NEUTRALIZING GRAVITY
How fast should the earth be spinning to neautralize gravity?
If we can get centripetal accelration from earth's rotation to equal the gravitational pull (g=9.8m/s^2), we have, basically, neturalized gravity. Let's go through this ROMANTIC calculation. (go to end if you are not interested in the calculation)
a(centripetal) = r x w^2 (w = angular velocity)
since r and w are perpendicular,
a= rw^2
we want this a to equal g
9.81 = rw^2
r = radius of earth = 6.4 x 10^6
plugging everything in, we get
9.81 = (6.4x10^6) w^2
so w = 0.00124 rad/s
T = 2*Pi/w
T = 5067 s = 1.4 hours
SO IF THE EARTH COMPLETES ITS ONE ROTATION IN 1.4 HOURS INSTEAD OF ITS USUALL 24 HOURS, WE DON'T HAVE ANY GRAVITY TO WORRY ABOUT. THAT MEANS EARTH SPINNING ABT 20 TIMES FASTER THAN ITS CURRENT RATE.
If we can get centripetal accelration from earth's rotation to equal the gravitational pull (g=9.8m/s^2), we have, basically, neturalized gravity. Let's go through this ROMANTIC calculation. (go to end if you are not interested in the calculation)
a(centripetal) = r x w^2 (w = angular velocity)
since r and w are perpendicular,
a= rw^2
we want this a to equal g
9.81 = rw^2
r = radius of earth = 6.4 x 10^6
plugging everything in, we get
9.81 = (6.4x10^6) w^2
so w = 0.00124 rad/s
T = 2*Pi/w
T = 5067 s = 1.4 hours
SO IF THE EARTH COMPLETES ITS ONE ROTATION IN 1.4 HOURS INSTEAD OF ITS USUALL 24 HOURS, WE DON'T HAVE ANY GRAVITY TO WORRY ABOUT. THAT MEANS EARTH SPINNING ABT 20 TIMES FASTER THAN ITS CURRENT RATE.


