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
'Watch the stars, and from them learn. To the Master's honor all must turn, Each in its track, without sound, Forever tracing Newton's ground.'