On the Moon with it’s lesser gravity, would a heap of loose granular material (sand) form a pile with the same steepness as here on Earth? Or would it be either higher or flatter?
<blink>WAG ALERT!!
WAG ALERT!!
WAG ALERT!!</blink>
I would guess that gravity notwithstanding, the lack of a real atmosphere (well, one as dense as ours) means the surface of the moon has a lot of turbulance which would flatten any sand and make mole hills out of mountains. Please note, I’m talking sand (dust), not rock.
Yer pal,
Satan
Yeah, I’d say it’d be steeper. Even on earth though, all loose material slopes don’t break at the same angle. Sand is different than other junk. Chemical composition of sand makes a diff and any “additives,” such as clay. Moisture content. Prevailing winds. But as to just gravity, yeah.
Here’s my WAG:
Hmm i thought i saw a show about the moon where they said the “sand” was really more like little glass beads than sharp angular grains (they said that was partly why the astronauts tripped a lot in that file footage). If that is true, the rounder grains wouldn’t hold together as well as they would if angular and lead to more smoother piles of sand. From the footage I have seen of the moon it looks like the hills have fairly shallow slopes and are rounded.
I would think that gravity makes the difference in how high you can pile things here and on the moon. With lower gravity, i would be you can have higher and steeper hills before the materials slide off.
‘The beginning calls for courage; the end demands care’
Given equivalent (i.e., same shape) sand particles you would see the same geometries for piles here and on earth. While less than on earth, the moon still exerts significant gravitational attraction. A sand grain on the pile is going to tumble when it reaches a position where its center of gravity is no longer supported. That will be the same on the moon and earth. So, no, you couldn’t make sand piles with steeper sides on the moon.
Naturally occurring shapes are affected by forces in addition to gravity, such as wind.
I vote with beatle. If you lean over backwards too far, gravity will make you fall over. The point at which you lose your balance should not change. Your ability to regain your balance and the speed with which you fall will probably be altered, but I leave that to a physics major.
(Um, I’m new here. What’s a WAG?)
Look up “angle of repose” somewhere. It has to do with friction vs gravity. If gravity lessens (as on the moon) and friction remains the same, I think your pile of sand would be taller.
eace,
mangeorge
Work like you don’t need the money…
Love like you’ve never been hurt…
Dance like nobody’s watching! …(Paraphrased)
How’d I make that little square? Supposed to a “P”. I’ll try again.
Peace,
mangeorge
Work like you don’t need the money…
Love like you’ve never been hurt…
Dance like nobody’s watching! …(Paraphrased)
mangeorge:
I did as you suggested and looked up ‘angle of repose’ on the Web. This is the only definition I found, but other sites also implied gravity was considered as a completely separate factor from angle of repose, in geological considerations, leading me to conclude that the angle of repose of any granular substance is not a function of the amount of gravity to which it is subject. See:
http://www.apnet.com/inscight/06181997/angle-o1.htm
, which quotes the AP Dictionary of Science and Technology as saying:
"angle of repose
Engineering
• the steepest angle of a surface at which a mass of loose or fragmented material will remain standing in a pile on a surface, rather than sliding or crumbling away; the angle will vary according
to the composition of the material. Also, angle of rest."
Thus I’d say that the lesser gravity of the moon would not permit a taller pile of a given amount of sand on the moon than would be the case on earth.
Ray
>>>(Um, I’m new here. What’s a WAG?)
<<<
wag means: Wager A Guess
Ray;
I’m thinking of the experiment with the weight on a board. you raise one end of the board till the weight slides, which gives you the angle of repose. I’m going on memory here, and physics was never my strong point. So my assumptions may be way off.
BTW; gravity must be a factor. Otherwise, the sand would never fall.
Peace,
mangeorge
Uh, WAG usually means “Wild ass guess”
Satan wrote:
Turbulence occurs in a fluid, and there is no atmosphere (fluid) on the moon for turbulence to form. The sand pile should sit there for millenia while a sand pile in the Sahara is eroded and rebuilt many times a year.
mangeorge writes:
In this experiment, the angle would be the same on the earth as on the moon because the force of gravity along the board is
mg * sin (angle)
where g = acceleration due to gravity.
and the force of friction keeping the weight in place is
N * coefficient of friction
where N is the normal force:
mg * cos(angle)
So, put it all together and we have:
mg * cos(angle) * frictioncoef. = mg * sin(angle)
Thus the mass and gravitational force cancel out and you are left with the tangent of the angle being equal to the coefficient of friction.
Gravity is by no means required to be a factor other than in the sense that it must be non-zero for there to be a sand pile, but as you can see it cancels out. Remember the feather and hammer trick from the http://www.straightdope.com/ubb/Forum3/HTML/003200.html faked moon landings.
That is only in this particular experiment. I’m guessing it applies to the sand pile case, but there may be other esoteric factors I’m not thinking of. Time to sleep…
[Note: This message has been edited by Nickrz]
Nope. Steeper on the moon. Definitely.
Take it to the extreme – what would be the angle of repose if there were no gravity?
BTW, John McPhee wrote an excellent article for the New Yorker a few years back about the steepness of the hills around L.A. The hills are steeper than the angle of repose. I.e., the hills are going to lose their load, sooner or later. Big brush fires, not uncommon there, burn off the vegetation and then the rain comes and loosens everything up. Result: an E-ticket ride at no charge. McPhee’s thesis was that in spite of the obvious danger and historical precedents, people keep building higher and higher up on the hills. They put in little catch basins and retaining walls and firmly believe it won’t happen to the.
“And comb London’s teeming millions for him? Had we but world enough and time.”
Dorothy L. Sayers
Murder Must Advertise
That last sentence would be more intelligible if the last word were “them”.
Sorry.
pluto writes:
For a moment, let’s focus on the weight-on-the-board experiment.
I’m interested in the thinking behind your ‘Definitely’, because the angle is quite definitely not going to be steeper on the moon - unless someone can point out a flaw in my math.
The extreme of 0 gravity doesn’t make sense. You (pluto) are looking for the answer “90 degrees” but at 0 gravity there would be no angle of repose - not 90 degrees, but no angle - you’d have a weight floating in the general vicinity of a board. For the sand pile, you’d have a sphere of sand floating in space, slowly spreading out if it can’t self-gravitate. If you look at the above math you’ll see that if g=0 the equation is a tautology and predicts nothing.
The appropriate extremes would be near-infinite gravity and near-zero gravity. The reduction or increase in downforce on the weight is exactly balanced by the reduction or increase in friction between the weight and the board.
Now for the OP: How would the weight-on-the-board be different from the sand pile? This is a question I honestly don’t know the answer to, but am NSWAGing it would be the same. Any other thoughts on the sand pile?
-
I admit my first answer was a “C’mon, that’s obvious!” reply, rather than a well-reasoned argument.
-
I agree, douglips, that your analysis is correct if you are considering, say, a block of wood sliding down an inclined plane. The tangent of the angle at which the block of wood starts to slide is equal to the coefficient of friction. (The static coefficient, actually. Once it starts sliding the friction coefficient usually gets smaller.) The angle is independent of the force of gravity. And you’re right – no gravity makes the question nonsense.
-
I don’t agree that the angle at which the whole pile starts to slide, as a unit, is the same as the angle at which the grains in the pile start to slide. So I don’t think your analysis is necessarily applicable.
-
I now think this whole thing is a lot more complicated than I thought at first.
-
I’m going to go to the library on Monday and see if I can find some answers.
“And comb London’s teeming millions for him? Had we but world enough and time.”
Dorothy L. Sayers
Murder Must Advertise
pluto said
pluto,
I’m not sure we’re all working on the same question. The OP said “On the Moon with it’s lesser gravity, would a heap of loose granular material (sand) form a pile with the same steepness as here on Earth?” I think between my response and douglips’ we’ve addressed the relevant considerations.
Brief Infomercial
If you’re finding this thread hard to read, just copy it all, or relevant portions, and paste into a text reader. I think, douglips, something about the way you brought in some of your response causes the horizontal scroll problem. How did you do it?
Back to the OP. The extreme case of 0 gravity is not under consideration. Neither is the case of lithified sandstone. The hills around LA are not just piles of loose granular material. Up there, your dealing with rocks, not sand grains. Many similarities, but not the same thing. The way I read the OP is, “could you get a steeper grade on a stack of marbles on the Moon?” And the answer is no.
The same. Definitely.
Having perused a couple of soil engineering texts and wading through terms like critical void ratio and negative dilatancy, I’ve come to the conclusion that piles of sand would look the same on the moon or the earth for the reasons listed above, notably by douglips.
The governing equation is called “Coulomb’s equation” (yes, it’s the same Coulomb), described in 1773, and it describes the shear strength of soils. In English:
shear strength = cohesion + (normal stress on shear plane * friction angle).
For sand and, in general, non-organic dry materials (e.g. gravel) the cohesion is zero and when the shear stress equals the shear strength the sand shears (slips) at the friction angle, which, in this case, is the angle of repose. The derivation of the equation is based on analysis very similar to that described in this thread.
However, if the soil is cohesive, that is, if the soil sticks to itself (because it’s damp or sticky, etc.) then the piles can become steeper in a smaller gravity field. But that’s not what was posited in the OP.
Thanks, Nickrz.