Course content > Kinds of Forces > Gravitational Forces
Prerequisites
In our discussion of flat-earth gravity we learned that the reason all objects accelerate the same (as long as air resistance doesn't have a significant effect) when they are unsupported --dropped or thrown-- near the surface of the earth is that the force of gravity is proportional to the object's mass. It seems like the earth is pulling down on each piece of the object with the same force, so bigger objects are pulled down with a bigger force. If we can ignore air resistance then the only force on an unsupported object is its weight so
a = Fnet/m = -mg/m = g
Since the force of the earth pulling on the object is mg, then g = W/m has units of N/kg, which conveniently for describing falling bodies, turns out to be the same as the units of acceleration, m/s2.
If we are going to restrict all our considerations to motion near the surface of the earth, the above discussion is sufficient. We can just leave "g" as a fixed number associated with the force of gravity. But if we are going to do rocket science and do interplanetary travel, then we have to take into account the idea that the force of gravity changes with position. We would have to allow g to be a function of position: g(x,y,z), where (x,y,z) represents the point in space where we are measuring a gravitational effect. Since the gravitational force have a direction, we should really write these as vectors:
This is a funny kind of mathematical object -- a vector function of a vector. We call it a field and we refer to the vector function, g, as the gravitational field.
What it means mathematically is that we have three functions: the x component of g, the y component of g, and the z component of g, and they are each functions of the three variables that determine the position we are looking at:
What it means physically is:
If we have a gravitational field it means that if we put an object of mass m at the position labeled by the vector, r, then the object feels a gravitational force equal to mg.
One way of representing this is to put an arrow at every point in space pointing in the direction of the gravitational force an object would feel if it were placed there. This allow us to focus not only on the force that the object feels right now (as we should to calculate its motion right now according to Newton's 0th law), but on the force it would feel when it moved somewhere else.
This is really overkill if we are only going to be talking about gravitational forces near the surface of the earth, since we already know the direction of the force. But this same idea -- the idea of a vector field -- is of great value and importance when we are talking about electric forces. This is because electric forces are humungously stronger than gravitational forces. In many of the situations we will consider there are many objects creating electric forces and the net electric force is changing dramatically, both in magnitude and direction, over very small distances (such as near the surface of a molecule). "Field" is a tough concept so it really helps to try to make sense of it in the case of gravity where it is much simpler. There are a couple of "dangerous bends".
Dangerous Bend #1:
The direction of the gravitational field does NOT represent the direction in which an object moves when it is experiencing the forces of that field.
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Dangerous Bend #2:
Despite the name "field", the gravitational field does NOT refer to the region of space in which the gravitational force is significant. It refers to the set of vectors at each point in space.
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Dangerous bend #1 is confusing because if an object is not moving and there are no other forces on it, then it will move (accelerate) in the direction of the arrow. But since it is the net force that goes with the object's acceleration, if there are any other forces on it, it won't accelerate in the direction of the field arrow. Also, if it already has a velocity, the force corresponding the the field arrow will point in the direction of the acceleration -- the direction in which the velocity is changing, not in the direction of the velocity itself.
In the figure below, we show a representation of a gravitational field near the earth and a thrown ball. The little arrows pointing down indicate the direction of g at their center point. We haven't shown an arrow at every point (the entire picture would be red!) but you should imagine that there is an arrow of the gravitational field at every point.
If we take a ball (blue) and throw it in the direction of the blue arrow, clearly, the direction of motion of the ball is not in the direction of the field.
Dangerous bend #2 is tempting because we are using the word "field" which in common speech represents an "area" (could be a region in space but also could be in some kind of symbolic space, like "the area of your specialization"). But our fields don't represent this (they have the wrong units, for one), but a set of vectors -- really everywhere in space. For forces like gravity and electricity, the influence of a source of these forces falls of like the square of the distance, so it never really goes away. While there are times when we can ignore a gravitational or electrical field because their sources are too far away or because (in the electrical case) there are opposite charges cancelling it, but there we say "the field is weak or negligible". We do NOT say "we are out of the field."
Joe Redish 10/4/13
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