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

'The mathematician's patterns, like the painter's or the poets, must be beautiful; the ideas, like the colours or the words, must fit together in a harmonious way. Beauty is the first test: there is no permanent place in the world for ugly mathematics.'