## A little Backstory…

For April Fool’s Day this year, Swungover became “David Bowie’s Guide to Space-Gravity,” and allowed me to ask a question I’ve had for awhile to our awesome math-science community. Feel free to add your two cents in the comments section.

Over the last year, I’ve asked this question to a few very intelligent people, including rocket scientists, and have gotten different answers, so I was interested to see if there was a general agreement of one. During one of the conversations, I was making a point and somehow found myself using the words “space-gravity,” which cracked myself up and several others. We almost immediately agreed it was the kind of gravity David Bowie would specialize in if he were a mastermind physicist. Hence, the title. (Aside from the fact that we here at Swungover love David Bowie and would be honored to take over his physics blog.) Also note: I’m not the best at explaining all the information physicists need to have in order to answer it. So, I think I’ll go back to writing about swing.

Here’s the post:

## A Space-Gravity Conundrum

Hello Cosmonauts. I’m currently busy finishing up my new theory of quasar periodicity and just need to finish laying down the vocal tracks for it. Still, though, it’s gonna be a pretty alligator cookie-crumble space-race, man. So I’m handing over the column to fellow blogger Bobby White of Swungover, who had a bit of a physic puzzle for you die-hard Space-Gravity scientists. — d.b. (phd)

Hi! My name’s Bobby White, and I’m honored to be presenting one of Dr. Bowie’s weekly puzzles. When I’m not listening to Doctor Bowie’s earlier works and studying Literary Physics (a special branch of physics that deals with measuring space-time in literary worlds, especially Beckett) I am a swing dance instructor. And so, I love to think about simple physics.

I am also a bit of a sci-fi buff, and one day was thinking about 2001: A Space Odyssey and had a question that I, being a Literary Physicist and not a Reality Physicist, could not come up with an answer to. I also asked a few rocket scientist friends the question, and they couldn’t agree on an answer (I really did). So, I began thinking to myself, “Where can I find a large group of math-science people whom I could ask?” Well, the swing dance community, of course, and Dr. Bowie’s revered online scientific journal here. So, here’s the puzzle:

RIDDLE ME THIS: In the film 2001, there is a space ship that creates artificial gravity by rotating a cylindrical hull. Because of centripetal effect, people would be pushed against the hull, giving it a fake gravity. Now, we know that such a thing is feasible, except that in reality, the hull would have to be more than two miles in diameter to make the artificial gravity safe for humans.

But we don’t actually need that for today’s question, which is more theoretically based. Let’s assume, measurement-wise, that the ship in 2001 is possible — that the center the ship orbits around is only ten or so feet above the astronaut’s heads. Now, let’s say a man standig on the “ground” in the spaceship tosses a ball towards the center (the space directly above his head).

WHAT COURSE WILL THE BALL TAKE? WHAT WILL HAPPEN TO THE BALL?

NEW! The Shana Challenge:
“Now an exercise for the reader: all of these explanations involve a static coordinate system with the spaceship rotating around one axis. But what if the coordinate system stayed with the astronaut. In other words: what is the path of the ball from the astronaut’s point of view?” –Shana Worel

Also, if you were to run on this spaceship, as Dave does in the film, what would happen in a real life situation? Because, when you technically run, both feet are off the ground at some point. Would that alter running? How?

UPDATED EXPLANATION (10:34 a.m.): In a real life situation like this, it’s not gravity, but centripetal(?) inertia giving the effect of gravity. In the film, because it’s filmed on earth, and the 1960s didn’t have the budget to make a space station (silly 1960s) to film it in, it appears to be real gravity, just always pushing straight away from the center of the craft. If you’d like to chose one or the other to base your answer on, I’m cool with that.

Also, there has been some questions raised about whether you’re inside the space craft, or outside of it, looking in, and possibly dying of suffocation and struggling to hold onto consciousness.

1. April 1, 2012 4:14 am

It’ll depend on how hard it is thrown. It can either describe a straight line to the ‘ceiling’ above Dave’s head, or varying degrees of parabolic arc (some of which would allow Dave to catch the ball again), or any number of forms of loops around the axis of rotation (imagine a spirograph), or even end up floating at that very axis of rotation.

2. April 1, 2012 4:17 am

It’ll depend on how hard it is thrown. It can either describe a straight line to the ‘ceiling’ above Dave’s head, or varying degrees of parabolic arc (some of which would allow Dave to catch the ball again), or any number of forms of loops around the axis of rotation (imagine a spirograph), or even end up floating at that very axis of rotation.

April 1, 2012 4:36 am

From the POV of Dave, if he throws the ball toward the exact center in a perfectly straight line, but with insufficient force to reach an obstacle or the center, it will come right back to him in a perfectly straight line. If he throws it off center, it will follow an arc similar to if he were throwing it at an angle towards the ground on earth, the angle being the 180° compliment to its ascent.

For example, if there’s nothing in the middle and the axis is outside the round chamber, and Dave throws the ball exactly straight towards the center, with just enough force for the ball to slow nearly to a stop (as if it would in a gravitationally normal situation), but not all the way halt at the center, it will pass through the center and then act as if it’s been dropped straight down from the “sky” to the ground on the exact opposite side of the chamber as Dave.

If Dave throws the ball forward a bit, the ball will follow a sharper arc to land forward in the chamber than gravitationally-induced parabolic, because from each point forward, the ball’s frame of reference shifts its centripetal momentum. So, any trajectory besides straight towards the center is effectively like throwing the ball at the ground.

One more bit: the ball would thus act like Dave. When Dave runs along the inside of the chamber, he “throws himself” upward a little bit, and then he “falls back down” to the inside of the chamber, a little bit in front of where he started. But although the exertion is similar to running up a hill, from Dave’s point of view, it feels like he’s staying in place and his movements are making the ground under his feet pass faster or slower. Dave is equivalent to the ball. If he could jump so high that he passed the center of the chamber, he would suddenly feel oriented upside down and fall up onto his head. As long as he jumps less than that, he just always feels like he’s going straight up and down and the ground is treadmilling below. The ball would “feel” the same way.

April 1, 2012 6:02 am

The trick is to ignore Dave and think about what happens from the point of view of a fixed observer looking down at the ship.

Suppose instead of throwing the ball upwards, Dave merely releases the ball. Let’s orient ourselves so that Dave is standing straight up at the moment he releases the ball.

From the point of view of the center of the ship, the ball’s instantaneous velocity is horizontally to the right (parallel to the x-axis) when Dave releases it. So the ball travels to the right and eventually hits the wall on that side.

It’s no different than if the ball were attached to a string tied to the rotating center of the ship and we cut the string. And that’s no different than if you’re swinging a ball on a string around your head (on earth) and you let go — the ball travels in a straight line tangent to the path it had been taking.

So now suppose Dave throws the ball upwards. You might think that because Dave is rotating as he throws the ball, he somehow imparts a “curving force” onto the ball. But remember Newton’s second law: f=ma means objects accelerate only when you apply a force to them. “Accelerate” here means “change directions” — in particular, an object doesn’t follow a curved path if you’re applying no force to it. So once Dave releases the ball, it’s going to travel in a straight line (from the point of view of our observer, anyway — Dave’s reference frame is accelerating, so Newton’s laws don’t apply in their vanilla form).

So when Dave throws the ball, he imparts an upward force on it. So now the velocity of the ball is determined by the rightward component (due to the rotation) and an upward component due to the throw. So the ball travels up and to the right and eventually hits the wall of the ship somewhere on that side.

In particular, it’s impossible for the ball to get “stuck” in the middle of the ship, or loop around, or take a curved path, from our observer’s reference frame, anyway. This is a simple matter of: Once the ball is released, there is no force acting on it. So it travels in a straight line until it hits something.

What all this looks like from Dave’s perspective is left as an exercise to the reader.

April 1, 2012 6:03 am

In the diagram above, the ball will travel along a chord of the circle upward and to the right.

Recall that if you have an object traveling in a circular path and then release it from that path, it will fly away along a tangent to that path at whatever velocity it had when released. Now, imagine if instead of throwing the ball in the diagram above, Dave simply released it–it would travel in a horizontal line toward the right side of the diagram along a tangent of the circle described by Dave’s hand at whatever velocity it had when Dave released it.

When Dave tosses it toward the center of the craft, he is imparting a “second” velocity to the ball–straight “up.” If you remember vectors, you can imagine one arrow pointing to the right from the point of release, and a second pointing up from the end of the first arrow (the length of each vector being proportional to the velocity the ball had in that direction at the moment of release). The ball’s path will be along the diagonal that forms the third side of a triangle with these two vectors as its two other sides.

The ball will not arc nor will it stop because it is not being accelerated once it is released (unlike a ball on earth which is constantly being accelerated by gravity). See the attached diagram. The blue circle represents the ball’s path if Dave holds it without releasing it.

April 1, 2012 6:04 am

Here’s the diagram:

April 1, 2012 6:55 am

What is the difference between how hard the ball is thrown? Once it leaves the hand, is it in zero gravity?

April 1, 2012 7:15 am

“Zero gravity” doesn’t mean much without a reference frame.

For example, you experience “zero gravity” when you’re in the Tower of Terror (an elevator in freefall). But of course gravity still exists — that’s what’s pulling the elevator down. So we need to distinguish the “zero gravity” reference frame *inside* the falling elevator from the earth-gravity reference frame *outside* the elevator.

From the point of view of someone standing outside the ship, the ball is in “zero gravity” the entire time. (That is, there are no external forces acting on the ball.) From the point of view of an ant on the ground of the ship, the ball is always “in gravity”. From Dave’s point of view, well, the problem was set up such that centrifugal effects are a poor approximation of “gravity” for him.

April 1, 2012 1:43 pm

The premise of the situation is that we accept whatever it takes to make the behavior we see inside the chamber is being done, regardless of how it’s explained in the film. So, the behavior we see in the chamber tells us that everything in the chamber acts as if there is gravity pulling you perpendicularly towards the inner surface of the chamber from every point.

The responses above that seem to be relying on the idea that this is a mere spinning cylinder are invalid if their explanations wouldn’t match up with how it looks when Dave is walking along the inside of the chamber, even if they would be valid in a real-life spinning room in space.

Now, if Mr. White doesn’t care for an explanation that also validates the human movement that we see in the film, then it’s just a non-gravitational situation that’s wholly explicable with a closed centripetal chamber.

April 1, 2012 2:33 pm

So…my set up is a little too ambiguous…. How about both answers?

In real life, it’s not gravity, but centripetal inertia giving the effect of gravity. So how would the ball react in that situation (I believe we already have some answers for that)

In the film, because it’s filmed on earth, and the 1960s didn’t have the budget to make a space station (silly 1960s), it appears to be real gravity, just always pushing straight away from the center of the craft. How would the ball react differently in that situation?

9. April 1, 2012 5:48 pm

I am no physicist, but it seems to me that we must remember the set up is only creating a force that replicates gravity. it does not create some gravitational field inside the cylinder or anything like that. It is completely reliant upon the affect of the pull. When the ball is thrown, it is affected by two different inertial forces I think. But once it is let go of,I don’t know what will happen exactly, but I don’t think it will continue to feel any additional pull beyond the intertia already applied to it. So, it will not hit the center and just float in space. I don’t think so anyway.

April 1, 2012 7:33 pm

I am so glad that you’ve asked this question. I’ve seen it attacked twice in sci-fi and both times it was unsatisfying. Both authors answered the question using the Coriolis effect. The ball was, in both cases, uncatchable by the thrower as it took a parabolic course due to Coriolis force. This never seemed right since Coriolis seems to me to work only under gravity and there is no gravity in space, only an imitation of gravity.
The other difference is that the two sci-fi scenarii were set on particular decks of larger stations whereas as here we’re looking at the single deck of a smaller station. Here the ball could go straight through to the far side of the station.
I will be thinking on this, but first must dance.

April 1, 2012 9:31 pm

No, Coriolis doesn’t require gravity; what it requires is differential rotational speeds. On Earth, it’s the result of the differing distances of points on the surface from the axis of rotation (not the center of the Earth). The north and south poles don’t move at all, while a point on the equator has to cover 25,000 miles every day, or over 1000 miles per hour (that’s why virtually all satellites are launched eastward and the Ariane launches from only 5 degrees N in French Guiana — it substantially reduces the amount of additional velocity needed), and a point at 45 degrees north or south is moving at about 700 miles per hour.

So if you have a mass of air which is stationary *with respect to the ground* at the equator, and push it due north, it will find itself moving faster toward the east than the ground it’s passing over, which makes it appear to curve eastward.

Similar thing with a ball in the space station. If for convenience we have a station with a twelve-foot radius and drop the ball from a height of six feet above the deck, the ball has a spin-ward speed of only half that of the deck when released — and will still only have half the spin-ward speed (neglecting the effects of the station’s co-rotating air) when it hits the deck. That’s what causes the curvature of trajectories as seen from the rotating reference frame of the station. How much the impact point diverges from directly “beneath” the drop point depends on how long it takes the ball to traverse the distance since the difference in rotational speeds will be multiplied by the total time.

April 1, 2012 10:22 pm

Ralf, That totally makes sense! Why is this just now making sense? Maybe I just was never ready to hear it before.
So in the case of the station, the ball tossed upward would have a greater rotational speed when thrown then the surrounding “higher” space into which it was thrown. But how would that look to an observer? wouldn’t that just improve the illusion of gravity?

April 1, 2012 10:01 pm

The dance practice I was supposed to do at 5 is rescheduled for 7. So I have time to think of physics now instead of later. However, after reading the responses above, it seems that Paulidin and Mobilhomme have it down. I doubt that I can add anything to this as far as physics goes.
I may not be a physicist but I am a linguist. So just for clarity I’d like to put in that Centripetal force isn’t really what we’re talking about here but rather Centrifugal force. Fugere means to run from, as in “Vampirus est! Fuge!” – “It’s a vampire! Run!”. Petere means to go to, as in “Starbucem peto. Visne aliquid?” – “I’m going to Starbucks. Want something?”
So the apparent gravity here is centrifugal force rather than centripetal, since it moves things away from the center and not toward. Then again maybe we’re talking about the force of the hull against the feet of the astronauts which IS centripetal.
But back to physics – it just occurred to me that if Dave threw the ball directly against the rotation of the station at just the right speed he could exactly cancel out the “gravity” and put the ball into a sort of internal orbit. From an external POV the ball would stop and hang in space. But from within the station the ball would fly indefinitely around the loop of the station. Or would it? What do Paulidin and Mobilhomme say on this?

• April 3, 2012 3:25 pm

I thought about mentioning the terminology issue. Bobby’s probably afraid of half-informed physics students (or teachers!) who claim that centrifugal force doesn’t exist. But it does exist, as defined. And it is defined in laymen’s terms as the appearance of an outward force when dealing with a rotating reference frame. The reason for using the word at all is that it’s much more specific than just saying “inertia.” “Inertia causes the Earth to bulge at the equator” doesn’t tell you how it happens. “Centrifugal force causes the Earth to bulge at the equator” immediately informs you that this bulging is due to a rotation (in this case, the rotation of the Earth).

April 2, 2012 4:42 am

Let’s model the spaceship as the unit disk in the Cartesian plane: i.e., label a fixed point as “the origin”, assign it the coordinates $(0,0)$, and then every point in the plane is given by a pair of coordinates $(x,y)$, where $x$ is the horizontal displacement from the origin and $y$ is the vertical displacement.

Then the unit disk is the set of all points $(x,y)$ satisfying $x^2+y^2 \leq 1$. Place Dave at the bottom point, or $(0, -1)$. If we fix Dave in place, and consider the spaceship from his point of view, then a rotation of the ship by a speed of $V$ (let’s say the outer edge moves at speed $V$) has the effect of imparting to every object in Dave’s field of view an acceleration which is NOT constant.

Contrast this with the usual model of gravitation on the surface of the Earth, in which objects experience a constant downward gravitation of 9.8 meters per second per second. In the spaceship model, objects experience an acceleration whose magnitude is $V^2$ times the distance from the center of the ship, and whose direction is radially outward from the center.

We can express this acceleration in the form of a differential equation: if $p(t)$ represents the position of an object in the spaceship at time $t$ which does not experience any force other than the rotation of the ship, then $p$ satisfies the equation
$p''(t) = V^2 p(t)$.
This is a simple example of a differential equation (“simple” means it’s one of the first you’d encounter in an introductory course) called a constant coefficient differential equation. It has a simple solution (same meaning of “simple). If Dave throws an object upward from his location at $(0, -1)$, with initial velocity $W$, then the coordinates of the object at time $t$ are given by
$p(t) = (0, \frac{W-V}{2V}e^{Vt} - \frac{W+V}{2V}e^{-Vt} )$,
where the $e$ bit refers to the exponential function, and $e ~ 2.718$.

What does this mean? If $W < V$, i.e. the object is thrown slower than the rotation of the ship, then the multiplier of $e^{Vt}$ is negative, and hence the ball will return to Dave. If the object is thrown faster than the rotation of the ship, then the multiplier is positive, and hence the ball will bounce off the ceiling. If the ball is thrown upward at exactly the speed of rotation, $W=V$, then the position of the ball simplifies to
$p(t) = (0, -e^{-Vt})$,
which means that the ball will approach the center point, and will appear to Dave to hover there.

An extra comment: the position of the ball can be approximated by a Taylor series — that means the vertical component can be written as an infinite polynomial. If we write out the first few terms of this series (a second college course in calculus will teach you this), we find that the vertical component of $p(t)$ is
$p(t) = -1 + Wt - \frac{V^2}{2}t^2 + \frac{WV^2}{6}t^3 - ...$.
For small $t$ values (meaning small time values, or directly after Dave throws the ball), the first three terms of this series give a good approximation to the position of the ball, so that
$p(t) ~ -1 + Wt - \frac{V^2}{2}t^2$.
This last expression is exactly the motion of an object thrown vertically on the surface of the Earth, if gravitation were given by $V$.

April 3, 2012 12:10 am

I’m skeptical of this interpretation.

Ignoring these differential equations, I thought we established that, from the point of view of a stationary observer outside the space ship, the ball travels along a cord of the circle (i.e, the ship) and eventually hits the wall. This happens no matter how hard the ball is thrown.

How do you reconcile this with your claim that the ball can hover in place, from Dave’s point of view? Surely if the ball eventually hits the wall, it does not appear to be hovering in place.

13. April 2, 2012 8:29 pm

Bobby, the reason why it matters how hard the ball is thrown is that the center of the spaceship* has a non-zero dimension. The question is “what course will the ball take,” so bouncing off the center is an option. Since it’s a vector problem, if the velocity it’s thrown is sufficiently high compared to the velocity of spinning, it could hit the center.

But barring that, the simple explanation (which is redundant due to mobilhomme’s explanation) that I would offer is based on Newton’s 1st law of motion: an object in motion stays in motion unless acted upon by an outside force. On Earth, this outside force would be the acceleration of gravity, but in space, there isn’t anything acting on the ball until it bounces off a wall. The astronaut is being acted on by the spinning of the spaceship, which is why she is moving in a different path.

Here’s an analogy: imagine that the spaceship is a tether ball that is spinning around a pole. The rope is constantly accelerating the ball toward the center, thus making it move in a path around the pole. Now imagine that you cut the rope while the ball is moving. It travels off in a path tangent the circle. If you hit the ball toward the center at the same moment you cut the rope, it would travel diagonally across the circle, with its path determined by how fast it was spinning combined with how hard it was hit (the velocity that it leaves the hand).

Incidentally, dynamics was always my least favorite subject. Thanks for bringing it up again Bobby.

*it feels super awkward using this word, since no one in aerospace calls something a spaceship unless it’s on a cartoon. :)

14. April 2, 2012 8:34 pm

Now an exercise for the reader: all of these explanations involve a static coordinate system with the spaceship rotating around one axis. But what if the coordinate system stayed with the astronaut. In other words: what is the path of the ball from the astronaut’s point of view?

Did I mention that I hate dynamics?

15. April 2, 2012 8:55 pm

BTW: here’s the answer key. I had to look it up so I could concentrate on work.

scitation.aip.org/getpdf/servlet/GetPDFServlet?filetype=pdf&id=PHTEAH000042000007000423000001&idtype=cvips&doi=10.1119/1.1804661&prog=normal

And for the record, I love the term “space gravity” :)

April 2, 2012 9:15 pm

wait a minute. Wouldn’t the astronaut and ball both travel in a line perpendicular to a line drawn from the astronaut’s feet to the center? Would not the centrifical force keep them moving in that direction unless a string tied to them or the disc beneath them continued to change their direction?

So, if you suddenly let go of the ball, wouldn’t it continue to move in that direction until it once again made contact with the disc, thus appearing as if it fell towards the “ground”?

If so, then if you threw it up in the air, it would have to directions, up and perpendicular to that line. So, wouldn’t this make it effectively, go at an angle between the two, thus arching up then towards the ground, despite having been thrown at what the astronaut perceived as straight up towards the center?

And, what if the ball were thrown past the center, almost to the other side of the arc, or actually hitting the other side? How would that be perceived by the astronaut? I guess, unless it were huge, any observer would still perceive that it is not a flat surface, but would see it for what it is, a curved ring he was bouncing a ball off the center of.

If my logic is faulty here, please let me know why. Am I missing something?

April 3, 2012 6:14 pm

I think that you’re right Mateo. It seems to me that the ball would follow an arc in toward the canter and lateral in the direction of rotation. This arc would eventually meet the surface of the craft. But it seems that from the point of view of the thrower that “arc” would not look like an arc at all but rather the ball going straight up and back down just like gravity.
I agree with you but we might be off. That’s what we’re discussing.

April 3, 2012 6:42 pm

One thing that occurs to me is that we are assuming that the astronaut will observe as if he is on the earth. Not necessarily so. depending on size, he may be well aware he is on a rotating disc. As I recall, in 2001, it was quite clear that they were on a rotating disc. I don’t know that an outside observer would necessarily be so different from inside.

April 3, 2012 6:46 pm

Of course, the questions that must be on everyone’s minds is, if a lindy hop leader does a hollywood style slotted swing out in the direction opposite to the rotation, would the follower want to come back in with normal directional force or lighter or heavier than usual? And would it make a difference if they were using counter balance?

Also, if a leader were to toss his partner into the air, while standing on the disk, would she come straight back down, or would he have to compensate for the arc when catching her.

How would bpm affect everything.

April 3, 2012 7:11 pm

Wait just one minute. I can’t believe how wrong I was. Due to centrifical force, the ball would always be directed away from the center, which would make it feel like towards the ground (gravity) that is the whole point of the set up. However, if the ball is thrown in the air with enough force to overcome the “gravity” just as you would on earth, you would be arresting that motion, and applying an opposite force equal to the “gravity” and greater, which would project the ball into the air. Depending on the atmosphere and the amount of inertia, it would continue in the opposite direction towards the center (or maybe not quite towards) and would eventually hit the disk, or, it might slow due to resistance of the atoms and molecules in the artificial atmosphere, and come to a rest in the middle somewhere floating in the zero gravity that exists in the whole place.

maybe this has already been said more clearly than I could.

In this artificial atmosphere you could have a couple of interesting phenomena. One would be, a ball could be thrown straight in the air, and it would arc back to the ground in a spot possibly quite different from the thrower, since the throw would continue to revolve around the center, and the ball would not. The other would be, you could take a ball, apply a little force, and throw it in the air, and watch it float up a foot or two, then float in the air for a second, then move laterally, not quite parallel to the perceived ground, and continue to hover in the air, but not infront of the astronaut.

Is this all accurate, or am I way off again?

April 3, 2012 7:25 pm

Going back to the first post, Greg seems to have gotten it right. But when you say Spirograph, do you mean if it hits the disk, then bounces off?

Then we would need to adjust for the material of the ball, and again, we must take into account any artificial atmosphere that would affect the intertia of the ball. Interesting stuff.

April 5, 2012 12:56 am

It has occurred to me that I misspoke earlier and that a few other people are off the mark as well. I said that the ball would arc. But of course it wouldn’t. It would go along a chord of the circle as Mobilhomme said. I am just so entrenched in Earth thought that it’s difficult to think of a thrown object NOT arcing.
The funny thing is to think of it as the same but backward. On Earth the ground is roughly straight but gravity curves the space so that the ball’s path is curved. On the space craft there is no gravity to curve the space but the “ground” is curved.

But here’s my other thought. A few people have said that the ball could be thrown with just the right velocity to reach the center and hover. This just makes no sense to me for two reasons.
First, there’s no force to account for this slowing and stopping. the ball follows a chord. At some point the chord MUST intersect the circle. There seems to be no other option.
But second, even if the ball could hit the center and hover, I really can’t see how that’s possible in this scenario. When the ball is thrown it will follow two vectors, one “up” toward the center from being thrown, and another laterally due to rotation. Assuming that there is at least some rotation, the second vector must be non-zero. That is to say there must be some angle to the chord such that it cannot be a diameter. The harder Dave throws the more the chord will approach the center but it can never actually intersect the origin.
Now if Dave were to throw up AND SLIGHTLY BACK against rotation, then I suppose we could get the ball to intersect the origin. However I still see no way to account for hovering. The closest to hovering is the internal orbit that I mentioned earlier, achieved by throwing the ball directly against rotation with just the right velocity to cancel out the rotation.

It also occurs to me that there’s a sweet spot where Dave’s push and the push of the rotation are equal. The chord described by that sweet spot will give the maximum possible hang time.
This is analogous to a 45degree angle on Earth. Throwing at less than 45 gives a small arc and thus less distance. Throwing at more than 45 gives a narrow arc and again less distance. Throwing at exactly 45 degrees gives the greatest distance to the throw.
Of course, on Earth we can increase hang time by adding more force. We could even put enough power behind a throw to launch a ball into orbit in theory. But aboard the craft, that’s not really possible. If we throw weakly we make a small arc. If we throw harder we make a bigger arc with more hang time. But we hit a point where we are out-throwing the rotation BUT because the “ground” is curved WE”RE STILL THROWING AT THE “GROUND”. So beyond a certain point we start to get less rather than more hang time for each bit of throw force.

I think that hovering in the station is just as unlikely as hovering over the Earth. Even in geosynchronous orbit an object never hovers. It simply falls continuously at a rate equal to the Earth’s rotation.

April 24, 2012 8:04 am

Following jlebar and mobilhomme and Shana:

You can see from mobilhomme’s diagram that the ball will make an arc from Dave’s point of view since it rises above the radius and falls back down to it.

If x is down, y is right, R is the radius from which the ball is thrown, w is the angular velocity, and v is the initial velocity of the ball, y = wRt and x=R-vt. Converting to radial coordinates gives

r = x^2 + y^2 = sqrt( ((wR)^2 + v^2 )t^2 + R^2 -2Rvt ).

That looks good since when t=0, r=R, and as the t^2 term takes over, it’s going to take over, so that gives a nice looking (upside-down) trajectory. As R gets big and w is small,

r = R – vt + (1/2)Rw^2

This is pretty neat, since, for a regular ball thrown in the air, h(t) = vt-R-(1/2)gt^2, where g is the acceleration due to gravity. (The minus sign is because r is measured from the center of the spaceship.) So, again, when the spaceship is big and the rotation isn’t too fast, the motion is about the same as a regular ball, from Dave’s perspective, with g = Rw^2. And Rw^w is the centripetal acceleration, which corresponds to the apparent acceleration due to Dave’s artificial gravity.

April 24, 2012 8:07 am

Typos:

That should be r = R – vt + (1/2)Rw^2t^2
and, in the last sentence: And Rw^2…

May 12, 2012 5:34 pm

Nice problem set-up, SomeGuy. With the addition that the proper condition for your approximation seems to be v << wR to get the units right, your argument satisfies a freshly minted physics masters student. For the less mathematically inclined, the condition v << wR means that the ball is thrown much more slowly than the human is being moved by the cylinder (as observed from outside the space station).

In the opposite case, where the ball is thrown much faster than the human moves due to the rotation of the cylinder, the ball appears to run away from the thrower down the cylinder as it "falls" back to the edge of the cylinder. It would be like throwing a ball straight up in the air but watching it arc and land four feet away. You'll have to trust intuition and limiting the limiting case of an infinitely fast throw on this one since the math can't be approximated as it can in the slowly thrown ball case. (Or you could write a program for it, using the first equation SomeGuy showed for r).

P.S. The handle is from my undergraduate days, not just made up for this post.