You have likely read in books that tides are mainly caused by the Moon. When the Moon is high in the sky, it pulls the water on the Earth upward and a high-tide happens. There is some similar effect causing low-tides. They also say that the Sun does the same as well, but has smaller effect compared to the Moon.

Here's my question: Why is the Moon the major cause of tides? Why not the Sun? The Sun is extremely massive compared to the Moon. One might say, well, the Sun is much farther than the Moon. But I've got a simple answer: Just substitute those numbers in $a=\frac{GM}{d^2}$ and find gravitational acceleration for the Moon and then for the Sun (on Earth, by the way). You'll find something around $3.38$ $10^{-6}$ $g$ for the Moon and $6.05$ $10^{-4}$ $g$ for the Sun - I double checked it to make sure. As you can see, the sun pulls about $180$ times stronger on the Earth. Can anyone explain this? Thanks is advance.

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    $\begingroup$ You may want to check out this section. $\endgroup$
    – HDE 226868
    Commented Nov 29, 2014 at 15:51
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    $\begingroup$ Related: physics.stackexchange.com/q/111685/16660. $\endgroup$
    – Wouter
    Commented Nov 29, 2014 at 17:08
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    $\begingroup$ It's not correct to say that the Moon causes tide by "pulling the water upwards". It pulls the water both on the near and far side of the Earth. But it pulls more on the near side, and it's that difference that matters. To make a long story short, it's not $m/r^2$ that matters, but it's derivative, i.e. $m/r^3$. If you calculate that you'll see that the Moon has a stronger effect than the Sun. $\endgroup$
    – Szabolcs
    Commented Dec 1, 2014 at 16:34
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    $\begingroup$ You will find every other person answering you about how Moon is causing the tides by means of differential and integral and what not. But nobody will be able to answer how a force of the order of 10^-5 N actually cause the tides. That is because it does not. $\endgroup$
    – user124734
    Commented Aug 28, 2016 at 6:05

5 Answers 5


What is important for tidal forces is not the absolute gravity, but the differential gravity across the planet, that is, how different is the force of gravity at a point on the Earth's surface near the sun relative to a point on the Earth's surface far from the sun. If you compare it with the moon, the result will be that the tidal force from the sun is about 0.43 that of the moon.

Suppose two different bodies in the sky that have the same apparent size. Because the mass M of the object will grow as $r^3$ (because $M=4/3\rho\pi R^3$ and $R=\theta r$), the gravitational force will actually grow linearly with $r$, where $r$ is the distance and $R$ is the radius of the object. So if two bodies have the same apparent size (such as the Moon and the Sun) and the same density, the tidal force would be the same. The density of the moon is about 2.3 times larger than that of the sun, that is why the tidal force is larger by that factor.

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    $\begingroup$ This explanation about having the same apparent size seems very confusing. I'd say "[...] the volume $V$ of the object will grow as $r^3$ - - the force grows linearly with $r\rho$", so there'd be no need to first assume (without saying it explicitely before making the calculations) that the Sun and the Moon had the same density. $\endgroup$
    – JiK
    Commented Nov 30, 2014 at 23:19
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    $\begingroup$ I liked the picture and I'm sorry it was edited out again. $\endgroup$
    – Floris
    Commented Dec 2, 2014 at 0:07
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    $\begingroup$ I think there is a gap in the reasoning in this answer. You start by saying the absolute gravity is not the relevant quantity, and then in the second paragraph you nonetheless talk about how the force (itself) grows with $r$. I think what is missing is that the differential of a $r^{-2}$ force fields is proportional to $r^{-3}$ and that this precisely cancels the factor $r^3$ by which mass increases under constant apparent size and density assumptions. Therefore such objects should have tidal effect independent of $r$. Then finally it is their density ratio that distinguishes Sun and Moon. $\endgroup$ Commented Dec 2, 2014 at 12:57
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    $\begingroup$ @MarcvanLeeuwen Thanks! I agree with you, the grammar is not good, and/or the reasoning is not straight. But I think people understood what I meant. I am going to improve the writting, due to the large number of votes. $\endgroup$
    – user65081
    Commented Dec 2, 2014 at 13:03
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    $\begingroup$ I've already written an answer to a closely related question that shows a different, more explicit comparison between the Lunar and Solar Tides. Since it might be of interest to readers of this answer, I'll link it here: physics.stackexchange.com/a/111695/16660. $\endgroup$
    – Wouter
    Commented Jun 30, 2016 at 14:41

Tides are caused by the gradient of the gravitational field - so the tidal "force" experienced drops with the third power of the distance.

This means that the relative strength of the tides should go as

\begin{align} \mathrm{ratio} & = \frac{M_\mathrm{moon} \cdot D_\mathrm{sun}^3}{M_\mathrm{sun} \cdot D_\mathrm{moon}^3} \\ & = \frac{7 \cdot 10^{22}\cdot (1.5 \cdot 10^{11})^3}{2\cdot 10^{30}\cdot (3.7\cdot 10^{8})^3} \\ & = 2.3 \end{align}

So although the sun is more massive, its greater distance makes its tidal force about 2.3x weaker than the moon's - consistent with your number (and my round numbers...)

Following a suggestion by @wolprhram jonny, if you assume a certain angular size $\alpha$ of the sun/moon (they are both about 0.5° across as seen from Earth), you can rewrite the above equation by first replacing mass with density times volume, then rearranging: \begin{align} \mathrm{ratio} & = \frac{(\rho_\mathrm{moon}r_\mathrm{moon}^3)\cdot D_\mathrm{sun}^3}{ (\rho_\mathrm{sun} r_\mathrm{sun}^3)\cdot D_\mathrm{moon}^3} \\ & = \frac{\rho_\mathrm{moon}\alpha_\mathrm{moon}^3}{ \rho_\mathrm{sum} \alpha_\mathrm{sun}^3} \end{align}

So when the apparent angle in the sky is the same, the tidal forces scale with the density of the objects. Interesting and unexpected result.

  • $\begingroup$ Great hint. Now it's clear; As you and @wolprhram jonny pointed, absolute gravity field is not important, and gradient of gravitational field matters. It makes sense; I had to tell myself: If it's all about force of gravity, the earth ground would fall freely as well as ocean water, and thus no tide would happen. Thanks for the crystal clear answer. $\endgroup$ Commented Nov 29, 2014 at 17:50
  • $\begingroup$ Could the destiny therefore be measured based on the tides? $\endgroup$ Commented Dec 1, 2014 at 16:07
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    $\begingroup$ @PyRulez yes you could estimate density (not destiny!) from this effect. You would want to measure changes in gravity rather than "tides" but it would be possible. $\endgroup$
    – Floris
    Commented Dec 1, 2014 at 16:26
  • $\begingroup$ But the tides is cumulative over a lot of water. I was wondering if tidal effects would allow us to determine the mass of the moon/sun in kilograms to greater accuracy, and therefore measure G accurately? $\endgroup$ Commented Dec 1, 2014 at 16:58
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    $\begingroup$ @PyRulez - no, the motion of the water is only loosely related to the force and phase of the tides (it is really a wave phenomenon that is driven by the tidal forces). See David Hammen's very complete answer on the topic $\endgroup$
    – Floris
    Commented Dec 2, 2014 at 0:06

The highly upvoted answer is right but to make things much simpler:

Tides are based on the change in gravity, not the gravity. That means they drop off at the cube of the distance rather than the square of the distance like gravity itself does. Thus the object with the most gravity isn't necessarily the one that causes the most tide.

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    $\begingroup$ "The highly upvoted answer" -- which one? Votes change over time. If you need to refer to a specific answer, link to it. But I don't see what you're actually adding, here. Floris's answer already mentiones the third-power law in its very first sentence. $\endgroup$ Commented Nov 30, 2014 at 14:24
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    $\begingroup$ What's added: Brevity. $\endgroup$
    – RobertB
    Commented Dec 1, 2014 at 4:43
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    $\begingroup$ @RobertB Brevity and a lack of math that makes posts harder to read for those not used to dealing with such things. I'm trying to answer for the average man, not the scientist. $\endgroup$ Commented Dec 3, 2014 at 3:12

As stated in other answers it is how much the gravitational force is different by on opposite sides of the earth that creates the tides.

You can still show this using $a=\frac{GM}{d^2}$ but you need to consider the difference, not the absolute force on the earth.

The sun while much more massive is just far enough away that it is getting to a much flatter part of the hyperbola.

Everything is better with graphs

Maths and Graphs

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    $\begingroup$ Welcome to Physics. We use MathJax to render LaTeX output here. You should change the equations in your image to LaTeX like so: $$a_{moon}(r) = \frac{G \cdot M_{moon}}{(r + d_{moon})^2}$$ This has the benefit of making your answer more searchable. $\endgroup$
    – pho
    Commented Nov 29, 2014 at 22:58

Behind any theory, the field reality indicate a stronger lunar tide for:

-oceanic waters (the lunar influence is detected as M2 and K1 tidal components; the tidal component is defined by the frequency of a tidal oscillation; the frequency is dependent on the relative movement of the implied celestial bodies (Eartth, Moon/Sun).

-earth's crust (the same tidal components; the tidal response is no longer influenced by coastal morphologies, but by the local mass change caused by oceanic loading and crustal deformation induced by oceanic loading) http://en.wikipedia.org/wiki/Earth_tide

-inland groundwater and rivers (the description "inland" was used because the coastal groundwater and rivers are influenced by the oceanic input; the K1 and M2 are much weaker when compared to their oceanic equivalents vecause many other strong cycles interfere, such as the day/night cycle) http://www.nature.com/srep/2014/140226/srep04193/full/srep04193.html

  • $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. $\endgroup$
    – Kyle Kanos
    Commented Jan 10, 2015 at 14:51
  • $\begingroup$ Updated with explanations. $\endgroup$
    – Andrei
    Commented Jan 10, 2015 at 15:07
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    $\begingroup$ This still doesn't answer the question. At least, not explicitly/clearly $\endgroup$
    – Jim
    Commented Jan 10, 2015 at 15:11

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