Consider a planet of mass $m$ like the Earth orbiting a star of mass $M$ like the Sun. According to Newton's law of gravity, the force on the Earth is

$$F = \frac{G M m}{r^2},$$

where $r$ is the distance from the Sun. The force pulls the Earth toward the Sun, which we'll suppose is sitting at rest at the origin.$^2$ From the force $F$, we get the gravitational potential energy $U$:

$$U = - \frac{GMm}{r}.$$

The total energy is the sum of the kinetic energy of the Earth plus the potential energy $U$. The kinetic energy is of course $K = \frac{1}{2} m v^2$, where $v$ is the speed of the Earth. The speed has two components, the radial speed $\dot r$, which is due to the changing radial coordinate, and the angular speed $r \dot \theta$, which is due to the changing angular coordinate.$^3$ (The dots stand for the rate of change of whatever's underneath.) Then the total energy is

$$E = \frac{1}{2} m \dot r^2 + \frac{1}{2} m r^2 \dot \theta^2 - \frac{GMm}{r}.$$

The motion due to a force like $F$ that only depends on the radial coordinate $r$ and not on $\theta$ is very special: it means that the angular momentum $L$ of the planet is constant:

$$L = m r^2 \dot \theta = \mathrm{constant}.$$

I'll explain why $L$ is constant in a future video; you can go ahead and check it for yourself by taking the rate of change of $L$ and plugging in the $F = ma$ equations. The fact that the combination $r^2 \dot \theta$ never changes means that as $r$ gets bigger the angular speed gets smaller, and vice-versa. So the angular speed is biggest when the planet is closest to the star, and smallest when it's farthest away.

We can use the fact that $L = m r^2 \dot \theta$ is constant to write $\dot\theta = \frac{L}{mr^2}$. Then if we plug this into the expression for the total energy, we can get rid of all the $\theta$'s and write the energy entirely in terms of $r$:

$$E = \frac{1}{2} m \dot r^2 + \frac{L^2}{2mr^2} - \frac{GMm}{r}.$$

That means we effectively have a one-dimensional problem for the coordinate $r$, subject to the effective potential

$$U_\mathrm{eff} = -\frac{GMm}{r} + \frac{L^2}{2mr^2}.$$

This is a huge simplification! It allows us to understand the qualitative shape of the orbits of the planet just by sketching a picture of the potential $U_\mathrm{eff}$:

On the left is the graph of $U_\mathrm{eff}(r)$ for some non-zero value of $L$. This picture is incredibly useful, because you can understand what's going to happen to a mass that's moving in this potential just imagining that it's a hill that the particle is sliding along.

The shape of the orbit will depend on the energy, represented by the horizontal dashed line. When the energy is at the minimum of the effective potential, the particle will just sit there at the bottom of the hill! That means that $r$ is constant. The planet isn't sitting still—remember that it has angular momentum $L = m r^2 \dot \theta$, and so the angular coordinate is changing. So when the energy is at the minimum of the effective potential, the planet will orbit the star in a circle!

Now what happens when we increase the energy a bit? You can see for yourself by dragging the slider that controls the energy in the animation at the top of the page. If you imagine the particle sliding along the hill with energy like this, it will just oscillate back and forth between the two turning points, where the horizontal dashed energy line intersects the effective potential. That means that $r$ will oscillate back and forth between its smallest value on the left (called the perihelion) and its largest value on the right (the aphelion).

This motion corresponds to an ellipse! This is the generic behavior for a planet orbiting a star. The planet oscillates back and forth between its point of closest approach to the star and its farthest distance, all the while swinging around in the $\theta$ direction because of the non-zero angular momentum.

I haven't proven here that the orbit is an ellipse—I'll show you how to derive that in an upcoming YouTube video (you can subscribe if you're interested in seeing that!). But the point is that just by sketching the picture of the effective potential we can understand at a glance what the qualitative features of the orbit will look like with almost no work at all. Actually solving the equations to determine the explicit orbit is a lot more work.

Finally, what if you drag the energy all the way up past $E>0$? If you release the particle from that high up on the hill, it won't oscillate back and forth at all—it'll just slide down the hill and then coast back up as it travels all the way out to $r \to \infty$. So a "planet" with $E > 0$ isn't really a planet at all, in the sense that it isn't trapped in an orbit around the star. It's like a comet that speeds into the solar system, sling-shots around the star, and then flies off on its way, never to return again. The shape of this trajectory is a hyperbola.

Actually, there's one more case to mention here—the crossover point when $E = 0$. In this case, the mass does escape out to infinity, but with no extra kinetic energy to spare. You can show that the shape is a parabola. So a mass moving around a star can realize all the different conic sections depending on how much energy it has!