How does the temperature of the triple point of water depend on gravitational acceleration? Suppose I do two experiments to find the triple point of water, one in zero-g and one on Earth.  On Earth, water in the liquid or solid phase has less gravitational potential per unit mass than water in the gas phase.  Therefore, the solid and liquid phases should be favored slightly more on Earth than in zero-g.
In a back-of-the-envelope calculation, how does the temperature of the triple-point of water depend on the gravitational acceleration and, if necessary, on the mass of water and volume and shape of container?
Edit
Let's say I have a box in zero-g.  The box is one meter on a side.  It has nothing in it but water.  Its temperature and pressure are just right so that it's at the triple point.  All the water and ice and steam are floating around the box because it's zero-g.
Now I turn on gravity.  The liquid water and ice fall to the bottom of the box, but the average height of the steam remains almost half a meter above the bottom of the box.  So when gravity got turned on, the potential energy of the ice and liquid water went down significantly, but the potential energy of the steam didn't.  Doesn't this mean that once gravity is turned on, water molecules would rather be part of the ice or liquid phase so that they can have lower energy?  Wouldn't we no longer be at the triple point?
Several people have posted saying the answer is "no".  I don't disbelieve that.  Maybe the answer is just "no".  I don't understand why the answer is no.  Answers such as "No, because gravity doesn't affect the triple point," or "No, because the triple point only depends on pressure and temperature" simply restate the answer "no" with more words.
 A: There is no difference; phase transitions does not change gravitational potential.
A: To expand on mpq and Greg's remarks.
A triple point is a unique (pressure,temperature) pair for each material (and where appropriate group of phases) where three phases of matter can coexist in equilibrium.
If you are not at the right pressure, there is no temperature where this occurs, and if you are not at the right temperature there is no pressure where this occurs.
In that sense the question is simply based on a misconception.

Several posters have tried to address the effects of changing gravitation on the locally experienced pressure due to a atmosphere column. That's fine as far as it goes, but it doesn't really go to the question.
Diddling the local gravity will change the pressure naturally experienced due the local atmosphere but will not affect the parameters of the triple point.
A: This is perhaps similar to what mbq meant, but I will elaborate.
The T-p phase diagram of water tells us, for a given temperature and pressure, what phase we will get if we have a bunch of that substance.  If I apply different pressures to a bottle of water, I am moving around in the p-direction of the T-p plane.  I am not changing the pressure of the triple point of water, just changing the pressure of that particular bottle of water! Similarly, if a tank of water is in a gravitational field, it affects the pressure.  In fact, it leads to different pressures at different locations of the tank.  It could lead some parts of the tank to freeze, for example.  But it does not in any way change the triple point of water itself, which is an intrinsic property of that substance.  So I would say that the question is ill-posed.  It might be better to ask: what will happen to a tank of water at a given temperature and density if we now apply a gravitational field?
A: I don't see how to derive triple point directly, but let me talk about something similar and see where it leads.
For ideal gas atmosphere you'll find out that $g$ works as an inverse temperature. That is, $0K$ is the same as infinitely strong gravity because both keep the gas at the ground.
The real gases exhibit phase transitions, so we should promote ideal gas to something like van der Waals gas. This should not change the above correspondence of $\beta = {1 \over k_B T}$ and $g$ much. Therefore, I would imagine that one would get a phase transition at high $g$ with constant $\beta$. Or in other words, that in the presence of gravitational field, higher temperatures are enough for the phase transition. Or in other words yet, we have another coupling $g$ that tells system to cluster besides the usual thermal coupling $\beta$.
Now, this carries over (by a huge and completely unjustified extrapolation) to the triple point of water. I'll try to provide some calculations later.
A: Back of the envelope calculation:


*

*gravity will manifest itself by increasing the pressure by a factor of $mg/A$ where A is the surface of the container with water at the triple point. So basically the effect is shifting the triple point phase diagram down. So the triple point would be at the same temperature with a lower pressure.


$$p_{measured} = p_{gas} + p_{gravity}$$


*

*Assuming a regular shape we can say that $p_{gravity}=\rho g h$ where $\rho$ is the density of the gas at our triple point pressure and h is the typical height of a column of gas.


$$p_{gravity}=\rho g h$$


*

*to calculate $\rho$ we can use the ideal gas law: 


$$\rho = \frac{Nm_w}{V} = \frac{p_{gas}m_w}{kT}$$


*

*we can now calculate what fraction of pressure is due to gravity


$$\frac{p_{gravity}}{p_{gas}}=\rho g h/p_{gas}=\frac{m_w g h}{kT}$$


*

*using the following values we can calculate this ratio


$m_w = 3 \times 10^{-26} \mathrm{kg}$
$g = 9.81 \mathrm{m/s^{2}}$
$h = 1 \mathrm{m}$
$k = 1.38 \times 10^{-23} \mathrm{J/K}$
$T = 273.16 K$
$$\frac{p_{gravity}}{p_{gas}}=0.000078$$
This means that you need a value for g one thousand times stronger to see a large effect.
