I get that Earth's mass is very large, so its acceleration is very tiny. But wouldn't the acceleration accumulate over a period of time and become noticeable?

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    $\begingroup$ Related: physics.stackexchange.com/q/454332/2451 and links therein. $\endgroup$
    – Qmechanic
    May 25, 2019 at 13:15
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    $\begingroup$ Ah, but part of the time the moon is ahead of the earth, part of the time it's behind the earth, and part of the time it's on one side or the other of the earth as the earth makes its annual trip around the sun (which is also moving, but ignore that for now). Things tend to cancel out. If they didn't the moon would have long ago smashed into the earth, we wouldn't exist, and this discussion would have never(*&(*&(*NO CARRIER $\endgroup$ May 27, 2019 at 3:28
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    $\begingroup$ @BobJarvis I will upvote that if you post it as an answer. It's far simpler than any of the other answers here. The moon is sometimes on one side and sometimes on the other, so the acceleration "accumulates" to zero change in velocity. $\endgroup$ May 27, 2019 at 6:59
  • $\begingroup$ Question title has been edited to add an additional question that is not in the body, and was not present when the majority of answers were given. Have edited to revert. $\endgroup$
    – Brondahl
    May 28, 2019 at 18:54

7 Answers 7


It seems you have the same misunderstanding like most people have before fully understanding Newtonian physics. They think: Only the moon rotates around the earth, and the earth stands still. But this is wrong.

Actually the earth does accelerate towards the moon, in much the same way as the moon accelerates towards the earth.

And that's why not only the moon, but also the earth rotates around their common barycenter (the $\color{red}{+}$ in the animation below), albeit with a smaller radius.

(animated image from Wikipedia: Barycenter - Gallery)

Edit (in reply to question asked in comment, now moved to chat):

The attractive force is pointing vertically down to the center of the earth. It has no horizontal component. Therefore this force adds no horizontal speed to the moon's movement. The moon had already a horizontal speed since its creation billion years ago. The attractive force acts only vertically. Therefore the moon's path is a curve bending towards the earth, instead of just a straight line.

The same applies to you when standing on the earth. The attractive force adds no horizontal speed to your movement, And since you had no horizontal speed from the beginning, it stays like this.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    May 27, 2019 at 1:37
  • $\begingroup$ Is this the same true for relation between the Earth and the Sun? $\endgroup$
    – kukis
    May 27, 2019 at 15:39
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    $\begingroup$ @kukis In principle yes. But it is more complicated, because there are more planets besides the earth. So you need to consider the barycenter of the sun and all planets together. $\endgroup$ May 27, 2019 at 15:52
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    $\begingroup$ An animated picture must be worth 10,000 words. $\endgroup$
    – nigel222
    May 28, 2019 at 9:05
  • $\begingroup$ This does not answer the question. An interesting fact but not the reason why the Earth is not pulled towards the Moon. That's explained by @Bob Jarvis' comment on the question. The moon is "behind" the Earth half of the time so that would pull it back again. $\endgroup$
    – Jacques
    May 28, 2019 at 14:04

It does: both the Earth and the Moon accelerate towards each other all the time. As you say, the acceleration of the Moon is significantly larger than that of the Earth. Both bodies, therefore, end up following trajectories for which the acceleration is always towards the other: those trajectories, of course, are the orbits they take around a common point, which is the barycentre of the Earth-Moon system. This is the centre-of-mass of the system, and for a two body system of masses $m_1$, $m_2$ the distance of the centre of the $i$'th body from it is given by

$$d_i = r\frac{m_j}{m_i + m_j}$$

where $r$ is the separation of the centres of the bodies, $i,j\in\{1,2\}$ and $j\ne i$.

If you take the Earth-Moon system and assume the orbit is circular (which is a pretty good approximation I think), then for Earth we get $d_1 \approx 4671\,\mathrm{km}$, which means the Earth is orbiting (and hence accelerating towards) a point about that far from its centre. This point is inside the Earth, sine the radius of the Earth is about $6371\,\mathrm{km}$.

By contrast, for the Pluto-Charon system, the barycentre is outside Pluto, and the bodies can be clearly seen to be orbiting a common central point: the Wikipedia page for barycentre has a rather nice little animation made up of images from New Horizons which shows this.

'Accumulation of acceleration'

I wanted to address the other notion, that acceleration somehow 'accumulates'. This is true, in the sense that the velocity of something is the integral of its acceleration over time:

$$\vec{v}(t) = \vec{v}(t_0) + \int\limits_{t_0}^t \vec{a}(\tau)\,d\tau$$

But the critical thing here is that $\vec{v}$ & $\vec{a}$ are vectors, which means we can arrange life so that this integral does not become very large, even if the magnitude of the vectors is never zero (indeed, even if it is constant).

So the obvious example is to think about an acceleration like this, in cartesian coordinates in 2 dimensions:

$$\vec{a}(t) = (a\sin\omega t + a\cos\omega t)$$

We can integrate this to get $\vec{v}(t)$ (dropping the constant of integration which we can safely do as it involves a change of frame):

$$\vec{v}(t) = \left(-\frac{a}{\omega}\cos\omega t, \frac{a}{\omega}\sin\omega t\right)$$

And we can integrate again to get the position, $\vec{p}(t)$, again dropping the constant of integration which corresponds to a choice of origin of coordinates:

$$ \begin{align} \vec{p}(t) &= \left(-\frac{a}{\omega^2}\sin\omega t, -\frac{a}{\omega^2}\cos\omega t\right)\\ &= -\frac{1}{\omega^2}\vec{a}(t) \end{align} $$

Well, now, this is motion in a circle, of course, and importantly, the magnitude of the acceleration, $|\vec{a}(t)| = a$: it's constant. And the direction of $\vec{a}(t)$ is always towards the centre of the circle.

This is what is going on in orbiting systems: the bodies are continually accelerating towards the barycentre of the system, and if the orbit is circular, then they never get any closer to it, and the magnitude of the acceleration is constant (if the orbits are elliptical then they do approach & recede from it, and the magnitude of the acceleration varies over time).

  • $\begingroup$ If the Earth and the Moon accelerate towards each other all the time, how are we losing it at ~4cm a year? Supposedly the sun will eat us before it ever does, but if that wasn't in the cards, at some point it will do nothing except accelerate away from the earth. No? $\endgroup$
    – Mazura
    May 25, 2019 at 22:39
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    $\begingroup$ @Mazura: No, it will never do so. Other things being equal the Earth-Moon system will eventually become tidally locked at which point the orbital radius will no longer increase (and in fact will very slowly start to decrease due to gravitational radiation). $\endgroup$
    – user107153
    May 26, 2019 at 1:40

The Earth's acceleration is towards the moon (ignoring forces from other things such as the sun). This is because the force of gravity between them is in the same direction of the geometric line connecting them, and according to Newton's second law the acceleration due to a force is in the same direction as that force. Just considering an Earth-Moon system, they both orbit about the center of mass of the system. So both bodies would be undergoing acceleration, where the acceleration of one body is towards the other.

I'm unsure what you mean about accumulation of acceleration. Forces determine acceleration. Acceleration isn't something that accumulates over time. It's just a result of the current net force acting on an object.

  • $\begingroup$ Acceleration will be tiny because of the huge mass of the earth. But in this case there is a constant gravitational force exerted by the moon on the earth. Since the moon exerts this gravitational force on the earth all the time won't this tiny acceleration increase and hence add up to a noticeable acceleration enough to move the earth. This is what I meant. Of course now I understand that the earth does accelerate towards the moon thanks to a number of answers that people gave. $\endgroup$
    – Ali
    May 26, 2019 at 8:04
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    $\begingroup$ @Ali The Earth does move: just because the acceleration is directed towards the moon this doesn't mean the trajectory must be. The direction of motion is the direction of the velocity, not the acceleration. $\endgroup$
    – gented
    May 27, 2019 at 9:43

The acceleration of the Moon due to the force of the Earth is perpendicular to the velocity of the Moon. That's why the path of the moon is a circle. The same is true concerning the acceleration of the Earth due to the force of the Moon. The acceleration of the Earth is perpendicular to its velocity. Hence it does not "accumulate"; the Earth follows a circular path, as does the Moon.

In fact, the Moon does not orbit the Earth. It orbits the common center of mass of the Earth-Moon system. The same is true of the Earth; it orbits the common center of mass. However, the center of mass of the system lies inside the Earth, so the radius of the Earth's orbit is much smaller than the radius of the Moon's orbit, and is ignored for many purposes. It is detectable, and must be taken into account when doing precise astronomical calculations.

While it is true that the orbits are actually elliptical, that fact has no bearing whatsoever on the question of whether or not acceleration accumulates.

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    $\begingroup$ "The acceleration of the Moon due to the force of the Earth is perpendicular to the velocity of the Moon. That's why the path of the moon is a circle. " Not true. The orbit is elliptical with an eccentricity of greater than 0.05, which is not insignificant. Therefore the acceleration is not perpendicular, but has a changing angle. $\endgroup$
    – Bill N
    May 25, 2019 at 13:49
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    $\begingroup$ The orbit of the Moon (and all the planets) is an ellipse: this was known to Kepler even before Newton. $\endgroup$
    – Tom
    May 26, 2019 at 0:25

Why doesn't the Earth's acceleration towards the Moon accumulate to push the Earth off its orbit?

Because the Earth doesn't orbit the sun, the center of mass of the Earth-Moon system does. Earth and Moon in turn orbit this center of mass.

Orbits are a consequence of motion, which is aptly measured by kinetic energy. In order to change an orbit, kinetic energy must be expended in order to slow down, speed up, or redirect the orbiting entity. The orbiting of Earth and Moon is largely conservative, and does not produce nor consume energy, so they cannot alter their orbit.

But wouldn't the acceleration accumulate over a period of time and become noticeable?

No. When you accelerate in your car, you apply acceleration. When you brake, you apply more acceleration (you can prove this by having your friend sit in the car with an accelerometer). The result is not that your car goes really fast, the result is that your car is stationary. Direction of acceleration also matters.

For orbiting bodies, the acceleration is such that it is always orthogonal to velocity, so it only changes the direction of movement, never its speed. It also happens to be quite predictable, so the directional change results in the orbiter going in a circle.

The acceleration imparted on the Earth by the Moon will not make it collide, because the direction of the acceleration is always towards the Moon. Earth already has a large velocity orthogonal to that direction (ie. it is flying "past" the Moon) so the acceleration can only curve its trajectory in a circle.

Generally, you can think of orbits as falling towards an object, but constantly missing.

  • $\begingroup$ Nice answer, but acceleration is only exactly orthogonal to velocity for the entire orbit if the orbit is a perfect circle. In an elliptical orbit, it's orthogonal when the orbiter is at the apsides, i.e., when it's nearest & furthest from the barycentre. $\endgroup$
    – PM 2Ring
    May 28, 2019 at 10:19

The moon is actually moving away from the Earth; 4 billion years ago it was much closer.

The moon raises tides which have the effect of slowing down the rotation of the Earth, so that the daylength is now much longer than it used to be, but some of the energy which has been lost by the Earth was captured by the moon, and it has boosted it into a higher orbit. It is still receding from us at the rate of a few centimetres per year.

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    $\begingroup$ This is not a helpful thing to bring up in this context. The OP needs to understand the basics of orbital mechancs before the long-term evolution of the system can be anything other than a confusing distraction. $\endgroup$ May 25, 2019 at 17:23
  • $\begingroup$ I know what you mean & it was something I already understood. Detecting the movement of the larger body is one of the means by which astronomers find exoplanets. $\endgroup$ May 26, 2019 at 6:32

The Earth is in free fall. We only experience acceleration because we are not at its centre, by tidal forces, and because of its rotation.


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