<p>The Earth and Moon are in orbit about each other, which means that the centrifugal (outward inertial) force <span class="math-container">$M_iV_i^2/d_i$</span> is balanced with the centripetal (real inward force), ie <span class="math-container">$GM_{\oplus}M_{\mathrm{Moon}}/(d_\oplus+d_{\mathrm{Moon}})^2$</span>, where <span class="math-container">$d_\oplus$</span> is distance from Earth to Earth-Moon center of mass (c.m.) and <span class="math-container">$d_{\mathrm{Moon}}$</span> is the Moon's c.m. distance.  Thus, most of the centrifugal force of the motion about the c.m. is canceled. Since we reside at one Earth radius from the center of the Earth, the cancellation is not exact, and the remainder is exactly what we call the Lunar Tide, i.e. the familiar tidal force of the Moon which causes most of the ocean tides (there is also a component from the orbit about the Sun).</p>

<p>The tidal acceleration is described in <a href="https://en.wikipedia.org/wiki/Tidal_acceleration" rel="nofollow noreferrer">Wikipedia page</a>.</p>

CC BY-SA 4.0
2019-03-12T20:33:47.383Z
[This](https://astronomy.stackexchange.com/questions/11882/what-causes-the-antipodal-bulge/11914#11914) is an answer I posted about anti-podal bulge Sept 2015.
BillDOe
CC BY-SA 4.0
2019-03-12T20:19:37.360Z
2019-03-12T20:19:37.360Z
eshaya
<p>Something interesting from <a href="https://en.wikipedia.org/wiki/Orbit_of_the_Moon#Path_of_Earth_and_Moon_around_Sun" rel="nofollow noreferrer">Wikipedia</a>:</p>

<blockquote>
  <p><a href="https://i.sstatic.net/1haFN.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1haFN.png" alt="enter image description here"></a>
  Scale model of the Earth–Moon system: Sizes and distances are to
  scale. It represents the mean distance of the orbit and the mean radii
  of both bodies.</p>
</blockquote>

<p>Earth orbits the common barycenter of the Earth–Moon system in about <strong>28 days</strong>, the same time the Moon orbits Earth (or more accurately, the common barycenter of the Earth–Moon system which is below Earth surface). It's estimated that the centrifugal force caused by this should be minimal on Earth surface, and only observable by a very large scale (e.g., the tides).</p>

CC BY-SA 4.0
2019-04-18T02:47:58.257Z
2019-04-18T02:47:58.257Z
Rockallite
<p>The Earth-Moon system is best thought of as the Moon orbiting the Earth, but not around the Earth's centre, rather their barycentre (the common centre of mass).  This means that the Moon orbits the Earth but pulls the Earth around a little so that the Earth has a translational motion around the barycentre as motion in a circle (but not a rotation/revolution about that point).  If the Earth and Moon both revolved around a common centre then they would be mutually tidally locked and both would show each other their same face forever, with each not changing position in the sky of the other at all (as is the case for Pluto and Charon, which are a binary dwarf planetary system).  This can be appreciated in animations contained in an article by Paolo Sirtoli at <a href="https://www.vialattea.net/content/tides-and-centrifugal-force/" rel="nofollow noreferrer">https://www.vialattea.net/content/tides-and-centrifugal-force/</a>.</p>
<p>A centrifugal force is a fictitious force that is introduced into a non-inertial reference frame to make sense of observations for someone inside that non-inertial reference frame.  So by definition, there is no centrifugal force that can be measured.</p>
<p>If the Earth's centre is to be considered as a reference frame then it is non-inertial because it is being pulled on by the moon, and so to make it quasi-inertial, a fictitious inertial force (since the word centrifugal implies rotation, which is not happening here) is introduced, which is everywhere, i.e. uniform, a force that opposes the Moon's pull as at the centre of the Earth.  The superposition of the Moon's varying gravity force across the diameter of the Earth and this fictitious inertial force produces the classic tidal force diagram that you see, that looks to produce the two tidal bulges, on the near and far side of the Moon.  In reality the tidal force is so small in comparison to the Earth's gravity force (by some ten million times) that it is not this force that causes the rise in the tides, rather the forces at ~45 degree offsets that serve to slide the water surface perpendicularly to the gravity force - here the small tidal force has no opposing gravity force - so as to collectively push up a noticeable tide at the near and far sides.</p>
<p>Alternatively, the reference frame can be thought of as the barycentre, which is truly fixed in relation to the inertial stars (ignoring other influences).  In this case, the reference frame is truly inertial, but the Earth's centre actually translates about the barycentre and so a real centripetal force has to be taken into account, pointing towards the barycentre.  Now, if you analyse the motion of the translating Earth (centre) about the barycentre carefully, you will notice that all points on the Earth are moving in a circular motion about their own (bary)centres, with a single radius value, and that these forces all point towards their own (bary)centres in a direction towards the distant Moon, i.e. it is a uniform centripetal force that acts in the region of the Earth, but towards the Moon.  This too is evident in Paolo Sirtoli's article.  This centripetal force is provided by the Moon's varying gravitational force across the Earth's diameter and when the two are compared, the result is the same (net) tidal forces acting in the same direction to produce the near and far side bulges (via the forces acting at the ~45 degree offsets, as before).</p>
<p>Now, you can consider a frame of reference in which things rotate around the barycentre, and so a centripetal force arises again, which might throw up a difference between the near side and far side centripetal force, which you might think might affect the tides.  However, you need to 'convert' your barycentre reference origin to the centre-of-earth reference origin, and when you do this you find that the centripetal forces are transformed into uniform forces around the centre of the Earth, and so no differential arises that could affect tides.</p>
<p>You may note that the Earth's gravity and axial rotation play no part in tides, - these forces are from within the Earth and so cannot exert a force on the Earth.</p>

CC BY-SA 4.0
2021-07-29T17:46:22.283Z
2021-07-29T17:46:22.283Z
JerryFrog
<p>That the Earth and Moon orbit about their center of mass from the perspective of an inertial frame of reference is a bit irrelevant. One thing that is quite relevant is that gravitational force is undetectable by a local measuring device. For example, people standing still on the surface of the Earth do not feel gravity. They instead feel the normal force pushing them up, away from the center of the Earth. The gravitational force on astronauts in the International Space Station is about 90% of what they experience on the Earth's surface, but they feel none of that.</p>

<p>Another relevant factor is that the Earth as a whole, along with objects on the surface of the Earth, accelerate toward the Moon (and the Sun, and Jupiter, and Venus, and ...) gravitationally. The gravitational acceleration of those surface-bound objects toward those other bodies is not exactly the same as is that of the Earth as a whole.</p>

<p>The difference between these accelerations results in a force that can be measured. This is the tidal acceleration. An extremely sensitive scale will show that you weigh slightly more when the Moon is on the horizon than when it is directly overhead. For a 61 kg person, this difference in weight between the Moon being on the horizon vs directly overhead is about 10<sup>-4</sup> newtons.</p>

<p>Compared to the ~600 newton weight of that 61 kg person, this is a very small effect. This very small effect, along with an even smaller effect from the Sun (roughly half), are however responsible for the tides in the oceans.</p>

CC BY-SA 3.0
2018-03-05T18:52:09.560Z
A quote of the difference in KG to a human would have been awesome, I figure its in 100dths of grams.
bandybabboon
CC BY-SA 3.0
2018-03-05T19:19:43.120Z
@com.prehensible -- Weight is measured in newtons, mass in kilograms. The latter is invariant, ignoring relativistic effects. The mass of a 61 kg person is 61 kg on the surface of the Earth, 61 kg on surface of the Moon, and 61 kg in the International Space Station (where the person in question is essentially weightless.)
David Hammen
CC BY-SA 3.0
2018-03-06T05:30:28.247Z
Not sure if you want to add a link to [this answer](https://physics.stackexchange.com/a/141981/83380) or not, but I've used it several times myself.
uhoh
CC BY-SA 3.0
2018-03-07T10:58:35.780Z
@com.prehensible -- the answer to your question is "one one-hundredth-thousandth of a kg, or about 10 milligrams" (61 kg * (1e-4/600))
Peter Erwin
CC BY-SA 3.0
2018-03-07T11:42:10.737Z
@PeterErwin - No it's not the answer, at least not in Newtonian mechanics, where mass is invariant. The zenith angle of the Moon does not change a person's mass. It does however affect a person's weight.
David Hammen
CC BY-SA 3.0
2018-03-05T18:08:38.077Z
2018-03-05T18:08:38.077Z
David Hammen
<p>I just realized the Earth is not a stationary object with the moon orbiting around it.  As shown in <a href="https://upload.wikimedia.org/wikipedia/commons/5/59/Orbit3.gif" rel="noreferrer">this minimalist animation</a> from Wikipedia, the Earth actually orbits a common barycenter with the moon.  If this was a carnival ride, people sitting in the moon and on the far side of the earth would certainly feel the centrifugal force acting on them.</p>

<p><a href="https://i.sstatic.net/PiKVD.png" rel="noreferrer"><img src="https://i.sstatic.net/PiKVD.png" alt="enter image description here"></a></p>

<p>However, this is a very large system in reality, and the Earth also has quite a strong gravitational force felt on the surface anyway.  So when it comes to accurately measuring gravity or doing experiments that rely on gravity, how big of a factor is this centrifugal force?  Does the apearant force of gravity fluctuate from 9.9 m/s^2 to 9.7 or is it more along the lines of 9.800001 to 9.799999 (assuming the average is exactly 9.8, which is a simplification).  Or is there something I'm missing that means the force is non-existent?</p>

CC BY-SA 3.0
2018-03-05T17:30:31.193Z
Have you calculated the Earth's orbital velocity?   Tried to work out the formula?
userLTK
CC BY-SA 3.0
2018-03-05T18:11:16.053Z
@userLTK my assumption would be that these equations have been done and someone could point out a source that discusses it.  I couldn’t find anything good with searches because of the avalanche of basic “his is how the moon and earth orbit” articles.
JPhi1618
CC BY-SA 3.0
2018-03-05T20:23:50.477Z
Fair enough.   I was just suggesting a bit of effort before asking as a kind of stack-exchange recommendation.   (not just googling, but trying to work it out).   Granted, if you calculate the centrifugal force based on radius and velocity, you'd get an incorrect answer but it would be a start (objects falling through space, aka, in orbits don't experience a force acting on them).   It was just a suggestion to try to work out the math before asking.
userLTK
CC BY-SA 3.0
2018-03-05T20:31:02.930Z
I just want to add that a carnival ride, you don't really experience the curvature, what you experience is the added g-forces from the centrifugal force.     The Moon can't add g'forces in the direction of the center of the Earth, only away from, and even then effects are tiny compared to Earth's gravity.
userLTK
CC BY-SA 3.0
2018-03-09T00:07:08.327Z
It's big enough to cause the tide on the side of the Earth opposite to the Moon.
Florin Andrei
<orbit><tidal-forces>
Does the Earth orbiting around the Earth-Moon barycenter cause a measurable centrifugal force?
JPhi1618
# Does the Earth orbiting around the Earth-Moon barycenter cause a measurable centrifugal force?

(Asked by: username_0 on 2018-03-05)

I just realized the Earth is not a stationary object with the moon orbiting around it. As shown in [this minimalist animation](https://upload.wikimedia.org/wikipedia/commons/5/59/Orbit3.gif) from Wikipedia, the Earth actually orbits a common barycenter with the moon. If this was a carnival ride, people sitting in the moon and on the far side of the earth would certainly feel the centrifugal force acting on them.

[![enter image description here](https://i.sstatic.net/PiKVD.png)](https://i.sstatic.net/PiKVD.png)

However, this is a very large system in reality, and the Earth also has quite a strong gravitational force felt on the surface anyway. So when it comes to accurately measuring gravity or doing experiments that rely on gravity, how big of a factor is this centrifugal force? Does the apearant force of gravity fluctuate from 9.9 m/s^2 to 9.7 or is it more along the lines of 9.800001 to 9.799999 (assuming the average is exactly 9.8, which is a simplification). Or is there something I'm missing that means the force is non-existent?

**Tags:** orbit, tidal-forces

> **Comments:**
> **username_1**: Have you calculated the Earth's orbital velocity?   Tried to work out the formula?  
> **username_0**: @username_1 my assumption would be that these equations have been done and someone could point out a source that discusses it.  I couldn’t find anything good with searches because of the avalanche of basic “his is how the moon and earth orbit” articles.  
> **username_1**: Fair enough.   I was just suggesting a bit of effort before asking as a kind of stack-exchange recommendation.   (not just googling, but trying to work it out).   Granted, if you calculate the centrifugal force based on radius and velocity, you'd get an incorrect answer but it would be a start (objects falling through space, aka, in orbits don't experience a force acting on them).   It was just a suggestion to try to work out the math before asking.  
> **username_1**: I just want to add that a carnival ride, you don't really experience the curvature, what you experience is the added g-forces from the centrifugal force.     The Moon can't add g'forces in the direction of the center of the Earth, only away from, and even then effects are tiny compared to Earth's gravity.  
> **username_2**: It's big enough to cause the tide on the side of the Earth opposite to the Moon.  

---

## 4 answers

### Answer by username_7 on 2018-03-05. Score: 6 | Accepted answer

That the Earth and Moon orbit about their center of mass from the perspective of an inertial frame of reference is a bit irrelevant. One thing that is quite relevant is that gravitational force is undetectable by a local measuring device. For example, people standing still on the surface of the Earth do not feel gravity. They instead feel the normal force pushing them up, away from the center of the Earth. The gravitational force on astronauts in the International Space Station is about 90% of what they experience on the Earth's surface, but they feel none of that.

Another relevant factor is that the Earth as a whole, along with objects on the surface of the Earth, accelerate toward the Moon (and the Sun, and Jupiter, and Venus, and ...) gravitationally. The gravitational acceleration of those surface-bound objects toward those other bodies is not exactly the same as is that of the Earth as a whole.

The difference between these accelerations results in a force that can be measured. This is the tidal acceleration. An extremely sensitive scale will show that you weigh slightly more when the Moon is on the horizon than when it is directly overhead. For a 61 kg person, this difference in weight between the Moon being on the horizon vs directly overhead is about 10-4 newtons.

Compared to the ~600 newton weight of that 61 kg person, this is a very small effect. This very small effect, along with an even smaller effect from the Sun (roughly half), are however responsible for the tides in the oceans.

> **Comments:**
> **username_8**: A quote of the difference in KG to a human would have been awesome, I figure its in 100dths of grams.  
> **username_7**: @com.prehensible -- Weight is measured in newtons, mass in kilograms. The latter is invariant, ignoring relativistic effects. The mass of a 61 kg person is 61 kg on the surface of the Earth, 61 kg on surface of the Moon, and 61 kg in the International Space Station (where the person in question is essentially weightless.)  
> **username_9**: Not sure if you want to add a link to [this answer](https://physics.stackexchange.com/a/141981/83380) or not, but I've used it several times myself.  
> **username_10**: @com.prehensible -- the answer to your question is "one one-hundredth-thousandth of a kg, or about 10 milligrams" (61 kg * (1e-4/600))  
> **username_7**: @username_10 - No it's not the answer, at least not in Newtonian mechanics, where mass is invariant. The zenith angle of the Moon does not change a person's mass. It does however affect a person's weight.  

---

### Answer by username_3 on 2019-03-12. Score: 2

The Earth and Moon are in orbit about each other, which means that the centrifugal (outward inertial) force $M\_iV\_i^2/d\_i$ is balanced with the centripetal (real inward force), ie $GM\_{\oplus}M\_{\mathrm{Moon}}/(d\_\oplus+d\_{\mathrm{Moon}})^2$, where $d\_\oplus$ is distance from Earth to Earth-Moon center of mass (c.m.) and $d\_{\mathrm{Moon}}$ is the Moon's c.m. distance. Thus, most of the centrifugal force of the motion about the c.m. is canceled. Since we reside at one Earth radius from the center of the Earth, the cancellation is not exact, and the remainder is exactly what we call the Lunar Tide, i.e. the familiar tidal force of the Moon which causes most of the ocean tides (there is also a component from the orbit about the Sun).

The tidal acceleration is described in [Wikipedia page](https://en.wikipedia.org/wiki/Tidal_acceleration).

> **Comments:**
> **username_4**: [This](https://astronomy.stackexchange.com/questions/11882/what-causes-the-antipodal-bulge/11914#11914) is an answer I posted about anti-podal bulge Sept 2015.  

---

### Answer by username_5 on 2019-04-18. Score: 2

Something interesting from [Wikipedia](https://en.wikipedia.org/wiki/Orbit_of_the_Moon#Path_of_Earth_and_Moon_around_Sun):

> [![enter image description here](https://i.sstatic.net/1haFN.png)](https://i.sstatic.net/1haFN.png)
> Scale model of the Earth–Moon system: Sizes and distances are to
> scale. It represents the mean distance of the orbit and the mean radii
> of both bodies.

Earth orbits the common barycenter of the Earth–Moon system in about **28 days**, the same time the Moon orbits Earth (or more accurately, the common barycenter of the Earth–Moon system which is below Earth surface). It's estimated that the centrifugal force caused by this should be minimal on Earth surface, and only observable by a very large scale (e.g., the tides).


---

### Answer by username_6 on 2021-07-29. Score: 0

The Earth-Moon system is best thought of as the Moon orbiting the Earth, but not around the Earth's centre, rather their barycentre (the common centre of mass). This means that the Moon orbits the Earth but pulls the Earth around a little so that the Earth has a translational motion around the barycentre as motion in a circle (but not a rotation/revolution about that point). If the Earth and Moon both revolved around a common centre then they would be mutually tidally locked and both would show each other their same face forever, with each not changing position in the sky of the other at all (as is the case for Pluto and Charon, which are a binary dwarf planetary system). This can be appreciated in animations contained in an article by Paolo Sirtoli at <https://www.vialattea.net/content/tides-and-centrifugal-force/>.

A centrifugal force is a fictitious force that is introduced into a non-inertial reference frame to make sense of observations for someone inside that non-inertial reference frame. So by definition, there is no centrifugal force that can be measured.

If the Earth's centre is to be considered as a reference frame then it is non-inertial because it is being pulled on by the moon, and so to make it quasi-inertial, a fictitious inertial force (since the word centrifugal implies rotation, which is not happening here) is introduced, which is everywhere, i.e. uniform, a force that opposes the Moon's pull as at the centre of the Earth. The superposition of the Moon's varying gravity force across the diameter of the Earth and this fictitious inertial force produces the classic tidal force diagram that you see, that looks to produce the two tidal bulges, on the near and far side of the Moon. In reality the tidal force is so small in comparison to the Earth's gravity force (by some ten million times) that it is not this force that causes the rise in the tides, rather the forces at ~45 degree offsets that serve to slide the water surface perpendicularly to the gravity force - here the small tidal force has no opposing gravity force - so as to collectively push up a noticeable tide at the near and far sides.

Alternatively, the reference frame can be thought of as the barycentre, which is truly fixed in relation to the inertial stars (ignoring other influences). In this case, the reference frame is truly inertial, but the Earth's centre actually translates about the barycentre and so a real centripetal force has to be taken into account, pointing towards the barycentre. Now, if you analyse the motion of the translating Earth (centre) about the barycentre carefully, you will notice that all points on the Earth are moving in a circular motion about their own (bary)centres, with a single radius value, and that these forces all point towards their own (bary)centres in a direction towards the distant Moon, i.e. it is a uniform centripetal force that acts in the region of the Earth, but towards the Moon. This too is evident in Paolo Sirtoli's article. This centripetal force is provided by the Moon's varying gravitational force across the Earth's diameter and when the two are compared, the result is the same (net) tidal forces acting in the same direction to produce the near and far side bulges (via the forces acting at the ~45 degree offsets, as before).

Now, you can consider a frame of reference in which things rotate around the barycentre, and so a centripetal force arises again, which might throw up a difference between the near side and far side centripetal force, which you might think might affect the tides. However, you need to 'convert' your barycentre reference origin to the centre-of-earth reference origin, and when you do this you find that the centripetal forces are transformed into uniform forces around the centre of the Earth, and so no differential arises that could affect tides.

You may note that the Earth's gravity and axial rotation play no part in tides, - these forces are from within the Earth and so cannot exert a force on the Earth.


---

