# Owl City and the Slow Earth

I’d like to make myself believe
That planet Earth turns slowly…

– Owl City, Fireflies

If you’re had any exposure to pop radio over the last few months, you’ve heard this plaintive rumination about the earth’s rotation. The first time I heard it I thought it must be the Postal Service getting back together, but alas it was not so. Oh well, at least we can get some physics out of it.

Does planet Earth turns slowly? It depends on how you look at it. At the equator it’s roaring along at more than 1000 miles per hour. This is pretty fast by terrestrial standards, and perhaps what Owl City had in mind. But on the other hand its angular velocity is just one rotation per day, or a very sedate 0.000694 RPM. Kids sometime imagine the earth as a spinning basketball and wonder why we don’t fly off, and that’s pretty much why: spin a basketball at 0.000694 RPM and I bet any ants on its surface will keep their grip just fine.

But this suggests a question: what would count as “fast” for the purposes of our scientifically inquisitive musicians? I think a good threshold might be the speed at which the earth’s rotation would start throwing people off at the equator. More specifically, the point at which the gravitational force required to hold people to the earth’s surface is exactly equal to the centripetal force required to hold them to the circular motion.

To be clear, there is no magical centrifugal force throwing people off, it’s simply that rotation is not straight line motion, and motion in anything other than a straight line requires forces. When standing on the ground there’s two of them. There’s gravity pointing down and the force of the floor (the “normal force”) pointing up. Together, they have the equal the exact centripetal force needed to keep you in circular motion about the center of the earth as the earth rotates.

We may remember that centripetal force is just the radius of the circle times the square of the angular velocity, and we may remember Newton’s equation for the gravitational force. If we do, we know they’re equal at the point where people start floating off their feet due to the rapid rotation of the earth. Why not write this relationship down?

Is it tough to solve this for the angular velocity? Nope!

Here M is the mass of the earth, m is the mass of an object on it (it’s canceled out in the last expression), G is the universal gravitational constant, and r is the radius of the earth. Plug in and you’ll get an angular velocity of 0.0012395 Hz. This isn’t too enlightening since it’s in radians per second, but if we turn it into revolutions per day it’s more interesting. Doing this (check my work please!) and I get a value of 17.045 revolutions per day.

So if the world rotated 17 times faster – in other words, a “day” of about 85 minutes – people near the equator would in fact go flying off. So would everything else, leaving us with a planet rapidly diffusing in space. Seventeen doesn’t sound like too much of a huge number, but then remember that centripetal force is proportional to the square of the velocity. Seventeen squared is a much more hefty 289 times more force. I think it’s fair to say planet earth would not in fact be turning slowly in that case, and Owl City’s worry would be justified. But as it is, I’d say it’s slow enough.

1. #1 Cyrus Robinson
January 5, 2010

1nd!

Er… time (and Earth, apparently) flies when your… dissecting pop music for physics lessons.

2. #2 Bruce E
January 5, 2010

The fact that things like the shuttle and ISS in low Earth orbit have a period of 90 minutes is an excellent confirmation of you 85 minute figure for the surface of the Earth at the equator.

3. #3 Watts
January 5, 2010

I remember a problem in my mechanics class in which calculated the offset between true “down” (i.e, towards the center of the Earth) and what we thought was “down” due to the radial component of the centrifugal “force”. At our latitude (roughly 32 degrees), I recall it being surprisingly large, like 10-15 degrees. That would seem to indicate that the Earth spins quickly.

The Earth also spun a bit faster back during the dinosaur eras. I wonder if the slight (the question is, therefore, how slight?) difference in apparent weight enabled larger creatures to exist more easily.

4. #4 Chris
January 5, 2010

If the earth spun faster the equatorial bulge would be larger so the actual velocity for a given angular velocity would be greater and therefore people would fly off (aka orbital velocity at the surface of the planet) at a lower rotational speed.

As a silly example if the earth were puffed up to geostationary orbit size without changing its mass then there would be no difference between orbiting and standing on the surface.

So what’s the answer when you take the equatorial bulge into account? You can find the size of the bulge using hydrostatics but you probably can’t treat the earth as a single point mass any more. Say, what IS the near-field potential on the edge of a thin disc vs. a sphere of the same mass?

5. #5 Uncle Al
January 5, 2010

Take care with your “day.” You need a sidereal day of 86,164.09054 seconds not a tropical day of 86,400 seconds. The proper number is 0.000696 RPM. That carries through your gravity vs. centripetal force calculation, too.

1.74 solar-mass pulsar PSR J1903+0327 is in a 95.17-day 0.437-eccentricity orbit with its 1.05 solar-mass star companion. It has an equatorial spin greater than 11% of lightspeed. 27% vs. 1.4×10^(-4)% gravitational binding energy, 1.8×10^11 vs. ~30 surface gees, 2×10^8 gauss vs. ~5 gauss magnetic field, superconductive compressed neutrons and exotica vs. proton-electron plasma.

The binary is perfectly described by General Relativity. No observable property – even relativistically and quantum mechanically extreme – falsifies the Equivalence Principle. Physics has no chance of empirically overthrowing GR.

http://www.mazepath.com/uncleal/boojum.pdf
Chemistry has a phat one – in physics’ own apparatus.

One should not be parochial about spin. All the fun is in the footnotes.

January 5, 2010

“Every day the world turns faster
Turning faster than it seems.
Leave me here on the Street of Dreams.”

— Oysterband, Street of Dreams.

7. #7 Dante
January 6, 2010

“Doing this (check my work please!) and I get a value of 17.045 revolutions per day.”
I get 107.0928 revs/day: http://www.wolframalpha.com/input/?i=0.0012395+Hz+%2A+1+day

8. #8 Marcia Earth
January 6, 2010

I suppose the speed of the Earth’s rotation is relative to the number of days you think you have left.

9. #9 Uncle Al
January 6, 2010

The Earth’s rotation and its derived physics are relative to the fixed stars (sidereal day). A tropical day is a sidereal day extended by a factor of (1 + 1/365.2422) for the incremental orbital movement that returns the sun to the same spot in the sky.

Leap year every four years except for leap centuries, but counting leap centuries evenly divisible by 400.

10. #10 Bean
January 6, 2010

So what you’re saying is that Superman not only could not have turned time back to save Lois in the movie but also probably would have flung most of Earth’s population off into space?

That bastard!

11. #11 r4 kort
January 8, 2010

I believe that it is about a man who is stuck between being alseep and awake, and it causes him to hallucinate about fireflys. As the city growth have been flown so.

12. #12 melior
January 10, 2010

“Doing this (check my work please!) and I get a value of 17.045 revolutions per day.”
I get 107.0928 revs/day

You have calculated radians per day. Dividing by 2pi will give you the right answer in revolutions per day, 17.04.

13. #13 Carl Brannen
January 18, 2010

Your mention of Owl City got me to hunt around for it and it’s a wonderful tune. If I begin feeling wealthy, I’ll consider buying the album.

For several decades I kept more or less current in my musical taste but I also don’t like hearing a lot of DJ chatter on a radio so I ended up with a station that didn’t play much new stuff. Now I’ve switched to one that seems to play an eclectic mix from old to new and sure enough, they give Fire Flies air time.

Thanks.

14. #14 Nikhil
December 31, 2011

Thanks exactly what i was looking for. Had solved this during my school days. But at 30 u dont remember much of that.