# Center of mass of the earth and the moon

1. Jul 15, 2009

### Labrodor

The center of mass of the earth is approximately at the center. The center of mass is also where the force vector of the gravitational force is pointing. (down that is).

Now the total gravitational force excerted on me should be the sum of the force excerted by the earth plus anything else, especially the moon. With the sum of these two vectors being what I feel as "down", should the direction of down change as the moon's position in the sky changes?

I thought I'd do a quick calculation and hear if you guys think it makes sense.

Approximate average distance from the surface of the earth to the center of the moon is
378 000 km. It has a mass of 7.34*10^22 kg.
The gravitational constant is G = (6.67*10^-11)*(m^3)*(kg^-1)*(s^(- 2))

Which should give me the acceleartion:
a_moon = G * Moon_mass / distance to moon^2
a_moon = (6.67*10^-11)*(m^3)*(kg^-1)*(s^(- 2)) * 7.34*10^22 kg / ((3.78*10^8 m)^2)

a_moon = 3.426x10^-5 m/s^2

With Earth's acceleration at 9.8 and the moon at the horizon we can do some quick trigonometry on the vectors. With the two vectors and their sum a_moon_vector + a_earth_vector forming a triangle, with the sum equal to the hypotenuse, we get an angle A which is the angle between the direction of felt down and the direction towards the centre of the earth.

Tan A = a_moon/a_earth
A = Tan^-1 ( 3.42*10^-5 / 9.8) = ca 2*10^-4 degrees

Is this enough to measured? Since we've got the ratio of the lengths in the triangle, we quickly find that if we have a pendulum hanging from a 300 meter long string (say, from the eiffel tower) it will deviate by 1 millimeter. That is, a change in 2 millimeters as the moon passes from one side of the horizon to the next.

2. Jul 15, 2009

### D H

Staff Emeritus
The center of mass of the Earth is exactly at the center of the Earth, by definition.

Only approximately so. The Earth is an oblate spheroid, approximately. For a true oblate spheroid, the gravitational force points toward the center of mass at the equator and at the poles. At intermediate points it does not.

A couple things wrong here. You are ignoring that the Earth itself is accelerating toward the Moon, and you are implicitly assuming that you can measure gravity. One at a time,

1. The Earth itself is accelerating toward the Moon. This means you need to subtract the acceleration at the center of the Earth toward the Moon from the acceleration at a point on the Earth's surface toward the Moon. If you don't do this you will come to the erroneous conclusion that the Sun should induce tides much higher than does the Moon. The measurable acceleration is much, much smaller than your 3.426x10-5 m/s^2. It is about 1.2x10-6 m/s^2

2. You can't measure gravity. You can measure all the forces acting on an object except for gravity. You would need a gravity shield to be able to measure gravity. There is no such thing as a gravity shield. A spring scale doesn't measure gravity; it measures the upward force exerted by the scale needed to keep you from sinking into the scale.

3. Jul 15, 2009

### negitron

No, it is not. The center of mass and the geometric center are in two different places because the density distribution is not homogenous.

Then explain what Cavendish did.

4. Jul 15, 2009

### Bob S

The Earth tides around the circumference of the Earth cause the ground under the particle accelerators at CERN (Geneva, Switzerland) to rise and fall about +/- 25 to 40 cm roughly every 12 hours. These tides are due to the roughly 24 hour period of the Moon, and Moon gravity. This is sufficient to require retuning of key accelerator components. See
http://accelconf.web.cern.ch/AccelConf/p93/PDF/PAC1993_0044.PDF

5. Jul 15, 2009

### D H

Staff Emeritus
http://earthobservatory.nasa.gov/Newsroom/view.php?id=32922 [Broken]
Scientists currently define Earth's center in two ways: as the mass center of solid Earth or as the mass center of Earth's entire system, which combines solid Earth, ice sheets, oceans and atmosphere. Argus says there is room for improvement in these estimates.

"The past two international estimates of the motion of the Earth system's mass center, made in 2000 and 2005, differ by 1.8 millimeters (.07 inches) a year," he said. "This discrepancy suggests the motion of Earth's mass center is not as well known as we'd like."

Argus argues that movements in the mass of Earth's atmosphere and oceans are seasonal and do not accumulate enough to change Earth's mass center. He therefore believes the mass center of solid Earth provides a more accurate reference frame.​

He measured the acceleration of his reference frame with respect to an inertial (stream of falling apples) frame. He inferred (and rightly so) that this acceleration resulted from the Earth's gravity.

Earth tides also affect the orbits of artificial satellites, and significantly more so than do the ocean tides. The Sun also contributes to the Earth tides, about half that of the Moon's contribution.

Last edited by a moderator: May 4, 2017
6. Jul 15, 2009

### negitron

No, you're thinking of Newton. This is Henry Cavendish.

7. Jul 15, 2009

### Creator

I hate to throw another wrench in your idea , Labrodor, but the center of mass of the moon doesn't coincide with its geometric center either. It is off by about 2 kilometers....which may be negligible in your calculation, but its just something to keep in mind.

..

8. Jul 15, 2009

### TurtleMeister

A torsion balance can be used to measure gravity.

9. Jul 15, 2009

### negitron

Right, hence my reference to Cavendish's experiment (which wasn't intended to measure G, although it ultimately had that effect).

10. Jul 15, 2009

### D H

Staff Emeritus
No, I'm thinking of Cavendish. You need to think of Einstein.

Cavendish type experiments measure a mechanical force or torque caused by the electrostatic force and a displacement. They do not measure gravity directly. There is no way to measure gravity directly. They infer the gravitational force based on the displacement and the measured mechanical force.

11. Jul 15, 2009

### negitron

Oh, please. You are engaging in nothing but semantic nitpicking at this point. You may as well say there's no way to measure voltage or pressure.

12. Jul 15, 2009

### Creator

13. Jul 15, 2009

### Staff: Mentor

Your calculation assumes a static case. The earth and moon are in orbit around their common center of gravity and I think such a variation would cancel out due to the orbit. When in orbit, you are in "freefall" torward the center of mass and not "feeling" any force in that direction.

14. Jul 15, 2009

### D H

Staff Emeritus
Re: Measuring gravity

Have you heard of the equivalence principle? Suppose you took a gravimeter on a spaceship to the middle of nowhere (no massive objects anywhere in sight). Now suppose the spaceship fires its engines to accelerate at 9.8 m/s2. The gravimeter will measure that acceleration even though you aren't close to any massive object. A gravimeter does not measure gravity. It measures acceleration with respect to an inertial frame -- an inertial frame in general relativity, that is.

The inability to measure gravity is critical for inertial navigation systems on aircraft and spacecraft. The accelerometers in an INS measure all forces acting on the vehicle but gravity. The vehicle's software must estimate the gravitational acceleration via some mathematical model in order to estimate where the vehicle is at some point in time.

15. Jul 15, 2009

### Creator

There are several flaws in your thinking here:

First, what about the sun? you have ignored its influence.

Second, ignoring tidal forces, either way there is no deflection because no matter what the gravitational source, you have to realize that the earth (and everything on it) is in free fall around that source (or the barycenter due to that source).

Lets use the sun ONLY (forget the moon) for simplicity, ignore TIDAL FORCES. Since the sun is so massive we can consider earth in free fall around it (without worrying about the barycenter). NOW; IGNORING TIDAL FORCES, this means that everything on earth is in free fall around the sun including the pendulum and so there is no differential force acting on the pendulum and the earth, so your answer is: No there will be no deflection from solar gravity alone (IF WE IGNORE TIDAL FORCES).

Likewise with the moon, except that we are in free fall about the earth-moon barycenter. So the pendulum again is in free fall also and no extra deflection due to lunar position as a result of earth rotation, (IF WE IGNORE TIDAL FORCES).

However, this ideal is unrealizable due to tidal forces which definitely will show up.

..

16. Jul 15, 2009

### Creator

Re: Measuring gravity

Earth to DH....
Like negitron said , YOU ARE NIT PICKING and missed the point all together ....we are NOT talking about going out on a space ship to take measurements, the question was about taking a gravity measurement from planet earth> Hello!
A gravimeter is named that precisely because it takes LOCAL gravity meansurements....Duh? It CAN be done and IT IS DONE ALL THE TIME.

P.S. Russ Watters ...you hit the nail on the head...my post ran over yours; missed it before I posted....sorry.

Last edited: Jul 15, 2009
17. Jul 16, 2009

### D H

Staff Emeritus
Re: Measuring gravity

The Earth is free falling toward the Moon, period. The equations for the tidal force do not care where the barycenter is. The barycenter comes into play in two ways. In addition to the Earth being in freefall toward the Moon (and Sun), the Moon (and Sun) are in freefall toward the Earth. Secondly, the barycenter makes for a handy reference for defining an inertial frame (in the Newtonian mechanics sense of an inertial frame).

Suppose you are observing a solar eclipse with the Sun and Moon directly overhead. Note that the Sun-Earth barycenter is above you but the Earth-Moon barycenter is beneath you. That is irrelevant when it comes to tidal forces. The tidal forces on you due to the Sun and the Moon are in the same direction: Up.

There is no reason to shout.

There is no reason to be rude, either.

The Earth rotates at one revolution per sidereal day with respect to inertial space. If a gravimeter measured gravity, how can you explain that the apparent centrifugal force that arises from the Earth's rotation is part of what the gravimeter reads? The gravitational force by the Earth on a vehicle in low Earth orbit is about 90% that of surface gravity. If a gravimeter measured gravity, how can you explain why won't it work in a spacecraft in low Earth orbit?

You pointed to the wikipedia article on gravimeters earlier. The start of the second sentence is key: "A gravimeter is a type of accelerometer". Accelerometers measure the acceleration of the accelometer's case frame with respect to a local inertial frame (general relativistic sense). An accelerometer (or gravimeter) on a non-thrusting spacecraft in low Earth orbit reads close to zero because the spacecraft is in freefall: It is a local inertial frame. A gravimeter on the surface of the Earth reads about 9.8 m/s2 because the gravimeter isn't in freefall. A falling apple is in freefall, and in general relativity, the falling apple serves as the basis for a local inertial frame.

18. Jul 16, 2009

### Labrodor

Thanks for several interesting replies, the technical level is a lot higher here than at comparable forums.

It's true that I ignored the acceleration of the earth. Two hypotheses seem to be offered in the discussion. That the pendulum will show a much smaller deviation or that the deviation will be exactly zero. The existence of tides makes me doubt that the sum of all graviational forces (mentioned and unmentioned ones) on the pendulum would stay unchanged over time.

However, the tides is a topic where I've tended to mess up the explanation, so I will not try to justify this hunch :-)

19. Jul 16, 2009

### BobG

The direction of gravitational force is not towards the center of the Earth (as other's have mentioned).

It does remain in a virtually constant direction provided you're remaining at the same location on the Earth.

Since your goal is to measure how the Moon changes the direction of down (regardless of what direction that may be), then Yes, that will work provided your pendulum is long enough that you can measure the deviation.

The main complication would be cancelling out the noise from the Sun and the planets since their geometric configuration will affect the direction of "down" as well. In practice, screening out the unwanted noise to measure the desired attribute is just part of the challenge (often the largest part of the challenge).