<p>That is because the moon attracts both the water <em>and</em> the earth. The gravity of the moon reduces with distance (by the inverse square law). So, the moon's gravity is </p>

<ul>
<li>Greatest at the point nearest to the moon (called sub-moon point)</li>
<li>Is lesser on the earth's center, and'</li>
<li>Is the least in the opposite side of the earth.</li>
</ul>

<p>Assume that earth is falling to wards the moon. Now, as the force on the point near the moon is the maximum and it has 'fallen' the maximum distance, while the earth 'falls' a lesser distance and the farthest of all, the antipodal point 'falls the least'.</p>

<p><a href="https://i.sstatic.net/CFfae.png" rel="noreferrer"><img src="https://i.sstatic.net/CFfae.png" alt="Tidal forces"></a></p>

<p>"<a href="https://commons.wikimedia.org/wiki/File:Tidal-forces.png#/media/File:Tidal-forces.png" rel="noreferrer">Tidal-forces</a>". Licensed under <a href="http://creativecommons.org/licenses/by-sa/3.0/" rel="noreferrer">CC BY-SA 3.0</a> via <a href="//commons.wikimedia.org/wiki/" rel="noreferrer">Wikimedia Commons</a>.</p>

<p>The above figure shows the forces acting on earth due to moon, The top figure shows the actual gravitational forces acting, while the bottom one shows the forces after subtracting the force on the earth itself. That should explain the antipodal bulge. See the wikipedia ariticle in <a href="https://en.wikipedia.org/wiki/Tidal_force" rel="noreferrer">Tidal forces</a></p>

CC BY-SA 4.0
2015-09-22T12:12:43.680Z
2018-11-06T07:08:30.660Z
2018-11-06T07:08:30.660Z
aeroalias
<p>I am probably going to get slammed for this, because it violates everything we were taught about tidal forces, but the antipodal tide is caused by the centrifugal force created by the Earth's rotation about the earth/moon barycenter, not differential gravitational forces.  While the moon's gravity is less on the side of the earth furthest from it, that force is still towards the moon, not away from it.<br><br>There is a good explanation of this centrifugal force explanation on NOAA's website <a href="https://tidesandcurrents.noaa.gov/restles3.html" rel="noreferrer">here</a>. I'll cite some pictures and pertinent text, but this site is a good read for anyone trying to understand the forces responsible for the Earth's tides.<br><br>Here is a diagram showing the earth and moon's movement around the system's barycenter:<br><a href="https://i.sstatic.net/1TxAd.jpg" rel="noreferrer"><img src="https://i.sstatic.net/1TxAd.jpg" alt="enter image description here"></a><br>And some of the pertinent text:<br><br><blockquote>The center of revolution of this motion of the earth and moon around their common center-of-mass lies at a point approximately 1,068 miles beneath the earth's surface, on the side toward the moon, and along a line connecting the individual centers-of-mass of the earth and moon. (see G, Fig. 1) The center-of-mass of the earth describes an orbit (E1, E2, E3..) around the center-of-mass of the earth-moon system (G) just as the center-of-mass of the moon describes its own monthly orbit (M1, M2, M3..) around this same point. </p>

<ol>
<li>The Effect of Centrifugal Force. It is this little known aspect of the moon's orbital motion which is responsible for one of the two force components creating the tides. As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force. And, since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon. This fact is indicated by the common direction of the arrows (representing the centrifugal force Fc) at points A, C, and B in Fig. 1, and the thin arrows at these same points in Fig. 2.</blockquote><br>And finally another diagram that the blockquote cites:<br><br><a href="https://i.sstatic.net/zVOzk.jpg" rel="noreferrer"><img src="https://i.sstatic.net/zVOzk.jpg" alt="Moon gravitational diagram"></a><br><br>Yes, one can probably find lots and lots of quotations declaring that the antipodal tide is caused because the moon's gravitational force is much less on the far side than on the near side, but that doesn't make it true. And I believe NOAA ought to be a pretty authoritative source. If they can't get it right...</li>
</ol>

CC BY-SA 3.0
2015-09-25T23:59:04.073Z
Without having ever worked it out, I'd wager this is equivalent to the answer of aeroalias, with it all depending on the reference frame you pick.  Aeroalias' answer is derived from the reference frame of the center of the Earth.  It appears that yours is derived from the reference frame of the barycenter.  Which still makes it a completely legitimate answer and derivation in its own right, to be clear.  I'm just not seeing anything in your answer that debunks the perspective in aeroalias' answer (though see my comment on the OP: the idea is more sensible on a planetary scale).
zibadawa timmy
CC BY-SA 3.0
2015-09-26T01:10:07.363Z
@zibadawatimmy, this explanation is, to me, more satisfactory than the gravitational difference explanation. And I think any explanation of the Earth's tides should include discussion of the planet's rotation around the earth/moon barycenter; it's there, it's real, and must have an effect on the planet.
BillDOe
CC BY-SA 3.0
2015-09-26T01:56:38.953Z
It's all relative.  If I'm right and they're the same answer in different reference frames, then I would prefer aeroalias' answer: it's really simple.  And if it's just a difference of frames, the gravitational difference explanation is *exactly the same as* the barycenter explanation, just in different coordinates.  I personally don't see a reason to prefer a barycenter perspective unless there's something else to gain.  If there were some other phenomena you were considering in conjunction with the moon's forcing function on the tides that is easier to deal with in the barycenter...
zibadawa timmy
CC BY-SA 3.0
2015-09-26T02:36:28.927Z
I disagree with some of the claims made. If we're in the barycentric inertial frame, then invoking centrifugal force doesn't really make sense because we're *not* in a rotating frame. The hypothetical tidal bulges would be as in **aeroalias**'s answer. ... Conversely, if we're in frame co-rotating with the Earth, then talk of centrifugal force on the Earth is equivalent to talk of the gravitational force due to the Moon by construction of the frame--so I don't think this answer's explanation is *wrong*, but it is needlessly confusing while misleadingly implying that it's The One True Way.
Stan Liou
CC BY-SA 3.0
2015-09-25T23:15:09.703Z
2015-09-25T23:15:09.703Z
BillDOe
<p>The bulges are caused by the fact that the earth's center is free falling  and the rest of the planet is not properly.  The vector force aplied by the Moon in a certain point minus the vector force aplied by the Moon on the Earth's center is not equal to zero. This difference causes the bulges. The Earth and Moon are orbiting the center of mass of the system. Consider someone in a box free falling. The person will not feel force acting. The diference between what the Moon's force on an Earth's point is and what the Moon's force on this same point should be is the cause of the bulges. When I said should be I meant that this point falls with the Earth but only the Earth's center of mass is free falling "correctly" with F =Mm/r^2</p>

CC BY-SA 4.0
2018-11-06T02:37:39.853Z
2018-11-08T21:49:17.273Z
2018-11-08T21:49:17.273Z
Anderson Alves
<p>The <a href="http://en.wikipedia.org/wiki/Tide" rel="noreferrer">Wikipedia article on tides</a> shows a diagram with a tidal bulge both towards the Moon and on the antipodal point (the Sun isn't included in the diagram). Surely that's wrong - what would cause the antipodal bulge?</p>

<p><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Tide_overview.svg/871px-Tide_overview.svg.png" rel="noreferrer"><img src="https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Tide_overview.svg/871px-Tide_overview.svg.png" alt=""></a>
<br>
<sup>Image courtesy of Wikipedia user Jhbdel under the <a href="http://creativecommons.org/licenses/by-sa/3.0/deed.en" rel="noreferrer">Creative Commons Attribution-Share Alike 3.0 Unported license</a>.</sup></p>

CC BY-SA 3.0
2015-09-24T09:05:45.187Z
For a detailed analysis of this diagram/effect with respect to oceanic tides, see [this Physics.Se Q&A](http://physics.stackexchange.com/questions/121830/does-earth-really-have-two-high-tide-bulges-on-opposite-sides).  The effect does make sense over the entire planet--rocky bits and all--but when restricted to just oceanic tides things don't work out quite so well.
zibadawa timmy
CC BY-SA 4.0
2018-11-09T17:43:48.193Z
I guess this seems surprising because it is tempting to imagine that the Earth and Moon are pinned in place as in an orrery.  https://en.m.wikipedia.org/wiki/Orrery  However, the real objects are not pinned in place,.
badjohn
<tidal-forces>
What causes the antipodal bulge?
rdt2
# What causes the antipodal bulge?

(Asked by: username_0 on 2015-09-22)

The [Wikipedia article on tides](http://en.wikipedia.org/wiki/Tide) shows a diagram with a tidal bulge both towards the Moon and on the antipodal point (the Sun isn't included in the diagram). Surely that's wrong - what would cause the antipodal bulge?

[![](https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Tide_overview.svg/871px-Tide_overview.svg.png)](https://upload.wikimedia.org/wikipedia/commons/thumb/e/eb/Tide_overview.svg/871px-Tide_overview.svg.png)
  
Image courtesy of Wikipedia user Jhbdel under the [Creative Commons Attribution-Share Alike 3.0 Unported license](http://creativecommons.org/licenses/by-sa/3.0/deed.en).

**Tags:** tidal-forces
> **Comments:**
> **username_1**: For a detailed analysis of this diagram/effect with respect to oceanic tides, see [this Physics.Se Q&A](http://physics.stackexchange.com/questions/121830/does-earth-really-have-two-high-tide-bulges-on-opposite-sides).  The effect does make sense over the entire planet--rocky bits and all--but when restricted to just oceanic tides things don't work out quite so well.  
> **username_2**: I guess this seems surprising because it is tempting to imagine that the Earth and Moon are pinned in place as in an orrery.  https://en.m.wikipedia.org/wiki/Orrery  However, the real objects are not pinned in place,.  

---

## 3 answers

### Answer by username_3 on 2015-09-22. Score: 25

That is because the moon attracts both the water *and* the earth. The gravity of the moon reduces with distance (by the inverse square law). So, the moon's gravity is

* Greatest at the point nearest to the moon (called sub-moon point)
* Is lesser on the earth's center, and'
* Is the least in the opposite side of the earth.

Assume that earth is falling to wards the moon. Now, as the force on the point near the moon is the maximum and it has 'fallen' the maximum distance, while the earth 'falls' a lesser distance and the farthest of all, the antipodal point 'falls the least'.

[![Tidal forces](https://i.sstatic.net/CFfae.png)](https://i.sstatic.net/CFfae.png)

"[Tidal-forces](https://commons.wikimedia.org/wiki/File:Tidal-forces.png#/media/File:Tidal-forces.png)". Licensed under [CC BY-SA 3.0](http://creativecommons.org/licenses/by-sa/3.0/) via [Wikimedia Commons](//commons.wikimedia.org/wiki/).

The above figure shows the forces acting on earth due to moon, The top figure shows the actual gravitational forces acting, while the bottom one shows the forces after subtracting the force on the earth itself. That should explain the antipodal bulge. See the wikipedia ariticle in [Tidal forces](https://en.wikipedia.org/wiki/Tidal_force)


---

### Answer by username_4 on 2015-09-25. Score: 8

I am probably going to get slammed for this, because it violates everything we were taught about tidal forces, but the antipodal tide is caused by the centrifugal force created by the Earth's rotation about the earth/moon barycenter, not differential gravitational forces. While the moon's gravity is less on the side of the earth furthest from it, that force is still towards the moon, not away from it.  
  
There is a good explanation of this centrifugal force explanation on NOAA's website [here](https://tidesandcurrents.noaa.gov/restles3.html). I'll cite some pictures and pertinent text, but this site is a good read for anyone trying to understand the forces responsible for the Earth's tides.  
  
Here is a diagram showing the earth and moon's movement around the system's barycenter:  
[![enter image description here](https://i.sstatic.net/1TxAd.jpg)](https://i.sstatic.net/1TxAd.jpg)  
And some of the pertinent text:  
  
> The center of revolution of this motion of the earth and moon around their common center-of-mass lies at a point approximately 1,068 miles beneath the earth's surface, on the side toward the moon, and along a line connecting the individual centers-of-mass of the earth and moon. (see G, Fig. 1) The center-of-mass of the earth describes an orbit (E1, E2, E3..) around the center-of-mass of the earth-moon system (G) just as the center-of-mass of the moon describes its own monthly orbit (M1, M2, M3..) around this same point.

1. The Effect of Centrifugal Force. It is this little known aspect of the moon's orbital motion which is responsible for one of the two force components creating the tides. As the earth and moon whirl around this common center-of-mass, the centrifugal force produced is always directed away from the center of revolution. All points in or on the surface of the earth acting as a coherent body acquire this component of centrifugal force. And, since the center-of-mass of the earth is always on the opposite side of this common center of revolution from the position of the moon, the centrifugal force produced at any point in or on the earth will always be directed away from the moon. This fact is indicated by the common direction of the arrows (representing the centrifugal force Fc) at points A, C, and B in Fig. 1, and the thin arrows at these same points in Fig. 2.  
   And finally another diagram that the blockquote cites:  
     
   [![Moon gravitational diagram](https://i.sstatic.net/zVOzk.jpg)](https://i.sstatic.net/zVOzk.jpg)  
     
   Yes, one can probably find lots and lots of quotations declaring that the antipodal tide is caused because the moon's gravitational force is much less on the far side than on the near side, but that doesn't make it true. And I believe NOAA ought to be a pretty authoritative source. If they can't get it right...

> **Comments:**
> **username_1**: Without having ever worked it out, I'd wager this is equivalent to the answer of aeroalias, with it all depending on the reference frame you pick.  Aeroalias' answer is derived from the reference frame of the center of the Earth.  It appears that yours is derived from the reference frame of the barycenter.  Which still makes it a completely legitimate answer and derivation in its own right, to be clear.  I'm just not seeing anything in your answer that debunks the perspective in aeroalias' answer (though see my comment on the OP: the idea is more sensible on a planetary scale).  
> **username_4**: @zibadawatimmy, this explanation is, to me, more satisfactory than the gravitational difference explanation. And I think any explanation of the Earth's tides should include discussion of the planet's rotation around the earth/moon barycenter; it's there, it's real, and must have an effect on the planet.  
> **username_1**: It's all relative.  If I'm right and they're the same answer in different reference frames, then I would prefer aeroalias' answer: it's really simple.  And if it's just a difference of frames, the gravitational difference explanation is *exactly the same as* the barycenter explanation, just in different coordinates.  I personally don't see a reason to prefer a barycenter perspective unless there's something else to gain.  If there were some other phenomena you were considering in conjunction with the moon's forcing function on the tides that is easier to deal with in the barycenter...  
> **username_5**: I disagree with some of the claims made. If we're in the barycentric inertial frame, then invoking centrifugal force doesn't really make sense because we're *not* in a rotating frame. The hypothetical tidal bulges would be as in **aeroalias**'s answer. ... Conversely, if we're in frame co-rotating with the Earth, then talk of centrifugal force on the Earth is equivalent to talk of the gravitational force due to the Moon by construction of the frame--so I don't think this answer's explanation is *wrong*, but it is needlessly confusing while misleadingly implying that it's The One True Way.  

---

### Answer by username_6 on 2018-11-06. Score: -1

The bulges are caused by the fact that the earth's center is free falling and the rest of the planet is not properly. The vector force aplied by the Moon in a certain point minus the vector force aplied by the Moon on the Earth's center is not equal to zero. This difference causes the bulges. The Earth and Moon are orbiting the center of mass of the system. Consider someone in a box free falling. The person will not feel force acting. The diference between what the Moon's force on an Earth's point is and what the Moon's force on this same point should be is the cause of the bulges. When I said should be I meant that this point falls with the Earth but only the Earth's center of mass is free falling "correctly" with F =Mm/r^2


---

