If you were on a planet how fast can it spin to cancel the affects of gravity ? so that you would float off ?
2006-10-27 02:44:47 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Astronomy & Space
A planet would need to spin fast enough for the surface velocity at the equator to be equal to the escape velocity. In the particular case of Earth, the escape velocity is 11186 m/s, while the surface velocity at the equator is only 465.11 m/s, very close to 1/24 the required speed. So the earth would need to spin about once an hour instead of once a day, to start throwing objects off at the equator.
2006-10-27 02:47:24 · answer #1 · answered by DavidK93 7 · 2⤊ 1⤋
Your answer is: velocity = squareroot(GM/r) (G is the universal gravitational contant = 6.67*10^-11, M = Mass of planet, r = radius of planet at equator) for a person standing at the equator of the planet for the effects fo gravity to be cancelled by the spinning of the planet. To be flung off the planet never to return any velocity faster than the velocity required to cancel the effects of gravity will do the job. Note that there is no intermediate. You are either standing on the surface of the planet being weightless or you are flung off never to return. A person would not just float on the surface they would be flung off. Note: This is the easiest way to calculate the velocity you desire. To find the velocity required for different latitudes would be quite involved.
2006-10-27 09:53:54 · answer #2 · answered by mg 3 · 1⤊ 0⤋
360mph speed of light . Each planet depends on inner core revolving at a different rate to the surface. The hotter the core grater the energy and therefore higher the gravity . Why do you think all the planets are circling the sun and not the earth.
2006-10-27 10:30:56 · answer #3 · answered by SUE N 2 · 0⤊ 0⤋
Earth would have to spin at about 17,694 MPH to make things on the equator weightless. That would make a day about an hour and a half hour long instead of the present 24 hour day.
Centrifugal acceleration is V^2/R. Acceleration of gravity on the surface of the Earth is 9.8 meters per second and Earth is 6,378,000 meters in radius at the equator. So V^2/6,378,000=9.8. Solve for V to get 7,906 meters per second, or 28,462 kilometers per hour or 17,694 MPH. Circumference of Earth is 2*PI*R=2*3.14*6,378 kilometers, or 40,054 kilometers. At 28,462 kilometers per hour it would take 1 hour and 24 minutes to rotate once.
2006-10-27 09:53:51 · answer #4 · answered by campbelp2002 7 · 0⤊ 0⤋
Are you talking about a centrifulgal force so strong that could cancel the force of gravity?
I think such a movement would crush us in the process.
2006-10-27 10:00:32 · answer #5 · answered by Just Me 2 · 0⤊ 0⤋
Well earth spins at apoximately 1000 mph.
24,901 around the earth and
24hrs in a day
2006-10-29 17:28:14 · answer #6 · answered by Anonymous · 0⤊ 0⤋
If that happened I would think the planet would explode.
2006-10-27 14:03:26 · answer #7 · answered by lisalau 5 · 0⤊ 0⤋
about 20,000 feet per sec
2006-10-27 16:24:18 · answer #8 · answered by garrbear99 1 · 0⤊ 0⤋