## Centripetal acceleration at nonzero latitudes

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selene
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### Centripetal acceleration at nonzero latitudes

If you're standing on the equator, then as the Earth rotates you move in a circle around the center of the planet, with gravity providing the centripetal acceleration.

But if you're anywhere other than the equator, you don't orbit the center of the planet. You travel in a circle around the nearest point on the Earth's axis (corresponding to your latitude).

Gravity is still pointed towards the center of the planet. The normal force holding you up is normal to whatever surface you stand on - if Earth were a perfect sphere this force would be straight "up", colinear with gravity. So we need another force angled somewhat towards the nearest pole in order to make the total force be towards the axis.

Friction is the obvious candidate for a force that adjusts to exactly what it needs to be to keep you moving at the same rate as everything around you. But I'm confused as to just how it manages to exert a force in the right direction.
If I'm at 40 degrees north latitude, then the ground beneath my feet is moving at roughly 350 m/s from west to east. If friction drags me along then it should exert a force in the west-to-east direction. This is tangential to the circle I want to be traveling in.

Where is the component of force that makes my actual acceleration 40 degrees north of down?

PM 2Ring
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### Re: Centripetal acceleration at nonzero latitudes

selene wrote:If you're standing on the equator, then as the Earth rotates you move in a circle around the center of the planet, with gravity providing the centripetal acceleration.

Sure, but the strength of gravity is much greater than the centripetal acceleration.

selene wrote:Gravity is still pointed towards the center of the planet.?

Almost true. It'd be true if the Earth were a perfectly smooth and homogenous sphere, but it's not.
However, the vector sum of the gravitational acceleration and the centripetal acceleration doesn't point directly at the centre of the planet, even if the planet were a perfectly smooth and homogenous sphere.

If we pretend the Earth is a perfect sphere, then to find the centripetal acceleration at latitude A:
ac = v²/r
v = 2*pi*r / T,
r = R * cos(A)
where R is the radius of the sphere and T is the rotational period.
So
ac = 4*pi²*r / T²
=4*pi²*R*cos(A) / T²
(4 * (pi^2) * radius of Earth * cos(40 degrees)) / ((24 hours)^2) = 0.0258391095 m / s2
The magnitude of the centripetal acceleration is much smaller than the gravitational acceleration, so it doesn't affect the direction of the combined acceleration very much.

gmalivuk
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### Re: Centripetal acceleration at nonzero latitudes

selene wrote:Gravity is still pointed towards the center of the planet. The normal force holding you up is normal to whatever surface you stand on - if Earth were a perfect sphere this force would be straight "up", colinear with gravity. So we need another force angled somewhat towards the nearest pole in order to make the total force be towards the axis.
The normal force is always straight "up", essentially because that's what "up" means. "Down" is the direction of local total acceleration, and "up" is the direction opposite that. There's no particular reason to expect that this would point directly toward the center of the planet.
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Drowsy Turtle
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### Re: Centripetal acceleration at nonzero latitudes

selene wrote:Gravity is still pointed towards the center of the planet. The normal force holding you up is normal to whatever surface you stand on - if Earth were a perfect sphere this force would be straight "up", colinear with gravity. So we need another force angled somewhat towards the nearest pole in order to make the total force be towards the axis.

Why? These are the only two forces that are actually acting on you, and they exactly cancel out.

selene wrote:Where is the component of force that makes my actual acceleration 40 degrees north of down?

You seem to have realised that, relative to a stationary observer at the equator, gravity has an apparent Southwards component, and also that the normal to this force acts in the opposite direction along the same line.

It stands to reason then, does it not, that the normal force will have a Northward component relative to an observer at the equator?
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selene
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### Re: Centripetal acceleration at nonzero latitudes

PM 2Ring wrote:the strength of gravity is much greater than the centripetal acceleration.

You are correct. I left out the normal force in mentioning the equator because I wanted to move on to the anywhere-else scenario, where I did include the normal force. My problem is with the direction, not the magnitude, of these forces.

Drowsy Turtle wrote:These are the only two forces that are actually acting on you, and they exactly cancel out.
...
It stands to reason then, does it not, that the normal force will have a Northward component relative to an observer at the equator?

The normal force cancels out most, but not all, of the gravitational force.

The vector sum of gravitational force and normal force is radial. That's the direction I'm calling straight down (or up). These two forces alone should produce an acceleration towards the center of the Earth. If they canceled perfectly, you wouldn't accelerate at all.
In fact you are moving in a circle of constant latitude. Your acceleration, and therefore the sum of all the forces acting on you, is towards the center of the circle. Which is NOT the center of the Earth, it's somewhere between that and the North Pole.

To make it clearer, let's use 89.9 degrees north instead of 40. You're just a few miles away from the North Pole. An observer on Polaris with a really good telescope watches you over the course of one day. In that time you trace out a circle around the Pole - or rather, around a point just beneath the pole.
Circular motion is caused by centripetal acceleration which is caused by centripetal force. Some force is pushing you towards the pole. It's certainly not gravity. If the normal force is not colinear with gravity due to the distortion of the Earth, then we're saying a perfectly spherical planet would be impossible to stand on. The net force isn't just "not quite" the same direction as gravity and normal force, it's 89.9 degrees away from their direction.

gmalivuk
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### Re: Centripetal acceleration at nonzero latitudes

No, the net force, the sum of gravity and centrifugal, is almost entirely dominated by gravity, and thus points quite nearly straight toward the center of Earth.

Things would maintain position just fine even on a completely frictionless part of Earth, as long as they started out not moving relative to the surface. The shape of the planet itself guarantees that.
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Qaanol
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### Re: Centripetal acceleration at nonzero latitudes

Selene, what you are missing is that the earth is not perfectly spherical: it is wider at the equator and shorter at the poles. Additionally, we define “down” as, essentially, the direction that an object released in freefall would accelerate. In other words, a surface is “level” if and only if an object resting on it feels no tangential forces.

If the earth were perfectly spherical (let’s say, made of rigid metal) and still rotated, then at non-zero latitudes we would say the surface is not level. If you poured water onto that spherical surface, the water would flow “downhill” toward the equator. If you set a marble on that spherical surface, the marble would roll toward the equator.

If you poured lots and lots of…let’s say wet sand, enough to cover the metal sphere with a thick layer, and you waited for the sand to come to rest, it would reach a shape much like the real earth. It would be broader at the equator and shorter at the poles. And then the surface would be level. If you put a flat sheet of metal somewhere on that surface, and put a marble on it, the marble would not roll.

Moreover, on the real earth, if you go somewhere at middle latitudes, and place a flat sheet of metal so that the surface normal points directly away from the center of the earth, that sheet will not be level. If you put a marble on it, the marble would roll off. If you poured water on it, the water would flow off.

On the actual earth, if you take a level surface, the direction “straight down” almost certainly does not point toward the center of the earth, because the earth is not spherical.

Gravity, to a good approximation, does point directly toward the center of the earth. And the normal force to the surface does point perpendicular to the surface. And these are not on the same line. The difference between them, depending on your coordinate system, is either a centrifugal force, or the net force causing circular motion at a constant latitude.

Don’t get caught up on the idea that centrifugal forces are “fictitious”. They absolutely are fictitious, but so is the “force” of gravity. The only “real” force present is the electromagnetic repulsion that is constantly accelerating you up out of the geodesic path you would otherwise follow (and the magnetic component of that may as well be considered fictitious as well.)
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PM 2Ring
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### Re: Centripetal acceleration at nonzero latitudes

Qaanol wrote:If you poured lots and lots of…let’s say wet sand, enough to cover the metal sphere with a thick layer, and you waited for the sand to come to rest, it would reach a shape much like the real earth. It would be broader at the equator and shorter at the poles. And then the surface would be level. If you put a flat sheet of metal somewhere on that surface, and put a marble on it, the marble would not roll.

In other words, the surface normal at any point on the spinning wet sand surface surface would be pointing in the opposite direction to the sum of the gravity vector and the centrifugal force vector. The actual Earth isn't exactly like this because it has lumpy bits, but it's fairly close to this equilibrium shape.

See http://en.wikipedia.org/wiki/Reference_ellipsoid
and http://en.wikipedia.org/wiki/Geoid

selene
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### Re: Centripetal acceleration at nonzero latitudes

I get it now, thank you!

I was stuck thinking it ought to be possible to rest a marble on a rigid spherical planet. But it makes perfect sense that no you can't, and that's why real planets aren't spherical.

Very good point about "down" being measured in a non-inertial frame, and thus being not the same as the "direction of gravity" I was picturing.