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  #51  
Old 04-16-2019, 04:20 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.
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.
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Old 04-16-2019, 04:27 PM
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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...
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.
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  #53  
Old 04-16-2019, 04:40 PM
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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.

https://en.wikipedia.org/wiki/File:G...ield_lines.svg
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Old 04-16-2019, 04:45 PM
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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.

https://en.wikipedia.org/wiki/File:G...ield_lines.svg
Thank you Naita
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Old 04-16-2019, 04:49 PM
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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.
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.
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Old 04-16-2019, 06:30 PM
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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.
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Old 04-17-2019, 08:28 AM
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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).
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Old 04-17-2019, 11:18 AM
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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.
Of course thereís a source. With regards to binary stars, there are two sources. No thereís no gravity being generated from the empty space thatís the centre of gravity between the two stars. The dimensionless point thatís the centre of mass of a planet isnít generating any gravity either. Itís the entire planet thatís 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 wonít 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.
  #59  
Old 04-17-2019, 11:40 AM
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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.
The force vector does that all by itself.
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Old 04-17-2019, 11:44 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.
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.
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Old 04-17-2019, 12:10 PM
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The force vector does that all by itself.
Right. And to have a vector, at least a Euclidean one, you have to have a starting point.

https://en.wikipedia.org/wiki/Euclidean_vector
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Old 04-17-2019, 01:03 PM
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Right. And to have a vector, at least a Euclidean one, you have to have a starting point.

https://en.wikipedia.org/wiki/Euclidean_vector
If you are putting a satellite in orbit around a non-spherical object, you need to know the gravitational force vector at every point around the object. So that's the starting point of the vector.

The vector field tells you for every point in space, "if the satellite is here, gravity is this strong and in this direction." That's exactly what you need to know to plot the trajectory of your satellite. It makes no sense to define "if the satellite is here, gravity appears to come from the this spot." Because that spot moves as soon as the satellite moves.
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Old 04-17-2019, 01:12 PM
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Of course thereís a source. With regards to binary stars, there are two sources. No thereís no gravity being generated from the empty space thatís the centre of gravity between the two stars. The dimensionless point thatís the centre of mass of a planet isnít generating any gravity either. Itís the entire planet thatís 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 wonít necessarily be the same as the centre of mass.
Yes, we agree on all that. My point is, the conventional and useful way to express this information is a vector field - ie by defining the total gravitational force (or acceleration) vector as a function of position. You COULD define the position of the apparent source of gravity as a function of position, but that has just as many dimensions and still carries less information, because you also need to supply the mass of the apparent source of gravity. It just makes things more complicated.
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Old 04-17-2019, 01:27 PM
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To the extent that a vector has a "starting point", it's just whatever point you happen to be using as an origin for the sake of convenience. The vector from (0,0) to (2,3) is the same vector as the vector from (-5,7) to (-3,10) or from (10,-12) to (12,-9) or whatever, even though they all have different "starting points".

In a vector field such as gravity, each vector will also have a location, but that location is the point where you're making the measurement, not some point remote from it.
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Old 04-17-2019, 03:19 PM
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Right. And to have a vector, at least a Euclidean one, you have to have a starting point.

https://en.wikipedia.org/wiki/Euclidean_vector
The most relevant starting point is the center of mass of the external object. If you draw it from your "center of gravity", how do you known what external point the force with that magnitude and direction applies to?
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Old 04-17-2019, 07:33 PM
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A plane (aircraft) in Spain lies mainly in the plane (2D space) of the plain (geographical feature).
I think I've got it.
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Old 04-18-2019, 05:51 AM
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The most relevant starting point is the center of mass of the external object. If you draw it from your "center of gravity", how do you known what external point the force with that magnitude and direction applies to?
Yes, I've stated many times that the position of the external object is a requirement of determining the force vector.

As to why the "centre of gravity" is the starting point of the force vector, that's (essentially) where the force is coming from. Of course any vector can be reversed by changing the start point and the sign of the magnitude. It's a question of perspective on pulling versus being pulled.
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Old 04-18-2019, 08:42 AM
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Yes, I've stated many times that the position of the external object is a requirement of determining the force vector.

As to why the "centre of gravity" is the starting point of the force vector, that's (essentially) where the force is coming from. Of course any vector can be reversed by changing the start point and the sign of the magnitude. It's a question of perspective on pulling versus being pulled.
You still seem to be under the impression that gravity has an apparent source point. It doesn't. If you measure the gravitational field at a specific point on or above the earth, you can't say "the apparent source of gravity is at these coordinates." All you have is a direction and a magnitude.

Gravity does not pull you towards a specific point in space. Gravity pulls you in a specific direction.
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Old 04-18-2019, 08:54 AM
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... or maybe you think it's useful to represent the gravitational field as being created by a point mass with constant mass, but whose apparent position depends on where you are. You could do that, but it's much more convenient to represent it as a vector field (direction and magnitude of gravity at each point in space).

Last edited by scr4; 04-18-2019 at 08:55 AM.
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Old 04-18-2019, 10:25 AM
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Originally Posted by Wrenching Spanners View Post
Yes, I've stated many times that the position of the external object is a requirement of determining the force vector.

As to why the "centre of gravity" is the starting point of the force vector, that's (essentially) where the force is coming from. Of course any vector can be reversed by changing the start point and the sign of the magnitude. It's a question of perspective on pulling versus being pulled.
Do you have any experience working with vectors?

Try drawing out these two situations illustrating the forces on a mass A approaching two much larger masses M1 and M2, with M2 twice the mass of M1. The initial situation has mass A a distance of 6 from both masses, and the masses are sqrt(72) apart, forming a right isosceles triangle.

For ease of explanation lets put A at the origin in a coordinate system and M1 and M2 at (0, 6) and (6, 0) respectively. The forces from M1 and M2 add up to a gravitational pull G, at an angle 26.57 degrees with respect to the X axis.

Using my approach you can draw this vector starting at A and making it any length you want. Using yours you now have to calculate the distance r you would have to get a pull of G from a mass of M1 and M2 and plot that point along the direction of the vector.

Now due to A having an initial velocity, at some later point in its trajectory, passes through (2, 1), which is on the line between the origin and your original "center of gravity", but by now we are closer to M2. Examining the change in distances we can determine that the gravitational pull is now 1.61 times the initial G, and the angle with the x-axis is now 13.39, so I draw the new vector, from (2, 1) with the correct angle and 1.61 times longer than my first vector.

Et voilŗ, I've drawn a useful representation of the forces on A at the two moments in time, whereas you still have to calculate the distance to this new center of gravity.

And with that extra work you are left with two diagrams that are less informative and intuitive to understand what forces influence A.

And I haven't even gotten into how much easier my approach is for solving these vector problems (at least partially) geometrically rather than algebraicly.
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Old 04-18-2019, 11:11 AM
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You still seem to be under the impression that gravity has an apparent source point. It doesn't. If you measure the gravitational field at a specific point on or above the earth, you can't say "the apparent source of gravity is at these coordinates." All you have is a direction and a magnitude.

Gravity does not pull you towards a specific point in space. Gravity pulls you in a specific direction.
Definition of apparent from Google: ď2. seeming real or true, but not necessarily soĒ.

Gravitational attraction comes from all the mass within a system. Itís accurate to say that an external object is accelerating towards every bit of mass within the system, but everywhere is not a direction. The actual direction of the gravitational attraction, aka the force vector, is from (or towards, depending on perspective) the centre of gravity. That is indeed an apparent point of attraction. The dimensionless point at the centre of the Earth isnít attracting external bodies towards the Earth, the entire mass of the planet is. But the direction of the acceleration is towards the centre of the Earth.

Are you saying that vector fields are a more accurate way of mapping the gravitational attraction, especially for a non-spherical body or group of bodies? I donít disagree with that, especially for an object in motion. But the cumulative effect of the vector fields, at least at a given moment in time, is going to be a force vector. And yes, the information provided by the direction and magnitude of that force vector is useful.
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Old 04-18-2019, 11:57 AM
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Definition of apparent from Google: “2. seeming real or true, but not necessarily so”.
In science, the term "apparent" refers to an observed quantity rather than an intrinsic quality. E.g. "apparent size" is the size of an object as seen by a specific observer, so the apparent size of the Moon as seen from the Earth is about 1/2 degree.

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Gravitational attraction comes from all the mass within a system. It’s accurate to say that an external object is accelerating towards every bit of mass within the system, but everywhere is not a direction. The actual direction of the gravitational attraction, aka the force vector, is from (or towards, depending on perspective) the centre of gravity. That is indeed an apparent point of attraction. The dimensionless point at the centre of the Earth isn’t attracting external bodies towards the Earth, the entire mass of the planet is. But the direction of the acceleration is towards the centre of the Earth.
Yes, but all you can say is it's towards the center of the Earth. It's equally true to say it's towards a specific piece of rock 10 feet below you. Or towards some point in the southern Indian Ocean (if you are in the US), or some star in Sagittarius. So why arbitrarily pick a point in that direction and call it the source of gravity, when you can just as correctly pick any other point in that direction?



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Are you saying that vector fields are a more accurate way of mapping the gravitational attraction, especially for a non-spherical body or group of bodies? I don’t disagree with that, especially for an object in motion. But the cumulative effect of the vector fields, at least at a given moment in time, is going to be a force vector.
I don't think you understand what a vector field is, because the phrase "cumulative effect of the vector fields" doesn't make sense. The vector field is the force vector at each point in space. That vector is already the cumulative effect of all masses. For every point, you measure/calculate the gravitational effect from all masses around it, add it all up and get a vector that represents the cumulative effect of all masses. That's the gravitational field vector at that point.

Last edited by scr4; 04-18-2019 at 12:00 PM.
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Old 04-18-2019, 12:26 PM
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... or maybe you think it's useful to represent the gravitational field as being created by a point mass with constant mass, but whose apparent position depends on where you are. You could do that, but it's much more convenient to represent it as a vector field (direction and magnitude of gravity at each point in space).
My mental model of gravity is as follows: gravity is a vector force, which all other forces aside would cause a previously immobile object to move in a particular direction at a particular speed (magnitude). For every particle in the universe there is a gravitational vector for every other particle in the universe. Distance drastically reduces the magnitude while mass increases magnitude. The gravity from an object is actually the aggregate of all gravitational vectors for every particle in that object; such that the gravity causing this apple to fall down is the aggregate of the gravity every particle in the earth exerts on the apple. There may be a 'heavy' building to the side of the apple but the forces pale in comparison and don't need to be calculated.

Also most particles are moving all the time, so it doesn't make sense to define a center of gravity with exact precision for more than one infinitesimally small moment in time. As particles move in the very next instant, their gravitational vectors and therefore the center of gravity shifts ever so little. This means you can't say "the gravitational center of the earth is about X from the geometric center of the earth" and use that to plot orbital paths. When the spaceship moves 100 kilometers away the gravitational center of the earth has moved also.

Right?

~Max
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Old 04-18-2019, 12:43 PM
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I am pretty sure that the "center of gravity" of a body subject to gravity traditionally refers to an imaginary point (usually) inside the body or nearby such that if suspended from that point there is no resultant torque. It does not refer to a point inside the Earth.

Note that this "center of gravity" may not exist, may not be unique, may depend on the orientation and distance of the body, etc.
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Old 09-10-2019, 10:35 AM
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I am pretty sure that the "center of gravity" of a body subject to gravity traditionally refers to an imaginary point (usually) inside the body or nearby such that if suspended from that point there is no resultant torque. It does not refer to a point inside the Earth.

Note that this "center of gravity" may not exist, may not be unique, may depend on the orientation and distance of the body, etc.
That's pretty much it. Let me add: Center of gravity is defined in terms of the attracting body, that is, the other body or bodies in the system. A standalone mass, without any other bodies nearby that it would be attracted to, doesn't have a center of gravity.

But a mass in orbit does. Then, the center of gravity for the object has a meaning, being the place where the net force from the attracting body (i.e. the Earth) applies. (And, it could cause the mass to rotate about its center of mass, if the net gravity force acts along a line that doesn't pass through the center of mass.)

Last edited by Limmin; 09-10-2019 at 10:36 AM.
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