jeffdanielk
Favorite Answer
The surface would have to move at orbital speed of 17500 miles per hour. Earth would have to rotate once ever 90 minutes I stead of 24 hours. If this was the case, we would be weightless on the surface. Any faster, and the matter that earth is made of would fly off into space. The planet would fall apart.
Michael
hahahaahahhahaha that would be awesome...lmao
daniel g
Well, it would depend on your latitude. At the equator, if earth surface velocity got near 17.000 MPH, you would go weightless. Faster would be negative G and you would be thrown off the surface. Now consider gravity, you would follow a trajectory not unlike throwing a ball into the air. Hit earth, the process repeats.
At the poles, days would be like 85 minutes, maybe get dizzy.
ANDRE L
In addition to the two good answers already posted, one must also take into account the drag from the atmosphere, in slowing down any objects flung off of the Earth.
In reality, no people or objects could land on such a body's surface, and it's debatable whether such a body could stay intact with such an insane amount of rotational spin.
billrussell42
centrifugal force would have to be greater than weight
mg = mrω²
ω = √(g/r) = √(9.8/6.37e6) = 0.00124 rad/s
0.00124 rad/s x 1 rev/2π rad = 0.0001974 rev/s
or 5066 sec/rev
or 1.41 hr/rev
or
earth circum. is 40,000 km = 1 rev
40,000 km/rev x 0.0001974 rev/s = 7.90 km/s
earth radius 6,371 km = 6.37e6 meters
Centripetal force f = mV²/r = mrω² ω is angular velocity in radians/sec 1 radian/sec = 9.55 rev/min m is mass in kg r is radius of circle in meters V is the tangental velocity in m/s = ωr f is in Newtons