I have always wondered and once I even got it, but then completely forgot. I understand that gravity causes high and low tides in oceans, but why does it occur on the other side of Earth?

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    $\begingroup$ A lot of tidal action is due to driven oscillations in the oceans (I'm not taking about the small surface waves). Think moving a box of water from side to side at a regular frequency, but on a much bigger and slower scale, with the Sun and the Moon doing the driving. $\endgroup$ – binaryfunt Jun 10 '14 at 15:15
  • $\begingroup$ Why not? Gravity doesn't stop halfway through the Earth. $\endgroup$ – user207421 Jun 11 '14 at 11:24
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    $\begingroup$ When I take a little boy by the hands, pick him up and twirls around, his arms are pulled horizontal; that is understandable, since I am pulling them. But his legs are pulled horizontal too, even though no one is pulling them. Likewise, he pulls my arms outward from my body, but somehow my coattails are also pulled outward from my back. $\endgroup$ – Beta Jun 11 '14 at 15:54
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    $\begingroup$ There are two good answers, and this explanation is short, so I'm making it a comment. Gravitation is a $1/r^2$ force. This means the Moon pulls the near side of the Earth away from the center of the Earth, and pulls the center of the Earth away from the far side of the Earth. $\endgroup$ – David Hammen Jun 26 '14 at 14:51
  • $\begingroup$ There's an animaton about these concepts here. $\endgroup$ – Pernk Dernets Oct 2 '17 at 13:40

Imagine that we have a very massive object in space. At some distance away (call it ten units) we release three tennis balls in a row:

Three tennis balls ten units away from a massive object

The tennis balls all fall towards the massive object. But because gravity goes like distance squared, the nearer balls feel a stronger attraction than the farther balls, and they move apart from each other:

Three tennis balls closer to the massive object

You're riding on the middle tennis ball. You feel like you're in free fall, in a good inertial frame. You look towards the heavy object and you see the leading tennis ball moving away from you. You look away from the heavy object and you see the following tennis ball moving away from you. The heavy object is pulling the three tennis balls apart.

Likewise, if you had three objects at the same distance falling towards the massive object,

Three tennis balls at the same distance of the massive object

you'd see them converge as they all fell along slightly different rays towards the same center. This gives the tidal compression. You can imagine the process of launching a whole constellation of tennis balls, choosing the center one as your "rest frame," and having their motions approximate the arrow pattern in Joshua's figure.

The situation stays essentially the same if you add angular momentum, except that then your tennis ball constellation doesn't crash onto the massive object.

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    $\begingroup$ I took the liberty of redrawing your ASCII art as images (mostly because I really liked the explanation and the illustration didn't go too well with it). The source files can be found here (Stack Exchange doesn't allow SVG uploads so if any changes are necessary to the images, the SVG source should make that easier). $\endgroup$ – Joey Jun 11 '14 at 8:22
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    $\begingroup$ The explanation for the tidal compression doesn't cut it. In fact, ignoring gravitational attraction among the tennis balls, they could all be circling in a common circular orbit, at the same speed but at slightly spaced out positions, without ever converging to each other. Indeed one could fill the whole orbit with $n$ equally spaced tennis balls at angles $2\pi/n$ from the orbit center, and they certainly would have no tendency to lump together. $\endgroup$ – Marc van Leeuwen Jun 11 '14 at 11:36
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    $\begingroup$ It is not "essentially the same if you add angular momentum" - it is critical to understanding how this really works. $\endgroup$ – Floris Jun 11 '14 at 14:36
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    $\begingroup$ Question: so angular momentum has nothing to do with the effect $\endgroup$ – Muhammad Umer Jun 14 '14 at 16:17
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    $\begingroup$ @MuhammadUmer There are two main effects. The big one is that with angular momentum the two bodies will not crash into each other. The smaller one is that the tidal stretching and compression takes energy, which comes from the difference between rotational and orbital periods. After a long time, two bodies with strong tidal interactions will orbit and rotate at the same rate, so that the same sides face each other. This is why we always see one side of the moon. $\endgroup$ – rob Jun 14 '14 at 19:15

First we must understand a little what is meant by "tide." A tide is the difference of gravitational force an object feels across its volume from another object. In the Earth's case the side closest to the moon feels a stronger force pulling it towards the moon than the center of the Earth does, while the side opposite the moon feels a force weaker than the center of the Earth feels. The picture below (taken from this site, which is a great reference as well, especially for explaining some misconceptions about the second lunar tidal bulge) shows this. The center of the earth feels a force toward the moon as calculated by Newton's Law of Gravitation:

$$F=G \frac{m_1 m_2}{r^2}$$

while the other areas of the Earth's surface feels a slightly different force from the moon than the center of the Earth does, as demonstrated by the arrows. The side closest to the moon feels an additional force by virtue of being closer to the moon, as demonstrated by the arrows pointing towards the moon, while the side furthest away feels a less strong force, represented by the arrows pointing away from the moon (here represented as a generic satellite).

The side closest to the moon has a tidal bulge because of the additional gravitational force pulling the sea level higher than the average level, while the side opposite the moon also has a tidal bulge by virtue of the lessened force of gravity it feels being further away from the moon. So, both bulges are caused by the moon; one side feels a greater attraction, while the other side feels a smaller attraction.

  • $\begingroup$ if the side far feels less gravity of moon then why does it bulge or why does it in fact go in opposite direction? Is the earth also being pulled..So bulge on far side is a place where would have been to some part if there was no moon.. $\endgroup$ – Muhammad Umer Jun 9 '14 at 13:50
  • $\begingroup$ @MuhammadUmer, look at the arrows on the picture. The lessened gravity on that side of the Earth leads to a force that points away from the surface of the Earth (when the main component of the moon's gravity that everything on Earth feels is taken out) the same as the increased force on the side facing the moon produces a force towards the moon away from the Earth. $\endgroup$ – NeutronStar Jun 9 '14 at 14:22
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    $\begingroup$ @MuhammadUmer, rob's answer below provides a good visualization of the effect. $\endgroup$ – NeutronStar Jun 9 '14 at 16:38
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    $\begingroup$ @Muhammad: You're forgetting that there's a big rightward arrow that all of these should be added to to get the actual gravitational attraction: none of the attraction is in the "opposite direction". The diagram above has simply subtracted off the big rightward arrow so we can more easily see the differences: in particular, how it looks from an earth-centered reference frame. $\endgroup$ – user5174 Jun 9 '14 at 19:06
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    $\begingroup$ @TheodoreNorvell, that is not correct. Centrifugal/centripetal forces from the Earth-moon system do not make significant contributions to the Earth's tides. $\endgroup$ – NeutronStar Jun 9 '14 at 19:12

The Earth is free falling towards the Moon. Because gravity decays with distance, the side near the moon wants to fall faster than the center of the Earth, while the other side tends to fall slower. So observed on the Earth, the other side "lags behind" and therefore we have high tide there.

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    $\begingroup$ Isn't this a website that is supporting this explanation? Why would another website be more/less credible than this one? I believe a better thing to ask is, "can you show the math or physical picture behind this answer?" $\endgroup$ – NeutronStar Jun 9 '14 at 4:21
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    $\begingroup$ @LDC3 While the explanation given here is not mathematical it is correct. You can also frame it in terms of orbital mechanics, of course, but that ends up sounding a little silly when applied to low angular momentum cases. $\endgroup$ – dmckee --- ex-moderator kitten Jun 9 '14 at 4:37
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    $\begingroup$ @Joshua I don't know user139981 and his statement didn't have any support. I agree that most web sites are also missing support for their statements. I asked user139981 to show a web site since I thought his statement was wrong, but I wasn't certain. His statement takes a different viewpoint of the situation, which you provided in your answer. $\endgroup$ – LDC3 Jun 9 '14 at 5:33
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    $\begingroup$ @Joshua: Too Long/Didn't Read. Usually written as TL:DR but now people have gotten even lazier and simply write TLDR $\endgroup$ – slebetman Jun 9 '14 at 14:37
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    $\begingroup$ @slebetman: Funny you say that; I think it may have originally been a semicolon. $\endgroup$ – Magus Jun 10 '14 at 19:08

Let us try to find the acceleration at points A and B with respect to the centre of the earth O due to the influence of moon and earth, as shown in the figure. enter image description here
O and X are the center of the earth and moon respectively. Let the radius of earth be $R_E$, distance between earth and moon be $d$, mass of earth and moon be $m_E$ and $m_M$ respectively. $O$ to $X$ is taken as the positive direction. I have assumed $R_E << d$.

Acceleration of point B $a_B$ is: $$a_B = -\frac{Gm_E}{R_E^2} + \frac{Gm_M}{(d-R_E)^2}$$ $$= -\frac{Gm_E}{R_E^2} + \frac{Gm_M}{d^2}(1+ \frac{2R_E}{d})$$
Similarly, $a_A$ is: $$a_A=\frac{Gm_E}{R_E^2} + \frac{Gm_M}{(d+R_E)^2}$$ $$=\frac{Gm_E}{R_E^2} + \frac{Gm_M}{d^2}(1-\frac{2R_E}{d})$$
And $a_O$ is: $$a_O = \frac{Gm_M}{d^2}$$

Thus, the accelerations of point A and B with respect to O are: $$a_{AO} = a_A-a_O = \frac{Gm_E}{R_E^2} - \frac{2Gm_MR_E}{d^3}$$ $$a_{BO} = a_B-a_O = -\frac{Gm_E}{R_E^2} +\frac{2Gm_MR_E}{d^3}$$
But now, we get $a_{BO}=-a_{AO}$, which means that on both the sides, water will be trying to move away from the centre of the earth, thus causing tides on both the sides of the earth.


I tried it this way. Consider two equal spherical with masses M, covered with a layer of water, circling around each other at a distance $l$ (measured between their centers). They have radius $R$ and angular velocity $\omega$.

Now let's look at one of them (the same applies to the other). For a point on the outside, (at distance $\frac 1 2 {l+R}$ from their CM) there are three forces:

$F_{cf}$, the centrifugal force directed away from the CM

$F_{gM}$, the gravitational force from the sphere we are looking at, directed towards the CM

$F'_{gM}$, the gravitational force from the other sphere, directed towards the CM

So for a test mass m:

$$F_{cf}=-\frac{mv^2}{{\frac 1 2}l+R}=-m{\omega}^2({\frac 1 2}l+R)$$

$$F_{gM}=m \frac{GM}{R^2}$$

$$F'_{gM}=m \frac{GM}{(l+R)^2}$$

Now we must find out if $F_{cf}$ pulls harder or less (or equal) on a test mass than (to) $F'_{gM}$.

We can write

$$M{\omega}^2({\frac 1 2}l)=\frac{M^2 G}{l^2}$$ so $$MG={\frac 1 2}{\omega}^2 l^3$$

Putting this in the expression for $F'_{gM}$ gives:

$$F'_{gM}=\frac 1 2 m{\omega}^2\frac{l^3}{(l+R)^2}$$

So we have to find out if ${\frac 1 2}l+R$ is smaller or bigger (or equal) than (to) ${\frac 1 2}{\frac{l^3}{(l+R)^2}}$

Now we can write:

$${\frac 1 2}{\frac{l^3}{(l+R)^2}}={\frac 1 2}{\frac{l^3}{l^2+2lR+R^2}}={\frac 1 2}{\frac l {1+{\frac {2R}{l}}+{\frac{R^2}{l^2}}}}$$

which is obviously less than ${\frac 1 2}l +R$.

So we can conclude that for a mass $m$, on the outside of the spherical mass, the outward directed centrifugal force is bigger than the gravitational force caused by the other mass. Which implies an outward bulge develops.

We can apply the same procedure for the inside (the side that faces the other mass M) of the mass M, and that will show that the gravitational force from the other mass is bigger than the centrifugal force, which implies an inward bulge (towards the other mass) develops. I took two equal masses for simplicity, but you can do the same for two different spherical masses, circling around their CM. The result will remain the same (two bulges in opposite directions if both are covered with water; so if the moon was covered by a sea, also on the moon two bulges, on the line connecting the earth and the moon, would appear).

Looking at one mass I only had to compare the outward centrifugal force with the gravitational force caused by the other mass, so I didn't have to care about the force caused by the mass we were looking at, because they are equal everywhere (in approximation) on the surface of the mass, as is the case for the centrifugal force caused by an eventual rotation of the masses.

I didn't calculate if the bulges were different in size, but I guess they are.


This is because the gravitational field of the Moon, like any object, is not uniform - in particular, closer to the Moon it is stronger, and further away it is weaker. With regard to the Earth, the side of the Earth that is closest to the Moon experiences a slightly stronger pull than the side further away, which effectively results in the Earth being "stretched" ever so slightly as the nearer side accelerates harder than the farther side in response - and when you stretch an elastic sphere it becomes an oblong with a bulge on each side and not just on one side like a pear (as I imagine you'd be thinking it would have to look.).

From the viewpoint of the Earth's center of mass, which is being accelerated by this effect and may be more natural for you, the reference frame change results in the side of the Earth closest to the Moon as experiencing a force toward it, and the opposite side experiences a "fictitious" force directed away, thus creating a stretch in both directions - this opposite fictitious force because that frame is not inertial, just as with driving in your car how there is a "fictitious" force when you slam the gas that wants to push you into the seat and throw the bobblehead off the dashboard.

In a uniform gravitational field, this effect does not occur. The difference in forces that produces the stretching is called, perhaps not unsurprisingly, a "tidal force".

Also, if one has read any pop sci books or seen movies about "black holes" and they've talked about getting "pulled apart like spaghetti" when you fall in because the force on your feet is higher than that on your head, that is exactly this effect but much more extreme - and conversely, this effect is a very, very incipient form of the "spaghetti pull" effect manifested over a far larger distance due to the much gentler gravitational gradient.


Another way of picturing the answer is by considering that earth rotates around the same point as does our moon. If the moon had the same mass as the earth this point would be halfway between the two bodies. As the moon is much lighter than earth the centre of rotation is much closer to our planet but it is outside earth's surface. Because earth rotates around this Centre of Gravity, all parts of it experience varying centripetal forces. Solid and liquid matter on the part of earth surface that is furthest away from the CofG have the highest forces, matter nearest the CofG has the least while in between there is a gradient of magnitudes. So water on the far side is forced out the most, on the nearside the least and the solid in between somewhere between the two. So two bulges, although not quite in line with the moon because of restrictions to water flow (friction). When the sun and the moon are on the same line as earth the tides are higher.

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    $\begingroup$ No, this is incorrect. Tidal forces are due to the differential of the gravitational field one object exerts on the other. Two equal mass objects orbiting around each other would exert qualitatively the same kind of tidal force fields (double bulge) upon each other as does the Moon on the Earth, even though in that case they would both be circling a centre-of-mass in the middle, with all "centrifugal forces" directed away from this point. $\endgroup$ – Marc van Leeuwen Jun 9 '14 at 15:14
  • $\begingroup$ Tides can be coherently explained in terms of orbital dynamics (this is the method preferred by most SF writers), but it has to be done carefully and not in terms of centripetal forces or centrifugal pseudo-forces. $\endgroup$ – dmckee --- ex-moderator kitten Jun 9 '14 at 15:18
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    $\begingroup$ Also the center of gravity of the Earth-Moon system is actually inside the Earth's surface. $\endgroup$ – Marc van Leeuwen Jun 11 '14 at 10:27

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