Right. And to have a vector, at least a Euclidean one, you have to have a starting point.
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.
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.
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.
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?
I think I’ve got it.
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.
… 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).
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 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.
Definition of apparent from Google: “2. seeming real or true, but not necessarily so”.
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 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. And yes, the information provided by the direction and magnitude of that force vector is useful.
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.
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?
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.
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
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.)