Why doesn't the Earth's acceleration towards the Moon accumulate to create noticeable motion of the earth,...





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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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    Related: physics.stackexchange.com/q/454332/2451 and links therein.
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    – Qmechanic
    May 25 at 13:15






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    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
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    – Bob Jarvis
    May 27 at 3:28






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    @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.
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    – Dawood ibn Kareem
    May 27 at 6:59










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    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.
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    – Brondahl
    May 28 at 18:54


















16












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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?










share|cite|improve this question











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    $begingroup$
    Related: physics.stackexchange.com/q/454332/2451 and links therein.
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    – Qmechanic
    May 25 at 13:15






  • 8




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    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
    $endgroup$
    – Bob Jarvis
    May 27 at 3:28






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Dawood ibn Kareem
    May 27 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.
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    – Brondahl
    May 28 at 18:54














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$begingroup$


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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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?







newtonian-mechanics forces newtonian-gravity reference-frames acceleration






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edited May 28 at 19:16









Brondahl

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asked May 25 at 12:09









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  • 2




    $begingroup$
    Related: physics.stackexchange.com/q/454332/2451 and links therein.
    $endgroup$
    – Qmechanic
    May 25 at 13:15






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    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
    $endgroup$
    – Bob Jarvis
    May 27 at 3:28






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Dawood ibn Kareem
    May 27 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 at 18:54














  • 2




    $begingroup$
    Related: physics.stackexchange.com/q/454332/2451 and links therein.
    $endgroup$
    – Qmechanic
    May 25 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
    $endgroup$
    – Bob Jarvis
    May 27 at 3:28






  • 2




    $begingroup$
    @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.
    $endgroup$
    – Dawood ibn Kareem
    May 27 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 at 18:54








2




2




$begingroup$
Related: physics.stackexchange.com/q/454332/2451 and links therein.
$endgroup$
– Qmechanic
May 25 at 13:15




$begingroup$
Related: physics.stackexchange.com/q/454332/2451 and links therein.
$endgroup$
– Qmechanic
May 25 at 13:15




8




8




$begingroup$
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
$endgroup$
– Bob Jarvis
May 27 at 3:28




$begingroup$
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
$endgroup$
– Bob Jarvis
May 27 at 3:28




2




2




$begingroup$
@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.
$endgroup$
– Dawood ibn Kareem
May 27 at 6:59




$begingroup$
@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.
$endgroup$
– Dawood ibn Kareem
May 27 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 at 18:54




$begingroup$
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 at 18:54










7 Answers
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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.



animation

(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.






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  • $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – tpg2114
    May 27 at 1:37










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    Is this the same true for relation between the Earth and the Sun?
    $endgroup$
    – kukis
    May 27 at 15:39






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    @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.
    $endgroup$
    – Thomas Fritsch
    May 27 at 15:52






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    An animated picture must be worth 10,000 words.
    $endgroup$
    – nigel222
    May 28 at 9:05










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    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.
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    – Jacques
    May 28 at 14:04



















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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



$$d_i = rfrac{m_j}{m_i + m_j}$$



where $r$ is the separation of the centres of the bodies, $i,jin{1,2}$ and $jne i$.



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:



$$vec{v}(t) = vec{v}(t_0) + intlimits_{t_0}^t vec{a}(tau),dtau$$



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:



$$vec{a}(t) = (asinomega t + acosomega t)$$



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):



$$vec{v}(t) = left(-frac{a}{omega}cosomega t, frac{a}{omega}sinomega tright)$$



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:



$$
begin{align}
vec{p}(t)
&= left(-frac{a}{omega^2}sinomega t, -frac{a}{omega^2}cosomega tright)\
&= -frac{1}{omega^2}vec{a}(t)
end{align}
$$



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).






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    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?
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    – Mazura
    May 25 at 22:39






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    @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).
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    – tfb
    May 26 at 1:40



















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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.






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    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.
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    – Ali
    May 26 at 8:04






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    @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.
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    – gented
    May 27 at 9:43



















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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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    "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.
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    – Bill N
    May 25 at 13:49






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    The orbit of the Moon (and all the planets) is an ellipse: this was known to Kepler even before Newton.
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    – Tom
    May 26 at 0:25



















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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.






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  • $begingroup$
    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.
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    – PM 2Ring
    May 28 at 10:19



















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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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    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.
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    – dmckee
    May 25 at 17:23










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    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.
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    – Michael Walsby
    May 26 at 6:32



















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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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    7 Answers
    7






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    7 Answers
    7






    active

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    active

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    active

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    81














    $begingroup$

    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.



    animation

    (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.






    share|cite|improve this answer











    $endgroup$















    • $begingroup$
      Comments are not for extended discussion; this conversation has been moved to chat.
      $endgroup$
      – tpg2114
      May 27 at 1:37










    • $begingroup$
      Is this the same true for relation between the Earth and the Sun?
      $endgroup$
      – kukis
      May 27 at 15:39






    • 5




      $begingroup$
      @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.
      $endgroup$
      – Thomas Fritsch
      May 27 at 15:52






    • 2




      $begingroup$
      An animated picture must be worth 10,000 words.
      $endgroup$
      – nigel222
      May 28 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 at 14:04
















    81














    $begingroup$

    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.



    animation

    (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.






    share|cite|improve this answer











    $endgroup$















    • $begingroup$
      Comments are not for extended discussion; this conversation has been moved to chat.
      $endgroup$
      – tpg2114
      May 27 at 1:37










    • $begingroup$
      Is this the same true for relation between the Earth and the Sun?
      $endgroup$
      – kukis
      May 27 at 15:39






    • 5




      $begingroup$
      @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.
      $endgroup$
      – Thomas Fritsch
      May 27 at 15:52






    • 2




      $begingroup$
      An animated picture must be worth 10,000 words.
      $endgroup$
      – nigel222
      May 28 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 at 14:04














    81














    81










    81







    $begingroup$

    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.



    animation

    (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.






    share|cite|improve this answer











    $endgroup$



    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.



    animation

    (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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jun 9 at 17:14

























    answered May 25 at 12:37









    Thomas FritschThomas Fritsch

    4,6382 gold badges17 silver badges27 bronze badges




    4,6382 gold badges17 silver badges27 bronze badges















    • $begingroup$
      Comments are not for extended discussion; this conversation has been moved to chat.
      $endgroup$
      – tpg2114
      May 27 at 1:37










    • $begingroup$
      Is this the same true for relation between the Earth and the Sun?
      $endgroup$
      – kukis
      May 27 at 15:39






    • 5




      $begingroup$
      @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.
      $endgroup$
      – Thomas Fritsch
      May 27 at 15:52






    • 2




      $begingroup$
      An animated picture must be worth 10,000 words.
      $endgroup$
      – nigel222
      May 28 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 at 14:04


















    • $begingroup$
      Comments are not for extended discussion; this conversation has been moved to chat.
      $endgroup$
      – tpg2114
      May 27 at 1:37










    • $begingroup$
      Is this the same true for relation between the Earth and the Sun?
      $endgroup$
      – kukis
      May 27 at 15:39






    • 5




      $begingroup$
      @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.
      $endgroup$
      – Thomas Fritsch
      May 27 at 15:52






    • 2




      $begingroup$
      An animated picture must be worth 10,000 words.
      $endgroup$
      – nigel222
      May 28 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 at 14:04
















    $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – tpg2114
    May 27 at 1:37




    $begingroup$
    Comments are not for extended discussion; this conversation has been moved to chat.
    $endgroup$
    – tpg2114
    May 27 at 1:37












    $begingroup$
    Is this the same true for relation between the Earth and the Sun?
    $endgroup$
    – kukis
    May 27 at 15:39




    $begingroup$
    Is this the same true for relation between the Earth and the Sun?
    $endgroup$
    – kukis
    May 27 at 15:39




    5




    5




    $begingroup$
    @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.
    $endgroup$
    – Thomas Fritsch
    May 27 at 15:52




    $begingroup$
    @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.
    $endgroup$
    – Thomas Fritsch
    May 27 at 15:52




    2




    2




    $begingroup$
    An animated picture must be worth 10,000 words.
    $endgroup$
    – nigel222
    May 28 at 9:05




    $begingroup$
    An animated picture must be worth 10,000 words.
    $endgroup$
    – nigel222
    May 28 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 at 14:04




    $begingroup$
    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 at 14:04













    14














    $begingroup$

    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



    $$d_i = rfrac{m_j}{m_i + m_j}$$



    where $r$ is the separation of the centres of the bodies, $i,jin{1,2}$ and $jne i$.



    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:



    $$vec{v}(t) = vec{v}(t_0) + intlimits_{t_0}^t vec{a}(tau),dtau$$



    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:



    $$vec{a}(t) = (asinomega t + acosomega t)$$



    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):



    $$vec{v}(t) = left(-frac{a}{omega}cosomega t, frac{a}{omega}sinomega tright)$$



    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:



    $$
    begin{align}
    vec{p}(t)
    &= left(-frac{a}{omega^2}sinomega t, -frac{a}{omega^2}cosomega tright)\
    &= -frac{1}{omega^2}vec{a}(t)
    end{align}
    $$



    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).






    share|cite|improve this answer











    $endgroup$















    • $begingroup$
      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 at 22:39






    • 3




      $begingroup$
      @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).
      $endgroup$
      – tfb
      May 26 at 1:40
















    14














    $begingroup$

    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



    $$d_i = rfrac{m_j}{m_i + m_j}$$



    where $r$ is the separation of the centres of the bodies, $i,jin{1,2}$ and $jne i$.



    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:



    $$vec{v}(t) = vec{v}(t_0) + intlimits_{t_0}^t vec{a}(tau),dtau$$



    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:



    $$vec{a}(t) = (asinomega t + acosomega t)$$



    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):



    $$vec{v}(t) = left(-frac{a}{omega}cosomega t, frac{a}{omega}sinomega tright)$$



    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:



    $$
    begin{align}
    vec{p}(t)
    &= left(-frac{a}{omega^2}sinomega t, -frac{a}{omega^2}cosomega tright)\
    &= -frac{1}{omega^2}vec{a}(t)
    end{align}
    $$



    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).






    share|cite|improve this answer











    $endgroup$















    • $begingroup$
      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 at 22:39






    • 3




      $begingroup$
      @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).
      $endgroup$
      – tfb
      May 26 at 1:40














    14














    14










    14







    $begingroup$

    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



    $$d_i = rfrac{m_j}{m_i + m_j}$$



    where $r$ is the separation of the centres of the bodies, $i,jin{1,2}$ and $jne i$.



    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:



    $$vec{v}(t) = vec{v}(t_0) + intlimits_{t_0}^t vec{a}(tau),dtau$$



    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:



    $$vec{a}(t) = (asinomega t + acosomega t)$$



    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):



    $$vec{v}(t) = left(-frac{a}{omega}cosomega t, frac{a}{omega}sinomega tright)$$



    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:



    $$
    begin{align}
    vec{p}(t)
    &= left(-frac{a}{omega^2}sinomega t, -frac{a}{omega^2}cosomega tright)\
    &= -frac{1}{omega^2}vec{a}(t)
    end{align}
    $$



    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).






    share|cite|improve this answer











    $endgroup$



    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



    $$d_i = rfrac{m_j}{m_i + m_j}$$



    where $r$ is the separation of the centres of the bodies, $i,jin{1,2}$ and $jne i$.



    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:



    $$vec{v}(t) = vec{v}(t_0) + intlimits_{t_0}^t vec{a}(tau),dtau$$



    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:



    $$vec{a}(t) = (asinomega t + acosomega t)$$



    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):



    $$vec{v}(t) = left(-frac{a}{omega}cosomega t, frac{a}{omega}sinomega tright)$$



    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:



    $$
    begin{align}
    vec{p}(t)
    &= left(-frac{a}{omega^2}sinomega t, -frac{a}{omega^2}cosomega tright)\
    &= -frac{1}{omega^2}vec{a}(t)
    end{align}
    $$



    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).







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 26 at 9:41

























    answered May 25 at 12:53









    tfbtfb

    19.4k5 gold badges39 silver badges60 bronze badges




    19.4k5 gold badges39 silver badges60 bronze badges















    • $begingroup$
      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 at 22:39






    • 3




      $begingroup$
      @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).
      $endgroup$
      – tfb
      May 26 at 1:40


















    • $begingroup$
      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 at 22:39






    • 3




      $begingroup$
      @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).
      $endgroup$
      – tfb
      May 26 at 1:40
















    $begingroup$
    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 at 22:39




    $begingroup$
    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 at 22:39




    3




    3




    $begingroup$
    @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).
    $endgroup$
    – tfb
    May 26 at 1:40




    $begingroup$
    @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).
    $endgroup$
    – tfb
    May 26 at 1:40











    8














    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$















    • $begingroup$
      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 at 8:04






    • 1




      $begingroup$
      @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 at 9:43
















    8














    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$















    • $begingroup$
      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 at 8:04






    • 1




      $begingroup$
      @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 at 9:43














    8














    8










    8







    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered May 25 at 12:25









    Aaron StevensAaron Stevens

    23k4 gold badges44 silver badges79 bronze badges




    23k4 gold badges44 silver badges79 bronze badges















    • $begingroup$
      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 at 8:04






    • 1




      $begingroup$
      @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 at 9:43


















    • $begingroup$
      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 at 8:04






    • 1




      $begingroup$
      @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 at 9:43
















    $begingroup$
    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 at 8:04




    $begingroup$
    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 at 8:04




    1




    1




    $begingroup$
    @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 at 9:43




    $begingroup$
    @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 at 9:43











    7














    $begingroup$

    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. 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.






    share|cite|improve this answer











    $endgroup$











    • 3




      $begingroup$
      "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 at 13:49






    • 3




      $begingroup$
      The orbit of the Moon (and all the planets) is an ellipse: this was known to Kepler even before Newton.
      $endgroup$
      – Tom
      May 26 at 0:25
















    7














    $begingroup$

    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. 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.






    share|cite|improve this answer











    $endgroup$











    • 3




      $begingroup$
      "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 at 13:49






    • 3




      $begingroup$
      The orbit of the Moon (and all the planets) is an ellipse: this was known to Kepler even before Newton.
      $endgroup$
      – Tom
      May 26 at 0:25














    7














    7










    7







    $begingroup$

    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. 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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 27 at 12:27

























    answered May 25 at 12:43









    garypgaryp

    17.7k1 gold badge32 silver badges69 bronze badges




    17.7k1 gold badge32 silver badges69 bronze badges











    • 3




      $begingroup$
      "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 at 13:49






    • 3




      $begingroup$
      The orbit of the Moon (and all the planets) is an ellipse: this was known to Kepler even before Newton.
      $endgroup$
      – Tom
      May 26 at 0:25














    • 3




      $begingroup$
      "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 at 13:49






    • 3




      $begingroup$
      The orbit of the Moon (and all the planets) is an ellipse: this was known to Kepler even before Newton.
      $endgroup$
      – Tom
      May 26 at 0:25








    3




    3




    $begingroup$
    "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 at 13:49




    $begingroup$
    "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 at 13:49




    3




    3




    $begingroup$
    The orbit of the Moon (and all the planets) is an ellipse: this was known to Kepler even before Newton.
    $endgroup$
    – Tom
    May 26 at 0:25




    $begingroup$
    The orbit of the Moon (and all the planets) is an ellipse: this was known to Kepler even before Newton.
    $endgroup$
    – Tom
    May 26 at 0:25











    3














    $begingroup$


    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.






    share|cite|improve this answer











    $endgroup$















    • $begingroup$
      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 at 10:19
















    3














    $begingroup$


    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.






    share|cite|improve this answer











    $endgroup$















    • $begingroup$
      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 at 10:19














    3














    3










    3







    $begingroup$


    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.






    share|cite|improve this answer











    $endgroup$




    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 27 at 3:35

























    answered May 27 at 3:29









    CathyCathy

    472 bronze badges




    472 bronze badges















    • $begingroup$
      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 at 10:19


















    • $begingroup$
      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 at 10:19
















    $begingroup$
    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 at 10:19




    $begingroup$
    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 at 10:19











    1














    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$











    • 6




      $begingroup$
      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.
      $endgroup$
      – dmckee
      May 25 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.
      $endgroup$
      – Michael Walsby
      May 26 at 6:32
















    1














    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$











    • 6




      $begingroup$
      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.
      $endgroup$
      – dmckee
      May 25 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.
      $endgroup$
      – Michael Walsby
      May 26 at 6:32














    1














    1










    1







    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited May 27 at 12:55









    Peter Mortensen

    2,0171 gold badge14 silver badges24 bronze badges




    2,0171 gold badge14 silver badges24 bronze badges










    answered May 25 at 12:55









    Michael WalsbyMichael Walsby

    1,4711 silver badge6 bronze badges




    1,4711 silver badge6 bronze badges











    • 6




      $begingroup$
      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.
      $endgroup$
      – dmckee
      May 25 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.
      $endgroup$
      – Michael Walsby
      May 26 at 6:32














    • 6




      $begingroup$
      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.
      $endgroup$
      – dmckee
      May 25 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.
      $endgroup$
      – Michael Walsby
      May 26 at 6:32








    6




    6




    $begingroup$
    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.
    $endgroup$
    – dmckee
    May 25 at 17:23




    $begingroup$
    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.
    $endgroup$
    – dmckee
    May 25 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.
    $endgroup$
    – Michael Walsby
    May 26 at 6:32




    $begingroup$
    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.
    $endgroup$
    – Michael Walsby
    May 26 at 6:32











    0














    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$




















      0














      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$


















        0














        0










        0







        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited May 29 at 1:06









        Buzz

        4,8304 gold badges19 silver badges30 bronze badges




        4,8304 gold badges19 silver badges30 bronze badges










        answered May 28 at 19:38









        my2ctsmy2cts

        7,8242 gold badges7 silver badges23 bronze badges




        7,8242 gold badges7 silver badges23 bronze badges


































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Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029