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#51




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Taking this situation and calculating a "center of gravity" based on treating the EarthMoon 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. 
#52




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__________________
What would Bugs Bunny say 
#53




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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 
#54




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__________________
What would Bugs Bunny say 


#55




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#57




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).

#58




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If the source of gravity is a nonspherical 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 nonspherical 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 nonspherical body/group of spherical bodies. It represents an apparent force that is actually the cumulative force of all sources of gravity. 
#59




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#60




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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. 
#61




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 
#62




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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. 
#63




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#64




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. 


#65




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#66




I think I've got it.

#67




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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. 
#68




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Gravity does not pull you towards a specific point in space. Gravity pulls you in a specific direction. 
#69




... 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; 04182019 at 08:55 AM. 


#70




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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 xaxis 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 voilà, 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. 
#71




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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. Are you saying that vector fields are a more accurate way of mapping the gravitational attraction, especially for a nonspherical 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. And yes, the information provided by the direction and magnitude of that force vector is useful. 
#72




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Last edited by scr4; 04182019 at 12:00 PM. 
#73




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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 
#74




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. 


#75




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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; 09102019 at 10:36 AM. 
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