- #1

IWantToLearn

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but if we took earth rotation into account then the normal force will be less by a value mv

^{2}/R

according to this relation

*N*= mg -mv

^{2}/R

Example : a 70 kg man standing on a bathroom scale on the equator , where the radius of the earth at the equator is R = 6378 km , and the time to complete one rotation is T = 24 hours

[assuming that the gravitational acceleration is constant everywhere on the surface of the earth and equals to g = 9.8 m/s

^{2}]

doing the mass : we found that the normal force (the reading of the scale in newtons)

equals

*N*= 683.64 N

while the gravitational force must be equal to mg = 686 N , this means that the reading is less than the actual value (2.36 N less)

if we did the same calculations near the north pole where the radius is R = 6,356 km

there will be less by a value (2.35 N less)

both readings are not accurate, i know that all scales are calibrated to indicate the mass

now my questions is :

1- These small deviations from the actual value, are they neglected?

2- If they aren't, and the manufacturers take into account the earth rotation and where in the earth you are, is that means that if bought a bathroom scale from london it will not be accurate for me to use it in kenya?

3- what about the change of the value of g itself from place to place on the surface of the earth?