L1, L2, and L3 are. L4 and L5 are (in practical effect) much larger than points, which is why actual L4 and L5 distributions are common in nature.
I was not aware of that.
Even the L1 point which would be at the center of gravity for two objects of identical mass? Or is it just called something else then?
penumbrage - You are corrrect. If the gravitational gradiant is high enough, you will get pulled apart in two separate directions. If it isn’t (which is what we would expect when significantly far away from objects with large masses), you’ll stay put.
The illustration using a body around a black hole compares the gravitational attraction at the top of the body, t[sub]1[/sub], with the gravitational attraction at the bottom of the body, t[sub]2[/sub], both points attracted by one center of mass very far away.
At the center of the earth a six foot body is being equally attracted by, roughly, 21,120,000 feet of mass on all sides in every direction. Any difference from one side of the six feet to the other is so minuscule that it can be completely disregarded.
Gravity is the weakest force in the universe. Enormous bodies of mass are required to produce the comparatively trivial tidal forces we witness on earth. If you reduce the scale, you reduce the tidal force to nothingness and, again, it can be disregarded.
As JWK already said, this isn’t the way Lagrange points work, and they have nothing to do with the way that gravity appears to cancel out at the center of a sphere.
That is metastable (i.e., it balances , but only as long as everything remains perfect), like L1, L2, and L3, but isn’t regarded as a Lagrange point because it’s trivial and obvious. The five Lagrange points require some actual math to solve.
However, there is no gradient anywhere in a hollow sphere. There is some while falling through a tube, but when one body is a man and the other is the Earth, we’re talking about a difference of about 0.06 gram at the surface, and less further in.
The ‘no gradient in a hollow sphere’ is what I’m trying to understand. Pardon my fuzzy thinking, poor communication skills and lack of formal physics training. Let me know where I’ve gone wrong here.
- An object at Earth/Moon L1 is being slightly stretched by the earth and the moon since half of it’s mass is closer to Earth than the L1 point and half is closer to the Moon.
- This is also true for the less stable situation of being centered between two equal masses. All that is required is the two gravitational fields be equal at the point in question. The only difference from (1) is that the gradients are equal instead of being biased towards the larger mass.
- Expanding on this in two dimensions, if it were four equal masses there would be two axes of stretching, five axes for ten masses, and for the extreme case of as many masses as you can fit into the same orbit we would be stretching the object in an entire plane.
- Expanding on this in three dimensions, a second filled orbit of masses (I know, it won’t work in space, just in our heads) would stretch the object in two planes and for the extreme case you would have a solid shell of masses stretching the object from all directions.
- Stack up various diameters of these shells until you have a sphere about 7900 miles in diameter, and while you’re at it cut a hole from surface to surface passing through the center.
Is this not an approximation of the Earth for the purposes of this thread and if it is, where did the streching force disappear to? Or is the ‘no gradients’ wrong and it should be ‘insignificant gradients’?
The stretching force disappears because there are equal masses on all sides pulling equally in all directions. The net effect is no effect. No stretching, no gradient.
Think of it this way. If you have a magnet you can pull a pin in the direction of the magnet. But what happens if you put magnets equally in all directions around the pin? It will stay still in the middle, because the various effects cancel out, leaving no net effect.
The difference is between cases where the canceling out is at only one point and cases where the canceling out is over an entire space. There is zero gravity throughout the interior of a hollow sphere. In the space between two equal masses, there is zero gravity only at the exact center.
It takes calculus to prove this properly, but, aside from theory, it is an elementary classroom demonstration to show that there is no electrostatic field inside a charged hollow sphere, and electric charges follow the same inverse-square law that gravity does.
So my chain goes bad at step 3, somewhere between ‘four masses pull you in two axes’ and the ‘extreme case’, which amounts to a solid torus of mass surrounding you, pulling you radially in a plane. I’m assuming that what holds true for the hollow sphere holds true for a cross section of the hollow sphere - that as long as you stay within the plane of the torus there is neither felt gravity or gradients, true?
Imagine a strain gauge in the middle of your needle measuring the tensile pull the two ends feel. Turn on two opposite magnets, let the needle align and the gauge reads 1mg. After you turn on all the magnets in the circle are you saying the strain gauge reads 0mg or are you saying it reads 1mg anywhere in the circle you point it?
It’s 0 as soon as soon as the opposing magnets are turned on. +1 + -1 = 0. It does not change after that as long as the all new magnets added are truly opposing.
No, it doesn’t, except, once again, at the very center point.
This is correct. The time dilation effects due to GR depend on the gravitational potential, not the gravitational field, and therefore would be greatest at the center of the Earth (where the potential is greatest, despite the field being zero).
The fact that it’s the potential that matters is also key in understanding why you age at the same rate anywhere on the surface of the Earth. The surface of the Earth (neglecting mountains and such) is at an equipotential of the gravitational and centrifugal forces: If it were not, then the parts which were at higher potential would slide downwards to the parts at low potential. Since all points on the surface are at the same potential, they all feel the same time dilation. Note that it’s not a matter of the GR effect canceling out the SR effect: In fact, I didn’t bother to do any explicit calculation of the SR effect. This is because SR is inherently included in GR, so the GR result (same dilation everywhere) automatically includes the SR contribution.
penumbrage, the exact cancellation only occurs for a complete spherical shell. Here’s a simple way to visualize why it occurs, without need for calculus:
Picture a point inside the spherical shell. It doesn’t need to be in the center; it can be anywhere. Now draw a pair of coaxial cones (of any opening angle, and with the axis in any direction), with the vertex at the point you chose. One end of the cone will slice off one piece of the sphere, and the other end of the cone will slice off another piece of the sphere. Which sphere-piece of mass pulls harder on the point? Well, one piece is going to be closer than the other, but that piece is also going to be smaller. In fact, the size of each piece is going to be proportional to the square of the distance from the point. This will cancel out the 1/r[sup]2[/sup] in Newton’s law, so both pieces of the sphere will pull equally hard on our point, but in opposite directions. Well, the whole sphere can be broken up into these pairs of conic sections, and the contribution of each pair of sections is also going to be zero. So the total force from the entire sphere will be zero.
Is the long axis of your needle perpendicular to your ring of magnets?
If so, your strain gauge would read 0mg with any opposite pair or any combination
of opposite pairs since the needle is symmetrical about the geometric point of balanced force. It might even read slightly negative as the field tries to pull both ends of the needle into the center of the plane.
But if the long axis of your needle is in the plane of the magnets, like how I was imagining the experiment, it is no longer symmetrical around the point of balanced force. The ends are available for magnet pairs to attract. Does my last question about the strain gauge reading make more sense now?
penumbrage, the exact cancellation only occurs for a complete spherical shell. Here’s a simple way to visualize why it occurs, without need for calculus:
Picture a point inside the spherical shell. It doesn’t need to be in the center; it can be anywhere. Now draw a pair of coaxial cones (of any opening angle, and with the axis in any direction), with the vertex at the point you chose. One end of the cone will slice off one piece of the sphere, and the other end of the cone will slice off another piece of the sphere. Which sphere-piece of mass pulls harder on the point? Well, one piece is going to be closer than the other, but that piece is also going to be smaller. In fact, the size of each piece is going to be proportional to the square of the distance from the point. This will cancel out the 1/r[sup]2[/sup] in Newton’s law, so both pieces of the sphere will pull equally hard on our point, but in opposite directions. Well, the whole sphere can be broken up into these pairs of conic sections, and the contribution of each pair of sections is also going to be zero. So the total force from the entire sphere will be zero.
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I think I get the 0g inside the hollow sphere. Start dead center, fire up your MMU and head for a point on the sphere. Intuition says the closer you get to that point the more the inverse square would attract you to that point, but in reality the closer you get more and more of the vectors of the sphere are behind you, canceling out the increase. Thanks for the conic visualization, however, I was very fuzzy about how the cancellation handled the huge inverse square spike of contact on the inner surface. That must be when the cone sweeps through the rapidly converging sphere just before you touch, at which point a few square feet around your boots cancels out nearly the rest of the whole sphere.
I even think I see why the sphere won’t scale down to a torus. Approaching a point in a sphere the geometry never changes but approaching a point on a torus the apparent thickness increases, changing the vectors. You only get the self canceling balance when completely surrounded.
What I’ve been trying to find out is if gravity gradients are also canceled out, if the magnitude of the gravitational field is a factor.
Whether there is any difference between being in a 10 foot hollow at the center of the Earth and being in a 10 foot hollow in the center of a sun sized sphere of neutronium.
I think it must work for all spheres - regardless of mass, size or wall thickness - otherwise there would have to be some point at which things suddenly stop working one way and start working a different way.
Irrelevant as long as the field is exactly balanced in all directions.
No, it would read 0. Until you understand and accept that basic fact which everybody is trying to explain to you, you can’t get any farther.
No. See above. The placement of the magnets is symmetrical by definition in this thought experiment. Just as the distribution of forces due to gravity is symmetrical in a sphere by definition.
And that is true for all spheres with symmetrical distribution of masses, no matter the size.
After thinking about it more I maybe see what is causing the confusion.
0g no like you feeling him!
If the gravitational field is zero everywhere inside the sphere, then it’s constant, so all of the derivatives of the field (i.e., the tidal forces) would also be zero everywhere inside.
I think that one question implied in that quote was: even though the net “acceleration” due to gravity is zero, are there any other implications from being at the exact center of an earth-mass sphere versus a neutron-star-mass sphere? (i.e. are you time-dilated differently?)
My guess would be no, they exactly cancel themselves out for all purposes, even time-dilation, since there is no imputed acceleration, unlike standing on an planet where the planet has to “accelerate” you away from the direction gravity wants to take you in.
You would get time dilation, since that depends on the potential, not the field. Of course, you’d never know it unless you compared to a clock outside the mass. Any local experiment would show no evidence of the mass around you.