Quick easy graviational problem

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Homework Help Overview

The problem involves calculating the rotational speed of the Earth required for a person at the equator to experience weightlessness. It is situated within the context of gravitational forces and centripetal acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to balance gravitational force with centripetal acceleration. There are attempts to clarify the concept of weightlessness and how it relates to the forces acting on a person at the equator.

Discussion Status

Some participants have provided guidance on setting up the equations necessary for the problem, while others are exploring the implications of increasing rotational speed on the forces involved. There is an ongoing exploration of the terms in the equations and their relevance to the concept of weightlessness.

Contextual Notes

Participants are working with specific values for the mass of the Earth and the radius, as well as the mass of the person in question. There is an acknowledgment of the complexity of the problem and the need for further clarification on certain concepts.

Demin
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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
 
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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?
 
dwintz02 said:
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
 
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?
 

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