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In 2008, NASA astronaut Clayton Anderson tossed a 635-kg tank of ammonia coolant overboard from the International Space Station (ISS). Subsequently the tank burned up in earth atmosphere as planned.

Obviously in the 1-g environment on Earth's surface, no human could have lifted the tank let alone tossed it. The astronaut was only able to do so because the tank weighed much less in the microgravity environment of the orbiting (free-falling) ISS.

Because I was asleep in much of physics class in high school, I am unable to work out a formula for how much weight mass a human can chuck overboard from the ISS unassisted. In particular, I have only the vaguest notions of inertia and action/reaction. (What I mean is, I can look up Newton's laws of motion but I am unable to use them to solve my question.)

In the picture within the article that I linked to in my first paragraph above, we see that the astronaut appears to have his feet strapped into a footrest at the end of a robotic arm.

Am I right that this is vitally important because otherwise he would not have a backstop to push off from?

Anyway, can you give me a way to calculate, using simple arithmetic, what is the heaviest object that an average-strength human can toss overboard from the ISS?

Also, why is not all trash discarded from the ISS by being thrown overboard manually or using a device instead of being packed into a space-craft that will depart from ISS and burn up in the atmosphere?

(A day later:) Thank you to all three who took the time to reply so far, I upvoted each of your Answers. I could not for the life of me single out one for "acceptance" so no green checkmark from me this time :)

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3 Answers

Although the object has no weight it still has mass.

With no friction the force is simply a question of how quickly you can accelerate it. It's basically the same as pushing an object on an ice rink.

So if you can provide a force of 800N, equivalent to lifting you own weight on earth ( eg. doing a pullup) you can accelerate the tank with a = F/m = 800 N/635 kg = a little over 1m/s^2 If you are in contact with it for 1 second it will leave the station with a (relative) velocity of 1m/s

The real force to deorbit it comes from the atmospheric friction on the large surface area.

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I understand (I think) the analogy to pushing an object on an ice rink: (almost) no friction there, no friction at all (except for a few stray air molecules) when tossing overboard objects from ISS into space. Also, thank you for noting the weight/mass distinction, which of course I managed to miss in my Question (duh...) However, if I understand you correctly then the absence of friction is the only difference here? The astronaut cannot lift the 635 kg tank any more than he could on Earth? –  Eugene Seidel Nov 27 '12 at 14:18
... and so the opposite force exerted on the ISS is also 800 N and if ISS mass is 450 tons then tossing the tank earthward accelerates ISS by 800 N / 450,000 kg = 0,00177... m/s² in the opposite direction? –  Eugene Seidel Nov 27 '12 at 14:39
@EugeneSeidel "lift" rather implies gravity - it doesn't really mean much in orbit. But if the tank was moving toward the astronaut at speed it would squash them just as effectively as on earth. Yes the speed, and so orbit, of the ISS is slightly changed in the opposite direction. This is what thrusters do, they "throw" gas at high speed in one direction and move the ISS in the opposite –  Martin Beckett Nov 27 '12 at 15:35
Yes, I never doubted that a person could get squashed by a heavy object in space just as on Earth. I take it, then, that the astronaut cannot bend down from where he is attached to the robotic arm, pick up and "raise" (in relation to his own longitudinal axis) the tank (what would be the technical term for this -- alter its vector?) by a meter using his muscle power alone. Although the tank is practically weightless, it still has the same inertia that it had back on Earth. But wait a minute, he was able to push it away. So... I thought I had understood it but now I am feeling confused again. –  Eugene Seidel Nov 27 '12 at 15:55
Nevermind it should not make any difference whether the astronaut pushes an object away from him or pulls it toward himself, the same formula as given by Martin above applies, so the answer to my question is, yes he can "lift" (= pull toward himself) the tank. –  Eugene Seidel Nov 27 '12 at 16:39
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I will first answer your second question: Waste is usually collected and then discarded in heavier, more massive objects (such as this tank or spacecrafts) as these can be easier tracked than smaller particles and can also be (ideally) ejected in such a way that they enter the atmosphere quickly anc hence burn up quickly.

If waste were to be simply ‘thrown out of the window’, it would circulate around Earth for quite a while – after all, the atmosphere up there is rather thin and will take a long time to slow the small waste particles down sufficiently that they enter lower regions and burn up in the atmosphere.

As even the smallest particles can be dangerous, both to the ISS and other spacecrafts, one usually tries to avoid polluting orbits.

Regarding your first question: The key equation is $\vec F = m \vec a$. On Earth, $\vec F$ is composed of a force caused by the human throwing something overboard, gravity and possibly friction and other forces (if you are flying a plane and want to throw something away):

$$\vec F = \vec F_{\textrm{grav}} + \vec F_{\textrm{other}} + \vec F_{\textrm{human}}$$

On the ISS, we usually¹ have $\vec F_{\textrm{grav}} \approx 0$, and other forces caused by friction are negligible as well, which then means that a human could accelerate any body to any speed (in the Newtonian limit) provided that he were able to exert even a small force sufficiently long.

However, there are two catches:

Firstly, the body accelerated by the human will exert a force on the human of equal magnitude but opposite direction, too - the footrest is essential, as you noted. It connects the human to the ISS and hence means that the inertia of the whole ISS resists the acceleration due to the reaction force of the body. This effect becomes obvious when you look at the ‘centre of mass frame’ rather than the frame of the ISS.

In this frame, take the ISS and the tank to be at position $0$ at $t = 0$, we will use a 1-dimensional coordinate system now. Subscripts denote the object having the corresponding property, superscripts the frame of reference.

The ISS now exerts a force on the tank, causing it to accelerate with an acceleration $a$ up to a velocity $v_{t}$ (an astronaut pushes the tank away, accelerating it by $a^{I}_{t}$ to speed $v^{I}_{t}$ from his point of view).

As you know, momentum has to be conserved, and while the total momentum before the push was zero (everything was neatly sitting at the origin), it is not anymore: the tank has momentum $p_{t} = m_{t} v$. Since

$$p_{\textrm{total}} = p_{I} + p_{t} = 0\quad,$$

we have

$$m_{I} v_{I} = p_{I} = -p_{t} = m_t v_t$$

and, as a result from a Galileo transformation:

$$ v_t - v_I = v_t^I $$

That is, while an astronaut may have the impression that the tank is flying away as a result of his action, the ISS is also flying away in the opposite direction. If $m_I \gg m_t$, $|v_I| \ll |v_t|$, but if $m_I \approx m_t$, $v_I \approx - v_t$. There is hence an upper bound on the mass an astronaut can throw away, given by how much you want the ISS to move itself ($m_I \approx 4.5 \times 10^{5}\textrm{ kg}$ according to Wikipedia).

Secondly, $F_{\textrm{grav}} > 0$ even at the ISS. But given the numbers in the footnotes and assuming $F_{\textrm{human}} = 500\textrm{ N}$, we have $m_{\textrm{max}} = 6.631\times10^3\textrm{ kg}$ (by setting the acceleration caused by the human equal to that caused by centrifugal/gravitational force).

Note that this calculation is strictly Newtonian and does not take into account any relativistic effects. So don’t throw stuff too fast!

[1] If you calculate the acceleration due to gravity at $6371\textrm{ km} + 400\textrm{ km}$ away from Earth's ($m_{E} = 5.9736 \times 10^{24}\textrm{ kg}$) centre, you will find $g = -8.69\textrm{ ms}^{-2}$. However, the centrifugal force experienced by the ISS due to its orbit ($v = 7706.6\textrm{ ms}^{-1}$) is $a = 8.7714\textrm{ ms}^{-2}$, resulting in $g_{\textrm{actual}} = 0.0754\textrm{ ms}^{-2}$. This would indicate that, from the point of view of an astronaut, it is actually easier to throw stuff up rather than down. Any comments on this?

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I believe I have understood all. I read "If mI≫mt, |vI|≪|vt|, but if mI≈mt, vI≈−vt" as, "If the mass of the space station is much greater than the mass of the tank, then the space station will be accelerated "up" (= away from Earth) by much less than the tank will be accelerated "down", which is not the case if the two masses are approximately equivalent: then, each will be accelerated in opposite directions by approximately equivalent amounts." (Why the vertical bars around vI and vt but not mI and mt?) ---- So, the answer to my first question is 6.631 tons (if human strength = 500 N). –  Eugene Seidel Nov 27 '12 at 15:12
I am still baffled as to why you equated the gravitational force to zero, then later calculated it. Then your calculation of $m_{max}$ has no clear meaning associated with it. What's even more confusing is $g_{actual}$ and your speculation that it's easier to throw things up. –  AlanSE Nov 27 '12 at 15:12
@AlanSE: I first set the gravitational force to zero to get an estimate based on the assumption that we only have to take care of inertia and momentum conservation. The calculation of $g_{actual}$ is based on the fact that gravity acts downwards (towards Earth) and the centrifugal force perceived by an astronaut on the ISS acts upwards (away from Earth). The combined effect of the two gives the "actual" gravitational force $g_{actual}$ (away from Earth). To overcome $g_{actual}$, an astronaut has to accelerate an object by $-g_{actual}$, which I can plug into $F_{human} = m_{max} g_{actual}$. –  Claudius Nov 27 '12 at 15:27
@EugeneSeidel Yes, you have understood correctly. $v_I$ and $v_t$ will have opposite directions, hence we have to compare their magnitudes if we want to compare them in a meaningful way ("slower", "faster"). $m_t$ and $m_I$ are masses of the tank and the ISS accordingly and hence always positive. Note that the restriction of 6.6 tons only applies if the astronaut wants to overcome the force of gravity (whatever that may be, I am not entirely sure myself and also appear to appear rather confused :)). –  Claudius Nov 27 '12 at 15:30
Re [1]: $g_{actual} = 0$, it's in free fall. You must be getting a non-zero value based on rounding error or something. Whatever microgravity there is would be due to tidal effects. –  Retarded Potential Mar 14 '13 at 20:33
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I struggled with this question. We have to be realistic. This is the altitude of the ISS over time:

ISS altitude

This reflects the operational range of the ISS, this is important because I find this deceptive:


Does this mean it burned up on the first pass? My conclusion is that this can not possibly be the case.

Consider an astronaut able to exert $(70 kg) ( 9.8 m/s^2)=686 N$ of force, over a length of $0.5 m$. That would result in a kinetic energy imparted of $KE=F d=343 J$. Assume gravity in the vicinity of the ISS to be roughly $9.0 m/s^2$, and use the very simple $m g h $ formula to find that, if we assume only circular orbits, he would only be able to change the altitude by:

$$h = \frac{343 J}{ 9 m/s^2 \times 635 kg} = 6 cm$$

We should also consider that orbital energy is $\frac{G M }{2 r}$, not just $\frac{G M }{r}$, so in reality it's $12 cm$, not $6 cm$. Also, we can consider that only the average radius of the radius would drop $12 cm$, so the lowest point of the orbit could conceivably change $24 cm$.

Compare a few centimeters to an ordinary orbital range of the better part of 100 km for the ISS. It simply doesn't work for the throw itself to make a significant difference. The best they could possibly hope for would be that the throw avoids collision with the ISS on a second pass. This is possible, not because of a few centimeters of movement, but because the period of the orbit can be changed with a throw, making it sure that it won't later risk collision.

The tank probably took many orbits to finally burn up in the atmosphere. Some quick searching confirms this.



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More realistically the astronaut didn't actually throw it - they were merely a clamp to hold it while the CanadaArm swung it away. Even if the arm could only exert a few 100N (without stretching the astronaut) it is 10m long so get's it at least that far away. –  Martin Beckett Nov 27 '12 at 15:57
@MartinBeckett Indeed, getting it away should be the main concern. I took the derivative of the orbital period w/r/t radius, multiplied by 12 cm and orbital velocity and got $3 \pi (12 cm) \sqrt{R/(MG)} (7.9 km/s)=1.3 m$ displacement per orbit. This, by itself, seems like too little for comfort. If I take $1 m/s$, which others have proposed for throw velocity, and multiply it by 90 minutes, I get over $5 km$. So maybe it's deceptive to think that just because something is moving away from you it won't come back and hit. Well at least it would be a relatively gentle bump. –  AlanSE Nov 27 '12 at 16:05
I think the important thing is to get it slightly below/behind and let orbital decay take care of it - you just need to be careful not to throw it into a higher/ahead orbit so it doesn't decay into you. There must be some space equivalent of "not spitting into the wind". .ps didn't know you could do tex in comments, is that new? –  Martin Beckett Nov 27 '12 at 16:07
"more than a year after an astronaut tossed it overboard" Wow, that comes as a real shock to me and greatly changes the mental image I had of the tank burning up soon afterward. (Although if I had done due diligence before posting the Question, I would have found that myself.) This confirms what Claudius had earlier written about why that is not commonly employed as a method for discarding trash. –  Eugene Seidel Nov 27 '12 at 16:13
@EugeneSeidel the speed at GSO isn't zero, it's about half that of low earth orbit (although you go around once in 24h rather than 90mins you have to go a lot further). The problem of de-orbitting stuff from GSO is that there is no drag, so you have to provide the full orbital speed (in the opposite direction) to get it to drop out of orbit. While in LEO anything will deorbit from atmospheric drag unless you keep boosting it higher. To deorbit the fuel tank they didn't need to do anything to it's velocity, they were just moving it slightly away from the ISS so it didn't hit it on the next lap –  Martin Beckett Nov 27 '12 at 16:28
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