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Old 04-14-2019, 04:44 PM
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Gravity, its the law


So I was thinking , in relation to the planet does gravity extend in a one dimentional plane at the equator or is that overly simplistic and its several circumferences from pole to pole, but smaller as you visualize the poles.
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Old 04-14-2019, 04:50 PM
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Your question makes no sense. The force of gravity at any point at or above the Earth's surface (ignoring effect of unevenness like mountains and valleys) points toward the center of mass and changes inversely with the square of the distance from that center of mass.
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Old 04-14-2019, 05:00 PM
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Gravity (at a sufficient distance) acts like it is originating from a point located at the (aptly named) center of gravity of the object. Anything that falls will attempt to fall towards that mathematical point.
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Old 04-14-2019, 06:22 PM
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Gravity (at a sufficient distance) acts like it is originating from a point located at the (aptly named) center of gravity of the object. Anything that falls will attempt to fall towards that mathematical point.

So from space, would that be the equator. Basically for purposes of simplifying it for me, if a rock traveling at no real great speed will be captured by Earth gravity and fall towards the center mass which would be equatorial ?
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Old 04-14-2019, 06:34 PM
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It's not on the equator. Stand up and look towards your feet: the point is down that way, at the centre of the earth.
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Old 04-14-2019, 06:53 PM
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So from space, would that be the equator. Basically for purposes of simplifying it for me, if a rock traveling at no real great speed will be captured by Earth gravity and fall towards the center mass which would be equatorial ?

No. Gravity works exactly the same way on the surface of Earth as it does in space. Newton did not notice an apple depart a tree and fly off towards the equator. Pick up an object near you, then drop it and observe it not heading towards the equator. Center of mass means just that, the center of mass, which for spherical objects is almost always going to be the geometric center. The equator is an imaginary line defined by the rotational axis of the Earth and points on the equator are no more significant than any other point on the surface of the Earth in respect to gravity--and points on the surface of the Earth are as far from the center of mass as you can get without leaving the surface.


tl;dr:All that gravity knows about is "down."

Last edited by Darren Garrison; 04-14-2019 at 06:55 PM.
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Old 04-14-2019, 07:06 PM
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No. Gravity works exactly the same way on the surface of Earth as it does in space. Newton did not notice an apple depart a tree and fly off towards the equator. Pick up an object near you, then drop it and observe it not heading towards the equator. Center of mass means just that, the center of mass, which for spherical objects is almost always going to be the geometric center. The equator is an imaginary line defined by the rotational axis of the Earth and points on the equator are no more significant than any other point on the surface of the Earth in respect to gravity--and points on the surface of the Earth are as far from the center of mass as you can get without leaving the surface.


tl;dr:All that gravity knows about is "down."
Thanks Darren
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Old 04-14-2019, 07:59 PM
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The only reason that the Earth has poles and that there is an equator is because of the spin of the Earth. However, the spin has nothing to do with the mass of the year. The Earth would have the same mass and hence the same gravity if there were no spin.
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Old 04-14-2019, 08:15 PM
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However, the spin has nothing to do with the mass of the year.

That isn't what time-space means.
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Old 04-15-2019, 01:31 AM
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Since these questions about gravity are factual in nature, this is better suited to GQ.

Moving thread from MPSIMS to GQ.
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Old 04-15-2019, 02:37 AM
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Gravity is a force that acts between all matter, all of the time.

An apple on your table is receiving a gravitational tug from the particles in your body, and a tug from the particles in your car parked outside. And you can imagine all parts of the earth are pulling on the apple; not just straight down but also north, south, east and west. When you add up all the little tugs, bearing in mind there's a trillion cubic kilometers of rock approximately "down", the resultant force is towards the Earth's center (of gravity).
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Old 04-15-2019, 07:08 AM
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What did we do before the Law of Gravity was passed?
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Old 04-15-2019, 07:12 AM
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To the OP, as others have explained, objects are pulled toward the center of mass. The equator is an imaginary line that we, humans, drew up to separate the top half of the planet from the bottom half. If an object were on a line extended from the equator (think like Saturn's rings), it would be pulled toward the equator; however, it if was over any other point on the planet, it would be pulled to the nearest point of planet & not towards the equator.
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Old 04-15-2019, 07:14 AM
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What did we do before the Law of Gravity was passed?
Oh, it was chaos! France and England couldn't agree on what was down, and any upstarts knight could change the direction things fell at a whim. Fortunately it was passed in 1107 and worked its magic quietly until it was discovered by Newton and used to develop modern science.
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Old 04-15-2019, 08:11 AM
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Actually, because the earth is not a perfect sphere, gravity is not towards the center of mass. In fact, it is very slightly offset towards the equator, because the earth has an equatorial bulge due to its rotation. This is enough to affect the orbits of satellites - namely, it causes nodal precession.
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Old 04-15-2019, 08:16 AM
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Well, when told in amazement that he was defying the law of gravity, he coyly responded with a twinkle in his eye, "I never studied law."
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Old 04-15-2019, 08:18 AM
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Actually, because the earth is not a perfect sphere, gravity is not towards the center of mass.

No, it is towards the center of mass--it is just that the center of mass may not precisely match the geometric center.
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Old 04-15-2019, 08:23 AM
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The general piece has been discussed but...you knew there was a but...it is possible to notice gravitation attraction that is not directed radially.

Being near a sufficiently large mass, say a mountain, can induce a small localized change in the path a particle would take. The Schiehallion experiment was the original experiment where the deflection due to a local mass was used to measure the density of the Earth.
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Old 04-15-2019, 08:29 AM
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No, it is towards the center of mass--it is just that the center of mass may not precisely match the geometric center.
No, it's not. See the link I posted. For a rotating body like the earth, the gravity is slightly towards the equator. The "center of mass" thing is only true for spherically symmetric objects.
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Old 04-15-2019, 08:34 AM
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The distance between Earth's center of mass and center of gravity is a couple of miles or less.
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Old 04-15-2019, 10:52 AM
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For someone standing on the Earth’s surface, isn’t the centre of gravitational influence closer than the Earth’s centre of mass? So the matter immediately underneath my feet has more gravitational effect on me than the matter on the other side of the Earth? Likewise, viewing myself at the top of the world (of course), doesn’t the hemisphere of my side of the planet exert a stronger gravitational influence on me than the far hemisphere?
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Old 04-15-2019, 11:10 AM
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Yes you may think of each little bit of Earth as having its own gravitational attraction on you. They vary according to the inverse square law (Newton -- ignoring Einstein). Those to your left pull you to the left and down, those to your right pull you to your right and down. For s symmetrical object, the left and right pulls cancel. Using Calculus also a Newton invention (at least partially), you can show that the total gravitational effect of a spherical object is the same as if all the mass were located at the center.

There is one exception to this. If you are inside the sphere, the net gravitational attraction is only that of all the mass located closer to the center than you. That's a scalar there not a vector. You can ignore the mass located beneath your feet on the other side of the earth that is farther from the center than you are. All those inside the Earth stories, like ERB's Pellucidar, get this wrong. You would not be attracted to the inside "surface" in a hollow earth unless the enclosure is off center.
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Old 04-15-2019, 11:12 AM
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For someone standing on the Earthís surface, isnít the centre of gravitational influence closer than the Earthís centre of mass? So the matter immediately underneath my feet has more gravitational effect on me than the matter on the other side of the Earth? Likewise, viewing myself at the top of the world (of course), doesnít the hemisphere of my side of the planet exert a stronger gravitational influence on me than the far hemisphere?
The Earth is not exactly spherically symmetric, but, as explained above, that is an excellent first approximation, so you are still being pulled more or less towards the center of mass. As for matter close to you versus farther away, the gravitational force is inversely proportional to the square of the distance between you and the mass so, yes, the chunk of ground you are standing on exerts a greater force than a same-sized chunk on the other side of the planet.

As for a bulgy planet, imagine for a moment standing on a dense but thin flat disc, somewhere away from the center. You would still be pulled downwards to some extent, so in that case the direction of gravity is not directly towards the centre of mass, with ensuing effects that scr4 mentioned.
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Old 04-15-2019, 11:20 AM
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It is a basic result of calculus that the direction of the gradient (gravity in this case) is perpendicular to the level surface. If the earth were a perfect ellipsoid, the gravitational force on a point north of the equator would point to some place south of the center of the earth and vice versa in the southern hemisphere. Draw an ellipse, pick a point on it and draw the perpendicular to see where, on the minor axis, gravity points. What happens as you approach a pole (an end of the minor axis)?
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Old 04-15-2019, 12:30 PM
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The Earth is not exactly spherically symmetric, but, as explained above, that is an excellent first approximation, so you are still being pulled more or less towards the center of mass. As for matter close to you versus farther away, the gravitational force is inversely proportional to the square of the distance between you and the mass so, yes, the chunk of ground you are standing on exerts a greater force than a same-sized chunk on the other side of the planet.

As for a bulgy planet, imagine for a moment standing on a dense but thin flat disc, somewhere away from the center. You would still be pulled downwards to some extent, so in that case the direction of gravity is not directly towards the centre of mass, with ensuing effects that scr4 mentioned.
The example I would offer is if someone was standing on one of the balls of a circus barbell shaped object (two balls connected by a shaft). The centre of mass would be in the centre of the shaft connecting the two balls. But the mass of the ball the person was standing on would have a stronger gravitational effect than the other ball. Therefore the centre of gravity, relative to the person, would be closer to his position than the centre of mass. In this example, gravitational down is not pointing at the centre of mass.

The same applies for a person standing on a sphere. The centre of gravity is much closer than the centre of mass. However, in this case gravitational down is the same. But if the person starts moving away from the sphere, as his distance from the sphere increases, the centre of gravity, relative to his position, moves towards the centre of mass.
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Old 04-15-2019, 03:39 PM
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The center of gravity of a barbell is still right in the middle. The center of gravity is not defined as the point to which things are attracted.
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Old 04-15-2019, 03:49 PM
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What did we do before the Law of Gravity was passed?
Just think! How easy it would be to launch a rocket if we repealed the law of gravity over just the Kennedy Space Center.
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Old 04-15-2019, 06:41 PM
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But then, we wouldn't even need a rocket. Just get the astronauts to walk onto the launch pad, and WAHEEEEEEE!
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Old 04-15-2019, 06:57 PM
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Oh, and since nobody else has mentioned it, what's a "one-dimensional plane"?
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Old 04-15-2019, 08:46 PM
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Oh, and since nobody else has mentioned it, what's a "one-dimensional plane"?
A line in the sky.
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Old 04-15-2019, 08:51 PM
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I think--taking another look at the OP--the question is about orbits, i.e. do they always circle around the circumference of the Earth or can the orbit be around a different cross-section, such as something orbiting the Tropic of Cancer. The answer to that is that an orbit must be around the circumference of the Earth, but it can be tilted at any angle relative to the equator.
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Old 04-15-2019, 09:05 PM
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So I was thinking , in relation to the planet does gravity extend in a one dimentional plane at the equator or is that overly simplistic and its several circumferences from pole to pole, but smaller as you visualize the poles.
A plain (not a plane) is, by definition, two dimensional, usually being 3 connected points. IIRC, if you connect two points that's one dimensional. Just saying. Gravity acts from the center of mass and can be thought of, in macro terms, as a point source, though I believe that the reality is that gravity varies across the globe, depending on the variable mass that mak sup the planet. So, more massive parts of the planet exert slightly more gravity than less dense parts. It's not something that is noticeable, but it is detectable. i'm not sure what any of this has to do with yoru original question though as I seem to have lost my Trane (not train) of thought...
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Old 04-15-2019, 09:33 PM
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A plane (aircraft) in Spain lies mainly in the plane (2D space) of the plain (geographical feature).
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Old 04-15-2019, 09:44 PM
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A point has no dimension.
If I have another point, and I am sure it's not the same point, I now have a dimension. Toward one of the points, from the other. And toward the other, from the one. They are both the same dimension, but in the opposite direction.

Of course if I am sure it isn't the same point, there must be a distance between them. That's why I chose to call it a dimension. Oddly enough, I now have an infinite number of points, between the two points, and beyond each in the direction away from the other. I can now assign what I call a distance, and can assign values to those distances.

The logical consequences are not characteristics of the universe, they are characteristics of my model of the universe, which I call geometry. I call it that so that you will think it does have characteristics of the world, which I can now measure.

Now I tell you I have another point, which I assure you is not the same point as either of the original two points. I also assure you that it is not any of the points I got from pointing from one of the original pointing toward the second of the original two. Since it isn't one of the original two, or any of the ones in the original dimension. I now have a second dimension.

I can keep this up. However, if I do, then each of the dimensions I create will have another dimension. Duration. If I don't have duration, it all goes away.

It turns out, I needed that dimension at the beginning. Even more oddly than all the above, I didn't need that dimension before that.

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Old 04-15-2019, 10:06 PM
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What did we do before the Law of Gravity was passed?
Certainly not hang anybody, that's for sure. They just float there.
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Old 04-16-2019, 04:38 AM
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The center of gravity of a barbell is still right in the middle. The center of gravity is not defined as the point to which things are attracted.
Then what is the correct term for the point within a body such as a sphere where gravitational influence, relative to a point on the surface of a body, is the same as the average gravitational influence of the entire body? Would it be the mean point of gravity? If it was a point between two bodies, it would be the point of gravitational equilibrium. I believe there would also be a point of gravitational equilibrium within a body, and that point would be the same as the centre of mass, but that’s not going to be a concept that applies to a point on the surface.

Suppose I’m standing at the North Pole and I’m calculating the gravitational effect on me of one ton of mass that’s immediately beneath me. It will be greater than that from one ton of mass at the South Pole. And the difference is not going to be linear, but proportional to the square of the Earth’s diameter. Somewhere in between the poles, there’s going to be a point where the gravitational effect on me of a ton of mass is equal to the average gravitational effect on me of all tons of mass of the Earth. Similarly, the gravitational effect on me (still standing at the North Pole) of the northern hemisphere is going to be proportionally greater than the gravitational effect of the southern hemisphere. Therefore that “average ton” point is going to be in the northern hemisphere.

I’m guessing the point is going to be (1 – 1/SQRT(2)) * Earth’s diameter directly south of the north pole, but I really don’t remember calculus well enough to back up that guess.

Last edited by Wrenching Spanners; 04-16-2019 at 04:39 AM.
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Old 04-16-2019, 06:43 AM
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Then what is the correct term for the point within a body such as a sphere where gravitational influence, relative to a point on the surface of a body, is the same as the average gravitational influence of the entire body?
I'm not sure why there would be any term at all for such a concept, nor even how one could rigorously define it (the direction to that point would be clear, but how far away?). If one wants to define the "center" of an object, one wants to define it for the entire object, not a separate center for every point of the object.

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Suppose I’m standing at the North Pole and I’m calculating the gravitational effect on me of one ton of mass that’s immediately beneath me. It will be greater than that from one ton of mass at the South Pole. And the difference is not going to be linear, but proportional to the square of the Earth’s diameter. Somewhere in between the poles, there’s going to be a point where the gravitational effect on me of a ton of mass is equal to the average gravitational effect on me of all tons of mass of the Earth. Similarly, the gravitational effect on me (still standing at the North Pole) of the northern hemisphere is going to be proportionally greater than the gravitational effect of the southern hemisphere. Therefore that “average ton” point is going to be in the northern hemisphere.
That "average point" is going to be the center of the Earth. The ton of rock right under your feet is closer than other tons, but there are more tons further away than close by. It was for solving this problem specifically that Newton invented calculus (though it's useful for many other problems as well).
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Old 04-16-2019, 08:06 AM
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That "average point" is going to be the center of the Earth. The ton of rock right under your feet is closer than other tons, but there are more tons further away than close by.
I’m still disagreeing with you. Let’s try a different model. Objects A, B, and C each have a mass of 1,000,000,000 kg and are positioned along an axis such that object A is at point 0, object B is 1000 metres from point 0 and object C is 2000 metres from point 0. For the sake of simplicity, let’s assume that objects B and C are supernaturally fixed in position, while object A is free to move.

Using Newton's law of universal gravitation, I can calculate the force pulling unfixed object A to fixed object B is 66.7 kg m / s^2 . The force pulling object A to fixed object C is 16.7 kg m / s^2 . The cumulative force is 83.4 kg m / s^2 .

The centre of mass of objects B and C is going to be 1500 metres from point 0. That’s not going to change whether I treat them as two distinct objects or as a set of objects. However, if I treat B and C as a set of objects, the “centre of gravity” of the set, relative to object A, is going to be different. The mass of the set is 2,000,000,000 kg. The force on object A is the same, 83.4 kg m / s^2 . But recalculating the distance r, the “centre of gravity” of the set is 1264 metres from point 0.

1) If “centre of gravity” isn’t the correct term for that point at 1264 metres from point 0, then I’m curious if anyone knows the correct term.
2) The mass of object A isn’t relevant to the calculation of the objects B and C “centre of gravity”. If I change object A’s mass to 1 kg, the calculation still returns 1264 metres from point 0. And the need for having objects B and C in a fixed position, at least before movement starts, virtually disappears.
3) I don’t see why this concept of the “centre of gravity” being in a different place than the centre of mass would be any different for someone on the surface of a planet than for three objects lined up along an axis. Note that I'm not discussing the planet as an independent object, I'm discussing the gravitational effects of the planets mass across distance relative to the position of the person on the surface.

Last edited by Wrenching Spanners; 04-16-2019 at 08:10 AM.
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Old 04-16-2019, 08:52 AM
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Iím still disagreeing with you. Letís try a different model. Objects A, B, and C each have a mass of 1,000,000,000 kg and are positioned along an axis such that object A is at point 0, object B is 1000 metres from point 0 and object C is 2000 metres from point 0. For the sake of simplicity, letís assume that objects B and C are supernaturally fixed in position, while object A is free to move.
That's an invalid model to examine Chronos' statement, which applies to the sphere of the Earth. For a very rough idea of why it doesn't work, think of the Earth as being two hemispheres where you are standing at the "top" of one of them. You're going to be pulled "straight down", but for the average point in the closer hemisphere a larger proportion of the force will be perpendicular to this pull and this perpendicular component will be cancelled out by the equivalent pull in the other direction at the other side of the that hemisphere.

You have the same effect for the further hemisphere, but the proportion between gravitational pull that adds up and gravitational pull that cancels out is different, and it turns out that for a perfect sphere it sums up to exactly the opposite of the effect of increasing distance.
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Old 04-16-2019, 08:55 AM
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What did we do before the Law of Gravity was passed?

Just floated around. It made implementing decisions awfully difficult. How they managed to get together to pass it remains a mystery.
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Old 04-16-2019, 10:28 AM
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That's an invalid model to examine Chronos' statement, which applies to the sphere of the Earth. For a very rough idea of why it doesn't work, think of the Earth as being two hemispheres where you are standing at the "top" of one of them. You're going to be pulled "straight down", but for the average point in the closer hemisphere a larger proportion of the force will be perpendicular to this pull and this perpendicular component will be cancelled out by the equivalent pull in the other direction at the other side of the that hemisphere.

You have the same effect for the further hemisphere, but the proportion between gravitational pull that adds up and gravitational pull that cancels out is different, and it turns out that for a perfect sphere it sums up to exactly the opposite of the effect of increasing distance.
Okay, I think I see your point, thank you. Iíve been assuming that because all of the horizontal components of the gravitational force vectors cancel out, I can treat a sphere as a column. However, the vertical components of the vectors are greater in the southern hemisphere than the northern hemisphere. So suppose I go 1000km vertically south from the North Pole, and then horizontally to the surface to point A. Someone else goes 1000km vertically north from the South Pole, and then horizontally to the surface to point B. The direct gravitational force will be greater from a mass at point A than a mass at point B. However, the vertical ratio (sine?) of the angle from point B will be greater than that from point A. So calculate the vertical component of the force vector from either point A or point B and they should be the same, correct?

Weirdly, since a thin column seems to have a centre of gravity north of the centre of mass (relative to a mass at the north pole) and a sphere has a centre of gravity at the centre of mass, that seems to imply that for a thin ellipse the centre of gravity is north of the centre of mass. Does that mean that for a wide ellipse, the centre of gravity is below the centre of mass? That would imply that for the Earth, which is a slightly wide ellipse, mathematically at least, relative to a mass at the North Pole, the gravitational centre of mass is south of the geometric centre. (Obviously thatís presuming the Earthís mass if symmetrically distributed, which of course it isnít. Iíve no idea on a planet-wide scale how proportionally big the symmetric discrepancies are.)
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Old 04-16-2019, 10:34 AM
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Originally Posted by Wrenching Spanners View Post
I

1) If ďcentre of gravityĒ isnít the correct term for that point at 1264 metres from point 0, then Iím curious if anyone knows the correct term.
.
There is no fixed point that acts as the "center of gravity" because the direction of gravity depends on where you are. If you draw a bunch of arrows (vectors) showing the gravitational field around a massive object , they don't all point to the same spot, unless the object is a point mass or is spherically symmetric.

If you are trying to define a "center of gravity" that is not a fixed point but dependent on where you are - why?? Gravity doesn't pull towards specific point. It pulls in a specific direction. The way to describe that is a vector, not a location.
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Old 04-16-2019, 11:01 AM
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Originally Posted by Wrenching Spanners View Post
Weirdly, since a thin column seems to have a centre of gravity north of the centre of mass (relative to a mass at the north pole) and a sphere has a centre of gravity at the centre of mass, that seems to imply that for a thin ellipse the centre of gravity is north of the centre of mass. Does that mean that for a wide ellipse, the centre of gravity is below the centre of mass? That would imply that for the Earth, which is a slightly wide ellipse, mathematically at least, relative to a mass at the North Pole, the gravitational centre of mass is south of the geometric centre. (Obviously thatís presuming the Earthís mass if symmetrically distributed, which of course it isnít. Iíve no idea on a planet-wide scale how proportionally big the symmetric discrepancies are.)
Your "center of gravity" comes from applying the laws of gravity, which apply to point masses and spherical objects, to non-spherical non-point masses. The resulting "point" doesn't have a meaning, other than it being on the line of the vector of gravity in that specific location, and it being the spot where a point mass of the same mass would give the same force. But that's not really a useful concept.
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Old 04-16-2019, 11:14 AM
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OK, so for irregularly shaped masses, the gravity vector at given points on the surface won't always point toward the center of mass of the object: take as an example Ultima Thule, where there are two lobes. If you mapped the orientation of the gravity vector at all points on the surface, the vectors would not all aim toward the same point. As you move away from the surface, the vectors converge so that at infinite distance they all point toward the center of mass. So what terminology is used to describe the direction of gravity at a point on the surface? I mean, it's 'down' by definition, but where is 'down' pointing? It's not toward the center of mass. Do we just call it 'downwards'?
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Old 04-16-2019, 11:19 AM
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There is a very slight tug in the direction of the equator due to your centrifugal momentum on the rotating earth. The perceived accelleration will be perpendicular to the axis of the earth, so at the equator it will tug you directly out into space, and but at other latitudes, it will be tilted towards equator. But this so small as to not be noticeable.

At the equator it is about .0053m/sec^2 directly out into space, the largest pull in the direction of the equator will be .0027m/sec^2 at a latitude of 45 degrees (just south of Portland Oregon). This compares to the gravitational acceleration which on the surface of the earth is 9.8m/sec^2.

Last edited by Buck Godot; 04-16-2019 at 11:19 AM.
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Old 04-16-2019, 11:29 AM
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Originally Posted by scr4 View Post
There is no fixed point that acts as the "center of gravity" because the direction of gravity depends on where you are. If you draw a bunch of arrows (vectors) showing the gravitational field around a massive object , they don't all point to the same spot, unless the object is a point mass or is spherically symmetric.

If you are trying to define a "center of gravity" that is not a fixed point but dependent on where you are - why?? Gravity doesn't pull towards specific point. It pulls in a specific direction. The way to describe that is a vector, not a location.
As I understand it, gravity is a property of mass and pulls in every direction. Gravitational force is the result of two or more masses/gravities pulling at each other. The vector of that force is going to be relative to the position of the mass that is the ďtargetĒ of the vector. The ďsourceĒ of the vector is going to be the centre of gravity of the other objects. For sufficiently distant objects (and I now understand individual symmetrical spheres) or sets of objects, the centre of gravity and the centre of mass are going to be the same. However, for close non-spherical objects, centre of gravity and centre of mass will be different points. And again, that centre of gravity is going to be relative to the position of the mass thatís ďobserving downĒ.

Iím trying to refine my understanding of the concept, so any feedback is appreciated.
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Old 04-16-2019, 11:34 AM
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Originally Posted by naita View Post
Your "center of gravity" comes from applying the laws of gravity, which apply to point masses and spherical objects, to non-spherical non-point masses. The resulting "point" doesn't have a meaning, other than it being on the line of the vector of gravity in that specific location, and it being the spot where a point mass of the same mass would give the same force. But that's not really a useful concept.
Well it's only in science fiction that space agencies are sending probes to binary stars. However, if science ever catches up to science fiction, then yes, I believe knowing the source point of the gravitational source vector will be a useful concept.
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Old 04-16-2019, 11:41 AM
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Missed edit window:

*gravitational force vector
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Old 04-16-2019, 12:02 PM
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Originally Posted by Wrenching Spanners View Post
Well it's only in science fiction that space agencies are sending probes to binary stars. However, if science ever catches up to science fiction, then yes, I believe knowing the source point of the gravitational source vector will be a useful concept.
There is no source! The direction of gravity isn't towards a specific point. It depends on where you are. There is no location you can define on an object and say the gravity of the object is always towards that point.

Last edited by scr4; 04-16-2019 at 12:03 PM.
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Old 04-16-2019, 12:43 PM
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Originally Posted by Jasmine View Post
Well, when told in amazement that he was defying the law of gravity, he coyly responded with a twinkle in his eye, "I never studied law."
But of course, ignorance of the law is no excuse.
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