3 ms·

About 2 million km if GNU Units is serving me correctly: You have: (1 gravity) / (1/1440 * rpm)^2 You want: million km * 1.854336 / 0.5392766 By com

About 2 million km if GNU Units is serving me correctly: You have: (1 gravity) / (1/1440 * rpm)^2 You want: million km * 1.854336 / 0.5392766 By comparison: You have: (0.5 gravity) / (1 rpm)^2 You want: m * 447.12962 (That's the scenario in the film here.) For 1 g at 1 RPM: You have: (1 gravity) / (1 rpm)^2 You want: m * 894.25925 And presuming 3 RPM is tolerable (a common assumption in early space station / space colony proposals): You have: (1 gravity) / (3 rpm)^2 You want: m * 99.362139 (Almost exactly 100m or 330 ft.)

dredmorbius 3y ago
If the Earth revolved at a rate of ~17 times per day, or once every 1:24:42, the centrifugal force at the equator should just match the pull of gravity. Also via GNU Units: You have: sqrt(1 gravity / earthradius) You want: revolution/day * 17.060434 You have: 1 day / 17 You want: hour; minute; second 1 hour + 24 minute + 42.352941 second Stack Exchange confirms this: <https://physics.stackexchange.com/questions/136486/how-fast-would-the-earth-need-to-spin-for-us-to-feel-weightless#136495 https://physics.stackexchange.com/questions/136486/how-fast-...>) That's about once every 84 minutes. Which, I recognise is also roughly the orbital period for low-Earth orbit, that is, an object falling around the Earth without ever striking the surface. Though I think that may be a coincidence. Compare geosyncronous orbit (~42,000 km from Earth's centre), with the radius at which there is 1g of centrifugal acceleration at 1 revolution per 24 hours, ~2 million km. Fun to play with however.
dredmorbius 3y ago
Just to clarify, I was solving for radius. Diameter would be 2x, and circumference would be 2pi larger. So 11.6 million km circumference for the 24-hour rotating ring. If you had a trans-ring commute, at 100 kph, that would take you about 6 years 7 months one way. A jet aircraft operating at nearer 1,000 kph would cut that to a far more manageable 7 months. Though I'd still use it as a reasonable argument for return-to-office. Ping times would still be about 38 seconds even for free-vacuum direct transmission, around the ring itself. Given lightspeed is reduced to about 66% of its in vacuo* value in fibre optics, you'd have almost exactly a 1 minute delay. Something to keep in mind during those Ring Zoom sessions.

dredmorbius 3y ago

If the Earth revolved at a rate of ~17 times per day, or once every 1:24:42, the centrifugal force at the equator should just match the pull of gravity. Also via GNU Units: You have: sqrt(1 gravity / earthradius) You want: revolution/day * 17.060434 You have: 1 day / 17 You want: hour; minute; second 1 hour + 24 minute + 42.352941 second Stack Exchange confirms this: <https://physics.stackexchange.com/questions/136486/how-fast-would-the-earth-need-to-spin-for-us-to-feel-weightless#136495 https://physics.stackexchange.com/questions/136486/how-fast-...>) That's about once every 84 minutes. Which, I recognise is also roughly the orbital period for low-Earth orbit, that is, an object falling around the Earth without ever striking the surface. Though I think that may be a coincidence. Compare geosyncronous orbit (~42,000 km from Earth's centre), with the radius at which there is 1g of centrifugal acceleration at 1 revolution per 24 hours, ~2 million km. Fun to play with however.

dredmorbius 3y ago

Just to clarify, I was solving for radius. Diameter would be 2x, and circumference would be 2pi larger. So 11.6 million km circumference for the 24-hour rotating ring. If you had a trans-ring commute, at 100 kph, that would take you about 6 years 7 months one way. A jet aircraft operating at nearer 1,000 kph would cut that to a far more manageable 7 months. Though I'd still use it as a reasonable argument for return-to-office. Ping times would still be about 38 seconds even for free-vacuum direct transmission, around the ring itself. Given lightspeed is reduced to about 66% of its in vacuo* value in fibre optics, you'd have almost exactly a 1 minute delay. Something to keep in mind during those Ring Zoom sessions.