How Long Must a Day Be for Weightlessness at the Equator?

[madis]

Hello, this is my first post here :)
The question I'm having problem with follows: how long would one day have to be so that there would be weightlessness on the equator? (the teacher said that the correct anwser would be roughly 5 minutes)

Solving this by myself I couldn't reach any other conclusion than:
gravitational force = centrifugal force => mg=mvv/r => g=vv/r => vv=gr
(v-velocity of any point on the equator, r- radius of the Earth, 6,4 million meters)
which means that: v=7920m/s => T=2*pi*r/v=5077s=84 minutes

What did I leave out of the account?

 

[Hootenanny]

Welcome madis.

An equation of the apparent strength of gravity is:
$$g_a = g - r\omega^2 \cos^2 \phi$$
Where $\phi$ is latitude (which would be zero at equator) and $g_a$ is apparent force of gravity, which in your case would be zero. I can show the derivation if you like.

 

[madis]

Thanks! Using this equation I still got 84 minutes for the solution so I guess this is the real right anwser...

 

[Hootenanny]

Yes, I get around 85 mintues also. I would be interested in seeing how your teacher calculated the five minutes, if you could post it here please