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?
Newtonian Mechanics – How to Explain Why Orbiting Objects Do Not Fall to Their Orbits
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Best 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).