What causes the centripetal acceleration of the earth?

[gsingh2011]

For a body to undergo uniform circular motion, a centripetal force which is perpendicular to the velocity at all times must be applied to the body. For the moon and the earth, the gravitational force exerted on the moon by the Earth causes the moon to move in a circle instead of continue on a straight path. A similar example is a ball on a string. When the ball is moving in a circle parallel to the ground, the centripetal acceleration is caused by the tension in the string. So what causes the centripetal acceleration of a person on the earth? The reason I'm confused is because the force of gravity on a person is exactly canceled out by the normal force. If it wasn't, then that person would either move up or move down. So if those two forces cancel out, what is the force perpendicular to our velocity that keeps us moving in a circle?

 

[tiny-tim]

hi gsingh2011!

… The reason I'm confused is because the force of gravity on a person is exactly canceled out by the normal force.

no it isn't …

although we usually treat the "laboratory frame" as inertial, the laboratory is actually rotating once every 24 hours, and therefore strictly speaking we should include a centrifugal force in that frame

that centrifugal force is the difference between gravity and N

however, when we measure the weight of something, we always actually measure N anyway, in other words the figure we use for g already has the centrifugal force subtracted from it!

ie, our 9.81 is really a combination of gravity and centrifugal force

 

[I_wonder_why]

Hi tiny-tim,

Is this a valid approach? -

The centripetal force applied to a person on the surface of the Earth must be supplied by gravity, the only inward-pointing force. However, this must be a NET inward force, meaning that the normal force from the Earth must be less than gravity by a value equal to the centripetal force (N=Fg-Fc).
You get the same result: "we always actually measure N anyway, in other words the figure we use for g already has the centrifugal force subtracted from it."

 

[tiny-tim]

Hi I_wonder_why!

The centripetal force applied to a person on the surface of the Earth must be supplied by gravity, the only inward-pointing force. However, this must be a NET inward force, meaning that the normal force from the Earth must be less than gravity by a value equal to the centripetal force (N=Fg-Fc).

This is very confused …

the problem is that you're using the term "centripetal force"

this is best avoided unless there's one and only one force with a centripetal or centrifugal component …

in this case, the centripetal acceleration is supplied by the NET force of gravity and N …

it would be more logical to call that net force the centripetal force, but it would be best to avoid the term completely …

in an inertial frame, you can say that the normal force from the Earth must be less than gravity by a value equal to the mass times the centripetal acceleration (N=Fg-mv²/r)

in the rotating frame, you can say that the normal force from the Earth must be less than gravity by a value equal to the centrifugal force (N=Fg-mv²/r)

 

[I_wonder_why]

"in an inertial frame, you can say that the normal force from the Earth must be less than gravity by a value equal to the mass times the centripetal acceleration (N=Fg-mv2/r) "

This is what I was saying, I just wanted to be sure that it's equally valid to look at this only using centripetal force (understanding that it's not a separate force but a net force of gravity minus N).

Thanks!