# The gravitational field

#### Prerequisites

In our discussion of flat-earth gravity we learned that the reason all objects accelerate the same near the surface of the earth is that the force of gravity is proportional to the object's mass. To make this connection between gravitational force and acceleration we had to assume that the net force (which leads to acceleration according to Newton's second law) is equal to the force of gravity - in other words our object of interest does not interact with anything else but the earth via gravity - it's not held up by a hand or slowed down by air resistance.

This page digs deeper into what "interacting via gravity" means. We introduce the concept of **vector field, **a conceptually difficult idea. While this is not really needed if all we are going to do is flat-earth gravity, it turns out to be absolutely essential when we are talking about electric forces. And since gravity is such a simple force (compared to electric forces), this is a good place to get a handle on a challenging concept.

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 or other interactions then the only force on our object of interest is its weight so in the up-and-down direction, Newton's second law for an object of mass $m$ is

$$a = F^{net}/m = -mg/m = -g$$

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/s^{2}.

Now let's improve our model from flat-earth gravity to round earth gravity. We know that the earth is round and we know that wherever we are on the earth, "down" points to the center. These are not all the same directions! But we also find that the magnitude of $g$ measured anywhere near the surface of the earth is the same. So to improve our model, we'll make our next model *surface-of-a-sphere *gravity. This means instead of being in a small region of the surface of the earth that we can treat as flat, we're going to consider anywhere near the surface of the earth.

In this model, we have to replace our constant $g$ by a vector $\overrightarrow{g}$ that points to the center of the earth. This isn't really enough notation. The direction of our arrow $\overrightarrow{g}$ depends on where we are on the surface of the earth so we have to write that it is a vector that depends on position:

$$\overrightarrow{g}(\overrightarrow{r})$$

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, $\overrightarrow{g}(\overrightarrow{r})$, **the gravitational field.**

What it means conceptually is pretty straightforward: Wherever you are in space, you can identify a vector $\overrightarrow{g}$ that has a direction and a magnitude. The gravitational force on an object of mass $m$ placed at that point is $mg$ and the force vector points in the direction of the field vector.

Mathematically things are more complicated: we have three functions: the x component of $\overrightarrow{g}$, the y component of $\overrightarrow{g}$, and the z component of $\overrightarrow{g}$, and they are each functions of the three variables that determine the position we are looking at:

$$\overrightarrow{g}(\overrightarrow{r}) = g_x(x,y,z) \hat{i} + g_y(x,y,z) \hat{j} + g_z(x,y,z) \hat{k} $$

Let us summarize this in the context of forces:

If we have a gravitational field$\overrightarrow{g}(\overrightarrow{r})$it means that if we put an object of mass $m$at the position labeled by the vector,$\overrightarrow{r}$, then the object feels a gravitational force$\overrightarrow{F}^{grav}= m\overrightarrow{g}(\overrightarrow{r})$.

One way of representing this gravitational field 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 moved somewhere else.

This mathematical representation is really only needed if either we need to be careful about the direction of gravity, or if the gravitational force is different than it is on the surface of the earth, so when we are launching rockets to the moon or considering the motion of planets. For this we need a model of the gravitational force in which the magnitude as well as the direction of the field change. This is Newton's Universal gravitation

Finally, thinking ahead, the concept of fields will also be very valuable for analyzing situations where our object of interest experiences interactions with multiple other objects. While we only worry about gravitational interactions with one object (the earth) in this class, we will encounter examples of electrical forces where our object of interest interacts with multiple other objects via electrical forces. A specific example is considering the electrical forces experienced by an atom of interest near the surface of another molecule. The idea of a "Field" itself is tough so let's try to make sense of it in this case where our object of interest only interacts with one other object.

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. It is important to remember that the field determines the gravitational force, but the net force is what determines acceleration. Furthermore, the direction of acceleration and the direction of motion can be very different as we will see below!*

if an object is not moving and there are no other forces on it, then it will start moving (accelerate) in the direction of the arrow. But since it is the *net force* that determines the object's *acceleration*, if there are any other forces on it, it won't accelerate in the direction of the field arrow.

It is also important to remember that the net force determines the acceleration, which is a change in velocity. If our object of interest already has a velocity, the force corresponding to the field arrow (assuming that there are no other forces on our object) will point in the direction of the **change** in velocity, 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 ball thrown at an angle. 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, the field will push the ball down, but clearly, the direction of motion of the ball will not be in the direction of the field.

**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. A field can have zero magnitude or be very small at some point (though we don't see this for gravity, it can be true for electric fields).*

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 rather, they are a set of vectors — they really exist everywhere in space. For forces like gravity and electricity, the influence of a source of these forces falls off like the square of the distance, so it gets smaller but 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 & Wolfgang Losert 10/4/13

#### Follow-on

Last Modified: February 4, 2019