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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?

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    $\begingroup$ 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. $\endgroup$
    – Innovine
    Nov 1, 2021 at 17:18
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    $\begingroup$ 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. $\endgroup$
    – Innovine
    Nov 1, 2021 at 17:20
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    $\begingroup$ This has happened at least once: youtube.com/watch?v=1vXdRUIZ_EM $\endgroup$
    – Bakuriu
    Nov 1, 2021 at 21:25
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    $\begingroup$ Of course it will stay in place, but only for as long as cows stay spherical. $\endgroup$
    – Andrei
    Nov 2, 2021 at 18:16
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    $\begingroup$ @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. 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 $\endgroup$
    – Innovine
    Nov 3, 2021 at 13:26

3 Answers 3

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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.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – called2voyage ♦
    Nov 3, 2021 at 15:05
  • $\begingroup$ ... and even if the ISS were in frictionless circular orbit, and the wrench were on the same orbit, tides would cause the distance between their two COMs to increase and decrease by small amounts. $\endgroup$
    – Eric Towers
    Nov 4, 2021 at 16:55
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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.

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    $\begingroup$ "The ISS is not, strictly speaking, in orbit" Ridiculous. $\endgroup$
    – Organic Marble
    Nov 1, 2021 at 17:48
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    $\begingroup$ "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". $\endgroup$
    – Woody
    Nov 1, 2021 at 18:13
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    $\begingroup$ Perhaps it might be clearer to say: "is not, strictly speaking, in a stable orbit"? $\endgroup$
    – spechter
    Nov 2, 2021 at 0:41
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    $\begingroup$ @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. $\endgroup$
    – spechter
    Nov 2, 2021 at 3:23
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    $\begingroup$ 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)? 😉 $\endgroup$
    – Matthew
    Nov 2, 2021 at 17:54
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Organic Marble
    Nov 8, 2021 at 13:22

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