Given a crewed satellite placed at a distance $r$ from the center of the Earth, with an initial velocity perpendicular to its position vector, the magnitude of the initial velocity that would allow it to maintain a circular orbit of radius $r$ is:
$$
v_0 = \sqrt{\frac{G M}{r}}
$$
Where $G$ is the gravitational constant and $M$ is the mass of the Earth.
My question is: if the astronaut inside the satellite were to exert some kind of force on it, wouldn't that cause a small change in the direction of its velocity vector, thus making the satellite break its circular orbit?
Best Answer
There is something which I do not like about the answers, and it has to do with the fact that you have a really good intuition here and the other answers are giving some specific exceptions to that intuition but not really directing you how to use it.
So my answer is instead something like, “It is a real satellite, it wasn’t in a perfect circular orbit to begin with.” So, we are physicists and we know that we are creating these glorious mathematically exact models of the universe: but part of the game of being a physicist is understanding that those models are generally only approximately true. I could quibble with this and say “oh, conservation laws, those are more than approximately true” but I hope you can see my meaning. The world has noise and we know about that. The actual satellite occasionally feels perturbations from the solar wind, from the gravitational force of the Sun and the Moon, from pieces of space dust and radiation pressures, all sorts of things like that.
And even with all of that, if it was approximately circular then this equation approximately represented its motion, and it is a useful tool in my toolkit.
What is happening is that you have an intuition you are building up called stability analysis. So if I have a normal standard pen sitting on my desk, there are several stationary configurations that it can inhabit. It can be laying on my desk at rest in a variety of ways. But there is one stationary configuration where, even though it is in a proper state of force-balance and all that, you pretty much never see: where the pen is perfectly balanced upon its tip. What makes that configuration different?
It is that all of the “nearby” configurations to that one, are unstable. It is that the world is noisy. All of these configurations where the pen is resting on its side on the desk, all of those are nearby other stable configurations and so the noise does not disturb us from our big set of stable situations. The one where the pen is balanced on its very tip, the noise will eventually disturb it and it will get worse and worse from there.
How do we measure “nearby”? We think about something called “phase space,” which combines the idea of being nearby in position but also nearby in momentum, and this lets us think of the two things that noise might perturb. And then it's a stable orbit if the nearby points of phase space also lead to stable orbits.
Space is not far, space is fast
To be in orbit, things need to move fast—so fast that the distance you fall by “falling down” gravitationally is the same as the distance the Earth’s surface falls away from underneath you by virtue of its curvature. So if you imagine a normal freefall parabola starting from a sideways motion of velocity $v$ up at radius $R$, you would say in Newtonian mechanics that it describes the point $(x, y)$ over time where $$y(t) = R-\frac12 g t^2,\\x(t) = v~t,$$ and $g = GM/R^2$ of course, and this would only be approximately correct for small deviations in $y \ll R$. Then you could solve for $t = x/v$ and describe this instead as the parabola $y(x) = R - g x^2/(2 v^2).$ Here we are imagining that the velocity is small enough that the ground does not ever “curve away”, we can treat the Earth as flat. But the Earth is not flat, and we might instead think about the circle of radius R, $y(x) = \sqrt{R^2 - x^2} = R\sqrt{1 - (x/R)^2}.$ Just a little bit of calculus later, you can find that for small $x$, we have $y \approx R - x^2/(2 R),$ and these are approximately the same line when $g/v^2 =1/R.$ This is the precise speed where that parabola is “falling down” just as much as the surface is curving away underneath it. And indeed if you substitute $g = GM/R^2$ you will find your formula, $v = \sqrt{GM/R}.$
But I wanted to put some numbers to this. This speed is something like 18,000 miles per hour or 29,000 kilometers per hour. It is a very fast speed.
How this all answers your question
The fact that space is fast has a really important consequence for this discussion: when you tweak the position by a few dozen meters or tweak the speed by a few miles per hour or so, you generally are not going to crash the satellite into the Earth. Crashing into the Earth requires removing thousands of miles per hour of speed from that satellite's orbit. The nearby orbits are not circular orbits but elliptical orbits in the perfect-$1/r^2$-force-law model, so they may approach closer or further from Earth on one or the other side; these are called the satellite’s “perigee” and “apogee” respectively. They do not maintain a constant speed but rather a constant angular momentum $L = m v r,$ so as they get further from the Earth (higher $r$) they move slower and as they get closer they move faster. But yes: eventually if they are perturbed enough, at their perigee they run into the Earth’s atmosphere which slows them down, and this causes their perigee the next orbit to be slightly lower which means that it hits more atmosphere and goes even slower, and so on, until it finally vaporizes from the heating of the air (or crashes into the earth if it is built in a way that it doesn't vaporize.)
In practice these drag forces can also motivate our satellites to have long-lived rockets on them and engage in station-keeping, which is an active coordination of rocket boosts designed to fix the difference between “where I am” and “where I want to be.” This can also be used on unstable orbits, in which case it is much like if I “help” my pen sit on its point by watching it very closely and whenever it starts to fall to one side I detect it extremely fast and give it a very precise “thwack” with my hand to knock it back up to the point of stability.
As a nice example of the latter, it turns out that the Earth-Sun system has several Lagrange points where the forces of the Sun and the Earth and the centrifugal aspects of co-orbiting the Sun with the Earth all balance out. The ones along the Earth-Sun axis are the “obvious” ones (of course, if the Earth pulls you one way and the Sun pulls you the other, at some point between them they should balance out and both be pulling you equally in either direction), but it turns out if you do the stability analysis you find out that these are unstable. (The ones on the far side of the Earth or the far side of the Sun are maybe less obvious, I will grant, but it's not too many equations to fall through to see that those must exist too.)
But, there are also two points, “ahead of us” and “behind us” by 60 degrees in the orbit respectively, which are stable. If you put a satellite there, it will stay there.
Think about why you might not want to put a satellite in such a position: There’s a lot of space dust in those spots! They are “vacuuming up” debris because they are stable. So you might prefer to do some active station-keeping to keep a satellite in one of these unstable positions: at least then you aren't running into space dust! This is becoming something of an issue for our current space environment, too: As we blast things off into orbit this region of phase space containing fast moving things in elliptical orbits that don't hit the atmosphere is progressively containing more and more junk, as the process by which things come out of this region of phase space is so slow. So we have to track all these little bits of space garbage and try to make sure that it doesn't hit our satellites—not fun!