<p>The tide would be locked in place, roughly, because the Moon always shows the same side to the Earth (tidally locked). But the height of the water at the location facing the Earth and the opposite side should be greater than the mean height, and at the limb, as we see it, the height will be lower than the mean but by half the magnitude of the high tide. </p>

<p>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:
$$
\rho(g_M - g_t)(h_t + r_M) = \rho g_M r_M
$$</p>

<p>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.</p>

<p>On the other hand, the water would begin boiling since it is at 0 pressure until a water atmosphere is created, but the moon is too small to hold an atmosphere for long and the whole ocean would be gone in a while.  </p>

CC BY-SA 3.0
2016-09-11T16:48:32.550Z
The height of the water relatively to what, to which level standard?
Incnis Mrsi
CC BY-SA 3.0
2016-09-12T20:03:04.647Z
Well, relative to "sea level", of course.  Sea level would be the level that the top of the water would reach with no external gravity source, if the entire body were covered in water.   That is, if there were no earth nearby the moon.
eshaya
CC BY-SA 3.0
2015-01-20T23:13:04.107Z
2015-01-24T05:05:41.953Z
2015-01-24T05:05:41.953Z
eshaya
<p>There would be a tide of approximately 28 days due to centrifuge effect,the water being free flowing,about the mid point centre of gravity of the 2 masses.
Our tides are monthly but appear daily due to earth 24hr rotation.Variation in sub-marine land masses and ocean baison cross flow have sub tidal effect that can lag or lead the moon phased tide.On a water depth of 2-4km it may,as an independent body, go towards the outside of the rotational system leaving one side dry so to speak.</p>

CC BY-SA 3.0
2015-01-13T20:56:41.023Z
2015-01-13T20:56:41.023Z
Colin Shorey
<p>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.)</p>

CC BY-SA 3.0
2015-01-10T16:28:38.657Z
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](http://en.wikipedia.org/wiki/Libration)).
fibonatic
CC BY-SA 3.0
2015-01-13T08:21:31.813Z
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.
engineer
CC BY-SA 3.0
2015-01-21T08:15:41.567Z
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.
Mitch Goshorn
<the-moon><gravity><earth><tidal-forces>
Tide on the Moon
Devgeet Patel
# Tide on the Moon

(Asked by: username_0 on 2015-01-09)

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.)

**Tags:** the-moon, gravity, earth, tidal-forces
> **Comments:**
> **username_1**: 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](http://en.wikipedia.org/wiki/Libration)).  
> **username_2**: 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.  
> **username_3**: 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.  

---

## 2 answers

### Answer by username_4 on 2015-01-20. Score: 1

The tide would be locked in place, roughly, because the Moon always shows the same side to the Earth (tidally locked). But the height of the water at the location facing the Earth and the opposite side should be greater than the mean height, and at the limb, as we see it, the height will be lower than the mean but by half the magnitude of the high tide.

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.

On the other hand, the water would begin boiling since it is at 0 pressure until a water atmosphere is created, but the moon is too small to hold an atmosphere for long and the whole ocean would be gone in a while.

> **Comments:**
> **username_5**: The height of the water relatively to what, to which level standard?  
> **username_4**: Well, relative to "sea level", of course.  Sea level would be the level that the top of the water would reach with no external gravity source, if the entire body were covered in water.   That is, if there were no earth nearby the moon.  

---

### Answer by username_6 on 2015-01-13. Score: 0

There would be a tide of approximately 28 days due to centrifuge effect,the water being free flowing,about the mid point centre of gravity of the 2 masses.
Our tides are monthly but appear daily due to earth 24hr rotation.Variation in sub-marine land masses and ocean baison cross flow have sub tidal effect that can lag or lead the moon phased tide.On a water depth of 2-4km it may,as an independent body, go towards the outside of the rotational system leaving one side dry so to speak.


---

