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I know that in order to travel in a circle I have to have a net centripetal force $F=mv^2/r$. I also know that my normal force and gravitational force cancel. How, then, am I traveling in a circle around the Earth as it spins?

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I don't understand this question (I'm way out of my zone here and I can't read latex). What do you mean by traveling in a circle? Are you asking why we don't fly off the Earth like a penny on a record player? Coming from the HNQ list, I thought this was going to be about the curved flight paths of airplanes. – Mazura Feb 1 at 21:33
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Just to add some concrete numbers here, another way to phrase this is $F/m = \omega^2 r$ where $\omega = 2\pi/\text{day}$ and $r = 6,378\text{ km}.$ Multiplying these by each other we find out that this quantity is only $0.0337\text{ m}/\text{s}^2,$ or about one-291th of the the acceleration due to gravity. Furthermore as we co-orbit the Sun there is another acceleration to orbit which is about one sixth of that, and then there's the acceleration to orbit the Milky Way which is positively tiny -- $2\cdot10^{-10}\text{ m}/\text{s}^2$ or so. – CR Drost Feb 1 at 22:04
up vote 28 down vote accepted

The answer is that the normal force and the force of gravity do not cancel each other out; the force of gravity is slightly stronger, and it's stronger by an amount equal to $m v^2 / r$.

In elementary physics problems we often ignore the rotation of the Earth to simplify the problem, since it is generally a very small contribution; we therefore conclude that $F_g = F_N$ for an object on a horizontal surface.

Here's the formal answer to your question. We can set up Newton's 2nd law like so:

$$F_\textrm{net} = m a$$

$$F_g + F_N = m a$$

We know from geometry that centripetal acceleration is given by $a = v^2 / r$, and it's pointing inward, which we'll call the "negative" direction, so

$$F_g + F_N = - m (v^2 / r)$$ $$- F_g = F_N + m (v^2 / r)$$

I've formatted the last equation to show that the force of gravity is opposite in direction and stronger than the normal force by an amount $m v^2 / r$.


Note: the actual centripetal force on the surface of the Earth actually varies in direction relative to the force of gravity as you move away from the equator. The angle of the ground with respect to gravity also changes, since the Earth is closer to a geoid than a sphere. The general situation at any latitude requires friction to resolve, as well as treating the forces as vectors, since they aren't collinear, but the general principle that the force of gravity is stronger than the normal force, and that extra force provides the centripetal acceleration that maintains our circular motion, is still valid.

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Unless they are vector equations ypur equations are only true at the equator where the centripetal acceleration is directed towards the centre of the Earth. – Farcher Feb 1 at 17:28
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That's a fair point, I'll note that in the answer. I don't think the level of the question warrants a vector treatment. – Brionius Feb 1 at 17:31
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WOW This is one of those answers that illuminate a corner that I didn't even know was dark. Thank you! – dotancohen Feb 2 at 16:07
    
And I believe that is the centrifugal force term that physics teachers constantly claim doesn't exist... =( – Izkata Feb 2 at 19:49

I also know that my normal force and gravitational force cancel.

But not exactly and that little bit of force produces your centripetal acceleration. It is one of the reasons that your weight as measure at the North Pole is greater than your weight when measured at the equator.

enter image description here

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"...your weight as measure at the North Pole is greater than your weight when measured at the equator." This seems counter-intuitive. The radius of the earth at the equator is greater than at the North Pole. So (assuming I am correct in that), it seems there is more mass directly below your feet at the equator. – Kevin Fegan Feb 2 at 1:45
    
Mass directly below your feet isn't the only mass pulling on you. The whole planet is doing so, but the sideways component of the forces os balanced and cancels out. (To a first approximation, anyway) – keshlam Feb 2 at 2:38
    
@KevinFegan: You're correct, but it's not the only issue at play here. There's a good answer here that goes into more detail. – MichaelS Feb 2 at 5:50
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The assumption was of a uniform spherical Earth. The actual shape of the Earth is such that the direction of $\vec g$ is approximately at right angles to its surface. A plumb line does not point towards the centre of the Earth but is approximately at right angles to the surface. – Farcher Feb 2 at 10:42
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@Kevin Fegan At the North pole you are closer to the centre of mass of the Earth so the attractive force is larger than at the Equator. – Farcher Feb 2 at 10:44

First of all, if you are at either pole, you are not traveling in a circle. Any place else, you would be traveling in a circle, with the largest being at the equator.
For you to understand what is happening, I will use two cases:
1 - No spin generated force.
When you are at a pole, you don't "fly out" into space, nor do you sink into the ground, therefore, the force of gravity on you (mg), is balanced by the force exerted on you by the ground(Fn). $mg = F_g = Fn$.
2 - Spin generated force.
When you are at the equator, there is a tangential velocity that creates a centripetal force, but since you stay in place, then the force of gravity must equal the ground (normal) force + the centripetal force $mg = F_g = Fn +\frac{mv^2}{r}$

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