Rotation Problem: Earth lengths/Apparent Weights

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The discussion centers on calculating the apparent weight of a 150 kg individual rotating around the Earth, with a radius of 6.40 x 10^6 m. The normal force was calculated to be 1.4649 x 10^3 N, which corresponds to the apparent weight experienced by the individual. The main question posed is how long a day would need to be for this apparent weight to be achieved, with the answer determined to be approximately 6.16 x 10^4 seconds, or 17.1 hours. The relationship between normal force and apparent weight is emphasized, as they are equivalent in this context. The calculations involve centripetal acceleration and the period of revolution.
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Homework Statement


A friendly Brazilian has a mass of 150 kg. Being in Brazil, he rotates in a circle around the center of the Earth once per day, The radius of this circle (which is essentially the radius of the Earth) is 6.40 x 10^6 m.

I have found that the normal force is 1.4649 x 10^3.

Homework Equations


ac = v2/r , centripetal acceleration
\SigmaF=ma
v = (2\pir)/T , T is period of one revolution


The Attempt at a Solution


I only got to mg - Fn = mac
I know what the apparent weight is but I'm not sure about the normal force. If I can find this, then I can then find ac, v, and then T to find about the length of day.
 
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What is the question?

ehild
 
Sorry forgive me. The question is: How long would a day have to be for the Brazilian's apparent weight to be 1.46 x 10^3 N?

Btw, the answer should be about 6.16 x 10^4 sec or 17.1 hours.
 
The weight is equal to the force the man presses a horizontal support - scales, ground, chair ... So its is the same as the normal force.

ehild
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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