View Full Version : Gravitational approximation

his_eminence

2004-Feb-09, 06:48 PM

I read a webpage here http://farside.ph.utexas.edu/teaching/301/lectures/node152.html that states that after considerable effort, Newton "proved that the gravitational force exerted by a spherical body (outside that body) is the same as that exerted by an equivalent point mass located at the body's centre." How did he prove this? How do you prove that the earth can be approximated by an equivalent point mass?

milli360

2004-Feb-09, 06:51 PM

Using another of his little inventions--the calculus. It allowed him to sum up the effects of all the small parts, and the result was the same as if the entire mass were concentrated at a point.

That's an approximation of course.

his_eminence

2004-Feb-09, 07:26 PM

Right...I know calculus, I'm just having difficulty coming up with the equations he used. I'm trying to derive for myself the idea that you can use a point mass for a large body.

Demigrog

2004-Feb-09, 09:03 PM

Multi-variable calculus, though Newton's form of calculus differs in form from modern calculus, I think. Simple shapes are easy; real world objects are hard.

A quick search on Google for +integral +"center of mass" yields a lot of math department websites, including:

http://math.etsu.edu/MultiCalc/Chap4/Chap4-3/part2.htm

Like milli360 said, this is only an approximation (like most of Newton’s laws seem to be). I'm not sure what level of error there is, though I can think of several things that would slightly alter the numbers. Slight differences in distance between points within two objects will alter the force of gravity between them. I would expect this to be most obvious with two large volume objects that are close together. General relativity would probably also come into play depending on relative velocities, acceleration, etc.

his_eminence

2004-Feb-09, 09:43 PM

The question came about when a friend and I were trying to develop an equation for the force of gravity as one goes into the earth (with the assumption of a linear increase in density as you go further inside). My friend surmised that there might be some error introduced by assuming the earth to be a point mass, so then we were trying to figure out how to determine this error, if it did exist.

To be more specific then (i.e. to exclude relativistic properties), we are dealing with two objects (earth and a person) with 0 relative motion, where the person is on the surface of the earth. If that person then digs into the earth (yes, we got on this from a discussion of The Core), what kind of gravitational effects can he expect to see. The point mass approximation works well for bodies whose distance apart greatly exceeds the radii of either body. So we were thinking, does this asumption hold true for (and if not, what is the error in making this assumption) bodies that are touching, or whose distance apart is equal to or less than the radii of either body?

swansont

2004-Feb-09, 09:52 PM

If the mass is uniformly distributed, this is simply Gauss's Law. It also tells you that any uniformly distributed mass outside of your radius does not contribute to the field.

You can account for known nonuniformities by looking at them as additions to a uniform background, which can then be reduced to the equivalent point mass.

Demigrog

2004-Feb-09, 10:06 PM

Well, if you’re trying to calculate force within the Earth, the point-source model isn’t going to work at all, as part of the mass of Earth would actually pull you away from the core as you got deeper in the planet. Also, Earth is not even remotely uniform in density, which would be a very big deal for gravity at points inside Earth.

The best way I can think of to calculate gravitational force within Earth is to integrate the force of gravity for all points relative to the point you’re interested in. Take the equation for the density of earth at any point within the sphere, substitute it into the equation for gravitational force between two objects (with the person/drill ship/whatever’s mass as the second object), and then integrate it for all points within the (approximate) sphere of Earth.

If you can write a computer program to do it, you could simplify things and estimate the answer with some degree of accuracy. Divide the Earth up into cubes, assign a density to each cube, and calculate the gravitational force vector between each cube and the drill/ship/whatever. Sum them, you’ve got an estimate. (In other words, calculate the integral instead of trying to solve some nasty multi-variable integral with a non-linear density function). The smaller you make the cubes, the more precise your answer is. The more accurately you assign the true density of Earth to the cubes, the more accurate the answer is. (note that I’m ignoring gravitational pull of the Moon, Sun, etc. as they are probably insignificant compared to the error in the density estimates).

My guess is you’re not that bored. Of course, seeing as I just spent half my day posting on BABB, maybe you are. It’d only take me a half hour or so, but my girlfriend is making lasagna for dinner!

milli360

2004-Feb-09, 11:45 PM

The question came about when a friend and I were trying to develop an equation for the force of gravity as one goes into the earth (with the assumption of a linear increase in density as you go further inside). My friend surmised that there might be some error introduced by assuming the earth to be a point mass, so then we were trying to figure out how to determine this error, if it did exist.

As swansont points out, there is a big error.

In a "perfect" sphere, with density varying as you describe, the effect is like a point mass--with mass equal to the mass "below your feet". All the mass at a greater radius doesn't contribute--including the mass on the far side of the Earth. That makes your example fairly simple.

In the real Earth, there are discontinuities--the biggest density jump is at the core-mantle boundary (CMB). The density distribution is such that the force of gravity (what you'd feel on your feet) is almost constant all the way down to the CMB, maybe increasing just a little. That's halfway to the center, about.

his_eminence

2004-Feb-10, 01:16 AM

Thanks for the replies all. I really think I should just go do the math to show this

daver

2004-Feb-10, 01:46 AM

Thanks for the replies all. I really think I should just go do the math to show this

The math isn't that hard--geometry is almost enough. Think cones (say you have a wall of density rho and thickness h and a flashlight whose apex angle is alpha--calculate the graviational attraction of the section of the wall illuminated by your flashlight).

AstroSmurf

2004-Feb-10, 08:42 AM

For a sphere of uniform density, the gravitational force decreases linearly with the radius as you descend. g(r) = g(R) * r / R in other words. It pops right out if you integrate over the total volume.

Cougar

2004-Feb-10, 03:04 PM

I could be wrong, but I thought Newton showed in his Principia that the gravitational force exerted by a spherical body (outside that body) is the same as that exerted by a point at the center of mass using nothing more than geometry.

Normandy6644

2004-Feb-10, 03:22 PM

I could be wrong, but I thought Newton showed in his Principia that the gravitational force exerted by a spherical body (outside that body) is the same as that exerted by a point at the center of mass using nothing more than geometry.

yeah, but it's really weird geometry. It would have been easier for him ti just show the calculus, but no one else really understood it at the time, and so he kept all his math in the (rather awkward at times) geometry.

jfribrg

2004-Feb-10, 04:32 PM

Obviously you can't treat an object as a point mass unless you are sufficiently far away. If the moon behaved as a point mass, we would only have one high tide per day.

Think of a small volume somewhere in the large object and somewhere away from the center of mass. Think of this force pulling on you. Now think of how this force would pull on you if if were at the center of mass. The difference is a function of the sine of the angle between these two points. Now the farther away you are from the object, the closer the angle (and therefore the sine of the angle) gets to 0. This means that when you are far enough away, the difference between the point mass calculation and the actual calculation becomes negligible.

Feynman's "Lectures in Physics" books (I think it is volume 1) talks about this kind of thing at length.

SeanF

2004-Feb-10, 04:47 PM

If the moon behaved as a point mass, we would only have one high tide per day.

How do you figure that?

Sam5

2004-Feb-10, 05:00 PM

Obviously you can't treat an object as a point mass unless you are sufficiently far away.

If I understand it correctly, it works something like this: If you are standing on the earth facing North, and if you split the earth down the middle, right between your two feet, to the front of you and the back of you, then all the half of the earth on your right side is pulling you down and to the right, while all the half of the earth on the left of you is pulling you down and to the left. The right and left pull cancel each other out, so you are pulled downwards as if all the mass of the whole earth were located at a point in the center of the earth. So, you can think of the whole mass as a point in the center, even if you are standing on the earth.

swansont

2004-Feb-10, 06:07 PM

Obviously you can't treat an object as a point mass unless you are sufficiently far away.

...

Think of a small volume somewhere in the large object and somewhere away from the center of mass. Think of this force pulling on you. Now think of how this force would pull on you if if were at the center of mass. The difference is a function of the sine of the angle between these two points. Now the farther away you are from the object, the closer the angle (and therefore the sine of the angle) gets to 0. This means that when you are far enough away, the difference between the point mass calculation and the actual calculation becomes negligible.

Feynman's "Lectures in Physics" books (I think it is volume 1) talks about this kind of thing at length.

(moon/ tide comment snipped)

This is why the "uniform density" caveat is so important - all of the contributions from different points end up averaging out. Under that condition (which applies to angular symmetry - the density can vary with r) you can treat the object as a point mass.

The problem is that the force is only given for one point, and varies with position - you can't assume that both objects are point masses.

jfribrg

2004-Feb-10, 08:49 PM

If the moon behaved as a point mass, we would only have one high tide per day.

How do you figure that?

I wish to retract that statement. #-o What I meant to say was that if the Earth were a point mass there would be only one high tide. The way it works, the high tide on the side farthest from the Moon is due to the fact that the pull of the moon is weaker than on the close side. Thinking about it though, it really isn't relavant to the question being asked. The tides would be the same if the Moon were a point mass. The differences that cause the tides relate to the location of the mass on Earth. As long as we assume a homogeneous (or at least symmetrical ) distribution of mass, the acceleration due to gravity behaves as a point mass located at the C.O.M.

I should have given my previous post a reality check before I sent it off. It's nice to know that folks actually read what I write.

SeanF

2004-Feb-10, 09:09 PM

If the moon behaved as a point mass, we would only have one high tide per day.

How do you figure that?

What I meant to say was that if the Earth were a point mass there would be only one high tide.

How do you figure that? :)

daver

2004-Feb-10, 09:17 PM

folks actually read what I write.

How do you figure that? :)

milli360

2004-Feb-11, 02:51 AM

The problem is that the force is only given for one point, and varies with position - you can't assume that both objects are point masses.

I don't know why not--if both objects are perfectly spherical.

Wait, I just got a headache...

swansont

2004-Feb-11, 11:24 AM

The problem is that the force is only given for one point, and varies with position - you can't assume that both objects are point masses.

I don't know why not--if both objects are perfectly spherical.

Wait, I just got a headache...

Assume you have two objects, A and B, that are spherically symmetric. The field at any given point in space can be calculated assuming point masses from Gauss's law. But the force of A on B can't necessarily assume B is a point mass, because there will be tidal forces on B from the variation of force with position.

jfribrg

2004-Feb-11, 03:24 PM

The problem is that the force is only given for one point, and varies with position - you can't assume that both objects are point masses.

I don't know why not--if both objects are perfectly spherical.

Wait, I just got a headache...

I suspect it depends on what you are trying to calculate. Perhaps you can treat both as point masses if you are only interested in the movement of the objects' centers of mass, but you can't treat both as point masses if the tidal forces need to be considered. If the object's mass is not symmetrically distributed, then you definitely can't treat both as point masses. For instance, the oblateness of the Earth is a factor in calculating the orbits of artificial sattelites. The oblateness may be approximately symmetrical with respect to the equator, but not with respect to the orbit of sattelites in low Earth orbit, such as Hubble and the ISS. The effect is small , but it accumulates and over time becomes significant.

milli360

2004-Feb-11, 05:01 PM

What I meant to say was that if the Earth were a point mass there would be only one high tide.

How do you figure that?

Good catch. Of course there would be no tide.

Assume you have two objects, A and B, that are spherically symmetric. The field at any given point in space can be calculated assuming point masses from Gauss's law. But the force of A on B can't necessarily assume B is a point mass, because there will be tidal forces on B from the variation of force with position.

As long as you still assume spherical symmetry, there is no effect. You can still assume each is a point mass.

But you know what happens when you assume--you make a mass of problems for you and me.

swansont

2004-Feb-11, 06:34 PM

Assume you have two objects, A and B, that are spherically symmetric. The field at any given point in space can be calculated assuming point masses from Gauss's law. But the force of A on B can't necessarily assume B is a point mass, because there will be tidal forces on B from the variation of force with position.

As long as you still assume spherical symmetry, there is no effect. You can still assume each is a point mass.

But you know what happens when you assume--you make a mass of problems for you and me.

But spherical symmetry is an assumption separate from a symmetric mass distribution of an unperturbed sphere. Saying you can treat the sphere as a point mass is not assumption, it is mathematically equivalent. But assuming that the objects won't deform under tidal forces is an assumption.

milli360

2004-Feb-11, 06:46 PM

But spherical symmetry is an assumption separate from a symmetric mass distribution of an unperturbed sphere. Saying you can treat the sphere as a point mass is not assumption, it is mathematically equivalent. But assuming that the objects won't deform under tidal forces is an assumption.

What's your point? (eh)

You can treat both as a point, or neither? That's kinda what I said.

swansont

2004-Feb-11, 09:03 PM

But spherical symmetry is an assumption separate from a symmetric mass distribution of an unperturbed sphere. Saying you can treat the sphere as a point mass is not assumption, it is mathematically equivalent. But assuming that the objects won't deform under tidal forces is an assumption.

What's your point? (eh)

You can treat both as a point, or neither? That's kinda what I said.

I just wanted to clarify that treating the symmetric sphere as a point mass is not an approximation or assumption. It follows from the statement that you have the symmetry.

milli360

2004-Feb-12, 10:21 AM

Ah. Point taken. When I said it was an approximation, I meant that in the sense that real-world bodies are not perfectly spherical.

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