*Michael Fowler 2/14/06*

Let’s begin with the *definition*
of gravitational field:

*The gravitational field at any point P in space is defined as the gravitational force felt by a tiny
unit mass placed at P.*

So, to visualize the gravitational field, in this room or on a bigger scale such as the whole Solar System, imagine drawing a vector representing the gravitational force on a one kilogram mass at many different points in space, and seeing how the pattern of these vectors varies from one place to another (in the room, of course, they won’t vary much!). We say “a tiny unit mass” because we don’t want the gravitational field from the test mass itself to disturb the system. This is clearly not a problem in discussing planetary and solar gravity.

To build an intuition of what various gravitational fields look like, we’ll examine a sequence of progressively more interesting systems, beginning with a simple point mass and working up to a hollow spherical shell, this last being what we need to understand the Earth’s own gravitational field, both outside and inside the Earth.

This is of course simple: we know this field has strength *GM*/*r*^{2},
and points towards the mass—the direction of the attraction. Let’s draw it anyway, or, at least, let’s
draw in a few vectors showing its strength at various points:

This is a rather inadequate representation: there’s a lot of blank space, and, besides, the field attracts in three dimensions, there should be vectors pointing at the mass in the air above (and below) the paper. But the picture does convey the general idea.

A different way to represent a field is to draw “field lines”, curves such that at every point along the curve’s length, its direction is the direction of the field at that point. Of course, for our single mass, the field lines add little insight:

The arrowheads indicate the direction of the force, which
points the same way all along the field line.
A shortcoming of the field lines picture is that although it can give a
good general idea of the field, there is no precise indication of the field’s *strength* at any point. However, as is evident in the diagram above, there
*is* a clue: where the lines are closer
together, the force is stronger. Obviously, we could put in a spoke-like field
line anywhere, but if we want to give an indication of field strength, we’d
have to have additional lines equally spaced around the mass.

The next simplest case is two equal masses. Let us place
them symmetrically above and below the *x*-axis:

Recall _{}. A point mass *m* = 1 at *P* will therefore feel gravitational attraction towards both masses *M*, and a *total* gravitational field equal to the *vector sum of these two forces*, illustrated by the red arrow in the
figure.

The fact that the total gravitational field is just given by
adding the two vectors together is called the *Principle of Superposition*.
This may sound really obvious, but in fact it isn’t true for every force
found in physics: the strong forces
between elementary particles don’t obey this principle, neither do the strong
gravitational fields near black holes. But
just adding the forces as vectors works fine for gravity almost everywhere away
from black holes, and, as you will find later, for electric and magnetic fields
too. Finally, superposition works for
any number of masses, not just two: the total gravitational field is the vector
sum of the gravitational fields from all the individual masses. Newton used this to prove that the
gravitational field outside a solid sphere was the same as if all the mass were
at the center by imagining the solid sphere to be composed of many small
masses—in effect, doing an integral, as we shall discuss in detail later. He
also invoked superposition in calculating the orbit of the Moon precisely,
taking into account gravity from both the Earth and the Sun.

*Exercise*: For the two mass case above, sketch the
gravitational field vector at some other points: look first on the* x*-axis, then away from it. What do the *field lines* look like for this two mass case? Sketch them *in the neighborhood of the origin*.

It is not difficult to find an exact expression for the
gravitational field strength from the two equal masses at an equidistant point *P*.

Choose the* x*,*y* axes so that the masses lie on the *y*-axis at (0, *a*) and (0,-*a*).

By symmetry, the field at *P* must point *along* the *x*-axis, so all we have to do is compute
the strength of the *x*-component of
the gravitational force from one mass, and double it.

If the distance from the point *P* to one of the masses is *s*,
the gravitational force towards that mass has strength _{}. This force has a
component along the *x*-axis equal to _{}, where _{} is the angle between
the line from *P* to the mass and the *x*-axis, so the total gravitational force
on a small unit mass at *P* is _{}directed along the *x*-axis.

From the diagram, _{}, so the force on a unit mass at *P* from the two masses *M* is

_{}

in the *x*-direction.
Note that the force is exactly zero at the origin, and everywhere else
it points towards the origin.

Now, as long as we look *only
*on the *x*-axis, this identical
formula works for a *ring* of mass 2*M*
in the *y*, *z* plane! It’s just a three-dimensional version of the
argument above, and can be visualized by rotating the two-mass diagram above around
the *x*-axis, to give a ring
perpendicular to the paper, or by imagining the ring as made up of many beads,
and taking the beads in pairs opposite each other.

*Bottom line*: the
field from a ring of *total *mass *M*, radius *a*, at a point *P* on the
axis of the ring distance *x* from the
center of the ring is

_{}.

*This is an optional section: you can safely skip to the result
on the last line. In fact, you will
learn an easy way to derive this result using Gauss’s Theorem when you do
Electricity and Magnetism. I just put
this section in so you can see that this result* can* be derived by the straightforward, but quite challenging, method of
adding the individual gravitational attractions from all the bits making up the
spherical shell.*

What about the gravitational field from a hollow spherical *shell *of matter? Such a shell can be envisioned as *a stack
of rings*.

To find the gravitational field at the point *P*, we just add the contributions from all the
rings in the stack.

In other words, we divide the spherical shell into narrow
“zones”: imagine chopping an orange into circular slices by parallel cuts,
perpendicular to the axis—but of course our shell is just the *skin* of the orange! One such slice gives a ring of skin,
corresponding to the surface area between two latitudes, the two parallel lines
in the diagram above. Notice from the
diagram that this “ring of skin” will have radius _{}, therefore circumference _{}and breadth _{}, where we’re taking _{} to be *very small*. This means that the *area of the ring* of skin is

_{}.

So, if the shell has mass _{} per unit area, this
ring has mass _{}, and the gravitational force at *P* from this ring will be

_{}.

Now, to find the total gravitational force at *P* from the *entire shell* we have to add the contributions from each of these
“rings” which, taken together, make up the shell. In other words, we have to *integrate* the above expression in _{}

So the gravitational field is:

_{}.

In fact, this is quite a tricky integral: _{}, *x* and *s* are all varying! It turns out to be is easiest done by
switching variables from _{} to *s*.

Label the distance from *P
* to the center of the sphere by *r*.
Then, from the diagram, _{}, and *a*, *r* are constants, so _{},

and _{}

Now _{}, and from the diagram _{}, so _{},

and, writing _{},

_{}

The derivation was rather lengthy, but the answer is simple:

**The gravitational
field outside a uniform spherical shell is GM/r^{ 2} towards the center.**

And, there’s a bonus: for the ring, we only found the field *along the axis*, but for the spherical
shell, once we’ve found it in one direction, *the whole problem is solved*—for the spherical shell, the field must
be the same in all directions.

Once we know the gravitational field outside a shell of matter is the same as if all the mass were at a point at the center, it’s easy to find the field outside a solid sphere: that’s just a nesting set of shells, like spherical Russian dolls. Adding them up,

**The gravitational
field outside a uniform sphere is GM/r^{
2} towards the center.**

There’s an added bonus: since we found this result be adding
uniform spherical shells, it is *still
true* if the shells have *different
densities*, provided the density of each shell is the same in all
directions. The inner shells could be
much denser than the outer ones—as in fact is the case for the Earth.

This turns out to be
surprisingly simple! We imagine the
shell to be very thin, with a mass density _{} kg per square meter of surface. Begin by
drawing a two-way cone radiating out from the point *P*, so that it includes two small areas of the shell on opposite
sides: these two areas will exert gravitational attraction on a mass at *P* in opposite directions. It turns out that they exactly *cancel*.

This is because the ratio of
the areas *A*_{1} and *A*_{2} at distances *r*_{1}
and *r*_{2} are given by _{}: since the cones have the same angle, if one cone has twice
the height of the other, its base will have twice the diameter, and therefore *four times the area*. Since the masses of the bits of the shell are
proportional to the areas, the ratio of the masses of the cone bases is also _{}. But the
gravitational attraction at *P* from
these masses goes as _{}, and that *r*^{2}
term *cancels* the one in the areas, so
the two opposite areas have equal and opposite gravitational forces at *P*.

In fact, the gravitational
pull from every small part of the shell is balanced by a part on the opposite
side—you just have to construct a lot of cones going through *P* to see this. (There is one slightly tricky point—the line
from *P* to the sphere’s surface will
in general cut the surface at an angle.
However, it will cut the opposite bit of sphere *at the same angle*, because any line passing through a sphere hits
the two surfaces at the same angle, so the effects balance, and the base areas
of the two opposite small cones are *still*
in the ratio of the squares of the distances *r*_{1}, *r*_{2}.)

This is a practical application of the results for shells.
On going down a mine, if we imagine the Earth to be made up of shells, we will
be inside a shell of thickness equal to the depth of the mine, so will feel *no* net gravity from that part of the
Earth. However, we will be* closer* to the remaining shells, so the
force from them will be intensified.

Suppose we descend from the Earth’s radius *r _{E}*

Then the fraction of the
Earth’s mass that is still attracting us (because it’s closer to the center
than we are—inside the red sphere in the diagram) is _{}.

The gravitational attraction
from this mass at the bottom of the mine, distance *r* from the center of the Earth, is proportional to mass/*r*^{2}. We have just seen that the mass is itself proportional
to *r*^{3}, so the actual
gravitational force felt must be proportional to _{}.

That is to say, the gravitational
force on going down inside the Earth is *linearly
proportional to distance from the center*.
Since we already know that the gravitational force on a mass *m* at the Earth’s surface _{} is *mg*,
it follows immediately that in the mine the gravitational force must be

_{}.

So there’s no force at all at
the center of the Earth—as we would expect, the masses are attracting equally
in all directions.