An astronaut is in orbit around the earth and takes out his bow and arrow. He shoots the arrow away from the earth. Assume that a line extending down the shaft of the arrow would meet the centre of the planet. What trajectory would the arrow take? Two possibilities that occur to me - the arrow simply travels away from the earth, but because it won’t exceed escape velocity, it will go into orbit at the altitude where it runs out of steam or the arrow will travel away from the earth, reach an apex, and then gravity will pull it back and it will resume orbit from the baseline orbital pathway the astronaut was on when he shot it.
The arrow doesn’t “run out of steam” - the initial trajectory after leaving the bow is part of a new elliptical orbit which will intersect the astronaut’s orbit twice per orbit.
Assuming that the spacecraft orbit is circular (so we don’t have to concern ourselves where is spacecraft is in orbit relative to its apsides) and the arrow is not provided enough impulse to escape the Earth’s sphere of influence (SOI), the arrow will enter into an elliptical orbit. In two dimensions it will periodically cross the orbit of the spacecraft, but reality is that it will likely be out of the ecliptic of the spacecraft orbit and depending on notational perturbations it may only actually cross the print of the spacecraft irregularly. It will not turn to its previous orbit or fall to Earth without other influences to apply additional impulse.
Stranger
Thank you!
Is it worth being so pedantic as to note that, by equal and opposite reaction, the astronaut’s own orbit is also altered?
There was a funny old sci-fi story where a stranded astronaut “got home” using that technique. (Meanwhile, the stranded cosmonaut manages the same thing by hurling javelins…)
Trinopus
Also note that the arrow’s new orbit will sometimes take it closer to the Earth than when it started. It’ll oscillate above and below its original height.
Agreed. The application of a single point delta-v will always result in changing the eccentricity of the orbit (hence, why I specified starting from a circular orbit for clarity). I was also assuming that the amount of change of kinetic energy added was insignificant compared to the existing total energy of the astronaut+arrow system before the arrow was fired, and thus, the amount of momentum change is negligible for the astronaut (at least as a first order calculation). If we assume that the momentum transfer to the arrow is significant, then we have to consider what happens to the astronaut himself, i.e. he also goes into an elliptical orbit that gets squashed in the opposite ordinate.
Now try this problem out for thought; what if the astronaut shoots the arrow as described by the o.p.; however, with the arrow tethered to him such that once it gets to a certain distance the tether goes taut and the momentum of the arrow is transferred to back the astronaut? (Assume the tether is completely inelastic, with no energy stored within the stretch or vibration of the tether for the purpose of discussion.) What now happens to the dynamics of the astronaut? Does this draw him up into higher orbit?
Stranger
While you’re mulling over SoaT’s question, a follow up of my own: what happens to the orientation of the arrow as it goes on this eliptical orbit? For roughly half the time is it traveling point first and half the time feather-end first?
I would think that one of the effects would be to temporarily increase Earth’s gravity over the astronaut-arrow system.
Suppose we had two astronauts of the same weight tied together with a huge cord, hovering stationarily above a atmosphereless planet, far enough away that the gravity effects are negligible. Then assume that the astronauts push off against each other hard enough with a long enough cord that when the cord stops, one astronaut is barely hovering above the planetary surface while the other is twice as far as the original pairing. The planet obviously is attracting the pair more than before, since the gravitational effects were originally negligible but now amount to at least 1/2 the gravity the pair would generate if both were on the surface.
I would assume that the same principle applies to objects of differing masses but I could be wrong.
Assuming it left the bow with exactly zero rotation of any kind in any dimension, it’ll be pointed the same direction (e.g. towards Andromeda galaxy) forever. Which means as it orbits it’ll be going point first or tail forst or anywhere in between.
Now in the real world it won’t be launched from the bow with exactly zero rotation. And the vacuum of orbital space isn’t perfectly zero. And the arrow will have some magnetic moment however small. etc. So soon enough it will be slowly tumbling & coning rather than pointing at Andromeda forever.
But if the essense of your question was whether the orbiting arrow would have any tendency to point in its direction of motion like it does in the atmosphere, … Well the answer is pretty much exactly No.
What would happen to the original bow astronaut? He’d exert some force pulling the bowstring back, but isn’t the force working mostly against the two ends of the bow? When he lets go, the two ends snap back, tightening the string. Since they’re opposite each other, does the bulk of the force point in opposite directions leaving little change to the floater?
I’m sorry I don’t have better words or more experience with bows and arrows to ask the question better, so if some physicist archer could describe the situation, that would be great.
Are they African or European astronauts?
IMHO, a very interesting question… For the record, IANA orbital physics/rocket scientist. Following Newton’s Laws of Physics (as far as my rudimentary knowledge of such, goes;)), wouldn’t the only influence upon the astronauts “orbit” be induced by the bowstring, as it finishes releasing it’s energy? (ie: when the arrow leaves the string)
Like I said, I’m not all that knowledgeable on “orbital dynamics” and such, I believe there are some SDMB’ers out there that are “rocket scientists”, or at the very least, are infinitely more knowledgeable on the subject. 
Perhaps one of them will grace us with their better understanding, of the problem. 
Pretty please?
The Duchy of Grand Fenwick has resumed its space program.
A perfectly spherical object (which would not have any directional bias with any gravitational gradients) would maintain the same orientation with respect to a fixed reference frame i.e. from someone on the surface of the Earth, the sphere would appear to rotate once every orbital period, even though an observer on the surface of the sphere would not perceive any of the inertial forces associated with rotating reference frames (Coriolis, Euler, and centrifugal forces). The reality is that without atmospheric pressure to stabilize the flight of the arrow via the fletching, minor perturbations like a slight angular momentum imparted during launch, solar pressure, et cetera would probably cause it to rotate or tumble, and precessional and nutational contributions will probably have it tumbling in an apparently erratic fashion.
Ludovic, an object or astronaut in stable low orbit is not “hovering stationarily…the gravity effects are negligible,”; it is instead falling around the Earth at a very high rate of motion, about 4.5 miles per second. It is attaining this speed, not the orbital altitude above the planet surface, into which most of the energy of the orbiting object is allocated, though in practice with conventional rocket space launch vehicles much of the energy is “wasted” in suspending the rocket as it attains the necessary orbital speed (gravity losses), transferred to the atmospheric as drag (acoustic and aeroheating losses), and lost in inherent thermodynamic inefficiencies (plume expansion and heat losses).
A hint to the question is to consider that energy (including that added to the system by launching the arrow) is conserved, as is momentum (of the astronaut-arrow-Earth system).
Stranger
Quoth LSLGuy:
There will also be tidal effects which will tend to keep the arrow oriented vertically. Point up or point down will both work about equally well for that, though.
Quoth Rhythmdvl:
This one’s easy, and we don’t need to look at any of the details of what the bowstring is doing. Momentum is conserved, so whatever momentum the arrow ends up with, the astronaut will end up with the same amount of momentum in the opposite direction. This is true no matter what the design of the bow, or how he fires the arrow, or if we replace the bow with a gun, or whatever.
I agree that I overlooked (or oversimplified) tidal forces. So let’s talk about them now. Assume we do launch the arrow pointed directly away from the Earth’s center with zero rotation in any dimension and we also ignore all other perturbations other than tidal force as it orbits a perfectly symmetrical & spherical Earth.
Positing starting from a shuttle-height orbit, the arrow’s orbit takes ~90 minutes. It might be 85 or 110 minutes, but it’s not 10 minutes, nor is it 4 hours. Tidal forces will try to rotate the arrow into a tail-towards-earth-center orientation. But compared to the fixed stars, that orientation is making a complete rotation once every ~90 minutes. Are the tidal forces really going to succeed at spinning up the arrow to also have a ~90 minute rotational period? And if so, how many millenia will it take?
My uneducated intuition is that at some intermediate rotation rate we’ll end up in a resonance where half the time the tidal forces are retarding the built-up rotation and half the time they’re advancing it. So the rotation rate stops growing right there. But as I say, that’s a WAG.
I can also easily intuit an infinite regress where for any given proposed resonance, there’s another high-order resonance that would have happened earlier at a lower rotation rate. By induction, therefore the original state of zero rotation is also tidally stable.
Comments, additions, debunks?
It’s actually pretty easy to figure out what the orbit would be from the vis-viva equation Vis-viva equation - Wikipedia and the facts that angular momentum is conserved and that shooting the arrow directly away from the Earth doesn’t change angular momentum.
I get (from the vis-viva equation, assuming that the arrow has tiny mass compared to Earth (duh!) and that the arrow has small initial velocity relative to a low-earth orbit velocity (~5 miles/second)) that the semimajor axis of the arrow’s new orbit is the astronaut’s orbit semimajor axis R+2RVarrow/(V-2*Varrow) where Varrow is the initial velocity of the arrow, and V is the velocity of the astronaut’s circular orbit.
I got a little tangled in my algebra figuring out the maximum distance the arrow gets to; if I feel like it, I’ll try again tomorrow (that’s where you need to use the fact that angular momentum is conserved)
I messed up - the new semimajor axis is R+R*(Varrow)^2/(V^2-Varrow^2).
The maximum distance from earth is approximately R*(1+Varrow^2/(V^2-Varrow^2)).
If I’ve made any more mistakes, I’ll find them only after sleep.
I believe this was mentioned upstream, but, as a first assumption of the force imparted “back” to the astronaut, shouldn’t the construction of the bowstring be important?
The bow frame is maximally loaded at two points. Isn’t the return energy dissipated quite a lot through the frame?
A fair amount of energy is dissipated (though of course modern bows are designed to minimize this), but momentum is never dissipated. If you know the velocity the astronaut gives the arrow, and you know the masses of both arrow and astronaut, then you have everything you need to calculate the astronaut’s resulting velocity.