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If an astronaut doing a repair outside the International Space Station (ISS) gently lets go of a tool into space, will it keep "flying" (orbiting) at the same spot relative to the ISS where it was left, forever? I'm not talking about throwing the tool in some direction, just gently letting go of it.

Since it was released from the same speed the ISS is circling Earth, and there is no air to cause friction and stop the tool, I wonder if it will remain at the very same position forever. If not forever, for how long, and why?

End Anti-Semitic Hate
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André Dias
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    Interestingly, if you push the object so it drifts away, exactly perpendicular to the stations orbital plane, it'll come back about 45 minutes later and you might be able to grab it again. Someone will eventually do a wireless yo-yo trick like this. – Innovine Nov 01 '21 at 17:18
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    If you released a tool when doing a spacewalk outside of a spaceship doing an interplanetary journey, it'd remain stationary (more or less) for a much longer time. Since its still in solar orbit it'll eventually drift, but over days or weeks, not minutes as on the ISS. An interstellar craft would have the tool alongside it for months or years. – Innovine Nov 01 '21 at 17:20
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    This has happened at least once: https://www.youtube.com/watch?v=1vXdRUIZ_EM – Bakuriu Nov 01 '21 at 21:25
  • Different but related (and lots of great answers and a few GIFs too) What kinds of things have been tossed out of the ISS? – uhoh Nov 02 '21 at 04:30
  • @Innovine Can you elaborate on that? I'm thinking if you push something so that it drifts away, it will take much longer than that to separate by one full orbit... – user3067860 Nov 02 '21 at 12:05
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    Of course it will stay in place, but only for as long as cows stay spherical. – Andrei Nov 02 '21 at 18:16
  • @user3067860 If you push the tool perpendicular to the stations orbit (the orbit normal), it enters into a very similar but differently inclined orbit. This orbit will intersect the stations orbit again on the opposite side of the planet, so you push the tool, it drifts away, slows down (relatively), and then gradually returns to you with the speed you threw it at. In practice it'll probably drift off due to lots of small influencing factors, but in general this is the behavior along the orbit normal. the objects orbit side by side, but diverge, and then reconverge a half-orbit later. – Innovine Nov 03 '21 at 13:09
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    @user3067860 You can see this effect in any orbit diagram showing a plane change manouver. When you release the tool with some velocity in the normal direction, you are at one node in the plane change. On the opposite side of the planet is the other node, where the tool returns to the station. https://tinyurl.com/2tw2btmx Here the ISS would move in the grey plane, while the tool moves in the yellow one. Should be easy to see how it departs, then returns again after a half orbit – Innovine Nov 03 '21 at 13:26
  • @Innovine Thanks! It all makes sense with the picture. – user3067860 Nov 03 '21 at 14:13
  • Bye bye expensive tools. https://www.newscientist.com/gallery/dn17038-hubble-space-tools/ – Liam Mitchell Nov 04 '21 at 04:18
  • From the perspective of the Earth, the lost tool is on a slightly different orbit. From the perspective of the space station, the tool will move in a circle, eventually returning to its original position. – Neal Nov 04 '21 at 12:38

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It will not. Once released, the object is in a very slightly different orbit from the space station.

If you put the object further out or further in from Earth than the space station it will be slightly further from, or slightly closer to Earth, while moving at the same velocity as the Space station, which will result in a slightly more, or less elliptical orbit, with a different semimajor axis and orbital period. As a result, it won't be able to remain stationary with respect to the space station

If you put the object ahead, or behind the space station in its orbit, if the space station's orbit is elliptical, the object will be moving too fast, or too slow relative to the space station's orbit to maintain exactly the same orbital path that the space station is travelling, and relative to the space station, the object will drift.

If you put the object perpendicular to the plane of the space station's orbit and release it, the orbit the object is in will have a very slightly different orbital inclination to the space station, and these orbits must cross, as such the distance between the space station and the object cannot remain constant.

Any position you release the object will be a combination of the above cases.

As a result, unless the space station is in a perfect, circular orbit, there is no place on its orbit where you can release an object and have it forever remain at a constant distance from the space station, without orbital adjustments.

JDługosz
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notovny
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Great question to illustrate small but measurable effects of orbital mechanics!

NO, even if both wrench and ISS are idealized point masses in the same circular orbit.

The mass of the ISS imposes a non-trivial gravitational acceleration on the wrench. If the wrench is dropped within EVA distance of the ISS, mutual gravitation will displace it on meter-scale distance within a few orbits. Surprise with such a small mass, but this is a 3body problem!

NO if the circular orbits are non-coplanpar since they would intersect.

NO, if coplanar eliptical orbits have the same period but different eccentricities. The pair would do a retrograde orbit around each other.

NO because of atmospheric drag. The ISS is in a slow spiral re-entry trajectory. The wrench, not so much due to its lower drag coefficient and higher sectional density.

The best you could do is place the wrench at the ISS center of mass (COM) which may be inside or outside the pressure hull. This would eliminate gravitational and tidal effects. Drag would still be an issue

A dated version of the COM location, for multiple conformations, is given in http://athena.ecs.csus.edu/~grandajj/ME296M/RevAB_Volume%20I%20Signed_updated.pdf. Too much information?

According to Scott Manly

a tool bag was dropped during an EVA on the ISS and it deorbited in about 9 months.
Woody
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    "The ISS is not, strictly speaking, in orbit" Ridiculous. – Organic Marble Nov 01 '21 at 17:48
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    "Orbit" as in "... especially a periodic elliptical revolution." (Oxford dictionary). ISS's trajectory is almost elliptical. And almost periodic... until it degrades and spontaneously re-enters. These details are essential to answering the posted question: if they are taken into account, the answer is "no". If not, the answer is "yes". – Woody Nov 01 '21 at 18:13
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    Perhaps it might be clearer to say: "is not, strictly speaking, in a stable orbit"? – spechter Nov 02 '21 at 00:41
  • @spechter: But then, what IS in a stable orbit, given sufficient time? Classic example: the radius of the moon's orbit is growing by 3-4 cm/year. – jamesqf Nov 02 '21 at 02:54
  • I guess it depends on the time frame you choose. The duration of the EVA? The lifetime of the spacecraft? The expected extinction of our species? I think in this problem the EVA duration is reasonable. Even in a single orbit, the hypothetical wrench will not "stay at the same spot WRT the ISS" – Woody Nov 02 '21 at 03:23
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    @jamesqf - Excellent point, well made, there is no perfect vacuum (etc) for an infinitely stable orbit. I think Bruce's first point is that the orbit noticeably degrades relatively quickly (days-months) unlike most natural celestial objects which we don't have to worry too much about over human lifetimes. – spechter Nov 02 '21 at 03:23
  • Or you could look at it as an n-body problem. All closed orbits in a 2-body problem are mathematically stable (absent of aerobraking). Not so if n>2. – Woody Nov 02 '21 at 03:26
  • Just for laughs, I calculated the gravitational acceleration due to the ISS mass. With some arbitrary assumptions about the location of the wrench when released, I came up with a displacement of about 1 social distance during a 90 minute orbit. – Woody Nov 02 '21 at 04:16
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    Oh, dear, is "social distance" a unit now? And (not that it matters overly much for your approximation) is that an imperial social distance (6ft), or a metric social distance (2m)? – Matthew Nov 02 '21 at 17:54
  • @Matthew And, worse yet, some countries/bodies just recommend 1 m or 1.5 m. I think this unit probably has more variants than the ton/tonne. :) – reirab Nov 03 '21 at 20:08
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If your tool has even the slightest relative movement to you, even just 1 mm/s, you can expect it to move away fairly quickly. 1 mm/s results in at least a 10 foot ellipse, also up to 50 feet per orbit drift. The figure below shows the orbit it will follow, relative to you, depending on the direction you send it. As there will likely be some "Posigrade"or retrograde component to it, it won't return directly to you.

It's an old picture I keep for reference because it's not simple to re-derive it. I'm sorry I don't know where it's from, probably screen shot from an old scanned NASA PDF. Google doesn't turn up any results.
enter image description here

This is all for perfect orbital mechanics, then there's still air friction, radiation pressure, etc.

tomnexus
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  • The last 2 cases in the diagram are essentially identical to one in JSC-1058 Rendezvous and Proximity Operations Handbook part 2 page 2-14. This document used to be available on James Oberg's website but it looks like the domain registration on that has lapsed :( I think the full diagram is from a JSC rndz training manual, I have a scan of what appears to be a newer version of this image that ISTR came from there. – Organic Marble Nov 08 '21 at 13:22