Cosmology has a reputation as being a very esoteric science; but some of it is actually quite accessible given the proper approach and the care to appropriately explain just a few terms and concepts. As an illustration, we use just a bit of simple physics, some high-school algebra, and a dose of careful thinking to derive the Friedmann equation for the expansion of the Universe.
The exposition below is my own; the line of argument is from Leonard Susskind's introductory cosmology lectures given at Stanford University.
In 1922, Alexander Friedmann derived an equation that describes the expansion of the Universe. What this means is that he developed a way of relating the size of the Universe to the amount of time the Universe has been expanding. Friedmann did this the hard way, by examining Einstein's equations of General Relativity (which is, essentially, a theory of gravity). There is an easier route to his equation, using gravity as described by Isaac Newton, coupled with the concept of the conservation of energy. It is that course we will follow here.
In order to mathematically model the Universe, we have to decide what our model looks like. Take the Universe to be a huge empty space filled with an even distribution of relatively small chunks of matter (call them galaxies). Draw a large imaginary sphere in that space, with some of the galaxies inside and some outside. Consider a small test mass m (say, a single galaxy) at the surface of that sphere moving away from the sphere's center.
At this point, let's talk about the gravity of a sphere. Newton showed that a symmetrical spherical body has a gravitational effect that is the same as if the entire mass of the sphere was concentrated at the very center. So, for our model universe, the evenly-distributed galaxies inside the sphere act gravitationally as a single mass at the sphere's center. Furthermore, since all of the galaxies outside the sphere are evenly distributed, all of their gravitational effects cancel out.
Newton's formula for the force due to gravity is $$F = -{{G M m} \over r^2}$$ where G is Newton's gravitational constant, M is the mass of the galaxies inside the sphere, m is the test mass, and r is the distance from the center of the sphere to the test mass. The "minus" sign comes from the convention that forces of attraction are negative, while those of repulsion are positive.
What is the mechanical energy of the test mass? (We are here not talking about the chemical energy in the molecular bonds of the mass, nor of the energy equivalent of the mass itself.) It is moving, so it has some kinetic energy, and it is in a gravitational field, so it has gravitational potential energy. The kinetic energy is $$E_k = {{m v^2} \over 2}$$ where v is the velocity at which the test mass moves. The gravitational potential energy of the mass at the surface of the sphere is the work it would take to lift the mass, against the force of gravity at the surface of the sphere, from the center of the sphere to its current position at r. Since work is force times distance, the potential energy U is $$U_g = -{{G M m} \over r^2} \times r$$ $$U_g = -{{G M m} \over r}$$
Energy is conserved, which means that it is neither created nor destroyed, though it may be converted from one form to another. Therefore, the mechanical energy of the test mass is constant, whatever its value turns out to be. So the energy of the mass m is $$constant = {E_k + U_g}$$ $$constant = {{m v^2} \over 2} - {{G M m} \over r}$$
Note that we've preserved conservation of energy; the test mass will slow down as it moves away from the sphere, losing kinetic energy. But that also increases its distance from the center of the sphere, which decreases its gravitational potential energy, so the difference between the two remains constant. (Another way to see this is to set the constant to 0 and then add the potential energy to both sides; it is easy to see that both quantities increase or decrease by the same amount. It may also help to realize that, since potential energy is negative, decreasing the potential energy increases the absolute value of that energy.)
Observe that the test mass is not going to change, so we can simplify by dividing it out of the equation and absorbing it into the constant (equivalently, we could set the value of the test mass to 1). $$constant = {v^2 \over 2} - {{G M} \over r}$$ We can do a similar thing to eliminate the division by 2; multiply both sides of the equation by 2 and absorb the multiplication into the constant, to yield $$constant = {v^2} - {{2 G M} \over r}$$
Let's introduce a bit of mathematical notation. In mechanics, physical quantities are quite often changing with time, so it is useful to have a compact notation for "rate of change with time". Physicists (and others) have borrowed Newton's original notation for this, which is the placement of a small dot over the quantity. Thus, if x(t) denotes a quantity x that changes as a function of time, the rate of that change is given as \(\dot x\). (We make use of another convention that if it is understood that x represents a function, the parenthesized list of independent variables is often omitted.)
Now set our model universe into expansion. Everyone knows what that means: "All of the galaxies are moving away from us." If that was the literal truth, it would be a remarkable fact indeed; the random motions of all the galaxies line up to make them move in a direction which is always away from our Milky Way galaxy. Further, as Edwin Hubble showed, the farther away a galaxy is, the faster it is moving away, in such a manner that a galaxy at twice the distance is moving away twice as fast.
The actual situation is at once simpler and more complex. Every galaxy is moving away from every other galaxy (actually, this is only true for clusters of galaxies; within a cluster, galaxies can and do collide, just as the Andromeda Galaxy is destined to eventually collide with the Milky Way). And every galaxy sees the same situation Edwin Hubble saw; the farther away a galaxy is, the faster it is receding.
The simplest way to explain this behavior is to posit that the galaxies themselves are not actually moving, but rather space itself is stretching, everywhere, at exactly the same rate (or, equivalently, that new space is being constantly created between every point of existing space). So a galaxy that is twice as far away has twice as much stretching space in between, so it appears to be moving away twice as fast. The rate of this stretching is named in honor of Hubble, and is known as the Hubble Constant. The value of the Hubble Constant has been measured (in fact, one of the primary missions of the Hubble Space Telescope was to determine the value), and is now known to be about 72 kilometers per second per megaparsec (km / sec / Mpc).
That's an odd-sounding set of units, but what it means is that for every megaparsec — million parsecs — of distance between two galaxies (a megaparsec is a unit that astronomers favor for measuring distances; it is about \(3 {1 \over 4}\) million light-years), 72 kilometers of new space is being created every second. Note, for later review, that the unit of the Hubble Constant is a velocity divided by a distance.
The rate of expansion of the Universe can potentially change with time, depending on its composition, conditions, and curvature (don't worry about what curvature is for now). So the Hubble Constant actually changes with time; the "constant" part is that it is the same everywhere in space at any given time. If you presume different values for composition and curvature, you get different values for the constant than another cosmologist at any given time in the Universe's past (or future). Additionally, the number 72 is, of course, dependent on the units; it would be different if given in miles, hours, and light-years, for example.
Cosmologists deal with all of this messiness by creating an abstraction called the scale factor, denoted by the symbol a(t). It's a function whose value has no units; this value is related to the size of the Universe at a given time. One way to think of it is to use a white magic marker to paint grid lines directly on space (space is black, so white shows up really well). Each galaxy is therefore located at a particular grid point. As the Universe expands by creating new bits of space, the grid lines move apart (because they are painted directly onto the old bits of space). Because the galaxies are also attached to the old bits of space, the coordinates of a galaxy on the grid do not change. What does change is the multiplier that specifies the amount of stretching that takes place as time passes — that's the scale factor.
Now consider two galaxies that started out at a distance apart denoted by X. Remember that this is in the coordinates of our magic-marker grid, so the coordinates won't change as the galaxies are pushed apart by the expansion of the Universe. Instead, the distance between the galaxies grows as the grid lines are spread apart by the scale factor a(t). Consider one of the galaxies as being at the center of the sphere, and the other on the surface of the sphere. So, writing the distance between the galaxies as D (the radius of the sphere), $$D = X a(t)$$ The velocity at which the galaxies separate is then $$\dot D = X \dot a(t)$$
We also have to deal with the fact that the sphere we started with is getting bigger with the expansion. Note that the number of galaxies inside the sphere does not change, because the space between the galaxies changes at the same rate as the sphere grows. So the mass density inside the sphere decreases as the Universe expands. If we multiply that mass density by the volume of the sphere, we get the total amount of mass inside the sphere. Adopting the standard symbol \(\rho\) for the density (which, remember, varies with time), we can write the mass as $$M = {{4 \over 3} {\pi D^3 \rho}}$$
Let's substitute these values into the energy equation above. D is r, the distance from the center (where all the mass acts as if it is concentrated), and \(\dot D\) is the velocity v at which the test mass (the other galaxy) is moving. $$constant = {{X^2 \dot a^2} - {{ 2 G ({{4 \over 3} {\pi D^3 \rho}}) } \over {X a}}}$$ Substitute once again for the D to get $$constant = {{X^2 \dot a^2} - {{ 2 G ({{4 \over 3} {\pi X^3 a^3 \rho}}) } \over {X a}}}$$ Simplifying, $$constant = {{X^2 \dot a^2} - {{8 \pi G X^2 a^2 \rho} \over 3}}$$
Remember that X, the coordinate difference of the galaxies, does not change; only the scale factor changes. This means that \(X^2\) is a constant. So, as before, we absorb that constant into the constant left-hand side by dividing the equation by \(X^2\): $$constant = \dot a^2 - {{8 \pi G a^2 \rho} \over 3}$$
Recall the Hubble Constant, which has units of velocity divided by distance. We have velocity in this equation as \(\dot a\) (the velocity of the expansion of space) and distance as a (the coordinate-independent distance scale over which that velocity is measured). The definition of the Hubble Constant is the velocity of the expansion of space divided by the distance over which that velocity is measured, so \(\dot a / a\) is the Hubble Constant. So let's rewrite our equation in terms of that, by first rearranging terms $$\dot a^2 = {{8 \pi G a^2 \rho} \over 3} + constant$$ and dividing by \(a^2\): $$\Big ( {\dot a \over a} \Big )^2 = {{{8 \pi G \rho} \over 3} - {k \over a^2}}$$ where k is the constant, which may be negative, positive, or 0, but we've chosen to define it such that we subtract its value here. (It turns out that k represents the curvature of the Universe, which is something we haven't defined here; but adopting the subtraction convention allows the value of k to be positive when the Universe is positively curved, etc., topologically speaking.)
What have we actually accomplished, beyond writing down a barely-comprehensible differential equation? Well, we have an equation that relates the scale factor of the Universe to the passage of time; in other words, this equation tells us how the Universe expands as it ages. It turns out to be possible to solve this equation for a and deduce some interesting facts about the Universe, but that's a subject for another essay.