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GrapesOfWrath
2001-Oct-24, 04:48 PM
This question has always bugged me, since the moon looks very rough to me, through a telescope.

Still, I checked with a technician at Brunswick (http://mentock.home.mindspring.com/ifaq.htm#IFAQ007), and as near as I can tell, the earth is as smooth as a billiard ball, but it is slightly (!) out of round.

Donnie B.
2001-Oct-24, 05:18 PM
From the peak of Everest to the deepest point in the Marianas Trench, we have a maximum deviation of just under 20km. Earth's radius is 6378km (equatorial), or 6357km (polar). This makes earth very flat indeed -- 20/6357 --> less than 1/3 of 1% deviation from perfectly flat.

However, your billiard ball is a little bit smoother, at slightly less than 0.2% (0.002in/1.125in radius).

If you go by sea level instead of crust surface, Earth is smoother, as Everest is less than half as high as the Marianas Trench is deep. Now your maximum deviation is about 9km/6357km = 0.15% deviation from flat.

I suspect that the moon is rougher than the Earth, but I don't have any numbers to back this up. But it has fewer weathering processes to even things out.

I just noticed that the difference in Earth's mean equatorial and polar radii is almost the same as the maximum deviation from Everest to Marianas... so even with the equatorial bulge, it's still damn flat!
_________________

My God! It's full of stars!

<font size=-1>[ This Message was edited by: Donnie B. on 2001-10-24 13:21 ]</font>

GrapesOfWrath
2001-Oct-24, 07:24 PM
On 2001-10-24 13:18, Donnie B. wrote:
However, your billiard ball is a little bit smoother, at slightly less than 0.2% (0.002in/1.125in radius).
You've confused diameter and radius. The .002 inch tolerance is on diameter, for the billiard ball. The radii of the billiard balls are not measured--only diameters. That's important for a couple reasons--it makes the tolerance tighter, and, because Everest and the Marianas Trench are not on a diameter, the surface relief is not actually as great as the differences due to oblateness.

Actually, now that I think about it, the profile of the full moon is pretty smooth.

Wiley
2001-Oct-24, 07:43 PM
So is the Earth smoother than the Moon?

The Moon does not have the weathering forces to smooth mountains, but it also does not have the volcanic and plate techtonic forces to create the mountains.

My initial thought is that the moon would be smoother. But I have no numbers to back such outearthish claim.

Mr. X
2001-Oct-24, 09:36 PM
Cool. I didn't know Planets (and satellites [natural]) could be smooth overall. Maybe if it's inaccurate it's damned cool to hear.

I guess that if you just look at the ground under your feet, to you the Earth is completely rough and you'd never know it is actually a sphere. Which I have been doing all my life. You lose the global pesrpective of things.

Well unless you have a way of being sure of walking in a straight line, dropping a quarter at the place you start walking and make your way to the same quarter again.

Hmm, somebody would surely pick up the quarter during that time. Oh well. It was worth a try. /phpBB/images/smiles/icon_smile.gif

Edited to add the (and satellites [natural]) thing, some of the people here can be picky when picky -> annoying!

<font size=-1>[ This Message was edited by: Mr. X on 2001-10-24 17:39 ]</font>

ToSeek
2001-Oct-24, 09:52 PM
This Website - http://website.lineone.net/~petergrego/mwjun98.htm - gives the impression that the tallest mountains on the moon are on the order of 1800 meters from base to top. Anyone want to do the math?

Phobos
2001-Oct-24, 10:35 PM
Planetary "smoothness" can actually be expressed by the simple equation "smoothness is proportional to mass".

In other words, the more massive a planetary object is, the closer it aproximates a smooth sphere.

To make the equation more accurate you would also need to factor in the amount of rotational momentum (how much energy is contained in its spinning). This is important because rotation is the principle cause for the bulging (centrifugal force around the equator cancels out some of the gravitational pull).

The only remaining major factor which afects the shape of a planet is the gravitational fields of neibouring objects (principally its moons).

Both Earth and our moon are subjected to contraction and bulging forces along the line which bisects both bodies centres of gravity. The effects are like both bodies being squeezed. The effect is more noticable on the moon, but can still be seen on Earth in the form of oceanic tidal motions.

Plate techtonics, and volcanic activity also have an effect upon the smoothness of solar objects, but the effect is significantly less as the planets get larger (and more massive).

Plate tectonics cause a resurfacing, but are also responsible for mountain ranges. However, where there is no plate movements you can end up with situations like at Olympus Mons on Mars where the volcanoe just keeps getting larger.

One other mechanism to consider is the effects of subsurface liquids such as those believed to be benief the surface of Europa. The presense of liquid water benief the frozen surface is the best explanation so far for this moon being so smooth.

Jeff

<font size=-1>[ This Message was edited by: Phobos on 2001-10-24 18:44 ]</font>

Wiley
2001-Oct-24, 10:38 PM
On 2001-10-24 17:52, ToSeek wrote:
This Website - http://website.lineone.net/~petergrego/mwjun98.htm - gives the impression that the tallest mountains on the moon are on the order of 1800 meters from base to top. Anyone want to do the math?


The Moon is 3476 km in diameter. This is about 0.1% maximum difference in the radius. (1.8 km / (3476/2) km )

I can't remember the above post, but the Earth is about 0.2%, right? The Moon wins.

Donnie B.
2001-Oct-24, 11:47 PM
In order to compare apples to apples, you'd have to know both the highest elevation and lowest elevation on the moon, and compare that to the Everest-to-Marianas figure for Earth.

There's no "sea level" on the moon, and sea level on Earth does not necessarily equate to the mean level of the crustal surface. If you don't accept that, remember that the sea level on Earth has varies markedly over geologic time. All sea level tells you is how much liquid water is present to fill in the low areas.

Donnie B.
2001-Oct-25, 12:05 AM
The .002 inch tolerance is on diameter, for the billiard ball.


You're right, I misread the reference from Brunswick. So let's run the numbers again:

Earth (using crust extrema) = 20km/(2*6357)km = 0.16% variation

Earth (using sea level to Everest) = 9km/(2*6357)km = 0.07% variation

Billiard ball = .002in/2.25in = 0.09% variation

Same result, but all three are about 2x smoother than I originally stated.

Or do you think the Earth numbers should use radius, rather than diameter? If so, then Earth turns out to be twice as rough as the cueball.

There is another factor to consider, however. The variation in Earth's surface is probably much more chaotic than the variation in a billiard ball. I wonder... if you made a cue-ball size scale model of the earth, would you be able to see the topography? Or perhaps feel it, with fingertips or tongue? The Himalayas would be rough like sandpaper, maybe, and the ocean plates smooth as a billiard ball. Midocean trenches might be like a faint scratch...

In any case, yes, planets are very smooth. Most of the 3D topographic maps and movies you see (of Earth and other planets) are grossly exaggerated in the z direction so they look kewler.

Another thing they often get wrong in SF movies and TV: the atmosphere is *much* thinner than they typically portray it. You don't see a fuzzy halo of atmosphere from orbit, and especially not from further away. After all, the bulk of the atmosphere is below 150km or so. That's 1% of the earth's diameter.

Ben Benoy
2001-Oct-25, 02:18 AM
I think that we don't have enough information concerning the method of measuring billiard balls for smoothness. It could be that I haven't spent enough time playing pool, but billiard balls always seem quite smooth to me. But the tolerance given allows for a billiard ball with a noticable seam.

That is, take a regular billiard ball, and now put a seam .002 inches (approximately the diameter of a human hair, according to this site (http://hypertextbook.com/facts/BrianLey.shtml)) around half of one circumference. Now, this seam should be pretty noticable, and I think that most people would be able to find it. But this ball still passes checks? Sounds fishy. (<font color="blue">Added in edit</font>) Also, what if you had a ball that was two hemispheres glued together, one with a larger diameter than the other, fudging on the "edges" to accomadate the measuring tool. Let one of the hemispheres be .004 inches larger than the other. Then any diameter you care to measure has a radius exactly the right radius for your ball, but there is a .004 inch seam all the way around the ball. (<font color="blue">End of addendum</font>)

Also, we don't need to look at the difference between Everest and Marianis, we need to know what the longest diameter is. For instance, the diameter which connects Everest to the other side of the world might be bigger than the tolerance. But maybe not. We basically don't have that kind of data. Also, this algorithm for smoothness assumes that the shape you're looking at is basically spherical. What is the "average diameter" of the Earth?

Whoever it was who asked whether we would be able to feel the topography of the Earth shrunk down to a billiard ball was on the mark, I think.

Regarding getting around the "average Earth diameter" question, I would suggest that we consider the Earth to have been mapped onto a sphere, preserving the distance between the ground and to local radius, but how do we assign this radius? Use the best fit ellipsoid? I think that this is the best option, but it's still a knotty problem.

Ben

<font size=-1>[ This Message was edited by: Ben Benoy on 2001-10-24 23:15 ]</font>

GrapesOfWrath
2001-Oct-25, 02:58 AM
On 2001-10-24 20:05, Donnie B. wrote:
There is another factor to consider, however. The variation in Earth's surface is probably much more chaotic than the variation in a billiard ball. I wonder... if you made a cue-ball size scale model of the earth, would you be able to see the topography? Or perhaps feel it, with fingertips or tongue? The Himalayas would be rough like sandpaper, maybe, and the ocean plates smooth as a billiard ball. Midocean trenches might be like a faint scratch...

I've played with billard balls like that. :)

Still, it is fairly easy to run the numbers. Sand (http://www.taunton.com/finewoodworking/pages/w00006.asp)paper (http://international.3M.com/intl/CA/english/centres/mfg_industrial/abrasives/pdfs/GradeComparisonChart.pdf) has grit on it, I guess 120-150 grit sandpaper has particles about 100 microns in diameter, or about one ten thousandth of a meter.

Wait, lets do this the other way. Everest is about 8km high, right? Let's assume it is a piece of grit on the earth (ignoring lateral extents of the mountains), what size grit on a billiard ball would correspond? My calculations say that the ball is 1/250,000,000 the size of the earth, so the grit would be 8000/250,000,000 meters, or 32 microns. That's 400 grit sandpaper. Pretty fine.

Ignoring the lateral extent is not an insignificant thing, also, since the mountain sits on the high Himalayan plateau, and land at sea level is a long way away. So, I'm not convinced that an Everest-size bump on the billiard ball would be noticed. I know it wouldn't affect my pool game.

GrapesOfWrath
2001-Oct-25, 03:13 AM
On 2001-10-24 22:18, Ben Benoy wrote:
Whoever it was who asked whether we would be able to feel the topography of the Earth shrunk down to a billiard ball was on the mark, I think.

I guess that would be me.


Regarding getting around the "average Earth diameter" question, I would suggest that we consider the Earth to have been mapped onto a sphere, preserving the distance between the ground and to local radius, but how do we assign this radius? Use the best fit ellipsoid? I think that this is the best option, but it's still a knotty problem.

The best fit ellipsoid is sea-level, for most intents and purposes. That is essentially how sea-level is defined for continental regions--for oceanic regions, it is the time-averaged (because of tides and waves) surface of the ocean.

Ben Benoy
2001-Oct-25, 03:16 AM
Cool, so most of the hard work is done. Still need to know what the Earth looks like on the antipodes, though...

GrapesOfWrath
2001-Oct-25, 03:40 AM
Ah, antipodes. My advisor's favorite subject.

Because the Pacific Ocean almost fills a hemisphere (just look at a globe sometime from that side), almost all land is antipodal to ocean. I think Santiago is antipodal to Xian, or something like that, though.

Although you could slap a human hair on a billiard ball, and call it good, I don't think that is going to occur in the manufacturing process of a billiard ball, nor in the process that created the surface of the earth. Although the Marianas trench is deep and linear, its lateral extent is such that it would not be like a scratch at all.

Ben Benoy
2001-Oct-25, 04:35 AM
On 2001-10-24 23:40, GrapesOfWrath wrote:
Ah, antipodes. My advisor's favorite subject.

)snip!(

Although you could slap a human hair on a billiard ball, and call it good, I don't think that is going to occur in the manufacturing process of a billiard ball, nor in the process that created the surface of the earth.


Even if it couldn't occur in the process of creating a billiard ball, any definition of "smooth" which allows it is flawed. Also, see the second example I added above (it's between the blue words. /phpBB/images/smiles/icon_biggrin.gif ), which is definately and noticably not smooth, and is a big enough "not smoothness" as to mess up your pool game, unless you're worse than I am...

Anyhoo, good call on the Pacific, didn't think of that. Now I want nothing more than to get a couple of billiard balls and a globe and start feeling them for smoothness. Somehow I don't think that a pool hall would take my licking balls too kindly. /phpBB/images/smiles/icon_eek.gif

Also, antipodes just popped up over here in the thread about gas giants. Ooh... ahh... Antipodes! (http://www.badastronomy.com/phpBB/viewtopic.php?topic=39&forum=2&1)


Ben

<font size=-1>[ This Message was edited by: Ben Benoy on 2001-10-25 00:36 ]</font>

GrapesOfWrath
2001-Oct-25, 09:28 AM
On 2001-10-25 00:35, Ben Benoy wrote:
Even if it couldn't occur in the process of creating a billiard ball, any definition of "smooth" which allows it is flawed.

Spoken like a mathematician. Hey, paisan!

The "definition of smooth" was not really a definition of smooth. It was more a definition of roundness rather than a lack of roughness. I'd heard for years that the Earth was as smooth as a billiard ball, or thought I had, but then I couldn't find anything to back up the claim. When I asked the engineer at Brunswick for their tolerances, he told me how they determined if a ball was ready for play.

They have a device, apparently, that rolls the ball between plates and the plates measure the thickness (diameter) of the ball. The tolerances are the ones he gave me.

As near as I can tell, if you plopped the famous blue dot down on the green, you wouldn't complain about its roll at all. In fact, the Billiard Congress of America (http://www.bca-pool.com/play/rules/equip.shtml) seems to have less stringent tolerances, increasing .002 inches to .005. Using that tolerance, as Karen, at the Straight Dope (http://www.straightdope.com/mailbag/mdownup.html) puts it, "When people hear the earth is an oblate spheroid, they tend to think 'hamburger bun' when they should be thinking "billiard ball.'"

Kaptain K
2001-Oct-25, 10:32 AM
When I asked the engineer at Brunswick for their tolerances, he told me how they determined if a ball was ready for play.

They have a device, apparently, that rolls the ball between plates and the plates measure the thickness (diameter) of the ball. The tolerances are the ones he gave me.
Dang it! You woke my "trivia magnet" again. Read in Scientific American (decades ago) that a tetrahedral trochoid (or any polyhedral trochoid for that matter) would pass this test.

GrapesOfWrath
2001-Oct-25, 11:22 AM
Yes. To tell the truth, I have not seen the device, and it may indeed be sophisticated enough that it would distinguish a ball from a trochoid.

Still, even as impressive as some of the Himalayan vistas seem to be in travelogues, the vertical relief is always greatly exaggerated by a bit of an optical illusion--and sometimes deliberately in making illustrations.

If I look at any photo of the earth from space, the edge looks sublimely smooth.

Vega115
2003-Nov-15, 08:28 PM
I guess its kinda like the study that showed that if you made a Pancake the size of Kansas, the state of Kansas would be flatter than the pancake.

Still...its AMAZING that we have a trench that is about 6 miles deep, and a moutain that height, and yet, the earth is still pretty smooth.

The Supreme Canuck
2003-Nov-15, 08:45 PM
Ha! Look at the ads in this thread!

20% off Billard balls!

Celestial Mechanic
2003-Nov-16, 04:54 AM
Planetary "smoothness" can actually be expressed by the simple equation "smoothness is proportional to mass". In other words, the more massive a planetary object is, the closer it aproximates a smooth sphere.
It might be more accurate to say "the more massive a planetary object is, the closer it approximates a smooth oblate ellipsoid. Jupiter and Saturn are much more massive than the Earth, and they are noticeably not spherical.

space geek
2007-Apr-14, 04:51 PM
:dance:If Earth would have been shrunk down to a size of an average billiard ball which is 2.25 in. @ 5.5 oz. (.16kg), which is 1/250,000,000 the size of Earth then some of Earths features as in Mt. Everest, which is 8.8km, may be impossible to feel on a billiard ball, an equation to show this is:


actual height of surface feature in km Model height of surface feature
----------divided by----------------- = --------divided by------------
Earth diameter in KM - 12756 km relief globe diameter in cm


SO.....:rolleyes:

8.8 km X - unknown
------- = ----------------
12756 km 5.715 cm - the same as 2.25 in

50.292 = 12756X :think:
so, X = 0.0039426152398......


So, what this equation is saying is that if Earth was shrunk down to the size of a billiard ball. Mt. Everest would be 0.0039 cm high, which may be impossible to feel. So, I'd imagine that Earth is pretty smooth....:dance:

space geek
2007-Apr-14, 04:54 PM
i hope that this is right... if it isnt, im sorry, im only in 8th grade....

Kaptain K
2007-Apr-14, 05:51 PM
Why did you dig up a three + year old thread?

space geek
2007-May-30, 08:29 PM
Because it sounded pretty interesting....

NEOWatcher
2007-May-31, 01:52 PM
Because it sounded pretty interesting....
I agree, but, there are also a lot of interesting things around here.

You may get some dirty looks by "bumping" a thread around here.

A lot of people look at this forum as a resource, and when an old thread is updated, it signals that there is something new going on, and not just someone who has their 2 cents. So, a bump is generally frowned upon.

Anyway; enough of my wanderings. It sounds like you are trying to confirm your understanding. Per an earlier post that said 32 microns, it seems close enough for me.

Welcome, and enjoy, and keep up the learning.

One last thing: Why did you mention the wieght of the ball?

elementary
2010-Jul-02, 04:39 PM
The statement is often made that "If the Earth were reduced to the size of a cue ball, it would feel just as smooth."

While it makes a great conversation opener at a cocktail party, it isn't quite true.

If the earth (radius 6378 km) is reduced to the size of a cue ball (diameter of 2.25 in which gives a radius of 2.85 cm) that is reducing it by a factor of 2.23E8. If Mt. Everest (height of 8848 m) is reduced by the same factor, it would be about 40 microns tall.

The width of a human hair ranges from 17 to 180 microns and even the finest hair can be felt on a smooth surface, so the Himalayas on our cue ball Earth would be felt as rough patches by a human fingertip. So would many other mountain ranges of a similar order of magnitude.

So the earth is not as smooth as a cue ball.

HenrikOlsen
2010-Jul-02, 09:55 PM
Note that this was a resurrection of an old thread.

There are multiple definitions of smoothness, one of them defines the maximum deviation from average height, others adjusts for the fact that "roughness" can be felt at very small height differences if the angle of the slope is large enough.
With the hair, the slope is very steep, which is why the hair is so easily felt, if at the other hand that same difference in diameter is the difference along the axes of an ellipsoid, no one would be able to feel it.

The Himalayas are spread over a fairly large area and Everest doesn't really poke that far above the surroundings. I doubt it would be possible to feel it on a cue sized earth.
Kilimanjaro on the other hand might just be noticeable as a feelable dot.

grapes
2010-Jul-02, 10:37 PM
The Himalayas are spread over a fairly large area and Everest doesn't really poke that far above the surroundings. I doubt it would be possible to feel it on a cue sized earth.
Kilimanjaro on the other hand might just be noticeable as a feelable dot.This wikipedia article about Kilimanjaro (http://en.wikipedia.org/wiki/Mount_Kilimanjaro)(hey, it says the first person recorded to climb it lived from 1871 to 1996, maybe there's something about it...), says that it is the highest free-standing rise (4.5km) from an encircled contour line that goes no more than fifty kilometers from the summit. In other words it is twenty times wider than it is high, and the elevation goes out ten for every one up.

So you'd have a very fine human hair-thick "bump" that was then feathered out in all directions. I'm not convinced that billiard balls don't have such "protruberences" :)

ETA: Welcome to BAUT, elementary!

neilzero
2010-Jul-03, 12:04 AM
We would expect the Moon and asteroids to have much less equatorial bulge than Earth because the radius is much smaller. While some asteroids have 5 to 10 hour days, the moon has a day 4 weeks long which makes for about 28 times less equatorial bulge. If 1800 meters is correct the moon is the same or less rough than Earth, by any analysis, possibly because the Moon does not have moving tectonic plates. Small asteroids and small moons on the other hand often have a large dimension more than twice the smallest dimension. This is likely because they have little tendency to assume sphereical shape because of their extremely weak gravity. Typical bearing balls are hundreds of times smoother than billiard balls = a rather irrelevant comparison. Neil