What happens if you jump on a "spin ship"?

If you're on a spin ship, or rotating wheel space station, and you jump, then you're no longer being accelerated by the rotation. What would happen?

First, I'll establish some terms to make discussing this easier. I've also popped in an image courtesy of 2001: A Space Odyssey to help with the visualization. Not the most realistic movie, but the visuals are helpful. Moving at or near the spinning surface in the direction of spin is spinward, moving up away from the surface toward the center of rotation is up, and moving down toward the surface away from the center of rotation is down.

If you jumped up, intuitively it seems you would keep your horizontal velocity and keep moving spinward until your feet reconnected with the surface. But if the radius of rotation is large, the surface would be very nearly flat and barely curved, so considering you wouldn't be "falling" back down toward the surface it seems you could coast floating just above the surface for quite a while before finally landing when the curvature catches up with you. In other words, "gravity" for objects in mid-jump would be lower than that for objects on the surface. Is that accurate? What would jumping on a spin ship actually be like?

• "Not the most realistic movie..." Really, in what way? Can you cite an example, I'm curious! Did Kubrick take care to get the strength of artificial gravity correct in “2001: A Space Odyssey”?
– uhoh
Jan 11, 2021 at 3:19
• Great question by the way! In addition to remembering that "gravity" decreases as one moves "up" towards the center of the hub though, we have to remember that our tangential or forward velocity doesn't decrease. My hunch is that our astronaut would land "forward" of were they jumped.
– uhoh
Jan 11, 2021 at 3:30
• Keep on mind that drum with larger diameter would need higher surface speed for the same amount of artificial gravity (angular speed will be lower, but not "translational" speed of the surface). So lower curvature is intuitively compensated by higher forward speed you are approaching it after jump. For the rest of physics, check Coriolis force. Jan 11, 2021 at 8:00
• Does this answer your question? Creating your own artificial gravity by running. (Part 1 - the basic idea) Jan 11, 2021 at 13:33
• @CarlWitthoft It's helpful for general information, but my question specifically concerns what happens when you jump, which is not discussed at all in that other question, though they do mention jumping as a form of exercise. So it doesn't answer the question, but I really appreciate the information! Jan 11, 2021 at 14:54

We must remember that there is no spoon gravity here and that as soon as our astronaut is no longer in contact with the floor they must have an essentially straight line trajectory.

Below is a simulation in an inertial frame moving towards Jupiter along with Discovery 1. Let's assume that the astronaut is initiall standing and has a tangential velocity equal to the $$\omega r_0$$ where $$\omega$$ is $$2 \pi / T$$ where the period of rotation $$T$$ is 20 seconds and $$r_0$$ is 5.3 meters from @OrganicMarble's SciFi SE answer to Did Kubrick take care to get the strength of artificial gravity correct in “2001: A Space Odyssey”?

If they jump "up" i.e. towards the center of rotation with a radial velocity of -1.2 m/s (I estimate 2.6 m/s is maximum possible) they will "land" about 3 seconds later, at about 1.7 meters in front of where they started, and almost flat on their face! (no, hat tip to @benrg for pointing out that to first order they'll continue to rotate about their centers of mass and land near vertically with respect to the floor underneath.)

They should either have their arms extended ahead of time to break their "fall", or jump with what on earth would feel like a partial backflip in order to land "upright".

update: I wondered what happens if one jumps harder. Here are the trajectories for several jump velocities in integer steps of 1.2 m/s (the 1.2 comes from my first attempt at a 3 second jump).

Counterintuitively at first, in this regime it seems the harder and "higher" you jump the faster you "come down" again!

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import solve_ivp

def deriv(t, X):
g0 = 9.80665 # m/s^2 exactly https://en.wikipedia.org/wiki/Standard_gravity
x, v = X.reshape(2, -1)
acc = -g0 * np.array([0, 1]) * 0 # times zero because there is no gravity!
return np.hstack((v, acc))

halfpi, pi, twopi = [f * np.pi for f in (0.5, 1, 2)]

r0 = 5.3 # meters = 35 feet/2  https://scifi.stackexchange.com/a/239904/51174
T = 20. # seconds https://scifi.stackexchange.com/a/239904/51174
omega = twopi / T # s^-1

v_tan = omega * r0

v_jumps = 1.2 * np.arange(1, 6)
X0s = [np.array([0, -r0, v_tan, v_jump]) for v_jump in v_jumps] # 2.6 m/s max from https://space.stackexchange.com/a/31729/12102

times = np.linspace(0, 3, 301)
t_span = times[[0, -1]] # first and last

for X0 in X0s:
answer = solve_ivp(deriv, t_span, X0, t_eval=times)

fig, (ax1, ax2) = plt.subplots(1, 2)

x, y, vx, vy = answer.y

cos, sin = [f(-omega * times) for f in (np.cos, np.sin)] # rotate backwards
xr, yr = x * cos - y * sin, y * cos + x * sin # rotating frame

xc, yc = [r0 * f(np.linspace(0, twopi, 201)) for f in (np.cos, np.sin)]

ax1.plot(xc, yc, '-k')
ax1.plot(x, y)
ax1.plot(x[::100], y[::100], 'o')
ax1.text(x[-1]+0.4, y[-1], 't=3.0 sec', fontsize=12)
ax1.text(x[-1]+0.4, y[-1], 't=3.0 sec', fontsize=12)
xo, yo = [r0 * f(omega*times[::100] - halfpi) for f in (np.cos, np.sin)]
ax1.plot(xo, yo, 'ok')
ax1.text(xo[-1]+0.4, yo[-1], 't=3.0 sec', fontsize=12)

ax1.plot([0], [0], 'ok')
ax1.set_aspect('equal')
ax1.set_title('r0 = 5.3 m, T_rot = 20 sec, v_jump = n times 1.2 m/s')

ax2.plot(xc, yc, '-k')
ax2.plot(xr, yr)
ax2.plot(xr[::100], yr[::100], 'o')
ax2.text(xr[-1]+0.4, yr[-1], 't=3.0 sec', fontsize=12)

ax2.plot([0], [0], 'ok')
ax2.set_aspect('equal')
ax2.set_title('rotating frame')

plt.show()

• Fantastic! And what happens to the distance travelled and air time as you increase the ship radius? Jan 11, 2021 at 17:55
• @TheEnvironmentalist What's not show here is for very small velocity jumps, air time also decreases, for 0.24 m/s jump velocity (1/5 of 1.2 m/s) is only 0.89 seconds. So in (this estimation of) Discovery-1's artificial gravity 3 seconds at 1.2 m/s is the maximum air time. A larger ship at the same level of simulated gravity would probably require a higher radial jump velocity to maximize air time. I'll bet there is an analytical solution out there somewhere that solves for $t_{max}$ as a function of $g_{sim}$ and $r_0$ and that might be the basis of a new reduced-gravity-sports question.
– uhoh
Jan 11, 2021 at 23:12
• For an ancient (well, 60 year old) illustration, see "Frames of Reference" at the 17:00 time hack This is a clear version : aeon.co/videos/… Jan 12, 2021 at 4:48
• Re "almost flat on their face" - their angular velocity is equal to the centrifuge's, so they'll remain vertical relative to the rotating frame, not the inertial frame. They will be tilted forward relative to the floor when they land, but not as much as you seem to suggest. Dec 3, 2022 at 23:41
• The higher you jump, the faster (harder) you land, just like on Earth. Outstanding graphics! Dec 5, 2022 at 13:03

On a sufficiently large spin ship the result would be indistinguishable from normal gravity, at least to human perception.

The idea behind a spin ship is simple: instead of gravity constantly pulling your body downward towards the ground as would happen on a planet, the ship rotates such that the floor is being constantly pulled "upward" towards you by the structure of the ship. This constant pull on the floor towards the ship's center is called centripetal force, and it's what prevents the floor from simply falling apart and flying off into space in a straight line, as would happen if the floor was not connected to the rest of the ship.

Like any other force, centripetal force causes the objects it pulls on (in this case, the floor beneath your feet) to accelerate. In the case of centripetal force this is known as centripetal acceleration, and on a ship designed for Earth-like gravity the ship will need to spin fast enough that the floor is being pulled towards the center of the ship at a rate of ~9.8m/s^2, the same rate at which objects accelerate towards the ground on Earth.

If you jump on a spin ship, you'll no longer be touching the floor and so will continue moving in a straight line "spinward". The floor however, will continue to accelerate towards the center of the ship as the station spins, making it feel as though you are "falling" back to the ground at a rate of 9.8m/s2, just as you would on Earth. This illusion of gravitational force is commonly known as centrifugal force1.

On a smaller ship (such as the one in 2001: A Space Odyssey) the illusion isn't perfect. As other answers have pointed out, there are additional factors at play in a spin ship which would lead to very noticeable, counterintuitive behavior for the occupants of such a vessel. Notably, varying "gravity" (centrifugal force) depending on altitude and your speed of travel in the direction of spin, plus Coriolis force2. However, the larger the ship's diameter the smaller and less noticeable those unwanted effects become on a human scale. Centrifugal force is proportional to radius (F = m*ω^2*r), so a large diameter spin ship doesn't need to spin as fast (in terms of angular velocity) to maintain Earth-like gravity, and small differences in altitude and speed don't matter as much when the ship is comparatively large. On a sufficiently large spin ship (a couple kilometers in diameter or so), these effects would be nearly imperceptible on a human scale, and jumping would feel identical to how it does on Earth.

1Not to be confused with the centripetal force discussed earlier. Centrifugal force is a fictitious but handy force that is used when we are looking at things from a rotating perspective (as a passenger on a spin ship would) but want to think of it as if we were stationary. Centripetal force is a real force that keeps the floor of our spin ship connected to the ship's center as it spins.

2Coriolis force is another fictitious force which would be felt when you move "up" or "down" in the spin ship. On larger ships it would be less noticeable at human scales, as the distances involved would be smaller compared to the radius of the ship.

• "floor is accelerated towards the center of the ship" ? Does it ever get there? Jan 11, 2021 at 22:02
• I've made a small edit to add a link or two. "floor is accelerated towards the center of the ship" is suboptimal. For myself, I was never able to figure out how to choose the right words to talk about what's happening in a rotating frame, so I always "hide" in an as-inertial-as-possible frame and language thereof. Great answer, +1!
– uhoh
Jan 11, 2021 at 23:07
• @uhoh When I talked about the floor accelerating I actually wasn't taking about centrifugal acceleration. The floor has to be accelerating in an inertial reference frame, otherwise it would just continue off in a straight line and the spin ship would fall apart. Trying to think of a better way to phrase that... Jan 12, 2021 at 0:12
• Yes, the radial spokes are constantly pulling each floor segment towards the center, otherwise they would go in a straight line. If we put a spring scale in each spoke we'd certainly measure a force; that's real, not fictitious. Hmm...
– uhoh
Jan 12, 2021 at 0:40

Look at this image, the sparks move away tangentially from the grinding wheel.

This movement is valid for the spin ship too, as long as it is possible inside the spinning ship. The jumping astronaut will move in tangential direction until he hits the floor again. I assume the circumference speed of the ship is much faster than the running speed of the astronaut.

• What would that mean for the actual movement of the astronaut? What would the body of the person inside the ship actually do? Jan 11, 2021 at 2:46
• Have a look to the excellent simulation by uhoh. The left sides of image 1 and image 2 shows the tangential movement.
– Uwe
Jan 11, 2021 at 10:42
• @TheEnvironmentalist that question can't be answered until you tell us what frame of reference you wan the answer in. In one perfectly valid frame the astronaut remains stationary (or at least their centre of mass does) from the moment they lose contact with the floor until the moment they hit the floor again. The ship is moving and spinning around them, Jan 16, 2021 at 15:56

All of the motion is relative to the central hub assuming the spacecraft is in a steady 0 G acceleration state. Running against the spin equal to the rate of spin would render one weightless$$^1$$.

"If the radius of rotation were large, the surface would be nearly flat"
"Gravity for objects in mid jump would be lower"

But a larger radius rotates faster$$^2$$ to get 1 G, and the "gravity" gradient would be less because there is more distance to the hub$$^3$$ if one jumped. The key here is every step running around a circle requires acceleration in a new direction. This is easily understood with banked turns here on earth. (One could actually try it in space with no cylindrical rotation at all).

"you could coast for quite a while just above the surface"

Because tangential velocity is not affected by a vertical jump, the floor would rise to meet you. There would be a slight loss of tangential velocity due to air drag.

Because "up" is toward the central hub, as long as the runner was slower than the rate of rotation, they would experience some centrifugal force. Ideally, from an outside observer, the cylinder could rotate like a second hand, with the runner trailing retrograde like a minute hand.

If stationary objects were 1 G, a moving object against the direction of rotation would be less than 1 G.

Now, how about running faster than the rate of rotation?

This would be difficult without magnetic or velcro shoes, as one would have to pass through the (net) velocity of zero G if running against the spin.

However, another possibility is running with the spin. Here, the faster you run, the "heavier" you are. So, the 2001 jogger could set static G's at a little less than one, then crawl down the ladder and happily take off on a nice 1 G jog. The person passing in the opposite direction may have a bit easier a run.

$$^1$$ in all cases the suspended runner is weightless, resistance is created by the floor rising to meet them unless they move forward with the same Tangential velocity as the cylinder rotates.

$$^2$$ F = m × V$$^2$$/radius

$$^3$$ there is no gravity, only tangential velocity and the vertical velocity of the jump.