# Tide on the Moon

If the Earth's Moon had a water ocean of depth 2-4 Km, how high would the tides rise due to the Earth's gravity? (Just a hypothetical question.)

• What do you mean by tides? Since our moon is tidally locked to the Earth, and thus Earth tides on moon will be stationary with respect to its surface (this is not 100% true since its orbit is slightly elliptical, which cause libration). Jan 10, 2015 at 16:28
• Slightly extending that question: What about tidal influence due to the sun? If the moon is locked to the Earth, it should experience tidal forces due to the changing distance to the sun. Jan 13, 2015 at 8:21
• The Sun's distance would not greatly affect tides, but its position would, as the Moon is not tidally locked to the Sun as it is to Earth. While the affect of the Earth would be rather consistent and may not qualify to the Earthlike concept of tides, the Moon would have Earthlike tides caused by the Sun, though perhaps not as pronounced - something which the current answers seem to have overlooked. Jan 21, 2015 at 8:15

Imagine installing a pipe with cross-section 1 cm$^2$ from the top of the ocean at its high point down to the center of the moon and then back up to the top of the ocean at a point where the net tidal field from the Earth is 0. Fill it with water until one side is equal with the water level on one side and its level on the other side will automatically be equal to the water level there. The weights of the water on both sides need to be equal if it is static. For now, we will ignore that g changes when one gets deep into the moon. Therefore, roughly speaking $\rho$gh should be the same in both sections of the pipe, and: $$\rho(g_M - g_t)(h_t + r_M) = \rho g_M r_M$$
where $g_t$ is tidal acceleration $$g_t(Moon) = 2\frac{GM_Er_M}{d^3}.$$ Subscript M means Moon and E means Earth and d is the separation between the Earth and Moon. Solving for h$_t$ gives: $$h_t = \frac{g_tr_M}{g_M - g_t} \sim \frac{g_t}{g_M}r_M$$ We could substitute values in here, but if we do this in comparison with the height of the tide on the Earth, 54 cm, we cancel out some of the error in the approximation and get an even simpler formula, namely, $$\frac{h_t(Moon)}{h_t(Earth)} = \frac{M_E}{M_M} \frac{g_E}{g_M} \Big(\frac{r_M}{r_E}\Big)^2$$ The mass of the Earth is 81.3 time greater, the acceleration at the surface of the Moon is 1/6.25 of Earth's g force, and the diameter of the Moon is 0.2725 of the Earth. So, that is 81.3*6.25*.2725$^2$ = 37.7 time higher than the Earth's tide or 37.7*54 cm = 20.4 meters. Note that the height of the ocean does not matter as long as it is considerably more than 20 meters.