Okay, I think I see your point, thank you. I’ve 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 that’s presuming the Earth’s mass if symmetrically distributed, which of course it isn’t. I’ve no idea on a planet-wide scale how proportionally big the symmetric discrepancies are.)
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.
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.
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’?
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.
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 “target” of the vector. The “source” 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 that’s “observing down”.
I’m trying to refine my understanding of the concept, so any feedback is appreciated.
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.
We need to deal with the gravitational pull of multiple object today as well. Send a probe to the Moon and you will need to know in what volume the Earth’s gravity dominate, where the Moon’s gravity dominates, and what the sum of the two is in the area in between. You calculate that, and you get a vector pointing from the probe’s center of mass in the direction it is being pulled. Your probe’s center of mass is the only point that is relevant to understand the effect of the vector.
Taking this situation and calculating a “center of gravity” based on treating the Earth-Moon system as a point source (or perfect sphere) doesn’t add anything helpful. You already have the direction of the force, and the distance to the fictional “center of gravity” is irrelevant. It’s an entirely useless additional calculation that you have to do on top of calculating the ever changing direction and strength of the total forces you experience.
My original question was regarding Earth and gravity, but for a visual aid how do you describe the rings of Saturn. That to me is what I would call a plane (potato/POtato), when I was trying to visualize the gravitic effect for Earth. So I get that when your on any point of the planet, or close to it, your constantly having gravity act on you. But there has to be a gradient for leaving Earth orbit when heading outbound.
Yes, the acceleration due to gravity at any point above the Earth is described by GM/r^2 where G is the gravitational constant, M is the mass of the earth and r is the distance to the center of the planet.
Like with any field it’s often visualized with field lines. Lines closer together means a stronger field. And when drawn on a flat piece of paper that field is drawn in a plane, but it’s a volume, not a plane.
The further from the Earth and the Moon an object is, the more that the effect of the Earth and the Moon on that object can be approximated by the gravity field of an imaginary point with the mass of both the Earth and the Moon, at the center of mass of the Earth and the Moon (which is, of course, quite near the center of Earth). But 1) at closer distances, it’s easier to deal with the Earth and Moon’s effect separately, and 2) (I suspect) by the time that an object is far enough that using the single object approximation is useful, the gravity of Sun and other planets are also in play, so why bother with the approximation.
Here is a quick and crude illustration of possible orbits–an “orbit” is really an object attempting to fall to the Earth’s center of gravity, but missing. So no “off-center” orbits.
Darren Garrison, I think you might have figured out what the OP was actually trying to ask, and answered it correctly. Good job (pending confirmation from Dacien, that is).
Of course there’s a source. With regards to binary stars, there are two sources. No there’s no gravity being generated from the empty space that’s the centre of gravity between the two stars. The dimensionless point that’s the centre of mass of a planet isn’t generating any gravity either. It’s the entire planet that’s doing so. With a planet, you can describe the gravitational force between the planet and an external mass as a force vector starting from the centre of the planet and pointing towards the external mass. The magnitude and direction of that force vector is going to be dependent on the location of the external mass, but the source point of the force vector will be fixed.
If the source of gravity is a non-spherical body, or a group of spherical bodies, then determining the force vector that describes the gravitational force is more complicated if the external mass and the source(s) of gravity are sufficiently close. At a distance, centre of mass and centre of gravity of the source(s) will be virtually identical. At close range, the location of the external mass is going to determine not only the magnitude and direction of that force vector, but also the calculated source point. That calculated source point will be dependent on the position and masses of the non-spherical body/group of spherical bodies. And it won’t necessarily be the same as the centre of mass.
Why is knowing the source point of that that calculated force vector useful? It describes the motion of the external mass towards the non-spherical body/group of spherical bodies. It represents an apparent force that is actually the cumulative force of all sources of gravity.
Hi Declan, hope you’ve got the gist of the (pretty good) answers so far. As Darren said, gravity points down, to the center of the earth. You can think of the earth as a sphere, even though it’s a little bit “squashed” by its own rotation, but close enough. In orbital mechanics (my area of PhD focus), you can think of a spherical mass as having all its mass concentrated at one point, at the center. And gravity gets less and less the further you are away from this point. Earth is almost a sphere, so this is pretty much how it works for Earth. So if you are a certain distance from the earth, no matter which direction, you should feel the same attraction due to gravity. (I am not talking about other effects that cause earth gravity to not exactly resemble a point mass, but those effects are small.)
Hey Chronos, you’ve just defined center of mass, not gravity. Center of gravity is where the net attracting force on an object is felt. Depending on the orientation of your barbell, the center of gravity might actually be behind the center of mass, which would eventually (and slowly) cause the barbell to rotate around its center of mass.
