Quick easy graviational problem

[Demin]

Homework Statement

How fast would the earth(Massearth =5.98X10^24 KG, Radius Earth = 6.4X10^6 m) have to spin on it axis so a 75kg person at the equator would be weightless.

Homework Equations

T^2= kr^3
Fg = G m1 m2 / r^2

The Attempt at a Solution

i know that first i need 2 find the W or the person then put the weight equal to some equation that i don't know =/ .. some 1 please help me ;P

 

[dwintz02]

Perform a force balance for a person standing on the equator and set that equal to the centripedal acceleration. Next, what does it mean to be weightless?

 

[Demin]

Perform a force balance for a person standing on the equator and set that equal to the centripedal acceleration. Next, what does it mean to be weightless?

weightless = u weight 0 N ;/ i still haven't figured it out .. do u mean

(G)(75kg)(5.98X10^24)/6.4X10^6 = V^2/r


? please respond

 

[dwintz02]

For a person standing on the Earth at the equator, how about:

GMm/r^2 - N = m*v^2/r

Sorry for the confusing post earlier, I meant mass times the centripedal acceleration. Remember the normal force (N) is still relevant here because the Earth is physically pushing the person away from the center. Ok now, which of the terms in my equation should go to zero as the Earth spins faster and faster (person becomes more and more weightless)? Or, if it helps, imagine the Earth is spinning just fast enough and the person's feet JUST leave the ground, then what term will be zero?