I saw a video today of a flash drive being thrown into space from the ISS (http://www.bbc.co.uk/news/science-environment-35487441).

What will happen to this flash drive? Will it burn up in the atmosphere, drift off into deep space or land back on earth?


2 Answers 2


According to Spaceflight Insider:

... After that, Volkov was the first to leave the module. He had with him a flash drive containing messages from last year’s 70th anniversary of Russia’s Victory Day. It was thrown retrograde from the space station so as to ensure that it will not collide with the station in future orbits before its orbit eventually degrades into the Earth’s atmosphere.

Spaceflight 101 adds that:

[the flash drive will] re-enter in a period of only a few months.

And includes this image of the release:

  enter image description here

Symbolic jettisoning of the video capsule containing messages from last year’s Victory Day celebration. Photo: NASA TV

As you can see, the flash drive was put into a sealed reflective bag similar to a candy bar wrapper. This will increase its surface area while not greatly increase its mass, making sure it loses its orbital speed and decays faster because of the nature of the drag equation:

$$a_D = - {1\over2}({C_D{A_v(t)\over{m_s}}})\ {pv_r}^2e_v$$

Where $a_D$ is total acceleration due to atmospheric drag (mind the sign, it is a deceleration), $p$ is atmospheric density (at ISS orbital altitude its density is around $10^{-8}\text{ Pa}$ or $1.45 \times 10^{-12}\text{ psi}$), $C_D$ is drag coefficient, $A_v(t)$ is the cross-sectional area of the object in the direction of travel, $m_s$ is the total object's mass, $v_r$ is the velocity magnitude relative to the ambient atmosphere, and $e_v$ is a unit vector in the relative velocity direction.

Once the flash drive loses its orbital speed and altitude, it will burn up in the upper Earth's atmosphere as it reenters it at at roughly 6.5 km/s, producing contact ionization heat due to friction with the atmosphere of over 6,000 °C (rule-of-thumb for peak shock layer temperature is as many Kelvins as speed of reentry in meters per second). And it takes a whole lot less of heat to completely melt and evaporate a typical flash drive reentering the atmosphere. Flash drive will start melting on its surface, fragment by ablation and outgas until it's completely gone. It might produce a small shooting star visible from the ground, if someone is lucky enough to see it, but that's about it.

Reflective bag (likely a metallized boPET film, better known as Mylar) that the flash drive was put in also increases its radar signature, so the object can be tracked by ground facilities like e.g. NASA's Haystack, and we can expect that its predicted reentry location will be announced beforehand. As technically a space launch, this object also, eventually, has to be registered with the UNOOSA (United Nations Office for Outer Space Affairs), as per the Launch Registration Convention which was signed and ratified by the launching nation (in our case Russia), and published in the OSOidx (Online Index of Objects Launched into Outer Space). And since it's not a classified object, it should eventually also find its way into the NORAD (North American Aerospace Defense Command) catalog and have published TLE, so the community at large can track it and predict its reentry.

  • $\begingroup$ I don't think the drag equation is the proper one to use at that altitude. It's derived from the terms in the Navier-Stokes equation and at 100km orbit, the mean free path is 1 meter, so the Knudsen number here is $K \approx 1/0.1 > 1$ if the packet is 10 cm. The fundamental assumptions used to derive that expression are violated. $\endgroup$
    – tpg2114
    Commented Feb 4, 2016 at 18:39
  • $\begingroup$ I am saying the package is small relative to the mean free path and so I don't expect the drag law to fit. Knudsen number affects both lift and drag since they both arise from viscous forces on a body. I did some additional reading and the approach used is to make $C_D$ a function of Knudsen number by experimental correlation to account for $\endgroup$
    – tpg2114
    Commented Feb 4, 2016 at 23:11
  • $\begingroup$ the molecular slip at the surface. So the equation does hold, but it's not under the same assumptions that happen in the continuum range and there are empirical correlations to make it work. It's active research to come up with a proper drag law that doesn't require those kinds of modifications. $\endgroup$
    – tpg2114
    Commented Feb 4, 2016 at 23:13

It will burn up in the atmosphere. Without and source of internal thrust it will decay into a lower orbit due to micro-drag forces, and it is too small to survive re-entry. There is no way it could land back on earth without a form of heat shield at the very least. Also from the video it appears that the Cosmonaut threw it against the direction of travel which would cause it to fall back to earth slightly quicker than if he had thrown it "in-front" of the ISS.

As a reference you can see what happened to this tool-bag in 2009 when a NASA astronaut accidentally let go of it while repairing one of the solar panels. It took about 9 months for the bag to de-orbit and burn up.

  • 2
    $\begingroup$ Quite likely he threw it like that on purpose or ISS could knock into it on next reboost. $\endgroup$
    – SF.
    Commented Feb 4, 2016 at 13:43
  • $\begingroup$ $100,000 for a tool bag?! $\endgroup$ Commented Feb 4, 2016 at 18:33
  • 4
    $\begingroup$ @LightnessRacesinOrbit I'm pretty sure that includes the cost of actually getting it to the ISS. Given it's 14kg, that's roughly in the ballpark of that number (10,000$/kg approx. cost) $\endgroup$
    – Ordous
    Commented Feb 4, 2016 at 19:55
  • $\begingroup$ @Ordous: Ah! That makes sense. Hehe. $\endgroup$ Commented Feb 4, 2016 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.