Centripetal acceleration on earth. The goal if this problem is to figure out how

Question

Centripetal acceleration on earth.
The goal if this problem is to figure out how strong the centripetal acceleration must be for us here in
Albany. For simplicity, we

What is the equation for A R ? You can approximate the latitude (indicated in the figure) as 42 degrees. Using the answer from (a), what is the centripetal acceleration of a person in Albany? (If you can't figure out the answer to (a), then just write this expression in terms of A R ). How fast would the earth have to be rotating (revolutions per hour) in order for the centripetal acceleration of a person in Albany to be equal to 1 g (9.8 m/s2)?

Explanation / Answer

a) Ra = Re cos(theta)

Ra = Re cos(42)

Ra = 6371E3*cos(42 degrees)=4.73E6 m

b) a = v^2/r = (2 pi r/T)^2/r = (2*pi*4.73E6/(24*60*60))^2/4.73E6=0.025 m/s^2

c) a= w^2 r

w = sqrt(9.8/4.73E6)=0.00144 rad/s =0.825 rev/hr

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