Space Station Gravity
The International Space Station (ISS) orbits the earth in a circular orbit at uniform speed.
Suppose that an astronaut releases a ball from rest with respect to the center of the ISS. In what direction will the ball appear to move immediately after release, according to the astronaut?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
10 solutions
Moderator note:
Fantastic explanation. Under the stated assumptions, there are no dissipative forces that contribute to orbital decay.
In practice, the ISS actually loses velocity due to drag with the very thin atmosphere that exists at the height of its orbit. The deceleration is very slow, approximately 2 × 1 0 − 6 m / s 2 according to this estimate but still requires the ISS to do a "re-boost" about once a month. It looks like fun .
Since you know it would fell to earth and the ball is moving, you should know when the ball is moving downwards, it will have smaller orbit and resulting an acceleration. link : Oberth effect
Log in to reply
The Oberth effect has to do with energy efficiencies in accelerating a spaceship in orbit. I don't see how it helps here.
The answer given for this question is incorrect, although the answer would be correct if the ball were released from an object with an orbit that is beyond the Earth's atmosphere (e.g. Hubble). Since the ISS' orbit keeps it within the technical atmosphere, it is not traveling in the vacuum necessary for the given answer to be true. The tennis ball will be slowed by atmospheric drag and its deceleration will ultimately cause it to fall to earth, although this would be such a slow process that it would not be immediately apparent.
Log in to reply
Right, the deceleration is so small ( 2 × 1 0 − 6 m/s 2 ) , that for all practical purposes, the ball would appear stationary to the astronaut.
As with many problems on this site, this one is poorly constructed.
Log in to reply
Care to elaborate? If the formulation needs improvement, I'll be happy to make the changes.
Log in to reply
You should specify that the ball is released inside the station, thus avoiding the dissipative forces that would cause its orbit to decay. You should also specify that the question is posed on a short time scale to avoid the snags associated with long-term considerations of this problem (like when the space station eventually has to use its propulsion system to correct its slowly decaying orbit, etc.).
Great question but the answer seems vague. I understand the ball is at rest relative to the iss but “the ball will remain where it is” seems wrong due to the ball traveking though spacetime, thus the ball not remaining where it is.
Log in to reply
Typically, "remains where it is" means that the spatial coordinate in the given frame of reference is constant, even though the object occupies various events in spacetime.
Wouldn't the orbit of the ball decay?
Log in to reply
Only if it were subject to dissipative forces, such as air drag and friction. The air inside the space shuttle does not affect the ball, however, because they are at rest relative to each other.
The answer is correct, in the first approximation. However, the more precise answer depends on where was the ball released relative to the center of mass of the space station.
If the ball is released at the location of the center of mass, it will stay there, floating without motion. However if it starts off-center, it will definitely move, and it is possible, for example, that in 1/4 of the period it will move to the center. This was discussed in a problem that I submitted earlier. In an other initial condition the ball is off center by 0.4m and will move away from the center, by 15m, in one orbital period. See this problem .
Log in to reply
Good point, I've updated the question so that it's released from the center of mass.
If there was an initial velocity given to the ball before releasing it, the case would have depended on the direction of the velocity vector. Since the ball was released from the ISS without this initial force, the ball would remain where it is however when seen from an external point of reference, although the ball is not moving but the ISS is, they seem to drift apart.
ISS is in a decaying orbit, but has fuel to maintain a constant orbit. The tennis ball does not have fuel, and its orbital decay will bring it closer to earth than the ISS. It doesn't really matter what time scale in which this happens.
The question is in which direction will the astronauts in the space station see it move? We know the ball will stay in its place because its space... so when the astronauts leave the ball in that moment they will start to see how it goes against earth but is only vision, we know it will remain in the same place or fall to earth due to its gravity. This problem should be reconsidered.
Log in to reply
"The ball will stay in place because it's space..." Be careful with that reasoning. The ISS is so close to the earth, that the earth's gravitational field is still 98% of what it is on earth. The ball is falling down.
The point of the problem is that both the ball and the space station are in free fall. Relative to each other, they are at rest. So the astronomers will see the ball remain in place, even though from the perspective of earth they are both in falling motion.
I have a question. When they released the ball, it could be thought of as an object separate from the ISS. The ball has a lot less weight than the space station does, and thus, it wouldn't need the same speeds to sustain the same orbit as the ISS. With the ball having the same speed as the ISS, wouldn't its orbit be different than that of the ISS? I could be horribly wrong in this conclusion, but I thought I would raise the question anyway. Hopefully, someone smarter than I can explain why this would or would not be the case.
Log in to reply
The speed needed to sustain orbit is independent of the object's mass.
The reason is the same as why light and heavy objects fall at the same rate (barring air resistance). Yes, more force is needed to accelerate a heavier object; but it also experiences more gravitational force.
If you are into the mathematical side of physics:
-
the force needed to maintain a circular orbit of radius R and speed v is F = m R v 2 ;
-
the gravitational force for an object at distance R from the center of the earth is ( M = mass of earth): F = R 2 G M m ;
-
equating these shows the condition for a satellite to maintain a circular orbit; note that the m cancel: m R v 2 = R 2 G M m , v = R G M .
Log in to reply
Thanks for helping me out with this. It highlighted a misconception I had. I was taking the fact that the gravitational force between two objects is determined using the masses of both objects, and applying it to this question. But as you said in the beginning of your comment: "The speed needed to sustain orbit is independent of the object's mass. The reason is the same as why light and heavy objects fall at the same rate" I was forgetting this concept. (Which looking back on it seems pretty silly of me) I was equating the gravitational force with the rate of acceleration when I shouldn't have been. Once again, thank you for correcting me, it helped a lot.
This is what I had understand so far. So imagine that I am in the ISS(imagine the room where many machines are at), as I release the ball, the ball will seem like remaining at the exactly same place where I released it. So it remained where it is, in my vision. Is this the scenario that is portrayed by the selection D?
For the ISS to hold an orbital course around the earth, it needs to be propelled. Which it is, ofcourse.(look at this article: https://www.edn.com/design/power-management/4427522/International-Space-Station--ISS--power-system ) And, the ball will only carry the force which the ISS moved it with. So upon releasing the ball, the ISS is accelarating furthermore and the ball carries only the force from the previous time interval. Which when differetiated, would turn out that they seem to move from one another.
Furthermore, I want to expand on the awnser. From Mr.Vreugdenhil "Viewed differently: The space station is stationary relative to a local inertial frame. In such a frame, there is no gravitational force." There is gravitational force. That's why the earth has the moon orbiting around it in the first place.
And, when a screw would fly off, on the ISS, that screw would orbit earth till it would be accelerated by the earths pull strong enough. (Based on my knowledge of the gravitational field and a documentary about ISS reparation procedures and development of tools for that purpose)
F=y * m' * m" * 1/r^2 y- gravitation constant r- radius between earth middle point and the body one of the m, is the mass of earth and the other is the mass of the attracted object
First cosmic speed is around 7.92 km/s, that is also the speed with which a body needs to move at to orbit.
I choose the awnser C because of the reasons above. And, i would like to hear what would you add/change.
Log in to reply
In principle, an object that orbits a star or a planet needs no ongoing propulsion. Once it travels at the right height, speed, and direction, it will keep moving without help. A simple example is the moon, which orbits the earth but has no propulsion.
The reason is the principle of inertia : if no forces are acting on matter, it continues to move at a constant speed in a straight line. Without any force acting on the ISS, it would continue at the same speed, but in a straight line.
But there is a force acting on the ISS, all the time. (And also on whatever screws might fly off!) This is the earth's gravitational pull, and the equation is as you describe. Because this force acts perpendicular on the orbit of the ISS, it causes neither speeding up nor slowing down . Rather, this perpendicular force continually changes the direction of the motion ( centripetal acceleration). This is precisely why the ISS travels in a circular orbit instead of a straight line.
Back to the ball inside the ISS: Both the space station and the ball experience centripetal acceleration due to gravitational force. Both will therefore orbit the earth in a circular orbit. If we were to remove the space station, the ball would remain in orbit just fine.
Log in to reply
I think also it would orbit around the earth if it had the speed needed, but because the ISS is propelled continuously it will move from the ball with an acceleration caused by a force. The force that the ISS is exposed to continuously, isn't affecting the ball anymore. Which in effect says that: F(inert)/t < F(inert)/t + d(F(accel)/d(t) I pressume it would look somewhat like this
Log in to reply
@Cnarfeł Anioł – "... because the ISS is propelled continuously".
That is not the case. About once a month the ISS uses its boosters to increase the radius of its orbit by a few miles. It needs to do this only because it still experiences some atmospheric drag; if it were higher above the earth, it wouldn't even need this.
Thus the orbit of the ISS is stable in the order of a few miles per week; and the same would be true by the ball. Again, the (main) force that keeps the ISS in orbit is the earth's gravitational pull only , and the same force would affect the ball.
Log in to reply
@Arjen Vreugdenhil – The question is if a ball is released into ISS and after a small time interval boosters are applied to bring back the station in its orbit. Will the boosters affect the motion of ball?
Log in to reply
@Anurag Srivastava – It depends on how the boosters are used.
If the boosters were used constantly to oppose exactly the (very slight) drag force on the ISS, then the ball would remain at rest relative to the ISS, because the ISS would remain stationary relative to its local inertial frame.
If the boosters are off, then the ball seems to drift forward/upward at an extremely low rate of acceleration while the boosters are off; the reason is that the ISS itself accelerates slightly in backward/downward direction.
If the boosters are turned on to make up for weeks or months of orbital decay in a few minutes, then the ball may noticeably accelerate backward/downward relative to the ISS, because now the ISS is accelerating slightly in forward/upward direction relative to its local inertial frame.
In each case, the ball remains at rest in the local inertial frame (which implies movement in a near-circular orbit around the earth).
I don't really get how this is supposed to work. I think guess and check works best for me. Now I kind of get though how it works. The ball stays in the same place because of the 0% amount of gravity. I would rate this problem an 8 because it took me a long time to get to the conclusion. At first I was kind of confused but then I got it.
Log in to reply
When analyzed from a normal perspective (e.g. in which the earth is at rest), it is not true that there is zero gravity. The ISS experiences about 98% of the gravitational force it would experience on earth. It is a common misunderstanding that "in space" means no gravity; but that is only true in deep space, far from any source of gravitational attraction.
Both the ball and the space station experience gravity, and that is precisely why both of them fall toward the earth. They fall at the same rate, so if you are inside the ISS and you see the ball fall alongside you, it appears as if the ball is at rest. The reason why neither the ISS nor the ball gets any closer to the earth is, that they have so much sideways speed that the falling motion becomes a circular orbit.
Why would the ball fall towards Earth that defies the laws of conservation angular momentum? In this case the ball already has the velocity to be in the free fall once it is released from the Space Station it would continue to have the same velocity. With that velocity and direction it would continue to rotate around the earth?
Log in to reply
It would travel in the same circular or elliptic orbit as the ISS.
The ball is in orbit, just like the station.
The problem should specify assumptions: the track of the station is not being adjusted by thrusters of any kind, atmospheric influence is being neglected, etc.
Could you explain how you used this law?
I know so,the ball must move with ISS.
This is a strange question. "In relation to ISS"??? ISS is moving with a speed of 7,66km/s relative to the earth surface. It does so because it would otherwise fall towards the earth. Now it is instead falling around the earth because the speed makes it "miss the earth". Releasing a ball at rest in respect to the ISS (Should be 7,66km/s in the negative direction, ending up at 0km/h relative to the earth surface) would of course make it look like it moves along C if you could even see something moving away from you at 7,66km/s. The ball would move along A though since it's no longer "missing the earth" when it's falling. It would probably fall slowly and with a speed of 7,66km/h ISS would be back where it released the ball in 92 minutes. Lets just hope the ball have had time to get out of the way until then. Otherwise we would have a pretty gruesome event happening!
Log in to reply
When the ball is released at rest relative to the ISS that means it's velocity is equal to the velocity of ISS and in the same direction as well. As the ball and the ISS have the same velocity and they are moving identically they will remain relatively at rest to each other.
Disagree. ISS is in CONSTANT orbit, thus it is not moving towards earth. It is using propulsion.
Log in to reply
Agreed, ISS is in a low orbit that experiences relatively significant drag and that requires maneuvers to maintain -- as such, the ball should move away from the Earth, correct?
Log in to reply
The ISS is constantly falling towards Earth, but it has sufficient horizontal energy that it keeps moving in a circular orbit. The speed of the ISS will decrease due to drag (but at a slow rate of 2 × 1 0 − 6 m/s 2 . So it does not have enough horizontal speed to keep in the circular orbit. This would cause it to move towards Earth.
The gravity is zero so it remains in the original position neither it falls or moves until a force is applied to move in a particular direction
This is not true. The ISS is just 250 miles above the surface of the Earth. The gravitational force there is 88% as strong as the gravitational force on Earth's surface.
I think that the solution is quite simple; for a given radius and velocity the centrifugal force is proportional to mass; the gravitational force is also proportional to mass. If these 2 forces are in equilibrium for the space craft, they will also be for the ball.
If the ball is going 0 m/s relative to the target then it wont move towards nor away from the target. A great way to learn orbital manoeuvres like this is in kerbal space programme. I highly recommend the game for anyone.
This is not true. The ISS is just 250 miles above the surface of the Earth. The gravitational force there is 88% as strong as the gravitational force on Earth's surface.
Stay in the same orbit as the ISS
Could you explain why it will stay in the same orbit as the ISS?
The ball at rest according to the astronauts will be at the same velocity of the space shuttle and the astronauts therefore being at rest according to the astronauts frame of reference.
If the ball is released when the astrounat is within the ISS it will stay within the ISS. The result will be same even outside the ISS. The ball tends to stay in motion with ISS.
Relevant wiki: Newton's First Law
The ball will fall towards the earth. However, the space station itself is also freely falling to the earth. Thus the ball and station are both free-fall projectiles with the same initial velocity, and therefore have the same motion. From the perspective of the space station, the ball will remain in place (D).
Viewed differently: The space station is stationary relative to a local inertial frame. In such a frame, there is no gravitational force.