Does the Earth orbiting around the Earth-Moon barycenter cause a measurable centrifugal force?

I just realized the Earth is not a stationary object with the moon orbiting around it. As shown in this minimalist animation 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.

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?

• Have you calculated the Earth's orbital velocity? Tried to work out the formula? – userLTK Mar 5 '18 at 17:30
• @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 Mar 5 '18 at 18:11
• 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 Mar 5 '18 at 20:23
• 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 Mar 5 '18 at 20:31
• It's big enough to cause the tide on the side of the Earth opposite to the Moon. – Florin Andrei Mar 9 '18 at 0:07

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.

• A quote of the difference in KG to a human would have been awesome, I figure its in 100dths of grams. – DeltaEnfieldWaid Mar 5 '18 at 18:52
• @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 Mar 5 '18 at 19:19
• Not sure if you want to add a link to this answer or not, but I've used it several times myself. – uhoh Mar 6 '18 at 5:30
• @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 Mar 7 '18 at 10:58
• @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 Mar 7 '18 at 11:42

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.

• This is an answer I posted about anti-podal bulge Sept 2015. – BillDOe Mar 12 '19 at 20:33

Something interesting from Wikipedia:

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).