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Engineering LibreTexts Tidal Forces

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    To summarize the considerations in the above four subsection: (i) it was explained that not only a vertical drop in gravitational field, but \textit{any} motion of a body under the influence of the force of gravity alone is classified by physics as a free fall -and (ii), that for the very same reason the satellite motion is nothing else than a special case of the free fall. Why so much attention has been devoted to these issues?, and to showing that satellite motion is just a special case of free fall? Because the next task is to explain the mechanism ocean tides -and it can hardly be done unless the readers has a good understanding of the mechanics of motion of satellites and planets (note that planets are also satellites - of the central star in the planetary system).

    The thing is that there are plenty of misconceptions about ocean tides. The explanation of the mechanism responsible for tides given in some college physics textbooks is partially incorrect, and in some it is even altogether incorrect. In Internet there is a number of articles discussing such misconceptions -for example, a thorough discussion presenting several ''wrong scenarios'' given in various textbooks can be found in this site.

    Most high-school students, or even elementary school students already know that the tides are caused by the the Moon gravity -which is correct. Some of them even know that there is also some weaker contribution from Sun, which is also true. But it's not enough to say only that in a college textbook -so the authors feel obliged to present a more detailed physical picture. And as is commonly known, ''the devil is in the details''. And this is exactly what happens.

    Bedeviled are the two tidal bulges that form on the oceans. One is on the side of the Earth facing the Moon. Here the explanations are ''less bedeviled'', saying that the bulge forms because the water in this bulge is closer to the Moon than the Earth's mass center, so it's attracted towards the Moon by a stronger force than that acting on the Earth's bulk. Earth is rocky and stiff, whereas water can flow - therefore it flows towards the stronger gravity and form the bulge.

    Quite embarrassing, though, is the formation of a similar bulge on the other side of Earth, the ''far side''. On the same token as before, here the Moon's gravity acting on things is weaker than that acting on the Earth's bulk. So, the fluid water should flow away and one would expect the formation of a ''dip'' rather than a bulge. It is quite common that for the formation of the other bulge authors blame ''the force of inertia''. Usually not going far into details -because the devil is in them, as has been said. In fact, it may be really difficult to explain why two different forces produce such a nearly-symmetric effect?

    The truth is that the two bulges are not produced by two different forces -there is a single force responsible for all tidal effects. Astronomers call it the ''tidal force''. Let's elaborate on the details. Consider a body of mass \(m\) and of small size experiencing gravitational attraction from another body of mass \(M\), \(r\) apart. The force of attraction, as we know, is given by the Newton's Law of Universal Gravitation:

    \[ F_{m\rightarrow M} = G\dfrac{mM}{r^2}. \]

    Now, suppose that the body of mass \(m\) cannot be thought of as a ''point-like object'', but is of finite size -call it \(D\). A closer analysis of the forces acting on \(m\) shows that in addition to \(F_{m\rightarrow M}\), another force emerges -one which tends to ''stretch'' the body, along the direction towards the mass \(M\). This force, called the ''tidal force'', is proportional to \(D\), and inversely proportional to \(r^3\):

    \[ F_{tidal} \propto \dfrac{D}{r^3} \;\;\; or \;\;\;F_{tidal}= k \dfrac{D}{r^3} ,\]

    where \(k\) is a certain coefficient depending on the body's shape. Anyway, no matter what the value of \(k\) is, because of the force's dependence on \(R^{-3}\), it grows quite rapidly when the distance between \(m\) and \(M\) decreases. Ocean tides on Earth are not the only known manifestation of the existence of tidal forces. Many other are observed in the Solar System. For instance, powerful tidal forces may occur in comets passing close to Jupiter. Comets are bodies consisting mostly of ''dirty ice'', which is a relatively soft material -in 1992, a Shoemaker-Levy Comet was ripped apart by tidal forces into several smaller parts during a close approach to Jupiter. The famous rings of Saturn are believed to made of debris of a moon that had been orbiting to close to the planet and was ''pulverized'' by tidal forces.

    As noted, the \(k\) coefficient depends on the shape of the body exposed to tidal forces -for a sphere it is different than for the cigar-shaped Oumuamua 750-by-115 foot object, a ''visitor'' from the interstellar space that zipped through the Solar System at the end of the 2017 year.

    In order to explain the origin of the tidal force, it may be instructive to calculate \(k\) for a simple object -we choose one whose shape is not necessarily similar to one of the existing celestial bodies, but such that it makes the calculations relatively straightforward. Namely, let's consider a spacecraft orbiting a planet of mass \(M\) along a circular trajectory of radius \(r\), measured from the planet's center. On board, there is a small spherical object of mass \(m\). Due to the weightlessness inside an orbiting spacecraft, we can imagine that the spherical object is suspended in mid-air inside the craft, so we can assume it orbits the planet independently, also moving on a circular orbit of radius
    \(r\) (see Fig. 2.2, plot (a)).

    This force -as has been discussed earlier -enforces the satellite motion, i.e., the free fall motion of the spherical object. The acceleration due to the force of gravity \(F_{gr}\) acting on the object is:

    \[ a_{gr} = \dfrac{F_{gr}}{m} = G\dfrac{M}{r^2}\]

    Figure \(\PageIndex{1}\):(a): A spherical object of mass \(m\) is inside a spacecraft orbiting a planet, but one can assume that it orbits the planet independently. (b): The ET-s manning (?) the spacecraft cut the object into two hemispheres of mass \(m/2\) each, and attach each one to one end of a long spring. Then they place one hemisphere at a distance of \(d\) above the spaceship, and the other at the same distance under it. They connect the other ends of the springs with each other. After that, the spacecraft flies away, leaving the ''object'' consisting of the two hemispheres connected by a spring of a \(2d\) length (shown as a thin line in the figure -the ''coils'' of the spring can be seen in the Fig. 2.4, which shows the ''object'' a blown-up scale). The mass center of the ''object'' is still on the original spacecraft's orbit of radius \(r\) So, now each hemisphere orbits the planet with the same angular velocity \(\omega\) as the craft had, but one moves along a trajectory of radius \(r-d\), and the other along a trajectory with radius \(r+d\). Because the ''lower'' hemisphere is closer to the planet than it originally was, the force of gravity acting on it (blue vector) increases -more than needed to hold it on a \(r-d\) orbit, tending to pull it down. This pull is compensated by the force of tension of the spring (the red vector). For the ''upper'' hemisphere, in turn, the force of gravity on the upper orbit is too weak to hold it on this orbit -the hemisphere would ''escape upwards'', if not pulled down by the spring's tension. So, effectively, one of the \(m/2\) masses pulls its spring's end down, the other pulls its spring's end up -producing a tension force in the spring. And this what we call the tidal force

    Figure \(\PageIndex{2}\): The ''object'' consisting of the two hemispheres and the spring shown here in a larger scale. The black dot in the middle of the spring marks the position of the ''object's'' mass center.

    Now, suppose that the spacecraft crew -not necessarily humans, they may be intelligent ET (extra-terrestrial) beings as well -cut the spherical object into two hemispherical halves of mass \(m/2\) each. The one part is attached to a long spring, and lowered from the spacecraft by a distance of \(d\). The other hemispherical part is attached to an identical spring, then raised, and positioned at a distance \(d\) above the spacecraft. Suppose that the other ends of the springs are connected, forming a single spring of length \(2d\).

    The spacecraft is not needed any more, let it fly away. Now the system of two hemispheres, \(2d\) apart, connected by the spring -as shown in the Fig. 2.3 - can be thought of a single elongated ''object'' of mass \(m\) (we assume that the mass of the spring is so small that it can be neglected). The object's mass center is still at a distance \(r\) from the planet center, so that the net force of gravity acting on the object is still the \(F_{gr}\) given by the Eq. 2.34, and the acceleration of the elongated object due to this force is the \(a_{gr}\) given by the Eq. 2.35.

    Let's now focus our attention on the ''upper'' hemisphere. Its mass is \(m/2\), and its distance from the planet center is \(r+d\), so the force of gravity acting on it -call it \(F^{upper}_{gr}\) -is:

    \[ F^{upper}_{gr} = G\dfrac{\dfrac{m}{2}\cdot M}{(r+d)^2}\]

    The acceleration \(a^{upper}_{gr}\) of mass \(m/2\) due to such force would be:

    \[ a^{upper}_{gr} = \dfrac{F^{upper}_{gr}}{\dfrac{m}{2}} = G\dfrac{M}{(r+d)^2}\]


    \[ G\dfrac{M}{(r+d)^2} < G\dfrac{M}{r^2}, \;\;\;{so~that}\;\;\; a^{upper}_{gr} < a_{gr},\]

    which means that the whole ''object'' is falling with acceleration \(a_{gr}\), yet the force of gravity \(F^{upper}_{gr}\) alone acting on the upper hemisphere is not strong enough for producing such acceleration. There must be an ''extra'' force \(F^{upper}_{extra}\) augmenting the downward pull, such that:

    \[ \dfrac{F^{upper}_{gr} + F^{upper}_{extra}}{\dfrac{m}{2}} = a_{gr}\]

    From which one obtains

    \[ F^{upper}_{extra} = \dfrac{m}{2} \cdot a_{gr} F^{upper}_{gr} = G\dfrac{\dfrac{m}{2}\cdot M}{r^2} G\dfrac{\dfrac{m}{2}\cdot M}{(r+d)^2}, \]

    which can be written in a more compact form:

    \[ F^{upper}_{extra} = G\dfrac{m\cdot M}{2}\left (\dfrac{1}{r^2} \dfrac{1}{(r+d)^2}\right )\]

    Now, let's switch our attention to the ''lower'' hemisphere. The force of gravity acting on it -call it \(F^{lower}_{gr}\) -is:

    \[ F^{lower}_{gr} = G\dfrac{\dfrac{m}{2}\cdot M}{(r-d)^2}\]

    The acceleration \(a^{lower}_{gr}\) of mass \(m/2\) due to such force would be:

    \[ a^{lower}_{gr} = \dfrac{F^{upper}_{gr}}{\dfrac{m}{2}} = G\dfrac{M}{(r-d)^2}\]


    \[ G\dfrac{M}{(r-d)^2} > G\dfrac{M}{r^2}, \;\;\;{so~that}\;\;\; a^{lower}_{gr} < a_{gr},\]

    which means that the force \(F^{lower}{gr}\) acting alone on the ''lower'' hemisphere would result in a higher downward acceleration than that of the object -therefore, to mitigate the \(F^{lower}_{gr}\) force and make the ''lower'' hemisphere to fall with acceleration \( a_{gr}\),
    there must be an ''extra'' force \(F^{lower}_{extra}\) pulling the ''lower'' hemisphere in the upward direction, such that:

    \[\dfrac{F^{lower}_{gr} F^{lower}_{extra}}{\dfrac{m}{2}} = a_{gr}\]

    From which we obtain:

    \[ F^{lower}_{extra} = F^{lower}_{gr} \dfrac{m}{2} \cdot a_{gr} = G\dfrac{\dfrac{m}{2}\cdot M}{(r-d)^2} G\dfrac{\dfrac{m}{2}\cdot M}{r^2} , \]

    which, again, can be written in a more compact form:

    \[ F^{lower}_{extra} = G\dfrac{m\cdot M}{2}\left (\dfrac{1}{(r-d)^2} \dfrac{1}{r^2}\right )\]

    The Eqs. 2.40 and 2.45 are not yet the final forms that we need. In order to simplify them even further, let's introduce an auxiliary parameter, \(\delta = d/r\). Doing some simple algebra, we get:

    \[ F^{upper}_{extra} = G\dfrac{m\cdot M}{2}\left (\dfrac{1}{r^2} \dfrac{1}{r^2(1+\dfrac{d}{r})^2}\right ) = G\dfrac{m\cdot M}{2r^2}\left(1 \dfrac{1}{(1+\delta)^2}\right )\]


    \[ F^{lower}_{extra} = G\dfrac{m\cdot M}{2}\left(\dfrac{1}{r^2(1-\dfrac{d}{r})^2}-\dfrac{1}{r^2} \right ) = G\dfrac{m\cdot M}{2r^2}\left (\dfrac{1}{(1-\delta)^2} 1 \right )\]

    To move further on, we would need to use a small calculus-based trick -in the case the Reader does not remember it, we present the ''recipe'' as a ''Mathematical Detour'':

    ****** Mathematical Detour ******

    Any continuous function \(f(x)\) can be expanded into a power series, also known as the Taylor series -which is a sum of the form:

    f(x) = f(0)& + &f'(0)\cdot x + \dfrac 12 f''(0)\cdot x^2 + \dfrac{1}{2\cdot 3} f'''(0)\cdot x^3 \nonumber \\
    &+ &\dfrac{1}{2\cdot 3 \cdot 4} f''''(0) \cdot x^4
    + \dfrac{1}{2\cdot 3 \cdot 4 \cdot 5} f'''''(0)\cdot x^5 +\ldots \;\;\;\;\;

    where \(f'(0)\), \(f''(0)\), \(f'''(0)\),... are the values of the first, second, third..... , and so on, derivatives of the \(f(x)\) function for \(x=0\).

    We want to apply such expansions to the functions \(1/(1+\delta)^2\) and \(1/(1-\delta)^2\) in the Eq. 2.46 and the Eq. 2.47, respectively.

    So, for the first of these functions:

    \[ f(x) = \dfrac{1}{(1 + x)^2} = (1+ x)^{-2}; \;\;\;\; f'(x) = (-2)\cdot(1+x)^{-3}; \]

    \[ f''(x) = (-2)\cdot (-3)\cdot(1+x)^{-4}; \;\;\;\;\; f'''(x) = 6\cdot(-4)\cdot(1+x)^{-5}; \]

    \[f''''(x) = (-5)\cdot (-24)\cdot(1+x)^{-6}; \;\;\;\; {and~so~on.} \]


    \[ f(0) = 1; \;\; f'(0) = -2; \;\; f''(0) = 6; \;\; f'''(0) = -24;\;\; f''''(0) = 120; {\rm~and~so~on.}\]

    Inserting the above numbers into the Eq. 2.48, one obtains:

    \[ \dfrac{1}{(1+x)^2} = 1 2x + 3x^2 4x^3 + 5x^4 -\ldots\]

    Next, we use the same recipe for the \(1/(1-\delta)^2\) function:

    \[ f(x) = \dfrac{1}{(1 x)^2} = (1x)^{-2}; \;\;\;\; f'(x) = (-2)\cdot(-1)\cdot(1-x)^{-3}; \]

    \[ f''(x) = (-2)\cdot (-3)\cdot(-1)\cdot(1-x)^{-4}; \;\;\;\;\; f'''(x) = 6\cdot(-4)\cdot(-1)\cdot(1-x)^{-5}; \]

    \[f''''(x) = (-5)\cdot (-24)\cdot (-1)\cdot(1-x)^{-6}; \;\;\;\; {and~so~on.} \]

    Now, the \(f(0)\) value and the derivative values we need are:

    \[ f(0) = 1; \;\; f'(0) = 2; \;\; f''(0) = -6; \;\; f'''(0) = 24;\;\; f''''(0) = -120; {\rm~and~so~on.}\]


    \[ \dfrac{1}{(1-x)^2} = 1 + 2x 3x^2 + 4x^3 5x^4 +\ldots\]

    ****** End of the Mathematical Detour ******,

    Based on the Eqs. 2.29 and 2.50, we can write:

    \[ \dfrac{1}{(1+\delta)^2} = 1 2\delta + 3\delta^2 4\delta^3 + 5\delta^4 -\ldots \]


    \[ \dfrac{1}{(1-\delta)^2} = 1 + 2\delta 3\delta^2 + 4\delta^3 5\delta^4 -\ldots \]

    Now, let's think for a moment about the \(\delta = d/r\) value. It is natural to assume that \(r\) is much larger than \(d\), so that the \(d/r\) ratio is much smaller than 1. A symbol for ''much smaller'' is ''\)\ll\)'', so that \(\delta \ll 1\).

    Well, if \(\delta \ll 1\), then, obviously, \(\delta^2 \ll \delta\), \(\delta^3 \ll \delta^2\), and so on. Therefore, the omission of all squares and higher powers of \(\delta\) will lead only to insignificant changes in the values of the right-hand sums in the two above equations. We can conclude that, to a very good approximation:

    \[ \dfrac{1}{(1+\delta)^2} \approx 1 2\delta \]


    \[ \dfrac{1}{(1-\delta)^2} \approx 1 + 2\delta \]

    Substitution of the Eqs. 2.53 and 2.54, to the Eqs. 2.46 and 2.47, respectively, yields:

    \[ F^{upper}_{extra} = G\dfrac{m\cdot M}{2r^2}\left (1 \dfrac{1}{(1+\delta)^2}\right ) =G\dfrac{m\cdot M}{2r^2}(1 1 +2\delta) =G\dfrac{m\cdot M}{r^2}\delta\]


    \[ F^{lower}_{extra} = G\dfrac{m\cdot M}{2r^2}\left (\dfrac{1}{(1-\delta)^2} 1 \right )= G\dfrac{m\cdot M}{2r^2}(1+2\delta 1 )= G\dfrac{m\cdot M}{r^2}\delta \]

    But, as was defined, \(\delta = d/r\), and by substituting this to the two above expressions, we obtain the final forms:

    \[ F^{upper}_{extra} = G\dfrac{m\cdot M}{r^3}d \;\;\;{and} \;\;\; F^{lower}_{extra} = G\dfrac{m\cdot M}{r^3}d\]

    It turns out that both ''extra'' forces are of the same magnitude (we cannot say that they are ''equal'', because the \(F^{upper}_{extra}\) vector is oriented down, and the \(F^{lower}_{extra}\) vector is oriented up). What is the origin of the two forces? Clearly, they are delivered by the stretched spring of \(2d\) length (if you grab a spring, right hand by one end, left hand by the other end, and you stretch it -each end will pull each hand ''inwards'', but each hand pulls the spring ''outwards'', with a force of the same magnitude -which is, of course, a manifestation of the Newton's Third Law of Dynamics). So, conversely, there is a force stretching the object, originating from the fact that the gravitational field which governs the motion of the object is not uniform -it exerts a stronger pull on one of the object's end than on the other end.

    In conclusion, the above analysis shows that an object of mass \(m\) and of finite size moving in the gravitational field of another object of mass \(M\), always experiences a force, tending to stretch it along the line connecting the mass centers of the two bodies. The force is proportional to the \(m\cdot M\) product, and to the object size - and inversely proportional to the \textit{\textbf{cube}} of the distance between the two objects. It's the famous ''tidal force'', as it is called by astronomers.

    For a ''dumbbell-shaped'' object consisting of two \(m/2\) masses, connected by a weightless ''handle'' of \(2d\) length, the tidal force is, as has been shown:

    \[ F_{tidal} = G\dfrac{m\cdot M}{r^3}d\]

    Based on this result, by performing a simple integration one can show that for a uniform rod of mass \(m\) and length \(2d\), the tidal tension force in the middle of the rod is:

    \[ F^{rod}_{tidal} = G\dfrac{m\cdot M}{2r^3}d \]

    As was mentioned earlier, the tidal force is generally of the form \(F_{tidal}= k\cdot GmMd/r^3\), where the coefficient \(k\) depends on the shape of the body experiencing the force.

    People may ask: ''OK, but it is a well established fact that it is the Moon which is responsible for the tidal effects occurring on Earth. And everybody knows that the Moon orbits Earth, not the other way. So, it seems to be something wrong with the above explanation.''

    No, nothing is wrong. Because it is not exactly so that only Moon orbits Earth. In fact, the two bodies form what in astronomy is called a ''binary system": both bodies revolve around a common mass center.

    So, the tidal force generated by Moon stretches the Earth along the line connecting their move mass centers. The \textbf{entire} Earth, not only its fluid part. With some simplification, Earth can be thought of as rocky sphere surrounded by an outer layer of water, i.e., the world oceans. The rocky core is also stretched -yet, since rocks are much less susceptible to stretching forces than fluids, the rocky part ''bulges up'' less than the fluid part. The average height of bulge on the rocky part is about 20 cm, while at an ocean, far from the shore, water raises, per average, about 60 cm, i.e., 40 cm relative to the ocean bottom.

    Why do we say ''average''? Well, the tidal effects on Earth are not only due to the Moon, but also due to the Sun. The Moon's mass is \(M_{M}= 7.35\times 10^{22}\) kg, and its distance from Earth is \(r_{M-E} = 3.84\times 10^8 \( m. For Sun, \(M_{S}= 1.989\times 10^{30}\) kg, and the Sun-Earth distance \(r_{S-E} = 1.496\times 10^{11}\) m.

    Figure \(\PageIndex{1}\): The effect of tides force on a planet with a rocky core (gray) covered by ocean (blue). Plot (a): No tidal force. Plot (b): the tidal force stretches both the rocky core and the ocean, but because rocks are stiffer than water, the core bulges up less than the ocean. Observers standing at towers set at the ocean floor (i.e., at the rocky core) do not see the core's motion -they move up and down together with the core's surface, and they cannot see the core's bulging. From their perspective, only the water level rises and falls. The same applies to observers standing on the ocean's shore, they are ''connected'' to the rocky core and they can only see the motion of waters relatively to the rocky core's surface.

    Therefore, if we consider the ratio of the tidal forces acting on Earth from the Moon and the Sun, only the above four parameters are important, because all other cancel out:

    \[ \dfrac{F_{due~to~Moon}}{F_{due~to~Sun}} = \dfrac{M_{M}\times r_{S-E}^3}{M_{S}\times r_{M-E}^3} = 2.185, \]

    meaning that about 68\% of the power of the tidal effects on Earth are due to Moon, and about 3 2\% are due to Sun. When Moon, Sun and Earth are aligned along a straight line (at Full Moon, or at New Moon), their contributions add up -the tide bulges are then the highest. They are called the ''spring tides'' (it's a tradition; no relation to the time of year). In contrast, when the directions of the Moon and the Sun viewed from Earth make a right angle -i.e., at the First Quarter Moon, or at the Third Quarter Moon -the effects from the Moon and Sun partially cancel out, and the tide bulges are the lowest. Such tides are called the ''neap tides''.

    For an observer on Earth, the two tidal bulges with the ''dips'' in between can be though of a wave traveling around the globe (actually, it's the bulge which always form below the Moon, and the impression that it ''travels'' from the East to West comes from the fact that the Earth rotates. Since the Earth spins and the Moon travels on its orbit in the same direction (counter-clockwise, if watched from above the North Pole). A lunar ''month'' -i.e., is the average period of the Moon's orbit with respect to the line joining the Sun and Earth -is about 29.5 days. So, every day the Moon reaches its highest point in the sky about 48 minutes later than the previous day. Therefore, the ''high tides'' occur not every 12 hours, but about every 12 hours and 24 minutes.

    As noted above, the tide's height at the middle of an ocean is only about 0.5 m. However, it is known that the height of a wave of water may significantly increase if the wave passes from a deep region to a much shallower region, or if it travels along a narrowing bay. Consequently, in some coastal regions on Earth the difference between the high tide and the low tide may be much larger, with the record value of 16 meters in the Bay of Fundy, Canada. Tidal Forces is shared under a CC BY 1.3 license and was authored, remixed, and/or curated by Tom Giebultowicz.

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