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The planet has a radius of 128 kilometres and the surface gravity is the same as Earth's. My calculations have discovered that the planet would have a density of about 50 times the Earth's: $$g_P=g_E$$ $$\frac{GM_P}{(r_P)^2}=\frac{GM_E}{(r_E)^2}$$ $$\frac{\rho_PV_P}{(r_P)^2}=\frac{\rho_EV_E}{(r_E)^2}$$ $$\frac{\rho_P\frac43\pi(r_P)^3}{(r_P)^2}=\frac{\rho_E\frac43\pi(r_E)^3}{(r_E)^2}$$ $$\frac{\rho_P(r_P)^3}{(r_P)^2}=\frac{\rho_E(r_E)^3}{(r_E)^2}$$ $$\rho_Pr_P=\rho_Er_E$$ $$\rho_P\times128\ km\approx\rho_E\times6371\ km$$ $$\rho_P\approx\frac{6371}{128}\rho_E$$ $$\approx49.78\ \rho_E$$ $$\approx49.78\times5.51\ g/cm^3$$ $$\approx274.29\ g/cm^3$$ $$=274290\ kg/m^3$$ I am aware that such a planet would be denser than the core of the sun. However, I am handwaving the problems, such as what the planet would be made of, etc. The density of the core is at around $320000\ kg/m^3$.

My question:

This planet has an ocean taking up roughly the same portion of surface area as the Earth's. How deep can it be? (The bottom must be liquid water) What else can be said about it?

Feel free to make a suggestion.

user_194421
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1 Answers1

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Assuming uniform gravity the ocean can be at least 60km deep at 0 degrees C increasing to around 128km at 220 degree C. So it depends on the temperature. Below these depths and temperatures the pressure is sufficient to turn water into solid phase VI ice and below 110km phase VII ice.

water phase diagram

https://en.wikipedia.org/wiki/File:Phase_diagram_of_water.svg

However gravity will not be uniform on such a small world. The increase in pressure will become much less the deeper you travel due to reduced net gravitational forces so it is reasonable to suppose that the ocean could remain liquid down to 128km even at 0 degrees C.

All of this is dependent on the properties of your hand-waved material allowing such a depth to be reached and not reacting with the water. It is also dependent upon any solutes in the water among other things.

Slarty
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