6
$\begingroup$

I've been reading about accelerometers. On earth, in a lecture demonstration, in a table-top experiment, they will generate a signal equal to the sum of the acceleration you'd like to measure plus the local acceleration due to gravity and all of the other accelerations of our laboratory frame.

Local gravity on the Earth's surface varies a few tenths of a percent; strongest at the North pole and weakest in equatorial mountains, due to variations in the distance from the center of the Earth and other mass-distribution effects.

There is also a variation in the centripetal acceleration due to the Earth's rotation from pole to equator, which points away from the Earth and further decreases local acceleration near the Equator. All together, it's about a 0.5% maximum difference.

There is a great inventory of these variations in this answer and I'd recommend a visit there if you are interested.

But my question is about daily variation in local acceleration due to the rotation of the Earth with respect to the Sun and Moon.

I was reading the document Precise Measurement of Mass, presented at the 60th Annual Conference of the Society of Allied Weight Engineers Arlington, Texas May 21-23, 2001 by Richard Boynton of Space-Electronics.

One statement on page 8 surprised me:

1.5 Tidal variations An object on the surface of the earth is attracted to every celestial body. Most of these masses are too far away to have any significance on weight, but the sun and the moon do have a significance. If you have a scale whose accuracy is 0.003 % or better, you will notice that the weight of an object varies as a function of the time of day. This effect is most pronounced during spring and fall when the sun and moon align. This produces the “neap tides” that often cause flooding of marinas at these critical dates. (emphasis added)

The figure of 0.003% is repeated twice more in other sections.

I have tried to reproduce this number, but it seems high by about two orders of magnitude.

The tidal effect I know how to calculate goes as follows. I like to imagine being an astronaut in the ISS, orbiting the Earth. At that low altitude Earth's gravity is almost as strong as it is on the surface, but an astronaut in the center of the ISS will not experience any acceleration with respect to the ISS since they are in the same orbit.

However if an astronaut moves to a point toward or away from the Earth, they will experience a tidal acceleration that points away from the center of the ISS, and which will be proportional in 1st order to the distance from the center. For a shift of $\pm \Delta r$ from the orbital distance $r_0$, the change in acceleration is:

$$\Delta a = \frac{-GM_E}{(r_0-\Delta r)^2} - \frac{-GM_E}{(r_0+\Delta r)^2}.$$

For $\pm$ 1 meter at the ISS's $r_0$ of say 6378+400 km, that's a $\Delta a$ of about 5E-06 m/s^2, or 5E-07 g.

Similarly, using the standard gravitational parameter $GM$ from the Sun and Moon of 1.327E+20 and 4.905E+12 and distances $r_0$ of 1.5E+11 and 3.8E+08 meters, and now using the radius of the Earth for $\Delta r$, I get $\Delta a$'s of 1E-06 and 2E-06 m/s^2, or 1E-07 and 2E-07 g.

When those add together, it's 3E-07 g, or 0.00003%, one hundred times smaller than the amount stated in the paper.

Question: Have I done my calculation correctly? Is the figure in the paper a factor of 100 too high?

note 1: 0.003% results if you make the change to percent (by multiplying by 100) twice.

note 2: FWIW, the paper points out that the company is selling scales with a precision of 0.002%.


enter image description here

$\endgroup$
  • 1
    $\begingroup$ they'd get the same error if they multiplied by g instead of divided by g to get the ratio, but yours looks right to me That paper needs a {{citation needed}}... $\endgroup$ – JCRM Sep 23 '17 at 7:06
  • $\begingroup$ @JCRM that's a good point, thanks! There may be other effects I haven't thought about as well, so the disagreement may not be arithmetic in nature. $\endgroup$ – uhoh Sep 23 '17 at 7:13
  • $\begingroup$ webcache.googleusercontent.com/… simplifies the equation for calculating the change in force to delta F = -4.GMmR/d^3 where m is the mass, R is the radius if the earth and d is the distance between the bodies; substituting F=ma, and your units d=ro, R=dr we get delta a = -4 GM dr/r0^3 whose values match yours. $\endgroup$ – JCRM Sep 23 '17 at 7:36
  • $\begingroup$ The german version of Wikipedia about tides or Gezeiten: "Die mittlere Erdbeschleunigung ist etwa zehnmillionenmal größer als die Gezeitenbeschleunigung". My translation: the Earth's gravity acceleration is about 10 million times bigger than the acceleration by the tides of the moon. $\endgroup$ – Uwe Sep 23 '17 at 9:46
4
$\begingroup$

The total gravitational pull of the Moon is $g_M = 3.3\cdot10^{-5} \frac{m}{s^2}$ or $0.0003\%$ of Earth's gravity. This is one order of magnitude smaller than the stated value.

At all times, most of this force is counteracted by the centrifugal force of the rotation of Earth around the barycenter of the Earth-Moon system, resulting in the small tidal force calculated by @uhoh.

In summary, even if we ignore the revolution around the barycenter of the system and make the Moon vanish instantaneously, the felt change in gravitational force would be $0.0003\%$ at most.

$\endgroup$
  • $\begingroup$ That's a good sanity check! I can't think of any other effects to consider, this seems to be pretty conclusive. Let's give it a few more days to see if anything else unexpected turns up, but I think this is pretty conclusive. Thanks! $\endgroup$ – uhoh Sep 23 '17 at 21:01
  • $\begingroup$ @MartinModrák No, you also have to take into account that Earth is in orbit around the Sun which just leaves tidal forces remaining. And these are about half the tidal forces of the Moon which I ignored because they are tiny compared to the whole pull of the Moon. $\endgroup$ – asdfex Sep 27 '17 at 11:10
4
$\begingroup$

From a book in french about tides: If a man with a mass of 102 kg is standing on earth, the constant gravitational force of Earth to the man is 1000 N, of the Sun 0.61 N or 610 mN and of the Moon 0.0034 N or 3.4 mN.
The maximal tidal variable force of the Moon is 0.000112 N or 112 µN and of the Sun 0.000052 N or 52 µN. The relation of 3.4 mN to 1000 N is 3.4 ppm (part per million) or 0.00034 %. But the tidal forces are much smaller: 112 ppb (part per billion) for the Moon and 52 ppb for the Sun.
The effect of the tides to the ocean is also very small far away from the coasts, about 0.7 m height variation. Compared to the Earth diameter of 6378 km it is only 110 ppb. Only the very small tidal forces would affect the scale. There are superconducting gravimeters able to measure such small variations. If the constant force of 3.4 mN were responsible for oceanic tides, the variations of water height would be incredible strong, about 31 times higher.

$\endgroup$
3
$\begingroup$

The reported difference of $0.003\%$ would be plausible, if the difference in measured weight (acceleration) was caused by the fact that the gravity of the sun acts in roughly the same direction as the Earth's gravity at midnight and roughly in the opposite direction at noon. But this is wrong.

Since the exact same accelaration affects the scale, the reading on the scale is unaffected by the alignment of the gravity of Earth and Sun (thx @asdfex for pointing it out).

$\endgroup$
  • 1
    $\begingroup$ That's what I thought first, too. But the Sun's gravitational pull is counteracted by the Earth being forced on its orbit, and accelerated. Only tidal forces remain, but they are minuscule. Maybe you are right, and this is the explanation for the value in the paper, but not actually correct. $\endgroup$ – asdfex Sep 27 '17 at 11:01
  • $\begingroup$ Maybe my intuition fails me badly, but I think that Sun's gravitational pull on the weighed object is not counteracted by the pull on Earth - how would that work? $\endgroup$ – Martin Modrák Sep 27 '17 at 11:16
  • $\begingroup$ Your scale gets accelerated by the same force as well, along with Earth. $\endgroup$ – asdfex Sep 27 '17 at 11:21
  • $\begingroup$ Oh I think I see the problem now - if it worked the way I expect, and Earth's weight was negligible, I would get forces appearing out of nowhere. Thanks for pointing it out, will edit my answer. $\endgroup$ – Martin Modrák Sep 27 '17 at 11:25
  • $\begingroup$ While an analytical, equal arm, or beam balance would not be.affected for the reason given; a spring, load cell, or torsion balance would be - and the weight of the pan and linkages would also be contributing to the variation (depending on how the scale were zeroed) $\endgroup$ – JCRM Sep 29 '17 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.