Sign up ×

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?

share|cite|improve this question
as per GR, gravity will eventually win over since the orbiting mass emits gravitational waves and loses energy. but this is very less so the eventual infall could take millions of years. Newtonian mechanics(NM) doesn't explain it. As per NM, it will be in perpetual motion. – guru Jun 29 '13 at 9:53
@guru I think "millions of years" is too short of a timescale for many systems to lose significant amounts of energy from gravity waves. Example: Earth has been orbiting the Sun for over 4 billion years. – Joshua Jan 27 at 18:29

3 Answers 3

up vote 15 down vote accepted

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 $(r - \frac{1}{2}at^2)\hat{i} + vt\hat{j}$ 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 $$ r_2 = r \sqrt{1 - (2\pi\frac{t}{T})^2 + \frac{1}{4}(2\pi\frac{t}{T})^4 + (2\pi\frac{t}{T})^2}$$ 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).

share|cite|improve this answer
Well said. @YWE, it might help to think of this in terms of Newton's Cannonball. – spencer nelson Feb 25 '11 at 21:06
I guess what I do not understand is why gravity doesn't slow the object down. It's an external force acting upon the object but the object's momentum does not change. – YWE Feb 25 '11 at 21:43
@YME: Gravity always acts perpendicularly to the current velocity, and as such it never reduces the magnitude of the momentum (which is a vector quantity), it does change direction, however. – dmckee Feb 25 '11 at 21:55
The momentum most certainly does change! The momentum at any point in a circular orbit is tangential to the orbit, and the change in the momentum (the acceleration) is towards the center of the orbit. – Colin K Feb 25 '11 at 21:55
Ah. Now I get it. A force applied orthogonally to an object does not change the magnitude of its momentum. I meant to say magnitude of momentum instead of just momentum. – YWE Feb 25 '11 at 22:05

The force of gravity has little to do with friction. As dmckee says, what is happening is that the body falls, but precisely because it has enough momentum, it falls around the object towards which it gravitates instead of into it. Of course, this is not always the case, collisions do happen. Also systems of astronomical bodies are complicated and the combined effect of the action of several different bodies on one can destabilize trajectories that in a simple 2-body case would be stable ellipses. The result could be collision or escape of the body.

In the 2-body case however, the crucial aspect of gravity which guarantees the stability of the system is the fact that gravitation is a centripetal force. It always acts towards the center of the other gravitating mass. One can show that this feature implies the conservation of angular momentum, which means that if the 2-body system had some angular momentum to begin with, it will keep the same angular momentum indefinitely.

(Extra note, even in the 2-body case, there can be collisions and escape to infinity, the first if there is not enough angular momentum (for instance one body having velocity directed towards the other body, like an apple falling from a tree), the other if there is too much angular momentum, resulting in parabolic or hyperbolic trajectories.)

share|cite|improve this answer

I always think of it this way: gravity and the centrifugal force generated by the orbiting object are exactly in balance. If you tie a rope to an object and spin it around you, the centrifugal force will pull on the rope. You pull back with the same force to keep the object orbiting around you. That's exactly what gravity does to orbiting objects.

You can also see that the speed of the object forces it into a particular orbit. If the object slows down, it will fall and either reach a new lower orbit, or crash into the object they're rotating around.

share|cite|improve this answer

Your Answer


