The acceleration of free falls in a gravitational field of moon we can discuss with using of Newton's law of gravity
(1) "g_\u0441=G\\frac {M_\u0441}{R_c^2} \\cong 3.7\\cdot 10^{-5} m\\cdot s^{-2}" , where "G=6.67\\cdot 10^{-11} m^3 kg^{-1}s^{-2}" - Newton's gravitational constant, "M_\u0441 =7.3\\cdot 10^{22} kg" - moon mass, "R_c=380\\ 000 km" - average distance from the center of Earth to the center of moon. If we consider a prob little mass at sublunar place on the Earth surface its distance form the moon center become smaller "R_{ps}=R_c - R_{\\oplus}" and we can understand from (1) that the gravity field will be grater "g_{ps}=G\\frac {M_\u0441}{R_{ps}^2} > g_c" . For opposite so called antipodal place the distance become grater then "R_c<R_{pa}=R_c +R_{\\oplus}" and the gravity field will be less "g_{pa}=G\\frac {M_\u0441}{R_{pa}^2} < g_c" . Thus at sublunar places the moon will attract the ocean water with increased force, and at antipodal places with decreased force. So we have two places with high tide on the Earth as shown in figure (https://en.wikipedia.org/wiki/Tide).
Answer: One expect the ocean to be not round, but will have two high tide on the Earth surface.
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