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?
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.
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.
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)?
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.
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.
Just think! How easy it would be to launch a rocket if we repealed the law of gravity over just the Kennedy Space Center.
But then, we wouldn’t even need a rocket. Just get the astronauts to walk onto the launch pad, and WAHEEEEEEE!
Oh, and since nobody else has mentioned it, what’s a “one-dimensional plane”?
A line in the sky.
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.
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…
A plane (aircraft) in Spain lies mainly in the plane (2D space) of the plain (geographical feature).
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.
Tris
So there. (see what I did, there?)
Certainly not hang anybody, that’s for sure. They just float there.
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.
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.
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).
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.
- 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.
- 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.
- 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.
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.
Just floated around. It made implementing decisions awfully difficult. How they managed to get together to pass it remains a mystery.