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.

  • 2
    $\begingroup$ What is the reason you changed your requirements from 64km to 128km? If you're not sure about your question it would be prudent to first put it into the question sandbox until it is properly fleshed out. $\endgroup$
    – dot_Sp0T
    Nov 5, 2017 at 12:34
  • 1
    $\begingroup$ @user_194421 If you will keep a direct link somewhere, you can delete it, edit when deleted, and undelete when ready. That way no one will be able to comment or vote (up, down, close votes) until you're ready. There are no votes on it now, but maybe you will want to consider this. $\endgroup$
    – Mołot
    Nov 5, 2017 at 12:43
  • 1
    $\begingroup$ "I am handwaving the problems, such as what the planet would be made of, etc." - this seems to be a bad fit for hard-science tag - it is impossible to find scientific papers on a celestial body with handwaved, impossible properties. $\endgroup$
    – Mołot
    Nov 5, 2017 at 12:50
  • 1
    $\begingroup$ we know the average total density of the mass of water and of the "useronium". How can we determine the amount of water if the "useronium" is purely handwave? At least give use what density you have handwaved for it... $\endgroup$
    – L.Dutch
    Nov 5, 2017 at 12:58
  • 1
    $\begingroup$ Freezing point of water with respect to pressure Will probably be useful to whoever has the patience to work out the math $\endgroup$
    – nzaman
    Nov 5, 2017 at 14:11

1 Answer 1


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


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .