Sciencemadness Discussion Board - Change in weight due to the moons gravity.

Change in weight due to the moons gravity.

wg48temp9 - 22-10-2019 at 09:31

I guess that the weight of say a calibration mass on the earth changes due to the moons gravity as the moon moves round the earth. That's what causes the tides.

But I can not find a description of the effect or its magnitude or alternatively if the weight does not change due to the moons rotation round the earth why it does not.

Putting it differently can I detect the moons motion by the changes in the measured weight of a test mass at my location and how many bits would I need on my digital scales? Hopefully it is less than 24.

PS Apparently for some reason, probably my finger trouble, my post was posted twice this one without the question. I have now added it and deleted the first post.

[Edited on 10/22/2019 by wg48temp9]

annaandherdad - 22-10-2019 at 09:36

Is this intended to be a question? If it is, then there is probably an answer (it depends on precisely what the question is).

RedDwarf - 22-10-2019 at 10:17

A quick back of the envelope(literally) calculation gives me a decrease of 3.34 x 10^-6 %
Gravity varies with the square of the radius and the acceleration due to gravity at the surface of the moon is 1.6 m/s ² compared to 9.8 m/s² on earth. The moon has a radius of 1700km and is about 384000km from the earth.

Feel free to tell me I (or my envelope) have screwed up the calculation, but I think that's it.

I suspect this effect would be completely outweighed by changed gravitational effect of the shifting position of the water on the earth.

Edit: missed a percent symbol in my excitement at having managed to get my superscripts correctly formatted!

[Edited on 22-10-2019 by RedDwarf]

wg48temp9 - 22-10-2019 at 10:24

Quote: Originally posted by annaandherdad

Is this intended to be a question? If it is, then there is probably an answer (it depends on precisely what the question is).

Sorry see my amended post.

Praxichys - 22-10-2019 at 10:26

I was interested so I broke out the calculator:

Standard 2-body gravitational equations were used. With a nominal distance of 384,400 km, the moon exerts approximately 0.00265N on an 82kg man. At 1g that's about ±270mg with the moon at zenith/nadir with respect to the local gravitational vector.

More accurately, the offset of the of the Earth/man system changes the results by 7.0% accounting for the diameter of the Earth, giving a more realistic +262mg/-280mg relative to your weight at nadir/zenith respectively, at distances of 390,771 km and 378,029 km, respectively.

Getting even more detailed, the orbit of the moon has a mean eccentricity of 0.0549, which gives a mean distance of 363,396 km at periapsis and 405,504 at the apoapsis. Adding this to Earth's 6,371 km radius, the maximum weight influence (moon at zenith and closest), occurring at 357,025 km would be -314mg, and the minimum weight influence (moon at nadir and farthest), occurring at 411,875 km would be +236mg.

EDIT: Now that the question has appeared I have calculated the variance to be 6.71x10^-7% for Earth objects, or roughly 0.0671ppb. By comparison, a local source (just spitballing) such as 5 cubic miles of water at an average distance of 5km exert almost double the force on an 82kg object, about 465mgf at 1g.

EDIT EDIT: @RedDwarf, can you check your calculations? Our results are similar but off by an order of magnitude. I have done it 3 times and I can't see an error...

[Edited on 22-10-2019 by Praxichys]

wg48temp9 - 22-10-2019 at 11:04

Quote: Originally posted by RedDwarf

A quick back of the envelope(literally) calculation gives me a decrease of 3.34 x 10^-6 %
Gravity varies with the square of the radius and the acceleration due to gravity at the surface of the moon is 1.6 m/s ² compared to 9.8 m/s² on earth. The moon has a radius of 1700km and is about 384000km from the earth.

Feel free to tell me I (or my envelope) have screwed up the calculation, but I think that's it.

I suspect this effect would be completely outweighed by changed gravitational effect of the shifting position of the water on the earth.

Edit: missed a percent symbol in my excitement at having managed to get my superscripts correctly formatted!

[Edited on 22-10-2019 by RedDwarf]

I have not checked your maths but I do suspect your missing a factor of approximately 2 due to the +/- effect of the moon being above my location or below it. Other than that your calculation was a smart way of getting an answer that did not need the mass of the earth and moon.

Praxichys: I have not checked your maths either, my mathcad is down and my fancy calculator needs new batteries if it still works. I prefer your answer as it suggests approximately 0.5g change in 82kg which is approximately only 18bits.

As check an 82kg man has a volume of about 82l, air is about 1kg/m^3 or 1g/l making the buoyancy correction of a man 82g. Significantly larger then the moon correction. Perhaps that's why the moon correction is not used.
I just realised the air density my change with the moon too.

With a lead test mass measured in temperature controlled vacuum chamber its apparently measurable with a 24bit scales and averaging over a day to reduce environmental noise.

PS: Actually it would have to be averaged over an hour, the earth rotates.

[Edited on 10/22/2019 by wg48temp9]

annaandherdad - 23-10-2019 at 08:32

You guys are missing the fact that the moon's effect on the earth is *tidal*, that is, the moon pulls on the earth as well as on your test mass, so the extra force the test mass experiences is proportional to the *difference* between the acceleration of the earth and that of the test mass, produced by the moons gravitational field. This introduces an extra factor of R/r in the computation of the force, and a nontrivial angular dependence, where R is the radius of the earth and r the distance to the moon.

Tidal forces explain why the bulge of ocean water points toward the moon on the side of the earth facing the moon, and points *away* from the moon on the opposite side.

However, you are correct to point out that the bulge in the ocean water has its own gravitational effect that cannot be neglected, and there is a bulge in the solid body of the earth, too. The problem is really self-consistent.

By the way, the centrifugal force due to the earth's rotation on its axis is much greater than the tidal force of the moon (but it is constant in time).

Sulaiman - 23-10-2019 at 09:31

2²⁴ = 16,777,216
the variation of gravity at fixed point on the Earth's surface near the equator due to the Moon is approximately 0.11ppm,
the Sun contributes 0.052ppm
so you should be able to see the effect ...
____________________________________________________________
but
you will have to compensate your balance for changes in buoyancy due to changes in atmospheric density,
which is a function of temperature and local climatic pressure,
with a small contribution from humidity.
also,
your electronics will have to be very carefully designed and compensated for temperature coefficients.
Typically you will have a voltage reference, a frequency reference, and amplifier offsets - that are all temperature dependant.
There is also the temperature coefficient of the current sensing resistor,
and the actual magnetic force compensation coil/magnet system are subject to temperature coefficients ... F=B.I.L
the magnet geometry/size is temperature dependant, as is the length of the wire.
These effects combined are typically much greater than the variations in gravity.

Note that a larger current is drawn from the supply when a heavier weight is measured, hence a change in internal heat generation.

If you are using a strain gauge with I-beam then there is virtually no chance of observing the effects of the Moon and/or Sun gravity.

aside: the limititations to accuracy for a well made pendulum clock are the Sun/Moon/tide changes in gravity.

wg48temp9 - 23-10-2019 at 14:42

Quote: Originally posted by annaandherdad

You guys are missing the fact that the moon's effect on the earth is *tidal*, that is, the moon pulls on the earth as well as on your test mass, so the extra force the test mass experiences is proportional to the *difference* between the acceleration of the earth and that of the test mass, produced by the moons gravitational field. This introduces an extra factor of R/r in the computation of the force, and a nontrivial angular dependence, where R is the radius of the earth and r the distance to the moon.

I had to think about that for a few minutes but I think your correct if you mean both the earth and the test mass feel the pull of the moon (actually a force between them). I do not know if the earth and moon are accelerating towards each other but they are in orbit about a common point and therefore are in free fall (weight less). I think that means they do not feel any acceleration.

Actually its their c of g's that are in orbit about the common point. But as the test mass is not at the c of g of the earth it will experience a different force depending on its distance from the moon's c of g.

The change in distance as the test mass at the equator as it rotates round the c of g of the earth is about 0.033, (earth diameter to moon distance) double that for the square effect 0.066. So only about 6.6% of the force between the test mass and the moon could be measured. In the UK where I am the variation will be about half that.

Correction: In the UK I think there is an other reduction of about half due to the misalignment of the moonsie its not in line with the vertical at the UK. That does depend on the inclination of the moons orbit relative to the spin axis of the earth.

[Edited on 10/24/2019 by wg48temp9]

wg48temp9 - 23-10-2019 at 16:17

Quote: Originally posted by Sulaiman

2²⁴ = 16,777,216
the variation of gravity at fixed point on the Earth's surface near the equator due to the Moon is approximately 0.11ppm,
the Sun contributes 0.052ppm
so you should be able to see the effect ...
____________________________________________________________
but
you will have to compensate your balance for changes in buoyancy due to changes in atmospheric density,
which is a function of temperature and local climatic pressure,
with a small contribution from humidity.
also,
your electronics will have to be very carefully designed and compensated for temperature coefficients.
Typically you will have a voltage reference, a frequency reference, and amplifier offsets - that are all temperature dependant.
There is also the temperature coefficient of the current sensing resistor,
and the actual magnetic force compensation coil/magnet system are subject to temperature coefficients ... F=B.I.L
the magnet geometry/size is temperature dependant, as is the length of the wire.
These effects combined are typically much greater than the variations in gravity.

Note that a larger current is drawn from the supply when a heavier weight is measured, hence a change in internal heat generation.

If you are using a strain gauge with I-beam then there is virtually no chance of observing the effects of the Moon and/or Sun gravity.

aside: the limititations to accuracy for a well made pendulum clock are the Sun/Moon/tide changes in gravity.

Yes there would be lots small effects to consider even thermoelectric effects. Input offset voltages can be almost eliminated with self zeroing op amps.
Weston reference voltage cells are defined to 7 significant figures though I suspect that some semiconductor references are as good as that these days.

I think 18 bits on the electronics is doable but as you suggest compensating for the buoyancy would be difficult but puting the sensor in a vacuum chamber would almost eliminate that effect.

If the data was accumulated over several months it may be possible to extract the tidal frequency from the noisy data using an FFT.

Unfortunately the effect pointed out by annaandherdad makes the change in weight much smaller by about 5 bits.

Of cause compared to detecting gravity waves it should be easy LOL

Twospoons - 23-10-2019 at 18:11

Quote: Originally posted by annaandherdad

Tidal forces explain why the bulge of ocean water points toward the moon on the side of the earth facing the moon, and points *away* from the moon on the opposite side.

I do wish they would stop teaching this overly simplistic tidal model in schools.
See
http://www.seafriends.org.nz/oceano/tides.htm

Where I live the simple model is clearly wrong : Tides on the east coast and west coast differ by 4 hours, yet the two coasts are only 50km apart.
This always bugged me in school, yet I didn't have an alternate explanation with which to challenge my teachers.

phlogiston - 24-10-2019 at 02:51

You should read the article "Detecting Extraterrestrial Gravity" by Shawn Carlson, in the great 'amateur scientist' series in Scientific American

It describes how to build a wonderful contraption that is able to detect directly the moons pull.

Carlson, S. (2000). Detecting Extraterrestrial Gravity. Scientific American, 282(1), 94–96. doi:10.1038/scientificamerican0100-94
(from scihub here: sci-hub.se/10.1038/scientificamerican0100-94)

[Edited on 24-10-2019 by phlogiston]

annaandherdad - 24-10-2019 at 06:27

The web article cited by Twospoons has a number of inaccuracies. For example, it says that the tidal bulge opposite the moon is hard to explain. On the contrary, the explanation was understood well by Newton. The author is also on the wrong track in trying to apply the water wave dispersion relation to the tidal bulge. Water waves travel by exchanging gravitational potential energy with kinetic energy, but the surface of the tidal bulges (on both sides of the earth) is a equipotential surface of the effective potential in the noninertial frame of the earth.

However, this article is correct in pointing out that there are serioius complications in the simple model of the two tidal bulges on opposite sides of the earth (with two more if you count the sun). The main ones are that the earth has continents and irregular ocean bottoms, and it is a flexible solid with a liquid layer, etc etc. The simple model of two tidal bulges (for the moon) would be correct if the earth were made entirely of liquid.

I can't access the Sci Am article cited by phlogiston from where I am now but will be very interested to read it when I get my hands on it.

By the way, on this general subject, it's worth while mentiioning that the tidal bulges do not align exactly with the earth-moon line, because of friction with the bottom of the ocean. Since the earth rotates on its axis more rapidly than the moon orbits the earth, the bulge actually leads the line joining the moon with the earth. Thus the gravitational attraction between the bulge of water and the moon has a component in the direction of the moon's motion, which slowly causes the moon's orbit to spiral away from the earth. In the process, rotational kinetic energy of the earth goes both into heat (friction of the tides), as well as the gravitational potential energy of the moon (which recedes from the earth), while angular momentum is conserved.

annaandherdad - 24-10-2019 at 09:08

Correction to my last post: As the moon recedes from the earth, its gravitational potential energy increases (as stated), but its kinetic energy decreases. Its total energy, kinetic plus potential, increases, since the gravitational potential energy is twice as big as the kinetic energy. The energy for this net increase comes from the rotational kinetic energy of the earth, which is decreasing. The effect is to move the moon further from the earth and to make the day longer. This process has been going on since the creation of the moon, which at an earlier stage in the history of the solar system was much closer to the earth.

wg48temp9 - 24-10-2019 at 10:18

Quote: Originally posted by phlogiston

You should read the article "Detecting Extraterrestrial Gravity" by Shawn Carlson, in the great 'amateur scientist' series in Scientific American

It describes how to build a wonderful contraption that is able to detect directly the moons pull.

Carlson, S. (2000). Detecting Extraterrestrial Gravity. Scientific American, 282(1), 94–96. doi:10.1038/scientificamerican0100-94
(from scihub here: sci-hub.se/10.1038/scientificamerican0100-94)

Great find phlogiston. I loved those articles in Scientific America. I did not know they were available at sci hub.

I am amazed that the moon's gravity can be detected so easily it also suggests how easily sub mg forces can be detected but then that is very similar to how modern analytical balances work. I have only had a quick look at the article. It claimed the temperature control circuit could hold the temperature to 1/100 C but it does not appear to have an integral term in its control loop so its performance could be significantly improved.

wg48temp9 - 24-10-2019 at 11:02

Quote: Originally posted by annaandherdad

Correction to my last post: As the moon recedes from the earth, its gravitational potential energy increases (as stated), but its kinetic energy decreases. Its total energy, kinetic plus potential, increases, since the gravitational potential energy is twice as big as the kinetic energy. The energy for this net increase comes from the rotational kinetic energy of the earth, which is decreasing. The effect is to move the moon further from the earth and to make the day longer. This process has been going on since the creation of the moon, which at an earlier stage in the history of the solar system was much closer to the earth.

Can you explain to me how the angular momentum of the earth moon system is conserved as the moon spiral out to a slower orbit, less angular momentum, and the earth rotation slows, less angular momentum too ???

[Edited on 10/24/2019 by wg48temp9]

annaandherdad - 24-10-2019 at 11:43

Yes, as the moon moves outward its angular velocity omega decreases but the radius r increases. The angular momentum is L= mvr where m is the mass and v the velocity, this is also L=m omega r^2. So you can see the two effects, omega goes down, r goes up, compete with each other.

But omega and r are related to each other by Kepler's laws, omega^2 = GM / r^3, where G=Newton's constant of gravitation and M = mass of earth (not to be confused with m=mass of moon). If you express L purely in terms of r, you find L is proportional to sqrt(r).

The total energy, by the way, is (1/2)mv^2 - GMm/r^2 which is proportional to -1/sqrt(r), after you use Kepler's law; it is negative but increases as r increases.

wg48temp9 - 24-10-2019 at 14:16

Quote: Originally posted by annaandherdad

Yes, as the moon moves outward its angular velocity omega decreases but the radius r increases. The angular momentum is L= mvr where m is the mass and v the velocity, this is also L=m omega r^2. So you can see the two effects, omega goes down, r goes up, compete with each other.

But omega and r are related to each other by Kepler's laws, omega^2 = GM / r^3, where G=Newton's constant of gravitation and M = mass of earth (not to be confused with m=mass of moon). If you express L purely in terms of r, you find L is proportional to sqrt(r).

The total energy, by the way, is (1/2)mv^2 - GMm/r^2 which is proportional to -1/sqrt(r), after you use Kepler's law; it is negative but increases as r increases.

Thanks for reminding me about the definition of angular momentum. Ok so perhaps angular momentum is conserved. Yes I agree the total energy must be constant.

However it still does not work for me as I thought the moon's angular velocity must speed up as its attracted to the tidal bulge of the earth that is is dragged ahead of the position directly below the moon by the higher angular rotation of the earth.

It probably the counter intuitive ways orbits change when an orbiting object is accelerated.

Sorry I should not have used "perhaps", angular momentum is conserved.

[Edited on 10/24/2019 by wg48temp9]

Putting it more succinctly, the attraction between the tidal bulge of the earth and the moon slows down the earth but instead of speeding up the moon it moves to a higher and slower orbit but an orbit with a higher angular momentum (the increase in height off setting the decease in orbital velocity) taken from the earth.

[Edited on 10/24/2019 by wg48temp9]

annaandherdad - 24-10-2019 at 20:39

The manner in which the moon moves into a higher orbit, due to a component of the force vector in the direction of its motion, is indeed a little counterintuitive. The force vector I am referring to is the one created by the bulge of ocean water on the earth, which is somewhat ahead of the line joining the earth and the moon.

Look at it this way. Suppose at one moment a small impulse were added to the moon, making it's velocity in its (originally) circular orbit slightly larger. This would mean the moon now had a velocity too large for the circular orbit it used to be in; so it will now be in an elliptial orbit, whose perigee is the radius of the original orbit, and whose apogee is slightly higher (reached on the opposite side of the earth). Now suppose on the opposite side of the earth, another small impulse is added to the moon, in the forward direction; this will smooth out the ellipse into a circle, in which now the radius is the same as the apogee of the previous elliptical orbit.

In this way, impulses in the direction of motion of the moon gradually move it to orbits of larger and larger radii.

It would be easier to explain with some pictures.

I forget the exact rate at which the moon is receding from the earth, but if I am not mistaken it is about 3cm/yr. This cannot be reliably extrapolated backwards in time over 4.6 billion years to find out what the moon was doing in earlier times, because the rate at which the moon recedes depends on the configuration of the continents (continential drift etc) and because when the moon was closer the tidal bulges etc were bigger. I mean that linear extrapolation won't work reliably over such long periods of time.

They can measure the current rate of recession of the moon using the cubical reflectors left on the moon's surface by the astronauts in the 1970's. It's laser ranging.

Yes, your final paragraph is exactly right.

[Edited on 25-10-2019 by annaandherdad]

wg48temp9 - 25-10-2019 at 01:08

In the improbable event that I am orbiting in a jet car I will have to remember if I put my foot on the gas I will slow down and if I but my foot on the breaks I will speed up and I thought reversing with a trailer was hard LOL

(the LOL indicates this a joke, I don't have enough money to buy a jet car LOL)

Sulaiman - 27-10-2019 at 00:44

My post above was based on memory,
here is one of the documents that I remember reading and just rediscovered,
that has real world data for the changes in g due to Moon, Sun, tides;
http://leapsecond.com/hsn2006/pendulum-tides-ch1.pdf

streety - 27-10-2019 at 04:06

Quote: Originally posted by phlogiston

The link referenced in the article is long-dead but wayback machine comes to the rescue again: https://web.archive.org/web/20010106055800/http://www.eden.c...

Quote: Originally posted by wg48temp9

In the improbable event that I am orbiting in a jet car I will have to remember if I put my foot on the gas I will slow down and if I but my foot on the breaks I will speed up and I thought reversing with a trailer was hard LOL

(the LOL indicates this a joke, I don't have enough money to buy a jet car LOL)

This video from Scott Manley suggests misunderstandings are common. It starts in the most relevant section but the whole video is worth watching:

https://youtu.be/i5XPFjqPLik?t=523

wg48temp9 - 27-10-2019 at 04:44

Quote: Originally posted by Sulaiman

My post above was based on memory,
here is one of the documents that I remember reading and just rediscovered,
that has real world data for the changes in g due to Moon, Sun, tides;
http://leapsecond.com/hsn2006/pendulum-tides-ch1.pdf

Thanks for the paper. When I searched I failed to find any reference to the effect though I may have missed one in the numerous bulging earth descriptions.

I would expect air buoyancy also effects pendulums.

Thinking about my 24 bit HX711 strain gauge bridge amplifier: by adding a stable fixed offset to the bridge or bridge inputs the resolution could be effectively increased significantly.

Ebay sells HX711 module and 4 strain gauges for only about three pounds.

See: https://www.ebay.co.uk/itm/1pc-HX711-Module-Amplifier-4-Pcs-...

Perhaps a semiconductor accelerometer could be used. Probably temperature controlled with serious filtering. In presence of stable random noise with an amplitude of several bits, averaging many measurements in effect produces more resolution.

[Edited on 10/27/2019 by wg48temp9]

A MPU-6050 GY-521 3 Axis Gyroscope + Accelerometer Module costing about three pounds looks like it has an accelerometer resolution of 16bits for +/- 2g, so only 14bits for one g.

[Edited on 10/27/2019 by wg48temp9]

Sulaiman - 27-10-2019 at 05:51

Quote: Originally posted by wg48temp9

Quote: Originally posted by Sulaiman

Strain gauges have quite significant hysteresis, linearity, creep and repeatability errors.
Check the specifications of your load cell carefully before buying.

Chemical balances that use strain gauges commonly go down to no less than 1mg resolution,
usually with several mg error.
This is why in my first post in this thread I wrote :
"If you are using a strain gauge with I-beam then there is virtually no chance of observing the effects of the Moon and/or Sun gravity."

0.1mg or less resolution digital balances are always of the electromagnetic force compensation type.

(my 220g x 0.1mg balance is only accurate to +/- 1mg absolute,
with 0.5 mg repeatability, and it is very temperature sensitive.
more expensive versions may have 0.1mg repeatability and 0.2mg linearity,
but check the price of suitable calibration weights ! )

Note that the HX711 datasheet indicates +/- 5ppm/^oC (+/- 6 bits per ^oC) offset

And there is no information on long term drift

I doubt that this IC is capable of detecting changes in g due to the moon.
(possibly excellent temperature and supply voltage control would give usable data if used with an excellent load cell)

wg48temp9 - 27-10-2019 at 08:59

Quote: Originally posted by Sulaiman

And there is no information on long term drift

I doubt that this IC is capable of detecting changes in g due to the moon.
(possibly excellent temperature and supply voltage control would give usable data if used with an excellent load cell)

Ignoring environmental vibration the load on the strain gauge/s is almost constant so hysteresis, linearity and repeatability are almost irrelevant.
Creep and drift will probably be much larger than the signal but it may be slower than the moon signal. Even changes in humidity may effect the strain gauges via the glue and insulation.

Environmental vibrations are likely to be much larger than the signal and in presence nonlinearities may not average to zero so nonlinearities will have an effect.

As you already suggested with the coil system changing current cause changes in temperature, the same must be true of strain gauges and also the silicon chip no matter how perfect overall temperature control is.

Taking the fft of the raw signal even with all its drift and noise the lunar signal may be detectable.

I guess I will have to try it to know.

Thinking about I live near a main road and near other houses so human and traffic noise will be high with frequency component very near to the moon signal frequency.

[Edited on 10/27/2019 by wg48temp9]

Experiment angular momentum of light form 1936

wg48temp9 - 26-11-2019 at 02:43

I found an interesting paper that may be of interest to others too.
Below is a snip of the apparatus used to measure the angular momentum of light directly. Its fascinating what can be achieved even back in 1936.
I wonder how the torque compares to the torque on a compass needle from the earth's magnetic field.

angmomlig.JPG - 42kB

Ref: