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.