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 v

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.