Is gravitational prospecting pseudoscience?

  • #1
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Main Question or Discussion Point

Is gravitational prospecting pseudoscience?

From the literature, it says that we can know what is beneath a site (say for prospecting for oil. metals, etc) from measuring the local "g" field/acceleration at various locations. Together with other information, we can know the mass distributions at the site below.

But according to gravity theory, the gravitational field and magnitude just above the surface of the earth would be always pointing directly towards the center of mass of the earth. We can assume that this CM relative to a location is a constant in general if we take into account purely gravitational forces between an object and the earth.

Small g is a constant at a location depending on mass m of an object, mass M of the earth and the distance r from the earth's CM. It is a constant in general.
 

Answers and Replies

  • #2
jtbell
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But according to gravity theory, the gravitational field and magnitude just above the surface of the earth would be always pointing directly towards the center of mass of the earth.
This assumes that the Earth's mass distribution is exactly spherically symmetric. This is true only as an approximation, when one wants to calculate g to only a few digits of precision.

The next level of approximation takes into account that the earth is not a sphere, but an oblate (flattened) ellipsoid, because of its rotation.

Then there are variations due to the irregular distribution of mass. You can see the results of measurements in the figure at the top of the following page:

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

This shows only variations over large regions. Some more discussion of these variations is at

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

and

https://en.wikipedia.org/wiki/Physical_geodesy
 
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  • #3
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This assumes that the Earth's mass distribution is exactly spherically symmetric. This is true only as an approximation, when one wants to calculate g to only a few digits of precision.
It is not so; spherical mass symmetry is not needed! It is a rigorous classical field theory principle and the application of the inverse square law of gravitation.

Starting from point mass particles, we assume gravitational forces obey the superposition principle. In chapter 1 of "Classical Mechanics, Hertbert Goldstein", you will find that the action of the earth's body on an object m above the earth is equivalent to the action of a mass M of the earth acting as a particle at the center of mass of the earth. This CM of the earth relative to any location - say New York city - is generally a constant (ignoring other non-gravitational factors). The "g" field and magnitude at any local postition is a constant and cannot tell anything about variation of mass distribution beneath a site.

Gravitational prospecting is clearly outside of classical mechanics. You have to reject the principle above if you accept gravitational prospecting.
 
  • #4
Vanadium 50
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Gravitational prospecting is clearly outside of classical mechanics. You have to reject the principle above if you accept gravitational prospecting.
I see. So you weren't really asking a question.

Do you think you could find a mountain with a gravimeter?
 
  • #5
phinds
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It is not so ...
Interesting. You join a physics forum filled with people who know what they are talking about and immediately tell a mentor that he doesn't know what he is talking about. Kind of makes one think you are not actually here to learn.
 
  • #6
davenn
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The "g" field and magnitude at any local postition is a constant and cannot tell anything about variation of mass distribution beneath a site.
this of course is garbage and if you believe that then you have some serious misunderstandings

you obviously didn't read the links by JTBell, for if you had and understood them, then you wouldn't make such silly comments

Please read the links and learn some real science


Dave
 
  • #7
sophiecentaur
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But according to gravity theory, the gravitational field and magnitude just above the surface of the earth would be always pointing directly towards the center of mass of the earth
Imagine a dumbell shaped planet. Its CM would be half way along the 'bar'. Take a point on the bar, nearer one ball then the other. Which direction would you expect the g force to act?
During the Moon Landings, were the astronauts attracted to the CM of the earth/ Moon or to the Moon's surface?
The CM is the centre of attraction only for objects a long way (infinity) from the body you are considering.
 
  • #8
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Imagine a dumbell shaped planet. Its CM would be half way along the 'bar'. Take a point on the bar, nearer one ball then the other. Which direction would you expect the g force to act?
During the Moon Landings, were the astronauts attracted to the CM of the earth/ Moon or to the Moon's surface?
The CM is the centre of attraction only for objects a long way (infinity) from the body you are considering.
Thanks. I actually found where I got things wrong. I did not read the links given by jtbell as I read some other explanations elsewhere and could not understand how a gravimeter could work.

I got the answer only after I read some other explanations about the force of an extended body on a point mass. Someone mentioned Newton proved that for a uniform spherical mass, it could be treated as a point mass at the center. In general, for an extended body, we have to do an integration and the g-field may not pass through the CM of the large body.

For the moon landing, I think the g-field still points approximately toward the center of the moon.
 
  • #9
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CoM isn't necessarily CoW.

For example : if a patch of sand on a beach contained a large number of mountain ranges compressed into black holes, the weight of your foot would perceptibly increase if you stepped on it (but a bird flying 500m overhead wouldn't notice).

As well, A (my) thread concerning mucking about at Earth/Moon L1.

And, as a side order, it's reasonable to assume that, in opposition to treating the Earth as a point source, as one moved closer to the center, net g would decrease.
 

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