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?

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    $\begingroup$ I have never seen $\gamma$ as a gravitational constant. Are you sure you don't mean $G$? $\endgroup$
    – jng224
    Commented Dec 16, 2020 at 12:49
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    $\begingroup$ @Jonas The book I'm reading used $\gamma$, but yes, I should have used $G$, thank you! $\endgroup$
    – Larpee
    Commented Dec 16, 2020 at 12:51
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    $\begingroup$ @Frank Don't put an answer in a comment. $\endgroup$
    – Bill N
    Commented Dec 16, 2020 at 14:22
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    $\begingroup$ Related: physics.stackexchange.com/q/556043/195139 $\endgroup$
    – Sandejo
    Commented Dec 17, 2020 at 0:46
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    $\begingroup$ @Jonas I was taught that it was $\gamma$ in Germany. Seems like that's the only country where you can use both G and $\gamma$ (from a quick glance at various Wiki pages). $\endgroup$
    – Rakete1111
    Commented Dec 17, 2020 at 11:09

7 Answers 7


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!


Yes you are completely right! The astronaut could apply a force onto the satellite and make it break its previous circular orbit! Suppose in fact the following scenario: the astronaut exits the satellite and then pushes it with their arms. The consequence of this would be both the astronaut and the satellite breaking their circular orbit.

And another amazing consequence would be that, despite both the objects breaking their circular orbits, their shared center of mass (the center of mass of the system: astronaut plus satellite) would still maintain the same perfect circular orbit! This is due to one of the fundamental theorems of Classical Mechanics:

The motion of the center of mass of a system cannot be effected by forces internal to the system itself, only by external forces.

But of course the astronaut alone cannot provide an external force since they are part of the system.

This is why astronauts in the space station don't have to worry about displacing their orbit around earth, the orbit of the center of mass is safe, in the sense that it cannot be altered by their actions, and so until they remain close to the space station there would be surely no problem. But suppose one of them pushes themselves away from the space station with all their force, then in principle they could alter the orbit of the space station, but in reality this is not a problem since the space station is way more massive than a human, and so the action of pushing themselves away would make almost no difference for the system as a whole, since their mass contributes so little to the position of the center of mass.

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    $\begingroup$ I'm not sure about "their shared center of mass would still maintain the same perfect circular orbit". The gravity of the Earth is an external force in this context. The field changes in a non-linear fashion along the line connecting two distinct points. I think this means that after the astronaut and the satellite separate, the sum of forces exerted by the Earth is different than a hypothetical force calculated for their center of mass. (cont'd) $\endgroup$ Commented Dec 17, 2020 at 0:33
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    $\begingroup$ (cont'd) Imagine the astronaut catapults himself forcefully enough to reach a hyperbolic speed. His trajectory is unbound and non-periodic. The satellite is massive and the impulse changes its orbit only slightly to some ellipse. The new orbit is bound and periodic. How can the center of mass "maintain the same perfect circular orbit" which is bound and periodic? $\endgroup$ Commented Dec 17, 2020 at 0:33
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    $\begingroup$ "cannot be effected" *affected $\endgroup$ Commented Dec 17, 2020 at 1:54
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    $\begingroup$ @KamilMaciorowski You are completely right. So the more interesting question is whether shifting masses, say, inside a hollow sphere could change the whole thing's orbit in an inhomogeneous gravitational field. The first intuition says "no" (because we are trained to look at it under the closed system paradigm) but the hull is transparent to gravity, so it's not at all "closed" in this context. $\endgroup$ Commented Dec 17, 2020 at 8:54
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    $\begingroup$ @Acccumulation Not necessarily. "Effect" is a verb that is somewhat synonymous to "cause". They could also mean "affect" as in "change". But it's not certain. $\endgroup$
    – Arthur
    Commented Dec 17, 2020 at 10:54

Until there is any external force on the satellite-astronaut system there will be no change in its velocity or orbit. Any other force arising within the system will be internal force and won't affect the system velocity. if astronaut applies force to the satellite the satellite will apply equal force to astronaut with net zero force on the system.

  • $\begingroup$ "Until there is any external force"?! The fact that the system is literally orbitting planet Earth means there's always an external force aplied: Earth's gravity. Otherwise poor astronaut and satellite sidekick will just drift off into space never to be seen again... $\endgroup$
    – Will
    Commented Dec 17, 2020 at 5:55
  • $\begingroup$ Well that force is already keeping the satellite in orbit otherwise there will be no question to ask. The question is whether any movement or force arising within the satellite will change its velocity and orbit. $\endgroup$
    – TKA
    Commented Dec 17, 2020 at 9:08
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    $\begingroup$ I agree with the answer, but IMO it should be added that this is only valid when the size of the satellite-astronaut system is small compared to the radius of the Earth, and the mass is small compared to the mass of the Earth. $\endgroup$ Commented Dec 17, 2020 at 21:38

Let's take a look at the equation :

$$v_o =\sqrt \frac{GM}{R}$$

The main thing to note here is that what you wrote with the above equation is derived for the center of mass of that satellite - astronaut system since the concept of center of mass is what makes us capable of applying Newton's laws to derive to these equations.

Newton's laws are only applicable for point masses and that's why you need to define center of mass for larger bodies to use Newton's laws on them.

Coming back to your question :

  1. Considering the satellite as your system

In this case your intuitions are absolutely correct. The push of the astronaut will be considered as an external force since it is not the part of the system and hence the satellite's center of mass will definitely deviate from the path .

  1. For both the astronaut and the satellite as a system

Then the velocity of the center of mass doesn't deviate. The astronaut pushes the satellite and the satellite pushes the astronaut and hence there is no external force on this system and hence the center of mass of this system doesn't experience any change.

Hope it helped 🙂.


Noumeno has a correct answer, but I wanted to add to it. In that answer, they point out that internal forces cannot affect the position of the center of mass of the system. However, it may not be obvious why they are "internal forces." In fact, they do not have to be!

The difference between internal forces and external forces is a choice made when formulating the problem. If we choose to treat "satellite and astronaut" as our system, we can go down the route of saying that the astronaut pushing on the satellite is an internal force, thus it cannot affect the trajectory of the center of mass of the whole system.

However, we can also choose to say we have two independent entities, a satellite and an astronaut, and it just so happens that the astronaut's position is inside the satellite. Now we can no longer claim the interaction is an internal force. Why? Because we have chosen to set up the problem such that these are now two separate objects interacting, externally. We will find that the result is exactly the same as if we thought of them as one system, but the math we use to get there is a little different.

When the astronaut pushes off of the side of the satellite, it does indeed break the circular orbit. It breaks it for both parties. Both parties are pushed into an elliptical orbit. Which orbits they end up in depends on which direction the astronaut pushed in (the 6 major directions are labeled prograde/retrograde, radial/anti-radial, normal/anti-normal based on the direction the satellite is traveling), but they'll both be elliptical.

Now, if that was the final interaction, that would be the end. However, it should be quite clear that, if that is the final interaction, that means the astronaut has jumped out of the satellite and is now floating away from it! And, if you were to actually plot out all of those potential orbits they could end up in, you would find that all of them show that the center of mass of "satellite plus astronaut" is following its original path. (consistent with the "internal forces" explanation). It just takes more math to prove this to be true.

However, inside the satellite, all good things must come to an end. Eventually the astronaut will gleefully impact into the other side of the satellite. This will perturb both of their elliptical orbits. Again, with a bunch of math, you find that if the astronaut is brought to a stop inside the satellite, those perturbations are exactly what is needed to put both of them back on a circular orbit.

So whether you think of them as internal or external forces, the result is the same. Thinking of them as internal forces, as Noumeno does, gets you to the answer really quickly. It's probably the best way to think about it. However, if you are uncomfortable with that approach (it feels a little handwavy), you can always treat the satellite and astronaut as two separate objects, and do all of the math to explore the elliptical orbits that can occur. You will, of course, end up with exactly the same answer. One approach is elegant, one is brute force with a bunch of extra math. But, because physics is consistent, both approaches yield the same outcome.

  • $\begingroup$ +1 nice answer, but regarding the "you would find that all of them show that the center of mass of 'satellite plus astronaut' is following its original path" bit: I think you might have overlooked one little monkey wrench here: escape velocity. What if the astronaut is ridiculously strong and kicks off against the satelite with such force that they propel their body over the escape velocity threshold (meanwhile dropping the satellite to a much lower orbit), hence setting off on a one-person one-way journey to outer space. The system is now broken, its center of mass on an ever-deviating path. $\endgroup$
    – Will
    Commented Dec 17, 2020 at 6:37
  • $\begingroup$ @Will That part will still hold, although my later claims that both parties will be in an elliptical orbit would fall apart -- the astronaut would enter a parabolic or hyperbolic orbit. But if you thought from a center of mass perspective, you'd still find the center follows the expected path. 'Course the center may not be a meaningful thing to look at, if the astronaut escaped. A bigger issue would arise if one of the orbits intersected the earth, as that would introduce additional forces... and probably a RUD (Rapid Unexpected Disassembly) $\endgroup$
    – Cort Ammon
    Commented Dec 17, 2020 at 8:59
  • $\begingroup$ In extreme cases, one may find that both the astronaut and the satellite end up on escape trajectories, in opposite directions. $\endgroup$
    – Cort Ammon
    Commented Dec 17, 2020 at 9:00
  • $\begingroup$ @Will I thought about this some more, and while the rules about internal/external forces still work, your escape velocity case is interesting because it is a three body problem. The nice elliptical/hyperbolic orbits are solutions to the 2 body problem. In the case of escape, we have to consider the transfer of momentum from the satellite and astronaut to the earth. The whole satellite/astronaut system can still be thought of as a unit, but that unit has to pull the Earth "off course" a little as it escapes. $\endgroup$
    – Cort Ammon
    Commented Dec 17, 2020 at 17:43

As others have mentioned, the equation that you're referencing is for the center of mass of the system, not some extended body representation. Also the mass of the satellite for a manned system will typically be much larger than the mass of the people manning it, so they can make only small perturbations to the system in any case.

Another point not yet mentioned, is that the formula is also only strictly true for a spherical Earth isolated from other astronomical bodies. In practice that is not real and the perturbations from these other factors will outweigh the effects of the people moving around. The Earth is not spherical and for a precise orbit determination you need to account for the gravitational pull of other bodies such as the Sun and Jupiter. At low orbits, there are effects from the upper atmosphere. At high orbits, there are effects from things like solar radiation pressure. So your perfectly spherical orbit is certainly doomed in practice, even if everyone onboard stays still.


If the astronaut is inside the satellite, then any force they exert on the satellite will have a reaction force that accelerates the astronaut. Eventually, the astronaut will hit the other side of the satellite, and exert an opposing force. So nothing inside the satellite can cause anything more than a transient effect on the orbit.

If the astronaut jumps out of the satellite, they won't be a significant fraction of the satellite, so the effect on the orbit will be minor. Moreover, there still won't be a permanent effect; both the satellite and the astronaut are now in orbit, and clearly their orbits intersect (the astronaut was previously inside the satellite), so they will collide further in their orbit. At that point, the astronaut will again exert a force that cancels out the force they initially exerted.

The phrasing of "breaking" circular orbit suggests that you think of the orbit as some sort of track that the satellite has to stay on, and maybe you even think that orbits have to be circular. No orbit is exactly circular. If a satellite's orbit is perturbed, it simply goes into a slightly different orbit, possibly more or less circular than its original. There are many satellites with orbits that are far from circular. If a satellite deviates significantly from its desired orbit, whether that's circular or not, they have thrusters to compensate.


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