Centrifugal Force: Earth Rotation & Its Effects

[Luca 123]

We know that the Earth is rotating, and its gravitational force is the centripetal force. So if I'm standing on the Earth, I'll feel 3 forces: Gravitational force, normal force and centrifugal force. However, the magnitude of the centrifugal force is equal to the gravitational force, so wouldn't the 2 forces cancel out, leaving the normal force to be 0? Why is the normal force I feel= the magnitude of the gravitational force, when I can feel the centrifuhal force?

 

[Dale]

However, the magnitude of the centrifugal force is equal to the gravitational force

That is not always true. In fact, the only time it is true is for an object in a circular orbit analyzed in the rotating frame where it is at rest.

 

[Luca 123]

That is not always true. In fact, the only time it is true is for an object in a circular orbit analyzed in the rotating frame where it is at rest.

So in this case, what would the centrifugal force be? I have normal force=mg=GMm/r^2, but the centrifugal force can't be zero right? So what went wrong in my calculations?

 

[Luca 123]

That is not always true. In fact, the only time it is true is for an object in a circular orbit analyzed in the rotating frame where it is at rest.

Also, what if we assume that the Earth is perfectly spherical, and since I'm standing still relative to the Earth, wouldn't the centrifugal force=GMm/r^2 ?

 

[Dale]

So in this case, what would the centrifugal force be? I have normal force=mg=GMm/r^2, but the centrifugal force can't be zero right? So what went wrong in my calculations?

I am not sure what case you are referring to.

If you are referring to an object in a circular orbit analyzed in the rotating frame where it is at rest then the normal force is 0. The gravitational force is mg directed inwards, and the centrifugal force is mg directed outwards.

 

[Luca 123]

I am not sure what case you are referring to.

If you are referring to an object in a circular orbit analyzed in the rotating frame where it is at rest then the normal force is 0. The gravitational force is mg directed inwards, and the centrifugal force is mg directed outwards.

However, in this case, wouldn't I feel weightless? Do I feel weight(i.e. normal force=/=0) because the Earth isn't perfectly spherical?

 

[Dale]

However, in this case, wouldn't I feel weightless?

Yes. You are in orbit and people in orbit feel weightless. Surely you have seen videos of astronauts in orbit.

Do I feel weight(i.e. normal force=/=0) because the Earth isn't perfectly spherical?

No. You feel weight because you are not in orbit.

 

[Luca 123]

Yes. You are in orbit and people in orbit feel weightless. Surely you have seen videos of astronauts in orbit.No. You feel weight because you are not in orbit.

I'm confused. Isn't the Earth rotating on its own axis, and we are rotating along with it? Doesn't gravity act as the centripetal force? The Earth itself is a rotating frame, so the laws of ratational motion should apply(?)

 

[Dale]

I'm confused. Isn't the Earth rotating on its own axis, and we are rotating along with it? Doesn't gravity act as the centripetal force? The Earth itself is a rotating frame, so the laws of ratational motion should apply(?)

Yes, all of this is true. What is NOT typically true is that the centrifugal force equals the gravitational force. That is only true for a circular orbit.

 

[Bandersnatch]

I'm confused. Isn't the Earth rotating on its own axis, and we are rotating along with it? Doesn't gravity act as the centripetal force? The Earth itself is a rotating frame, so the laws of ratational motion should apply(?)

Gravity acts as the centripetal force PLUS the extra overhead that tries to push you into the ground. But the ground resists, pushing back on your feet with the normal force - this is the force you 'feel'. You never feel gravity directly, it's approximately uniform on the scale of a human being. Unless you're being tidally stretched by a black hole, that is.

 

[nasu]

Also, what if we assume that the Earth is perfectly spherical, and since I'm standing still relative to the Earth, wouldn't the centrifugal force=GMm/r^2 ?

No. This is is probably the source of your confusion.
The centripetal acceleration of a point on the equator is about 0.03 m/s^2. Not even close to the acceleration of gravity.
The same value will have the centrifugal acceleration, in the non-inertial frame.
The centrifugal force is just a small fraction from the weight of an object.

 

[A.T.]

the magnitude of the centrifugal force is equal to the gravitational force

The Earth would disintegrate if it would spin that fast, which would solve your problems with the normal force.

 

[russ_watters]

Also, what if we assume that the Earth is perfectly spherical, and since I'm standing still relative to the Earth, wouldn't the centrifugal force=GMm/r^2 ?

That's the equation for gravitational force, not the equation for centrifugal/centripetal force.

 

[ashtadmir]

You are getting confused here about the concept of orbital velocity. When an object is revolving around the Earth without moving towards or away from its center then we can say that the centrifugal force is equal to the gravitational force.
But when you are standing on Earth then it is not the case, there is a centrifugal force acting on you then too but it's magnitude is very small, this magnitude is small because the tangential velocity you would have will be nowhere near the orbital velocity for a satellite near the surface of earth. Try doing the calculations for the orbital velocity near the surface of Earth and compare it to the tangential velocity of the surface of earth.

PS. you might want to study about geo stationary satellites. They remain stationary in the sky with respect to ground (the way you are stationary when you are standing still on ground). Geostationary satellites have same angular velocity as that of Earth about its axis and for it the equality of centrifugal force and gravitational force holds true.