## National Geographic SocietyIn partnership with

P.O. Box 98199
Washington, DC 20090-8199
800-647-5463

Lat/Lon: 38.90531943278526, -77.0376992225647

# Science Kids, Fictitious Forces, and Frictionless Surfaces

Posted by on March 12, 2013
Sid the Science Kid screenshot, originally from http://www.confessionsofapsychotichousewife.com/2010/10/sid-science-kid-party.html

SteelyKid has started to demand Sid the Science Kid videos, which of course we are implacably opposed to around here. One of the recent episodes available online was “Slide to the Side,” talking about friction. While this partakes a bit of the Feynman “Energy makes it go” problem, it was generally pretty good, and prompted a question from Kate that (combined with Rhett’s latest.

Kate and SteelyKid were talking about how friction keeps things from sliding, then Kate asked me whether it would be possible to keep anything at rest on a truly frictionless surface, due to the rotation of the Earth. My initial answer was that it wouldn’t be a problem, because the surface and whateveer object you were placing on it would initially be at rest with respect to the rotating surface of the Earth, and thus would continue to move together. It’s not like the Earth would spin out from under the object once you put it on a frictionless surface, or anything.

But after reading Rhett’s analysis of people flying off a spinning Earth, I want to change my answer slightly. As Rhett notes, if you’re going to talk about motion on the surface of the Earth, you need to talk about some fictitious forces that come in due to the rotation, and when you do it out carefully, I think you do get a slight force making things move, but not in the way you might expect.

The relevant force is the one Rhett discusses, the “centrifugal force.” This is a somewhat dangerous term to use in intro classes, because students want to invoke it for everything, and it’s not quite right. There’s no interaction that leads to a real force that would be measurable in a stationary lab frame; the “force” that appears occurs only when you transfer to a rotating frame, where things appear to start moving even without a specific interaction causing them to move. This is the basis of a bunch of amusement-park rides, and is what marks the rotating frame as non-inertial, that is, a frame of reference where Newton’s Laws of Motion don’t appear to work properly.

That said, there’s nothing inherently wrong about looking at the world from the perspective of someone in a non-inertial frame– we do it all the time, as beings riding on a spinning planet that’s orbiting a star. So let’s think about these forces, using the not-remotely-to-scale diagram below:

The black arc here represents a slice of the surface of the Earth, and the blue circles are spherical frictionless objects placed on the surface. If you look at the lower of the two objects, sitting on the equator, from the perspective of an observer who is also rotating with the surface of the Earth, this object experiences two forces: a gravitational force pulling the object toward the center of the Earth (represented by the blue arrow), and a “centrifugal force” which is directed outward, away from the axis of rotation (represented by the green arrow). These are exactly opposite one another, and the gravitational force is vastly bigger than the centrifugal force, so an object at the equator will fall straight down, and just sit on the ground nicely.

For the other object, at higher latitude, though, the situation is more complicated. The two forces aren’t quite in the same direction any more– the “centrifugal force” is still directly out from the axis, to the right in the figure, but the gravitational force is now down and to the left, toward the center of the Earth. This mismatch in directions makes all the difference.

To add these two forces together, we really need to break them into components. since the force of gravity is vastly bigger than the “centrifugal force,” we’ll leave it alone, and only think about what happens to the red arrow. We can break this into two pieces, one that goes along the same line as the gravitational force (the small blue arrow), and one perpendicular to it (the small green arrow). As with the object at the equator, the vertical component is more than balanced out by the gravitational force, so an object will still appear to fall straight down when dropped.

The other component, though, will be unaffected by gravity, and should look like a small force pushing the object toward the equator– south, in the figure above, assuming the usual America-centric view of the globe. If you had a truly frictionless object, and set it down on a frictionless surface in the middle latitudes, you would see it slowly drift to the south.

So, how big is this force? Well, if we call the latitude of the object (the angle measured up from the equator) $\theta$, the green arrow will have a magnitude given by:

$|F_{south}| = |F_{cent}|\sin \theta$

(a little bit of trigonometry there, but an easy bit), where $|F_{cent}|$ is the magnitude of the centrifugal force. this, in turn, is given by:

$|F_{cent}| = m \omega^2 R_{\perp} = m \omega^2 R_{Earth} \cos \theta$

where $\omega$ is the rate of the Earth’s rotation in radians/sec. The cosine creeps in there because the radius that matters is the radius away from the axis of rotation, which gets smaller as you move to higher latitudes. Putting these together, we find that:

$|F_{south}| = m \omega^2 R_{Earth} \cos \theta \sin \theta$

This will have its biggest value at a latitude of 45 degrees, coincidentally not too far from where I happen to be writing this, where the product of trig functions will be equal to exactly 1/2. Putting in numbers for everything (the Earth turns through 2 π radians in 86400s, the radius of the Earth is about 6,400,000 meters), we get a force equal to the mass multiplied by 0.017 m/s/s. Which means that an object placed on the surface of the Earth at that latitude should accelerate southward at 0.017 m/s/s, or about 2/1000ths the acceleration of gravity.

What’s that mean for your frictionless surfaces? Well, it suggests that an object placed on a truly frictionless surface ought to take about 11 seconds to slide 1m to the south. That’s actually a bit faster than I would’ve expected, which may indicate that I’ve made an arithmetic error someplace, but I’ve already spent more time on this than I should’ve, so I’ll leave that for people to point out in the comments.

Anyway, there you go. From SteelyKid to the Sid the Science Kid to Rhett the Science Blogger, your weird physics-y blog tidbit for the week.

For extra credit, calculate the oscillation frequency of a mass sliding back and forth across the equator on a smooth frictionless sphere the size of the Earth. this is, of course, anharmonic, but for a first pass you can use the small-angle approximation for the latitude.

For extra extra credit, add the Coriolis force to the problem. Send your extra credit problems for grading to Neil de Grasse Tyson, The Hayden Planetarium, 81 Central Park West, New York, NY, 10023.

(The Sid screenshot that’s the “Featured Image” for this post came from Confessions of a Psychotic Housewife.)

## Comments

1. NL
March 12, 2013

I think you mean “red”, not “green” when talking about the equator?

2. Eric Lund
March 12, 2013

You haven’t made any arithmetic errors that I can see, but you did assume a spherical earth. The actual shape is an oblate spheroid, so the normal force of the surface does not point exactly toward the center of the earth. Instead, it will have a slight poleward component, countering the equatorward component due to centrifugal force. I haven’t done the full calculation, so I don’t know if they exactly cancel, but I would expect them to be the same order of magnitude.

3. March 12, 2013

Yes, I meant “red” when talking about the equatorial point. This is what I get for spending so much time talking to the dog– my color vision gets all wonky.

I had meant to mention the shape of the Earth in this, but forgot when I started to close in on my self-imposed deadline. You’re probably right about the order of magnitude– the difference in radii is about 0.3%, and if you took the really naive approximation of saying there was an 0.3% axial component of the gravitational force, that would be bigger than the force above. It’s probably more complicated than that, but almost certainly on the same order of magnitude as the force described here.

Clearly, what we need is a perfectly frictionless surface so we can see which way, if any, an object slides on it. Somebody get on that.

4. CCPhysicist
March 12, 2013

And you forgot that the Earth’s axis is tilted.

If the earth were an inviscid liquid, it would take a shape that eliminates the force you are calculating. It isn’t, of course, but that is why the oblate shape of the earth works against the simplistic spherical argument.

And the statements above (and your analysis) ignore the moon and sun. Their gravity, normally manifest as tidal forces, act on that mass as well and the tilted axis gives a seasonal contribution.

5. Ian
March 13, 2013

Surely the oblate spheroidiness of the earth would exactly cancel the rotation.

The non-sphericity of the earth is determined by the exact same two forces as you are using to determine the object’s acceleration.

Or to put it another way. For the purposes of determining the shape of the earth, it is a fluid. A fluid in a steady-state must have no shear force. A particle of fluid (or a body floating on the surface) on the surface, feels gravity, buoyancy along the surface normal, and the centrifugal force. If the three did not cancel exactly, the surface particle would feel a sheer force. Since it is a fluid, it would deform ans so the shape of the earth would not be a steady-state. But it is, so the forces must cancel exactly.

6. Wilson
March 13, 2013

From the sublime to the ridiculous: Not having seen any episodes of Sid the Science Kid, why the implacable opposition? Bad science?

7. March 13, 2013

That was a joke. We’re happy to have her watch science-y shows– Mythbusters is one of the tv staples in our house.

I should’ve included an explicit sarcasm flag, because Internet, but I was writing in a hurry.

8. MobiusKlein
March 13, 2013

What about gyroscopic effects? An object on a frictionless surface will still be changing orientation as the Earth spins.
(except at the poles)

9. Wilson
March 14, 2013

Gotcha.

Yes, the Internet needs a sarcasm tag of some kind. Text communication in general does, but I guess the bulk of that is the Internet anyway.

Thanks for the clarification!

10. Clay B
United States
March 15, 2013

I think Ian is onto something there. The sliding particle is going “uphill” in terms of distance from the center of the Earth, so gravity is opposed to (cancels out?) the Equatorward push. At least my physics intuition says everything should cancel out.

11. thomas
NZ
March 15, 2013

I would have thought the oblateness and centrifugal components would have cancelled out exactly at the time the crust solidified. The earth was rotating faster then, so now, with a slower rotation, the oblateness should win and the object should slide away from the equator — but probably very slowly.

12. Ian
March 19, 2013

The solid crust is flexed by the moon’s gravity on a daily cycle. It is only a few 10s of km thick at most, compared to the 1000skm diameter of the earth. It will have zero rigidity over geological time periods.

Another way of thinking about the whole problem is to think of what mean down, and what means a flat, level surface. A flat level surface is one which has no net gravitational force, as resolved in the directions of the plane. When setting up this surface, assuming that you use a spirit level, you can’t tell which part of the forces are gravitational and which are centrifugal. So, almost by definition, you will put this surface down so that there is no acceleration along the surface.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 6, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9346328973770142, "perplexity_flag": "middle"}