Newton's Law of Gravitation

[ellipsis]

Just a quick question as to WHEN this law can apply to objects...

Is it applicable when two objects are REALLY close to each other? I saw this question asking what acceleration a person experiences when he is inside the earth. Is it even possible to calculate the acceleration?

[insane_alien]

the can be applied if the distance between the two objects is less than infinity.

 

and yes, you can calculate the force on an object below the surface of the earth. just put the mass of the earth BELOW that person into the equation and ignore the stuff above, that all cancels out.

[timo]

In the naive form of taking the total masses of the objects, the law applies to cases where the distance of the two objects is large compared to their size (and some important special cases). If the distance is not significantly larger than the size of the objects, then you have to use a slightly more sophisticated version than the one I assume you have in mind (the one where you only plug in the total masses of the objects): Basically, you divide the objects into smaller pieces (small enough so that they are small compared to the distance), calculate the forces on these smaller pieces and -if appropriate- add those forces up in a suitable manner to get the total forces on the two objects.

 

the can be applied if the distance between the two objects is less than infinity.

That should always be the case. Did you possibly mean zero?

[insane_alien]

i meant what i said. if there is somehow an object at a greater distance than infinity then something very wrong indeed has happened to the universe.

[John Cuthber]

There's something equally amiss with things if there are 2 objects separated by less than zero distance.

Technically, I think the latest theories say Newtonian gravity may break down at really small distances, (Planck length anyone?) but it still works fine for things as big as atoms.

[doG]

the can be applied if the distance between the two objects is less than infinity.

 

Did you possibly mean zero?

 

How could the distance between two objects be less than zero?

[timo]

Approaching zero lets the force term rise towards infinity, hence my idea that IA mixed that up.

[swansont]

In the naive form of taking the total masses of the objects, the law applies to cases where the distance of the two objects is large compared to their size (and some important special cases). If the distance is not significantly larger than the size of the objects, then you have to use a slightly more sophisticated version than the one I assume you have in mind (the one where you only plug in the total masses of the objects): Basically, you divide the objects into smaller pieces (small enough so that they are small compared to the distance), calculate the forces on these smaller pieces and -if appropriate- add those forces up in a suitable manner to get the total forces on the two objects.

 

Or, the short version:

 

It assumes point masses. If you can't reasonably assume that, you have to break the problem up into manageable pieces where the assumption holds.

[ellipsis]

To Atheist: That's what I was wondering, since, while I did not learn the more sophisticated version, I learned that the law cannot be applied to objects separated by a distance smaller than their sizes.

So I guess I can't really "plug in numbers" and calculate the acceleration!

Thank you all!

[D H]

Or, the short version:

 

It assumes point masses. If you can't reasonably assume that, you have to break the problem up into manageable pieces where the assumption holds.

 

More precisely, Newton's law of gravitation assumes spherical mass distributions. If the density of some object is a function only of the distance from the center of the body, the body looks exactly like a point mass for any point outside the surface of the body. If that is not true one must use the more general form of Newton's law of gravitation:

 

[math]\mathbf a (\mathbf R) =

-G\iiint \frac {\rho(\mathbf r)}{||\mathbf R - \mathbf r||^3}(\mathbf R - \mathbf r)\,d\mathbf r[/math]

 

This integral is an atrocious mess except for very simple shapes. So, "break the problem up into manageable pieces". A bunch of point masses, maybe? That is not what the people who work in the field of big, lumpy mass distributions (aka planets) do.

 

Quantum physicists typically do not "break up" the wave function into a finite number of pieces; they break it up into an infinite number of pieces instead! This turns out to be very manageable approach thanks to the use of orthogonal functions. The orthogonal functions geodesists use are the same ones quantum physicists use to "break up" the wave function: spherical harmonics.

 

Researchers at NASA and the USGS have developed a huge (360x360) spherical harmonic model of the Earth called EGM96 (Earth Gravity Model 1996). Models of lesser degree and lesser accuracy have been developed for the Moon and Mars.

 

Researchers at the University of Texas, NASA, and elsewhere are refining the the EGM96 model using a pair of orbiting satellites. This is the Gravity Recovery and Climate Experiment or GRACE for short. One of the refinements is looking for time variations in these terms. The time variations include short-term variations such those caused by tides, annual variations such as those caused by seasonal migration of mass away from / toward the equator, and even longer term variations such as those caused by plate tectonics.

 

The oceans are thin compared to the size of the Earth; how can tides affect gravitational attraction? Simple: The Earth is not a solid body; it is plastic. There are tides in the "solid" body of the Earth. While these solid body tides are smaller in amplitude than the ocean tides (30 centimeters versus 1 meter), the solid body tides affect the entire Earth. These solid body tides have an order of magnitude greater effect on orbiting bodies than do the ocean tides.

 

Edited to add:

 

Whoa! It looks the DoD is far ahead of NASA, the USGS, and tons of slaves graduate students at UT and elsewhere. The National Geospatial Intelligence Agency (acronym: WTF) has recently released a 2190x2159 spherical harmonics model of the Earth called the Earth Gravitational Model 2008 (EGM2008). Their web site: http://earth-info.nga.mil/GandG/wgs84/gravitymod/egm2008/index.html.

Edited June 14, 2008 by D H

[swansont]

Gauss's law means that a spherical distribution is the same as a point source; in some sense that's really mathematical semantics. —a uniformly distributed spherical mass is one where the point-mass assumption holds.

 

Another approach one can take for a lumpy planet (or equivalent problem) is solve for the ideal case, e.g. the uniform mass distribution, and then solve for the lumps(s). Not sure how often people do this in gravitation, but perturbation theory is common in QM.

[D H]

Gauss's law means that a spherical distribution is the same as a point source; in some sense that's really mathematical semantics. —a uniformly distributed spherical mass is one where the point-mass assumption holds.

One important difference between point masses and spherical masses: collisions. The probability of a collision between a pair of point masses is zero because the set of position/velocity states that lead to collisions has measure zero. Non-point masses have a non-zero chance of colliding.

 

Another approach one can take for a lumpy planet (or equivalent problem) is solve for the ideal case, e.g. the uniform mass distribution, and then solve for the lumps(s).

Relative to an oblate spheroid (rather than a sphere), here is the mass distribution for the Earth:

and for the Moon:

 

The Moon is very lumpy, and some have indeed suggested using a mass concentration model rather than spherical harmonics model for the Moon. The lumpiness of the Moon's gravity gave scientists a big surprise back in the early '70s. Apollo astronauts released a couple of little satellites to orbit the Moon as they departed the Moon. The second of these satellites started orbiting normally, then dipped toward the Moon, then pulled back away from the Moon, and then crashed into the Moon. The other satellite, released eight months earlier, eventually crashed into the Moon as well. For details, read this article.

 

Not sure how often people do this in gravitation, but perturbation theory is common in QM.

Perturbation theory is used in modeling earth orbits and solar system dynamics as well. For example, we use almost always use Newtonian mechanics rather than general relativity to model orbits. If relativistic effects must be included to satisfy accuracy requirements we still use Newtonian mechanics and treat the relativistic effects as perturbations to Newtonian gravity.

[DJBruce]

There was just a show on the history channel that touched on this topic. The said that of the coast of Puerto Rico Earths gravity is .3% weaker than normal.

[swansont]

This reminds me — D H, what's the reference for that cool figure of the moon's acceleration map?

[D H]

This reminds me — D H, what's the reference for that cool figure of the moon's acceleration map?

I got that map from this article:

http://science.nasa.gov/headlines/y2006/06nov_loworbit.htm

This is a fascinating article. Click on the lunar gravity image at the middle of the article and you will get the full image I posted.

 

Also see Konopliv et al, "Improved Gravity Field of the Moon from Lunar Prospector", Science 281 1476-1480 (1998).

[swansont]

Thanks!

[Pete]

More precisely, Newton's law of gravitation assumes spherical mass distributions.

Newton's law of gravitation F = Gm₁m₂/r² is the force between two point particles. The force between two objects is found by using the principle of superposition and thus integrating over the source body and the body which the source is acting on.

 

Pete

[swansont]

Newton's law of gravitation F = Gm₁m₂/r² is the force between two point particles. The force between two objects is found by using the principle of superposition and thus integrating over the source body and the body which the source is acting on.

 

Pete

 

 

But, as we've discussed, a uniform sphere acts like a point particle. That's what the integral reduces to.

[Mr Skeptic]

But, as we've discussed, a uniform sphere acts like a point particle. That's what the integral reduces to.

 

I had trouble believing that one. So I taught myself calculus from my dad's old textbook, and verified that it was indeed true. Actually, that was for electricity, but anything with an inverse square law would work.

[Pete]

But, as we've discussed, a uniform sphere acts like a point particle. That's what the integral reduces to.

 

The gravitational field of the sphere is the same as that of a particle. However the force on one sphere due to another is not the same as that of two point particles. This is because the sphere is an extended body and as such the tidal effects of one sphere will have an effect on the other sphere.

 

Pete

[D H]

The gravitational field of the sphere is the same as that of a particle. However the force on one sphere due to another is not the same as that of two point particles. This is because the sphere is an extended body and as such the tidal effects of one sphere will have an effect on the other sphere.

In short, center of gravity is not the same as center of mass. From The Feynman Lectures on Physics:

"In case the object is so large that the nonparallelism of the gravitational forces is significant, then the center where one must apply the balancing force is not simple to describe, and it departs slightly from the center of mass. That is why one must distinguish between the center of mass and the center of gravity."