1. Mick West

    Mick West Administrator Staff Member


    Source: https://www.youtube.com/watch?v=XkG1iFanoNo
    Cody Don Reeder, a science educator, did a nice experiment to show that gravity (and hence weight) changes based on your location. The two main things that affect gravity are latitude and elevation, and you should be able to detect these with a sensitive enough scale and a large enough change in altitude and/or location.

    Cody uses a 295g cube of non-magnetic tungsten, and a jewelry scale that weighs up to 500g with an accuracy of 0.030g (30mg). He weighs this in Logan UT, and Manchester NH.

    wieght varies by location Metabunk.
    The weight varies as expected. At higher altitudes you are further from the center of the Earth, so the gravity and weight decrease. At higher latitudes (more North) the effects of the spin of earth are less, and so apparent gravity is greater (plus you are closer to the center of the earth, due to the equatorial bulge).

    Going from Logan, UT to Manchester,NH is both a lower elevation and a higher latitude, so weight is expected to increase, which it does by about 0.22g, or 0.074%. With just an increase in altitude from 4,500 to 7,600 ft the weight decreases by about 0.06g, or 0.020%.

    We can calculate the expected weight, assuming the Earth is of even density, with this calculator, which also gives the equations used:
    https://www.sensorsone.com/local-gravity-calculator/

    (Note, the FAC value is approximating a section of the inverse square curve with a straight line. This is accurate for altitudes within the atmosphere)

    The theoretical local g values are given in the diagram above. The Manchester value increases by .052% (vs. .074% observed). The Higher altitude Logan value varies by about (.028%). The differences here are due to this being a theoretical local g.

    Actual local g varies based on the composition of the Earth, this can be local variations based on the type of rock nearby (see: Bouguer anomaly), but also deeper variations inside the Earth. However the direction and magnitude of the changes demonstrate both that gravity decreases as you get higher, and increases as you move away from the equator.
     
    Last edited: Jun 8, 2017
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  2. Mick West

    Mick West Administrator Staff Member

    The fact that weight decreases as you get higher is a bit a problem for the rather odd "density" theory of gravity. Air gets measurably less dense as you go higher. So if relative density is what makes things fall down then they would weigh more as they moved into less dense air. However they weigh less. They weight less because "density" has nothing to do with.
     
  3. Mick West

    Mick West Administrator Staff Member

    The scale used by Cody is a generic Chinese scale, costing about $10: http://amzn.to/2qRMNvh

    The accuracy listed is 0.01g. Reviews on amazon are generally positive, but say the scale might not be calibrated perfectly. However for the purposes of this experiment the only important figures are the relative weights. So long as the measurements are carried out with the same scale then it does not matter if there's a small calibration error.

    And if you want to repeat the experiment exactly, here's the tungsten cube for $75: http://amzn.to/2sLMliL

    Tungsten is great because:
    • It's not magnetic
    • It will not absorb water or change weight by chemical reactions
    • It's easy to carry around, as it's so dense.
     
    Last edited: Jun 5, 2017
  4. Z.W. Wolf

    Z.W. Wolf Active Member

    The first to measure a difference in the strength of the earth's gravitational field due to location was Jean Richer during an expedition to French Guiana in 1671-1673. He found that his seconds pendulum didn't keep proper time (as measured against the stars). It lost 2m 28s per 24 hours. He had to shorten the pendulum length by 2.8 mm in Guiana as opposed to its length in Paris.

    From this data and from his theory of gravity (the force of gravity decreases with the inverse square of the distance between objects) and the assumption of a rotating earth, Newton created a theory that the earth is an oblate spheroid, and that this shape was produced by the centrifugal force of the rotation.

    This led to the dispute between Newton and Cassini as to whether the earth is an oblate or prolate spheroid, which was only settled by a measurement of the meridian in Lapland and comparing it to the measurement of the meridian in France, showing that Newton was correct. A measurement of the meridian in Peru was further confirmation.
     
    Last edited: Jun 6, 2017
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  5. Mick West

    Mick West Administrator Staff Member

    A related type of thing was used by George Airy, with a pit and a pendulum:
    https://todayinsci.com/A/Airy_George/AiryGeorge-Pendulum.htm
    [​IMG]

    Measuring the difference in period of a pendulum on the surface and down the pit. Essentially calculating the gravitational constant.
     
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  6. Z.W. Wolf

    Z.W. Wolf Active Member

    One also has to ask what the gravimeters used by mining engineers and so on are actually measuring.

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


    Someone in the comments section of that YT video joked about buying gold somewhere where it's lighter and selling it somewhere where it's heavier. That's an old, old whimsical observation. But for that very reason, and others, precious metals are weighed on balance scales against reference masses.
     
    Last edited: Jun 6, 2017
  7. Mick West

    Mick West Administrator Staff Member

    Or digital scales calibrated with a reference mass. Like the one I just ordered for the scale I also just ordered.
    http://amzn.to/2sBdhCA
     
  8. NobleOne

    NobleOne New Member

    I wonder if weight changes at the same location over night since the object gets further away from the Sun.
     
  9. Mick West

    Mick West Administrator Staff Member

    Not easily measurably, and certainly not with this scale. Also the moon would have far more effect - still not measurable except with a super sensitive scale though. It's been done:
    [​IMG]
    Source: https://physics.stackexchange.com/q...easurements-relate-to-gravitational-wave?rq=1
    Source Article: http://iopscience.iop.org/article/10.1088/0026-1394/18/3/006/meta

    1µm/s^2 (micrometers per second squared) is about 1 ten millionth of g, which is around 9.81 m/s^, or 9,810,000 µm/s^2
     
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  10. Astro

    Astro Member

    Such a setup should be capable of detecting the Eötvös effect in a commercial airliner (assuming you could average out vibrations during level flight), should it not? Was that accounted for in the theoretic difference, or might that help explain the deviation from the theoretical difference?
     
    Last edited by a moderator: Jun 8, 2017
  11. Mick West

    Mick West Administrator Staff Member

    See the video from 12:55 onwards. He measured on the plane, and average it out later. I left it out as it's a bit complex, but it's really another good piece of evidence for both the curvature and rotation of the Earth.
     
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  12. 0x90

    0x90 New Member

    Provided I haven't messed up some mundane detail, when earth is at perihelion, the gravitational force between the sun and a kilogram mass fluctuates by 1.063e-9 newtons (~108 nanograms in earth gravity) between noon and midnight, and 9.614e-10 newtons (~98 nanograms in earth gravity) at aphelion. Between perihelion and aphelion, it fluctuates a far greater 3.967e-4 newtons (~40 milligrams in earth gravity).
     
  13. Mick West

    Mick West Administrator Staff Member

    I've got one of those scales. Next time I drive up into the Sierra's I could check the altitude variation out.
    20170608-160742-efff4.

    That's a polished stone ball which should be a nice consistent weight. It seems like you can get 0.01 to 0.04g variations just from handling it with bare hands. Need to keep it clean and use gloves.

    Ideally I'd have a 495g weight - the more mass, the more noticeable the difference in weight.
     
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  14. Astro

    Astro Member

    I'm thinking of trying that experiment on my next flight to try to capture the Eötvös effect. I see an identical looking scale for sale on ebay. Does it have an internal battery or do I have to supply external power for it?
     
  15. Mick West

    Mick West Administrator Staff Member

    It uses two AAA batteries. I bough my scale from Amazon ($10.10), the box is branded TBBSC TB011, but the scale itself is just the chinese generic. It does work well though. It powers off after a minute if the weight does not change, but I think the constant variation on the plane will keep it awake.

    I also got a 500g calibration weight ($11.58). That will be good for the ground based test, as I can calibrate the scale to 500g near sea level, and then that will give the maximum decrease in weight as I go higher. However you will probably want to use a lighter weight on the plane, as the scale stops with an error if you go over 501.00g momentarily.

    You will also want some method (like a bullseye level, real or virtual) of verifying the scale is level (on average), as (experimentally) around 1° tilt gives an error of 1g for my 500g weight.

    You want to be in straight and level flight, you should be able to verify this with a few minutes of video with the bubble not shifting significantly. Basically what Cody does.
     
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  16. NoParty

    NoParty Senior Member

    Hmmm...a polished stone that shouldn't be touched by human hands.

    Does this have something to do with debunking super powers?

    Screen Shot 2017-06-09 at 11.40.57 AM.
     
  17. Mick West

    Mick West Administrator Staff Member

    I wonder if you could get a measurable Eötvös effect in a car? There's some flat straight roads round here. What if I went 50 mph?

    https://en.wikipedia.org/wiki/Eötvös_effect
    20170609-113417-va25a.

    Using SI units, not sure what Ω would be in? radians per second? 7.2921150*10^-5 according to Wikipedia

    Assuming only east-west motion, so zero v, u = 50mph is 22.35 m/s, and I'm around 38.5°N latitude

    R = Radius of Earth = 6371000m

    ar = 2*(7.2921150*10^-5)*22.35*cos(38.5 degrees) +22.35^2/6371000 = 0.00262937596 m/s2 ??

    given g=9.81, so for a 400g weight, that would be .00262937596/9.81*400 = +/- 0.1 gram variation, easily detectable.

    Unless Ω is in rotations per second, which would make it 2pi smaller or about 0.016 grams, much harder.

    Looking at this paper:
    http://empslocal.ex.ac.uk/people/staff/gv219/classics.d/persson_on_coriolis05.pdf

    Which implies radians/second as
    2*(7.2921150*10^-5)*sin(43 degrees) ~= 10^-4

    So seems like a plausible experiment.
     
    Last edited: Jun 9, 2017
  18. Mick West

    Mick West Administrator Staff Member

    Well, just did a little exploratory experiment at 30mph, and I think this one would need a bit more sophisticated equipment. I used a weight combination of 400g (screws in a plastic jar, on a high friction Bench Dog). However the weight varies by around +/- 8g just from variations in the road surface. That makes a 0.1g variation difficult to detect without continuous data logging.
    20170609-131830-b9pxq.
    20170609-131005-a2mlu.
     
  19. angelote

    angelote New Member

    Hi,

    I would like to warn you will have different readings on the scale because the air density is not the same at sea level as inside the aircraft cabin.

    At sea level the air density is r0 = 1,2 kg/m³ (20 ºC; 101325 Pa; 50 %HR).

    Inside the plane the air pressure could be around 80 kPa, thus the air density would be r1 = 0,95 kg/m³.

    If “your weight” (495 g) has a density around 2500 kg/m³, this mean the volume is around 200 cm³. The buoyancy (in mass units, actually it is a force) at sea level would be b0 = r0·V = 240 mg. The buoyancy inside the plane: b1 = r1·V = 190 mg. Thus, the difference between the two weighing readings, due to air density, would be Dm = b1 – b0 = 50 mg. That means the weighing reading inside the plane will be 0,05 g higher. This is just 5 intervals (the difference in mass between two consecutive indications) of your indicating device. This is too little, you can neglected :)

    (For estimation of air density, see “Guidelines on the Calibration of Non-Automatic Weighing Instruments”, EURAMET Calibration Guide No. 18, Appendix A):
    https://www.euramet.org/publications-media-centre/cgs-and-tgs/
     
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  20. Mick West

    Mick West Administrator Staff Member

    Steel has a density of 8000 kg/m3, so the error there will be around a third that. Tungsten has a density of 19,250 kg/m3, so the buoyancy error would be around 1/8 your calculation, or <0.01g

    Assuming around 7,500 pressure altitude for the plane, the corresponding free air correction would be 0.38g

    It's something to consider, but at worst it's just going to decrease the gravity/height effect slightly, it's going in the opposite direction, so you can't mistakenly ascribe gravity effects to density here.
     
  21. Mick West

    Mick West Administrator Staff Member

    A related experiment:
    http://gnomeexperiment.com/

    The site has a map showing how much an approximately 308g gnome weighs at different points on the globe.