Does weight on earth vary as a function of earth-moon distance?

[Griffon]

Not at all .... or at least that's what I argued during a a discussion in the pub yesterday. Am I right?

 

The lunar distance varies by about +/- 6.5% between apogee and perigee. It was speculated during the pub discussion that weight should vary because gravitational force is inversely proportional to distance squared. In other words, if the moon were directly overhead one should be lighter, and even more so if the moon were closer.

 

I argued to the contrary that the earth and moon are in orbit about one another and in a state of 'free fall' relative to one another. With respect to the other body we are weightless regardless how near or far it is. Is this right? Or are there higher order effects?

 

PS I assume tidal forces, which depend on gravitational differential, will vary with lunar distance, but this (I think) is a different matter.

[swansont]

The effective value of g decreases when the moon is overhead as opposed to being on the opposite side of the earth, owing to the different distances. Alfred Loomis measured this in ~1930 by comparing the oscillation of pendulum clocks, whose period or frequency depends on the square root of g, with quartz crystal clocks, which have no such dependence.

[D H]

Am I right?

Yes and no. It depends entirely on what you mean by weight.

 

Legally, colloquially, and historically, weight is a synonym for mass. Ask someone how much they weigh and they'll give an answer in kilograms, stones, or pounds. Those are all units of mass, not force. (The pound is a unit of mass. The pound-force is a unit of force.) By this definition, yes, you're right, but you'll have to admit that this is a pretty lame justification for saying that you are correct.

 

To scientists and engineers, weight is a force. One way of defining weight is as the gravitational force exerted by the Earth on the object in question. This definition of weight tautologically removes the influence of the Moon. So once again you are right by this definition. However, this is once again a rather lame justification for your point.

 

So, let's extend that definition to the gravitational force exerted by everything in the universe. Per this definition, your weight does change based on the relative location of the Moon. The Moon makes you lightest when it is directly overhead, heaviest when it is directly underfoot. Not by much, a factor of ±3.4×10^-6.

 

Finally, let's use a definition of weight you can measure. (You can't measure gravitational force. You can only calculate it. Einstein's equivalence principle.) What you measure when you stand on a bathroom scale isn't the force of gravity. It's the sum of all forces *but* gravity. With this definition, the Moon makes you lightest when it is directly overhead or underfoot, heaviest when it is on the horizon. Now the variation is even less; tidal forces vary as 1/R³ rather than 1/R².

 

Either way, the right answer is "not by much". You said "not at all".

[Griffon]

The effective value of g decreases when the moon is overhead as opposed to being on the opposite side of the earth, owing to the different distances. Alfred Loomis measured this in ~1930 by comparing the oscillation of pendulum clocks, whose period or frequency depends on the square root of g, with quartz crystal clocks, which have no such dependence.

OK. That's interesting. I'd never come across Loomis.

 

So the moon does have an influence on g, which in turn affects weight.

 

I'm still intrigued though because ... Suppose you were in a spacecraft in an eccentric orbit about a planet. You would be weightless (not massless) irrespective of how near or far you were from the planet. What is the difference between this and being on the earth? I can't see how the moon being nearer or further away makes any difference to our weight, unless of course it's a higher order effect.

Edited March 16, 2013 by Griffon

[Enthalpy]

This has very practical implications for geologists, who use extremely sensitive gravimeters to guess what lies in the ground (in fact, they prefer to use differential apparatus to remove these perturbations). Their instruments show with great precision the variation of g (is that g?) over one metre height, for instance.

http://en.wikipedia.org/wiki/Gravimeter

 

Tidal force is the first perturbation for gravimeters and must be accounted for very precisely. Its influence exceeds by far the observed effects of deposits. Since Moon's distance varies a lot (+-6.5%) so does its effect (+-20%) and this change demands an accurate evaluation.

[Griffon]

The Moon makes you lightest when it is directly overhead, heaviest when it is directly underfoot. Not by much, a factor of ±3.4×10^-6.

How does one calculate that factor?

[D H]

How does one calculate that factor?

http://www.wolframalpha.com/input/?i=(gravitational+constant+*+moon+mass+/+(mean+earth-moon+distance)^2)+/++(standard+gravity)

 

That's not really the answer you want, however, particularly since you referred to astronauts being "weightless" in your previous post. You are thinking of weight in terms of "scale weight" (aka "apparent weight", aka "proper weight") as opposed to "gravitational weight" (or just "weight" to some).

 

You can't measure gravitational weight. You can measure scale weight. It's what your bathroom scale measures.

 

So let's go to that previous post of yours:

 

I'm still intrigued though because ... Suppose you were in a spacecraft in an eccentric orbit about a planet. You would be weightless (not massless) irrespective of how near or far you were from the planet. What is the difference between this and being on the earth? I can't see how the moon being nearer or further away makes any difference to our weight, unless of course it's a higher order effect.

The difference is that the spacecraft is in free fall. An object at rest on the surface of the Earth obviously isn't in free fall.

 

An accelerometer (a scale is essentially a one dimensional accelerometer) measures what is called "proper acceleration", acceleration with respect to a free-falling object. General relativity offers a very solid explanation why proper acceleration is so important, and why it is the best way to look at "weight". Scale weight is the product of proper acceleration and intrinsic mass.

[Griffon]

Yes, I am thinking of scale weight I believe. The weight you can measure, or feel if you like.

 

Aren't the moon and earth in free fall with respect to one another?

 

I can see that the force due to the moon on any point on the earth is going to be a function of the distance from that point to the moon's centre of mass. That distance varies of course, as will the force. But why isn't that force offset by an equal and opposite force due to orbital acceleration - at least to first approximation? After all, the moon is not in orbit of the earth. The earth and moon are in orbit of one another - specifically their barycentre.

[D H]

But why isn't that force offset by an equal and opposite force due to orbital acceleration - at least to first approximation?

 

That's a zeroth order approximation, not a first order. To first order, there's an additional term that is proportional to the ratio of the Earth's radius to the distance between the Earth and the Moon. This is the tidal force. It's very small, on the order of 10^-7 times the gravitational force exerted by the Earth, but it is very important. This tiny tidal force is what causes the tides.

 

In addition to this direct way in which the Moon affects your weight (scale weight), here's also an indirect path. That tidal force does more than cause the ocean tides. It also causes the Earth itself to change shape. These Earth tides mean that you are pushed away from / pulled toward the center of the Earth as the Earth changes shape, and that too changes your scale weight by a tiny, tiny amount that is observable in the high precision gravimeters that Enthalpy referenced in post #5.

[Griffon]

OK. Thanks.

 

Just to clarify - are you saying that the primary changes in scale weight caused by the moon are due to tidal forces?

Edited March 18, 2013 by Griffon

[Cryptonix]

Yes

[Griffon]

That results in g going through two maxima and minima per day as the earth rotates - much as there are two tides per day. So g is a minimum when the moon is both above our heads and beneath our feet if I understand this correctly. Not much though - ±1.5 × 10^−6 m/s^2 or so. http://www.trin.cam.ac.uk/clock/theory/rcl_report.pdf

Edited March 21, 2013 by Griffon

[fertilizerspike]

The effective value of g decreases when the moon is overhead as opposed to being on the opposite side of the earth, owing to the different distances. Alfred Loomis measured this in ~1930 by comparing the oscillation of pendulum clocks, whose period or frequency depends on the square root of g, with quartz crystal clocks, which have no such dependence.

 

Loomis proved only his obsession with time and that it is impossible to keep "perfect time", it is only possible to compare clocks to each other.

[swansont]

Loomis proved only his obsession with time and that it is impossible to keep "perfect time", it is only possible to compare clocks to each other.

 

He showed a timekeeping variation that was consistent with the magnitude and period expected from the variation from the moon. And all timekeeping is comparing clocks to each other, so that's no great revelation.

[fertilizerspike]

He showed a timekeeping variation that was consistent with the magnitude and period expected from the variation from the moon. And all timekeeping is comparing clocks to each other, so that's no great revelation.

 

You apparently believe he showed "variation that was consistent with the magnitude and period expected from the variation from the moon". Like all users of the forum I'm sure you're perfectly willing to provide the data he acquired. Please note that anecdotal stories about the evidence are unconvincing. Please provide the data, and then demonstrate that his data agrees with predictions based on whatever model you feel was used to make those predictions.

[John Cuthber]

If you want to look it up, here's a good start

http://www.nature.com/nature/journal/v130/n3273/abs/130124b0.html

[swansont]

The papers I read were Monthly Notices of the Royal Astronomical Society, Vol. 91, published in 1931: “The Precise Measurement of Time” by Alfred L. Loomis (p. 569-575) and “Time, Analysis of records made on the Loomis chronograph by three Shortt clocks and a crystal oscillator” by Brown, E. W. & Brouwer, D. (p.575-591)

 

It's interesting that you "know" the results were somehow flawed, implying familiarity with the experiment, and yet you require a citation.

[fertilizerspike]

The papers I read were Monthly Notices of the Royal Astronomical Society, Vol. 91, published in 1931: “The Precise Measurement of Time” by Alfred L. Loomis (p. 569-575) and “Time, Analysis of records made on the Loomis chronograph by three Shortt clocks and a crystal oscillator” by Brown, E. W. & Brouwer, D. (p.575-591)

 

It's interesting that you "know" the results were somehow flawed, implying familiarity with the experiment, and yet you require a citation.

 

I didn't ask for a citation, I asked for the data, which you apparently can not provide. Loomis came to many absurd conclusions, one being that his three pendulum clocks, which were essentially rooted in bedrock, phase shifted 120 degrees, were modulating each others' periods by transmitting inertia through the Earth. He was a rich guy who had a tiny handful of good ideas but mainly bad ones.

[John Cuthber]

Since the data is behind a paywall, it would be a breach of copyright and, therefore, probably illegal to paste it here.

Pleas don't ask us to break the law.

Instead, perhaps you might like to tell us hat you think caused the reported variation in the clocks.