (8) to get m as a function of a and you can simply throw in whatever surface acceleration you want and get the mass required, but just playing around with these (plugging 999.84 kg/m^3 for
rho at the triple point, 6.67384x10^-11 m^3 kg^-1 s^-2 for
G), I get that a 100,000km diameter blob of water under these conditions would mass 5.2352x10^26kg and exert a surface acceleration due to gravity of 13.98 m/s^2.
Way more than enough to hold down the measly 611.71 Pa partial pressure of water vapour at the triple point