How does Newtonian mechanics explain why orbiting objects do not fall to the object they are orbiting?

The force of gravity is constantly being applied to an orbiting object. And therefore the object is constantly accelerating. Why doesn't gravity eventually "win" over the object's momentum, like a force such as friction eventually slows down a car that runs out of gas? I understand (I think) how relativity explains it, but how does Newtonian mechanics explain it?

Answer

Newtonian mechanics explains that they do fall toward the object they're orbiting, they just keep missing.

---

Quick and dirty derivation for a circular orbit.

Let the primary have mass $M$ and the satellite mass $m$ such that $m \ll M$ (it can also be done for other cases, but this saves on mathiness).

Assume we start with an initial circular orbit on radius $r$, velocity $v = \sqrt{G\frac{M}{r}}$. The acceleration of the satellite due to gravity is $a = G\frac{M}{r^2}$ which means we can also write $v = \sqrt{\frac{a}{r}}$. The period of the orbit is $T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r}{a}}$.

Chose a coordinate system in which the initial position is $r\hat{i} + 0\hat{j}$ and the initial velocity points in the $+\hat{j}$ direction. Chose a short time $t \ll T$ and lets see how far from the primary the satellite ends up after that time.

If we have chosen $t$ short enough, we can approximate gravity as having uniform strength through the time period (and we shall show later that that is justified).

The new position is $\left(r - \frac{1}{2}at^2\right)\hat{i} + vt\hat{j}$ which lies at a distance $$ r_2 = \sqrt{r^2 - r a t^2 + \frac{1}{4}a^2 t^4 + v^2 t^2} $$ pulling our at factor of $r$ we get $$ r_2 = r \sqrt{1 - \frac{a}{r} t^2 + \frac{1}{4}\frac{a^2}{r^2} t^4 + \frac{v^2}{r^2} t^2} $$ and converting all the $\frac{a}{r}$ and $\frac{v}{r}$ terms into expressions of the period we get $$ r_2 = r \sqrt{1 - \left(2\pi\frac{t}{T}\right)^2 + \frac{1}{4}\left(2\pi\frac{t}{T}\right)^4 + \left(2\pi\frac{t}{T}\right)^2}$$ Finally, we drop the $(t/T)^4$ term as negligible and note that the $(t/T)^2$ terms cancel so the result is $$r_2 = r$$ or the radius never changed (which justified the constant magnitude for acceleration, and a small enough $t$ justifies both the constant direction and the dropping of the fourth degree term).