View Full Version : Classical Gravity Question: F = G m1 * m2 / r2 goes to infinity as r->0?

seanjmiller@yahoo.com

2011-Apr-14, 03:51 PM

Please explain to me where I am wrong.

I have a doughnut, a very large massive doughnut made of lead... Ok so it's not edible.

For calculating the force of gravity exerted by this doughnut I can replace it with a point located at its center. Correct? So as I approach this point the force of gravity will increase. As my distance to this point goes to 0, F = G m1 * m2 / r2 will go to infinity. Intuitively this is wrong. Why is this so?

Jeff Root

2011-Apr-14, 04:47 PM

You can only replace the doughnut with a point at its center for

measurements of gravity at a distance from the doughnut. The

simplification works very well-- though not perfectly-- for spheres,

less well for doughnuts, and even less well for more complicated

shapes. If you approach a sphere so closely that you are

inside the sphere, the force not only stops increasing, it begins

to fall. Same with the doughnut.

-- Jeff, in Minneapolis

macaw

2011-Apr-14, 05:10 PM

Please explain to me where I am wrong.

I have a doughnut, a very large massive doughnut made of lead... Ok so it's not edible.

For calculating the force of gravity exerted by this doughnut I can replace it with a point located at its center. Correct?

No, this is not correct. The correct way is to divide the donut into infinitesimal elements and to calculate the force due to each element. If you do that correctly, you will find out that the force in the center of the donut is precisely 0. Due to the symmetry, the forces balance each other.

So as I approach this point the force of gravity will increase.

No, see above.

As my distance to this point goes to 0, F = G m1 * m2 / r2 will go to infinity. Intuitively this is wrong. Why is this so?

Because you use intuition instead of math.

jfribrg

2011-Apr-14, 05:15 PM

You may want to check out Feynman's Lectures on Physics Volume 1 (http://www.amazon.com/Feynman-Lectures-Physics-Vol-Mechanics/dp/0805390464/ref=sr_1_4?ie=UTF8&qid=1302800868&sr=8-4). It has several pages describing (mostly in words with a few equations as well) what Jeff Root just said. The equations assume that you know some integral calculus, but even if you don't, the accompanying description alone is worth reading. As I said, it is only a few pages and can be easily read in the aisle of your local big chain bookstore.

tommac

2011-Apr-14, 06:23 PM

The problem is the shape. One can use the center of gravity as being the center at a distance but as you get closer to the center of gravity forces counteract each other in the center there will be a net force of 0 not infinity.

The max force would be at the surface of the outside of the doughnut.

However the gravitational potential would probably be a max at the center.

Please explain to me where I am wrong.

I have a doughnut, a very large massive doughnut made of lead... Ok so it's not edible.

For calculating the force of gravity exerted by this doughnut I can replace it with a point located at its center. Correct? So as I approach this point the force of gravity will increase. As my distance to this point goes to 0, F = G m1 * m2 / r2 will go to infinity. Intuitively this is wrong. Why is this so?

pzkpfw

2011-Apr-14, 07:42 PM

Another way to look at it is: If all the mass of that doughnut was all concentrated in one point, then F would be getting quite large at small distances... it's possibly black hole time.

swampyankee

2011-Apr-15, 01:05 AM

It's time for you to break out your calculus. I suspect that this is not a trivial calculation, and it may not even be possible in closed form--I'm too short for time to even try to find out--but the principle is that you would need to integrate dg resulting from each dm within the toroid to find the gravitational acceleration at any given point. When the point is sufficiently far from the toroid, with "sufficiently far" dependent on how accurately you need to know the gravitational acceleration at any given point, you can just use the basic g=GMr/|r|^3, and assume that M (the mass of your toroid) is spherical.

Have fun, and tell us how you fared.

(note that I'm using g as shorthand for F/m, where m is the mass of a test particle. Bold type denotes vector quantities)

Andrew D

2011-Apr-15, 05:39 AM

However the gravitational potential would probably be a max at the center.

The potential is max at r=\infty. Potential at the center is 0.

WayneFrancis

2011-Apr-15, 07:40 AM

Please explain to me where I am wrong.

I have a doughnut, a very large massive doughnut made of lead... Ok so it's not edible.

For calculating the force of gravity exerted by this doughnut I can replace it with a point located at its center. Correct? So as I approach this point the force of gravity will increase. As my distance to this point goes to 0, F = G m1 * m2 / r2 will go to infinity. Intuitively this is wrong. Why is this so?

Like I'm sure others have pointed out already, but I have not read their posts yet, this formula only works when you are outside of the object's surface. So if your doughnut is of major radius of 100km and a minor radius of 50km and your distance is 20km from the centre then this is outside of the applicability of that formula.

At R=0 the gravity would actually not go to infinity but go to zero but, again, you can't use that formula for that calculation. Use the wrong formula for a given and you'll get a wrong answer. Garbage in, Garbage out.

WayneFrancis

2011-Apr-15, 07:42 AM

Another way to look at it is: If all the mass of that doughnut was all concentrated in one point, then F would be getting quite large at small distances... it's possibly black hole time.

If all the mass of the doughnut was all concentrated in one point then it wouldn't be a doughnut :P

tommac

2011-Apr-15, 01:12 PM

???? gravitational potential .... shouldnt this be a max at the center? For a sphere it is the max in the center right?

Isnt that what this diagram shows:

http://en.wikipedia.org/wiki/File:GravityPotential.jpg

The potential is max at r=\infty. Potential at the center is 0.

seanjmiller@yahoo.com

2011-Apr-15, 01:16 PM

I am trying to model the attractive force of gravity in a computer program... Why? I'm bored. So I'm watching Stephen Hawking's 'Into the Universe' and there is a simulation that shows a bunch of marbles layed out on the floor of the cafeteria at Cambridge University. They were all equally spread out and nothing would happen because all of the forces would cancel out (this assumed an infinitely sized floor with an infinite number of marbles). If you removed some of the marbles the forces would be stronger and weaker in other areas and the marbles would begin to move. This was an explaination of the early Universe and the creation of galaxies/stars...

I was going to go into a long detailed explaination of what I have attempted, but instead I will pose a single question:

I have two masses, m1 and m2 seperated by some distance x. What are their equations of motion? You do not have to account for collisions, the 'particles' can pass right through each other. I would think that the masses seperated by x would accelerate towards each other and when they meet would start to decelerate until they are x distance apart again and this would repeat. For the more adventurous what is the equations of motion for the 3 'particle' problem.

Thanks in advance.

grapes

2011-Apr-15, 01:27 PM

???? gravitational potential .... shouldnt this be a max at the center? For a sphere it is the max in the center right?

Isnt that what this diagram shows:

http://en.wikipedia.org/wiki/File:GravityPotential.jpgTo me, the diagram looks like the lowest point is at the center.

I have two masses, m1 and m2 seperated by some distance x. What are their equations of motion? Keplerian ellipses, depending upon their initial velocity. If the velocities are too great, they might "escape" each other.

For the more adventurous what is the equations of motion for the 3 'particle' problem.An unsolved mathematical problem, in general! :)

seanjmiller@yahoo.com

2011-Apr-15, 02:14 PM

I am trying to model the attractive force of gravity in a computer program... Why? I'm bored. So I'm watching Stephen Hawking's 'Into the Universe' and there is a simulation that shows a bunch of marbles layed out on the floor of the cafeteria at Cambridge University. They were all equally spread out and nothing would happen because all of the forces would cancel out (this assumed an infinitely sized floor with an infinite number of marbles). If you removed some of the marbles the forces would be stronger and weaker in other areas and the marbles would begin to move. This was an explaination of the early Universe and the creation of galaxies/stars...

I was going to go into a long detailed explaination of what I have attempted, but instead I will pose a single question:

I have two masses, m1 and m2 seperated by some distance x. What are their equations of motion? You do not have to account for collisions, the 'particles' can pass right through each other. I would think that the masses seperated by x would accelerate towards each other and when they meet would start to decelerate until they are x distance apart again and this would repeat. For the more adventurous what is the equations of motion for the 3 'particle' problem.

Thanks in advance.

Strange

2011-Apr-15, 03:09 PM

The potential is max at r=\infty. Potential at the center is 0.

Hmmm...

http://en.wikipedia.org/wiki/Gravitational_potential

By convention, the gravitational potential is defined as zero infinitely far away from any mass. As a result it is negative elsewhere.

So, sounds like the potential is -something at the center.

Strange

2011-Apr-15, 03:14 PM

An unsolved mathematical problem, in general! :)

Indeed. You will have to model this with a simulation; taking small enough (?) time steps should produce something reasonably accurate.

You should just need F=ma and F=Gm1m2/r2 - you can then calculate the total force on each particle due to all the others and therefore its acceleration, at each time step, for every particle. Tedious and repetitive. Which is what we have computers for.

macaw

2011-Apr-15, 03:52 PM

I have two masses, m1 and m2 seperated by some distance x. What are their equations of motion?

See here (https://docs.google.com/viewer?a=v&pid=explorer&chrome=true&srcid=0BzsR_aYKdLVbZjE4YTkxYWYtMzcxZS00NjRmLWIwNDI tYTljMTNjOTE5YjNj&hl=en)

I would think that the masses seperated by x would accelerate towards each other and when they meet would start to decelerate until they are x distance apart again and this would repeat.

This is false, they end up colliding (see link cited above).

Jeff Root

2011-Apr-15, 06:40 PM

... there is a simulation that shows a bunch of marbles layed out on

the floor of the cafeteria at Cambridge University. They were all equally

spread out and nothing would happen because all of the forces would

cancel out (this assumed an infinitely sized floor with an infinite number

of marbles).

That's what I tried to argue, but Ken G (one of BAUT's real experts)

wouldn't go along with it. My argument was Newtonian and his was

probably relativistic, so that likely makes the difference...

... You do not have to account for collisions, the 'particles' can pass

right through each other.

If you treat them as points, and the points get very close together,

then you *will* get absurd interactions. If the points are exactly on

top of one another, you will either get infinite gravitational attraction,

or a division by zero error. So you'll need to test for it.

I would think that the masses seperated by x would accelerate

towards each other and when they meet would start to decelerate

until they are x distance apart again and this would repeat.

Yes. The general terms for it are "harmonic motion" or "oscillation".

It looks exactly the same as looking at two masses orbiting each

other edge-on to the plane of the orbits. Keplerian circular and

elliptical orbits are examples of harmonic motion.

-- Jeff, in Minneapolis

Andrew D

2011-Apr-16, 01:19 AM

Hmmm...

http://en.wikipedia.org/wiki/Gravitational_potential

So, sounds like the potential is -something at the center.

My interpretation was potential energy. Oops.

WayneFrancis

2011-Apr-16, 02:27 AM

...

I was going to go into a long detailed explaination of what I have attempted, but instead I will pose a single question:

I have two masses, m1 and m2 seperated by some distance x. What are their equations of motion? You do not have to account for collisions, the 'particles' can pass right through each other. I would think that the masses seperated by x would accelerate towards each other and when they meet would start to decelerate until they are x distance apart again and this would repeat. For the more adventurous what is the equations of motion for the 3 'particle' problem.

Thanks in advance.

For your experiment you have to think of what type of particle you are talking about.

If they can pass right through each other then they are bosons and as bosons they will be traveling at c and never slow down. If they are Fermion then they can't pass through each other so you'd get a collision. Congratulations you've designed a particle accelerator. If you have them miss each other by just a little bit what you've done is put them in orbit around each other.

Jeff Root

2011-Apr-16, 08:24 AM

His particles are almost certainly either stars or planets, since he's

working on a gravity simulator. Could be galaxies, though.

-- Jeff, in Minneapolis

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