Gravity (acceleration) on earth is roughly 9.8m/s/s. What is the
maximum CHANGE in gravity or acceleration due to the moon being
directly overhead vs the other side of the world, as the moon's own
mass will reduce the earth gravity by pulling upwards?

Can this cyclical change in earth gravity (due to the orbiting moon)
be actually detected (from noise) using commonly found
accelerometers...

...such as the ADXL311 seen at
http://www.analog.com/UploadedFiles/Datasheets/473532491ADXL311_0.pdf

A good gravity reference...
http://www.cepo.interacesso.pt/Artigos/Astrol/GCalcE.htm

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Clarification of Question by chrispeaks-ga on 22 Apr 2003 20:59 PDT

The resolution of the best Analog accelerometer sensor is around 2mg
or less, which I assume is 0.002m/s/s and the maximum acceleration
effect of the moon is 0.00000224N/kg (I think).

Not sure about the weight of the accelerometer sensor mass, or if this
is an issue.

It looks like you can't use these sensors to very roughly detect the
moon's orbit - though I am unsure about my units or physics here!

Hi Crispeaks

Answers is "yes". 
 Tidal effects are measurable and amount some
30 microgals, that is 30e-3 gals.

Here are the details:


First, let's define our units:

 'Gravity measures the Earth's gravitational field = 980 x 103
milligals = 980 x 106 microgals

Gravity anomalies are local changes in the Earth's gravitational field
 Objective: observe changes in gravity of 5 microgals.


Corrections to be applied:

Drift and Tidal = 20 - 50 microgals
Latitude Adjustment (40 degrees) = less than 1.5 microgals
Elevation Adjustment (free air correction)= 310 microgal/m = 3/1
microgal/cm
Accuracy of elevation measurement required: plus or minus 0.5 cm
Bouguer Correction = 25 micro gals/m = 0.25 microgals/cm "
http://www.epsci.ameslab.gov/etd/technologies/projects/esc/technologies/microgravity.html


 Here,  980 x 103 milligals = 980 x 106 microgals means
 980 E3 milligals = 980 x E6 microgals using the
Scientific notation
(1E3=1000, 1e-3 = 1/1000 etc) which is recommended.
http://www.galactec.com/timothy/index.php3?code=tj5e


Absolute measurement of gravity field and magnetic field
as function of time and location is used in mineral exploration.

Data are analysed statistically, basically by analysis of variance,
tidal effects are one of observable effects.

A quote:
' The following procedures are needed to convert raw data into
observed gravity values, and are
listed in order of application:

Calibration of the data,
Removal of tidal effects due to the passage of the sun and the moon
for each reading,
Averaging of repetitive values at each base and station occupation,
Removal of drift,
Calculation of gravity differences,
(Optional in some instances) Calculation of statistics to determine
precision, and, usually,
Conversion of relative values into absolute values..."
..
"Tidal fluctuations (changing "pull" of the sun and moon on the
gravity meter mass) can be as much
as  300 microgals; on May 14th at Pine Mountain Club, as shown here,
the value was a total range of
95  microgals, a very "quiet" day.
http://seis.natsci.csulb.edu/grannell/gravity1.html

Here is a
Primer on Gravity and Magnetics
http://www.gravmag.com/gmprimr.html

Tidal variations can also be detected by analysing the dynamics of
satelites:
http://pdsproto.jpl.nasa.gov/catalog/dataset/Results.CFM?resultsselbox=LP-L-RSS-5-GRAVITY-V1.0


Search Terms
tide gravity measurement
microgals, absolute gravitometry

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Request for Answer Clarification by chrispeaks-ga on 30 Apr 2003 19:22 PDT

Your answer is well researched and presented, however I wish to
clarify it. The question was can these forces be detected by
accelerometers, such as the ADXL311 (or the many similar sensors)
which list a range of 2G or 5G etc, and a resolution of '2mg'. Is this
2miligravs / 2microgals? Or 2milligrams of mass or what? The only
existing devices I found to detect these forces use springs (IE the
zero-length-spring) or mass suspended in magnetic force fields - both
combersom technology but delivers very precise results.

You stated your assessement of the problem objective to be: observe
changes in gravity of 5 microgals. Can I assume that the ADXL311 is
capable of detecting a change of 5 microgals - is this 5mg? In which
case the 2mg would be suitable. I am concerned that the UNITS involved
are incorrect somehow, and that mg <> Microgals.

Regardless, your material was most helpful and appreciated,

Thanks,

Chris

---

Clarification of Answer by hedgie-ga on 01 May 2003 04:13 PDT

Your additional information is helpful.
   The more you can tell us about your project,
  the goals and means, the better we can fit the
 search and answers to your situation.

 The sensor you mention is a low end device :

 the ADXL311 sells for $2.50 each
in quantities of 10,000.
http://makeashorterlink.com/?L11D24764

and your signal is buried in noise. For research you should
use statistical analysis of data and high quality sensors, 
if  the project is more than just  for fun or a hobby.

   Using the symbol g for gals is not the SI standard.
It  is confusing and  conflicts with use of g for grams.
But,  as you know,  some folks   follow
 "industry practice"  no matter what.

 In gal  units ( 1 g is 9.81 m/ s*s ) you have
for this sensor
  .. typical noise floor  300 micro g per root
 Hertz at 3V operation which signals
below 5 milli g to be resolved   ...

 Your signal  is 30 milli gals - so in principle you can
use it for this purpose.

  However, this sensor ADXL311 is  intended
for 'consumer products', like Virtual Reality goggles
or mass-produced  intrusion detection devices ,
rather than for gravitometry research.

 Therefore, you may want to look around.

 These two sites have  info on this and may
offer free expert advice:
   http://www.noiseboard.com/
  www.cis.ohio-state.edu/~duttap/nest/docs/LITeS-SENSORS-1.pdf       

To see what is available on the market, enter keyword
 'accelerometer' into this guide
      http://www.directindustry.com/

 High precision accelerometers are being developed,
 often with federal funding, as they have important,
 sometimes 'undisclosed' applications; The work with
sensitive charge amplifiers requires sophisticated skills
and calibrations. Lacking these,
 one can easy end up with spurious results.

So -  my recomendation is to select your sensors carefully,
do some reading (below), calibrate, check and doublecheck
and verify your data.


Here are references to some of the high-precison metrology studies.
http://makeashorterlink.com/?T11E21764

Good luck with your research .

hedgie

I think this question rating now takes you to a total of 50, and you
clearly deserve the stars bestowed upon you. Well done, well
researched and well presented. Attention to details such as even
shortening urls etc. Outstanding and gives me more confidence to
proceed with my project. As for Mr_Fluffy's comment, his input & URL
was valued and interesting, however the 'cheap' accelerometer is
exactly what I need for my special project!

Your second reference (CEPO) seems to be doing reasonably well on this
problem (given various unstated assumptions and limited precision
constants) right up until they look at the difference between the
acceleration due to moon overhead and the acceleration due to moon
opposite:
"gm(closest) = 3.48E-05 N/Kg
 gm(furthest) = 3.25E-05 N/Kg
 So the Moon’s gravitational pull varies by 2.24E –6 N/Kg ... "

This is wrong, because these aren't scalars, but vectors going in
opposite directions.  Thus, the difference between them is 3.48E-5 +
3.25E-5, or 6.73E-5 N/kg.  That is, the 3.48 pulls upwards on the
observer on the Moom side, and the 3.25 pulls downwards (towards feet)
on the observer opposite the Moon side.

This value is 30 times the erroneous value (2.24E-6 N/kg) shown in the
web site.

So, a 60kg person has a difference in pull of 6.73E-5 * 60 N = 4.04
E-3 N.
This would be equivalent to a weight on the person from a mass of m =
4.04 E-3 N / 9.80 N/kg = 0.412 g, not 0.01 g.

Continuing on to the centrifugal force section, the apparent
"centripital force" on an Earth-surface observer would be "upwards"
whether near or far from the Moon, so, rather than adding the two
numbers, these would be differenced (the opposite error to that
above).  I get slightly different accelerations than the web site
(1.17 and 7.91 E-5 N/kg), with a difference of 6.74 E-5 N/kg.

This last number is virtually identical to the acceleration difference
above.  Since the centrifugal forces are more "up" for opposite the
Moon, and the gravitational forces are more "up" for the Moon
overhead, they completely cancel each other out.

Therefore, I don't think it matters what accelerometer you use.

You are asking about detecting tidal forces.  The change in the value
of gravity you'll measure caused by the moon when it is directly
overhead is actually exactly the same as when it is directly opposite
(below your feet).  That's because when the moon is overhead it tries
to pull you off the surface of the earth, but the centrifugal force
trying to fling you off the earth because the earth is spinning around
the center of mass of the earth-moon system is at a minimum, while on
the other side of the earth, the moon's gravity is pulling you into
the surface of the earth, but the centrifugal force is maximum, and
the two effects cancel.  If you want to see a difference, you need to
measure when the moon is either directly overhead, or directly
opposite, at which times gravity will be least, and then when the moon
is just at the horizon (90 degrees from the other position), at which
time gravity will be greatest.

The decrease in gravity in the first case relative to what it would be
without the moon is 2GMR/D^3, and the increase in the second case is
GMR/D^3, where G is the universal gravitational constant, M is the
mass of the moon, R is the radius of the earth, and D is the distance
between the earth and the moon.  The total difference is 3GMR/D^3,
which is about 1.7 E -6 N/kg.  So the change is about 1 part in 6
million, and may be difficult to measure.  Even if you can measure it,
you will need to think about the position of the sun as well, since it
produces tidal forces about half as strong as those of the moon.

One thing you definitely can measure is the change in gravity with
latitude.  Because of the centrifugal force due to the earth's
rotation, gravity is stonger by about one part in 300 at the poles
than it is at the equator.

Litton makes a great inertial accelerometer that can do what you want.
The ADXL 311 is a tilt sensor for Christ's sake!!!! i.e. a crude
accelerometer. It will not measure micro g's (i.e. 1e-6 * 9.81 m/s/s.
Sincerely, a former accelerometer designer.

Motorola declined to develop an ultra-sensitive accelerometer I
invented due to the marketing department's assessment of a
non-existent usage. Furthermore, this patent is up for grabs due to
lapse of time. Nevertheless, the simple operation stems from the
infusion of acceleration-induced charge directly into the gate of a
MOSFET transistor. Identified by the patent number, 4,378,510,
Bennett, March 29, 1983, Here is a direct link copy/pasted from the
U.S. Patent Database:

http://patft.uspto.gov/netacgi/nph-Parser?Sect1=PTO2&Sect2=HITOFF&u=/netahtml/search-adv.htm&r=42&f=G&l=50&d=PTXT&s1=bennett.INZZ.&s2=motorola.ASNM.&co1=AND&p=1&OS=IN/bennett+AND+AN/motorola&RS=IN/bennett+AND+AN/motorola

Chris,

 Thanks for the rating and words of appreciation.
 Fedback matter to us and helps us to improve.

  We also got some valuable and relevant comments and I just
  want to add to them an elegant way of looking at this
  gravity +/-cetrifugal forces  --pulling this way and that way
  at different locations: 

 The search terms here is
 DECOMPOSITION OF FORCES 
and an example of interesting site which come up is:

http://kr.cs.ait.ac.th/~radok/physics/a6.htm

 In a nutshell: the complex motion of (e.g.) moon can be decomposed 
 into
 1) center of gravity of the moon-earth system (a mass point) orbiting the sun
 2) rigid  rotation of that system around that center of mass
 3) and (this one is of interest here) tidal forces - due to the gradient
    of the gravitational field around that center of gravity -
     which indeed are almost symmetrical.
     They depend on the 'direction' but not the 'sense' of the  direction
    (defined as an un-orinted ray) - leading to the factor of 2 
    (two senses of each direction)
    which appears repeatedly in physics. Obvious manifestation of this
    factor and symmetry is the fact  that we have TWO high tides a day as
    (in a less abstract way) is explained in the comments and  here:

http://www.physlink.com/Education/AskExperts/ae338.cfm



hedgie