calculate the centripetal acceleration for this situation (where R earth= 6.371x

Question

calculate the centripetal acceleration for this situation (where R earth= 6.371x10^6) this is non calculus physics ng Eartth Y: 1. Calculate the centripetal acceleration for this situation (where Rearth compare it to the acceleration due to gravity (g = 9.8 m/s2) ere Rea "6371 x 102 m) and -6.371 x 10 m) and earth 2. Image speeding up the rotation of the earth. Therefore, the tangential speed, v (or ) would increase. Looking at your equation of motion, what happens to the normal force as v gets bigger? Eventually you reach a speed which makes N-0. Physically, what does it mean if N becomes equal to zero? 3. Solve for the earth's minimum speed Vimin which will make you fly off the earth. Using Vmin ine the corres revolution. Saying it another way, this would be the new time for one day on earth. Determine this time. 78

Explanation / Answer

Image is not compete but I am answering whatever I understood from the question. If this is not correct thn please give me the link of complete question in comments.

I am assuming that a person is standing on surface of earth and was rotating with earth.

So speed_of_preson(V) = angular_speed_of_earth*Radius_od_earth

and angular_speed_of_earth =2*/(24*60*60) (because earth rotate 2 andle per day and 1day have 24hrs = 24*60 mins = 24*60*60 sec).

so V = 6.371*106*(2*/24*60*60) = 463.31 m/s

So accleration = V2/R = 463.31*463.31/6.371*106 =0.0336 m/s2 Answer

Which is very less then 9.8m/s2

Please comment if you have any doubt.

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