How fast would the Earth have to rotate so that it would neutralize gravity?

Asked by: Brad Nelson

Answer

In order to neutralise the acceleration due to gravity the centripetal acceleration needs
to be equal to the acceleration due to gravity:

Centripetal acceleration = 9.81 m/s^{2}

The centripetal acceleration is, a:

a=r x w^{2}

Where r is the Earth's radius (in our case the radius at the equator), and w is the angular
velocity.

Let a = 9.81 m/s^{2} and r = 6.4 x 10^{6} m

9.81 = 6.4 x 10^{6} x w^{2}

Therefore w = 0.00124 rad/s

This is how fast the Earth would need to rotate to get centripetal acceleration at the
equator equal to 9.81 m/s^{2}.

So if we use this value in this equation:

w = 2/T

Where w is the same as before, the numerator is constant, and T is the time for rotation or
the period.

If we put our value of omega (angular velocity) into the equation we find that T = 5074.99 seconds or 1.409 hours. This
means that the Earth would need to rotate with a period of 1 hour 24 minutes. This means
it would need to rotate approx. 20 times faster than it does now!
Answered by: Dan Summons, Physics Undergrad Student, UOS, Souhampton

'In a way science is a key to the gates of heaven, and the same key opens the gates of hell, and we do not have any instructions as to which is which gate.
Shall we throw away the key and never have a way to enter the gates of heaven? Or shall we struggle with the problem of which is the best way to use the key?'