There are lots of information about this topic, but I'm still greatly confused about something:

Astronauts in the ISS are in free fall all the time because they have only gravity acting on them, but when it comes to their floating motion I just cannot grasp the concept. How can a body be floating seems counter-intuitive to me.

I "concluded" from researching various texts around the Internet: because an astronaut is falling at the same rate with the ISS, it will remain stationary; likewise, if an astronaut jumped a meter in the ISS, his new "surface" will be that one meter above, so he will remain at that point, float in a way.

What I still do not yet understand are two things: Why did the astronaut really float? I mean is my conclusion correct, or am I utterly misguided? I drew my conclusion from the classic elevator example; if the cable of the elevator were to broke, you'd just float. This makes sense, because it's not the person's motion that causes the floating. But in the ISS, people can push themselves in a direction and move. What caused that direction? Did the astronaut push themselves from a place? Please explain this in a simple way from Newton's Laws.

My second question is that if the astronaut were in a spaceship orbiting the Earth, but that he would never hit the spaceship's parts from inside, and pushed himself in a direction, would he continue forever in that direction-we are assuming there are no obstacles. If so, as I learnt from the Newton's 2nd Law, there must be no forces acting on him. But there is gravity.

I read in places that this might be the centrifugal force, and if it is, would make much sense. But I wanna know why it is the case by the definition, "Because they are accelerating at the same rate."

  • 2
    $\begingroup$ Instead of "floating", what do you think should be happening instead? $\endgroup$ Apr 25, 2020 at 23:35
  • $\begingroup$ Related, possible duplicate: physics.stackexchange.com/q/56620/123208 But please let us know if the answers there (and on the various linked & related questions) do not adequately answer your question(s). $\endgroup$
    – PM 2Ring
    Apr 26, 2020 at 0:07
  • $\begingroup$ Why would they be floating in the first place? Is the conclusion I arrived at correct? $\endgroup$ Apr 26, 2020 at 2:01
  • $\begingroup$ Also, I couldn't find my answer in that link. $\endgroup$ Apr 26, 2020 at 2:02

3 Answers 3


There're two questions in what you've written. I'm answering the first one (about floating), since the second is conceptually different and should be in a different question.

Let's say you jump off a cliff. Would you say you're floating? The Earth's rushing towards you, so you might say "no", but imagine what it'd be like if we locked you in a box and threw the box off a cliff. Now you can't see the Earth rushing towards you anymore. Are you floating now?

That's what's happening on the ISS - the astronauts are falling, which is the same as floating in this context. Everything on the ISS is falling towards the Earth at the same speed, therefore they appear to be floating.

  • $\begingroup$ I asked the second question to understand the motion of the objects in the ISS. If indeed the astronaut can continue in a forever motion, what force compensates for the centripetal force, I was gonna ask. $\endgroup$ Apr 26, 2020 at 2:05
  • $\begingroup$ @Oregon.Vlogrago that should be a separate question. $\endgroup$
    – Allure
    Apr 26, 2020 at 2:27
  • $\begingroup$ Am I missing what "floating" really means in this case? I know they're all falling at the same rate, but I do not understand why falling at the same rate would make you float? Is it about frames of reference? $\endgroup$ Apr 26, 2020 at 11:13
  • $\begingroup$ Also, what do you think about my "conclusions"? $\endgroup$ Apr 26, 2020 at 11:13
  • $\begingroup$ @Oregon.Vlogrago what will you see if you're falling off a cliff inside a box? You won't see the items in the box hitting the bottom of the box, because the box itself is falling; you'll see them floating. I don't know what you mean by "conclusions". $\endgroup$
    – Allure
    Apr 26, 2020 at 11:24

My second question is that if the astronaut were in an infinitely huge spaceship orbiting the Earth, and pushed themselves in a direction, would they continue forever in that direction?

The equivalence principle is valid only locally. The required orbit velocity of an object at any place inside the ISS is pratically the same as that of the centre of gravity of the ship. So the relative velocity is the same and everything seems floating.

In a ship of 1000 km of diameter, the required orbit velocity of an object close to the region near the Earth would be greater than the ship CG, and the object would fall down in the ship's frame. The opposite for the region far from the Earth, and the object there would "fall up".


I "concluded" from researching various texts around the Internet: because an astronaut is falling at the same rate with the ISS, it will remain stationary;

In physics, the term geodesic path was introduced. All bodies in space that

  • move from the same point
  • in the same direction and
  • be at the same speed

will go the same path. An observer will notice that this path is curved. The curvature of the line depends on the distribution of the surrounding celestial bodies.

Near the Earth, the Earth has the main influence on the geodesic path. Up to a certain approximation, the influence of the sun and moon can be neglected. (In the reality of the space industry one could not). That is why the astronaut is floating together with the space station.

likewise, if an astronaut jumped a meter in the ISS, his new "surface" will be that one meter above, so he will remain at that point, float in a way.

This statement is not correct because one meter above the curved path is flatter.
In a space station of 100 m height, at 50 m height all bodies hover with the station, above them they hover to the ceiling (seen from the Earth) and closer to the Earth they fall to the floor of the space station.

How can you imagine that? Every change of the above parameters changes the path:

  • For a higher speed, the curve is flatter. So light is influenced by the earth only to a very small extent.
  • The greater the distance to the celestial bodies, the flatter the path becomes.

The latter is your case with the one meter jump.

I don't want to confuse you, but I must add that the disturbances between the bodies (in this case the space station and Earth) are interdependent. This cannot be measured because of the tiny mass of the station in relation to the Earth. But it is measurable for the moon-earth interaction. The orbit of the Earth is not only influenced by the Sun. The Moon makes the Earth wobble a little on its orbit around the Sun.

  • $\begingroup$ I don't wanna be rude, I really like your answer, but my physics knowledge is not enough to grasp geodesic path, let alone the concept I mentioned above. Can you explain it with the Newton's Laws? $\endgroup$ Apr 26, 2020 at 11:08
  • $\begingroup$ @Oregon.Vlogrago The earths gravitational attraction depends from the distance from the earth. That’s why for a given distance from earth the speed of all bodies rotating around earth is a constant. The centrifugal force is equal to the gravitational attraction F=mg. All bodies float together, as in your example with the station and the astronaut. At a smaller distance the speed for the circular motion must be higher, at a further distance lower. $\endgroup$ Apr 26, 2020 at 14:06
  • $\begingroup$ Your statement suggests that 2 objects, regardless of their mass, at the same distance from the earth, would orbit in the same speed. I don't think this must be true. $\endgroup$ Apr 26, 2020 at 15:32
  • $\begingroup$ So, when one of the masses is almost negligible compared to the other mass, as the case for Earth and Space ship or / and the astronaut one can approximate the orbit velocity v = SquareRoot(GM/r), where M is the (greater) mass around which this negligible mass or body is orbiting. The mass of tiny bodies is negligible. en.m.wikipedia.org/wiki/Orbital_speed $\endgroup$ Apr 26, 2020 at 15:43
  • $\begingroup$ What does this have anything to do with my question? $\endgroup$ Apr 26, 2020 at 15:44

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