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Asteroid gravity question, Which way is down?
David
post Feb 23 2007, 07:23 AM
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On a non-spherical asteroid or moon (which I suppose is most of them), is the direction of "down" simply toward the overall center of gravity of the entire body, so that, for instance, on a flattened spheroid, a wheeled rover near the equator would find itself on a downward slant, and could roll "downhill" from the equator all the way to one of the poles? Or on a dumbbell-shaped asteroid (assuming the center of gravity to be halfway between the two ends of the "dumbbell"), on proceeding from the ends of the asteroid to the center, it might find itself plummeting straight down as the slope of the asteroid momentarily coincided with a line drawn through the center of gravity? Or is it more complicated than that?
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tasp
post Feb 23 2007, 07:36 AM
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Keeping in mind any impacts on this object will tend to jostle loose surface debris 'downslope' over the centuries. Dramaticly weird gravity gradients relative to an appropriately smooth rover friendly expanse would be pretty unusual, I think.

2 monolithic asteroids in a very close orbit about each other might do some intersting things in the 'gap', but this is a rather idealized situation I suspect is rare in nature.
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AndyG
post Feb 23 2007, 09:53 AM
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QUOTE (David @ Feb 23 2007, 07:23 AM) *
On a non-spherical asteroid or moon (which I suppose is most of them), is the direction of "down" simply toward the overall center of gravity of the entire body, so that, for instance, on a flattened spheroid, a wheeled rover near the equator would find itself on a downward slant, and could roll "downhill" from the equator all the way to one of the poles?


This is a subtler question than I first thought!

Thought #1:

If we look at the gravitational attraction of an oblate spheroidal body as a point source acting at the centre of the body, with the surface being a non-massive, thin, skin - then there is a change of gravitational "dip" compared to the local surface "normal".

The maximum angle between dip and normal depends on the degree of flattening, and its location depends on the amount of sphericality of the body - it moves equatorwards the more flattened the body, and as you tend back towards a sphere, the latitude of maximum dip moves to eventually disappear at 45 degrees north when the body becomes a sphere. The dip-normal angle is, of course, always 0 degrees at the equator and at the poles.

Following thought #1, a smooth-skinned oblate spheroid would allow you (sitting just away from the equator) to roll ball bearings past the nearer pole and back again.

Thought #2:

As you move north on a non-spheroidal body, from the equator, you have what is effectively a mountain growing behind you: this will make the model of #1 incorrect - so I can't declare an oblate spheroid to act as a point source when I'm situated on the surface.

Thought #3:

Centripetal effects: a rotating oblate spheroid may well have enough angular velocity to cancel out the "to the pole" acceleration that #1 suggests.

...And then, like the sun rising over the tumbling surface of 202 Chryseďs, it struck me:

Thought #4:

Why is an oblate spheroid an oblate spheroid? Because it rotates (#3) and was once molten. It has the shape it has since the surface has set in the lowest potential energy state that it can. Surely that means that the local surface gravity has to be normal to the plane, no matter what the latitude?

I'm going to crunch some numbers later to prove #4 is correct.

Dumbells and contact asteroids would, indeed, have odd grav effects - but I wouldn't use the word "plummeting" with such low gravities!

Andy
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ngunn
post Feb 23 2007, 03:29 PM
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Just an anecdote. I was once on a gravity survey in Greece that required centimetre-accurate surveying-in of the station elevations. One particular survey traverse gave a significant mis-tie however often it was rechecked. This turned out to be due to a nearby mountain altering the direction of the local apparent vertical, exactly as in Cavendish's famous experiment to determing G.
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JRehling
post Feb 23 2007, 03:38 PM
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Imagine that a thread attached the Sun and the Earth, making them one body. The center of gravity is inside the Sun. Now, for someone on the Earth, do they feel gravity's pull only in that (the Sun's) direction? No. QED.

Gravitational fields are "lumpy" around irregularly-shaped bodies. The Sun-thread-Earth is a hyperbolic example to provide an existence proof of that without heavy math, but Eros and Phobos well qualify as well. And for that matter, the Moon and Ganymede, which have significant mass concentrations (mascons) that need to be mapped before they could/can be orbited efficiently.
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Bill Harris
post Feb 23 2007, 04:12 PM
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Look at the gravity and slope maps produced by/during the Hayabusa encounter last year. Some of the flattest-looking areas has a 35% apparent slope. On a small irregular body down ain't necessarily so.

--Bill


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centsworth_II
post Feb 23 2007, 06:18 PM
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QUOTE (Bill Harris @ Feb 23 2007, 11:12 AM) *
Some of the flattest-looking areas has a 35% apparent slope.


It would be interesting to see astronauts trying to manage that!
Like a tilted room.
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stevesliva
post Feb 23 2007, 06:49 PM
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QUOTE (JRehling @ Feb 23 2007, 10:38 AM) *
And for that matter, the Moon and Ganymede, which have significant mass concentrations (mascons) that need to be mapped before they could/can be orbited efficiently.

http://science.nasa.gov/headlines/y2006/06nov_loworbit.htm
http://science.nasa.gov/headlines/y2006/30nov_highorbit.htm

I was wondering when mascons would be mentioned.
QUOTE
The mascons' gravitational anomaly is so great—half a percent—that it actually would be measurable to astronauts on the lunar surface. "If you were standing at the edge of one of the maria, a plumb bob would hang about a third of a degree off vertical, pointing toward the mascon," Konopliv says. Moreover, an astronaut in full spacesuit and life-support gear whose lunar weight was exactly 50 pounds at the edge of the mascon would weigh 50 pounds and 4 ounces when standing in the mascon's center.
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David
post Feb 23 2007, 07:37 PM
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QUOTE (JRehling @ Feb 23 2007, 03:38 PM) *
Imagine that a thread attached the Sun and the Earth, making them one body. The center of gravity is inside the Sun. Now, for someone on the Earth, do they feel gravity's pull only in that (the Sun's) direction? No. QED.


The example is so unrealistic that it's hard for me to appreciate the point. The Earth is orbiting around the Sun, so they're not dynamically one body (and any "thread" connecting them would instantly snap); of course, close to the center of the Sun is the center of gravity for the whole Earth-Sun system, but on Earth we can practically ignore that when dealing with local gravity, because both we and the Earth share the same orbit.

A more realistic thought experiment might be: On a contact (or near-contact) binary, where the two objects are orbitally locked to each other and can be considered to be one body dynamically, is "down" towards the center of mass of each object considered separately, or is it toward the common center of mass of the two bodies?

If the latter, then one might expect the space between the two bodies to get filled up with dust, rocks, and other impact debris that gets freed from the surface of either body, producing an Itokawa-like agglomeration.
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dvandorn
post Feb 23 2007, 07:58 PM
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Rotation between the two objects in a contact or near-contact binary may occur around a mathematical "center of mass" of the two bodies. But I'm thinking that only actual *mass* can generate a gravitational field. The spot where their gravitational fields interact and counter-balance may define their common center of gravity, but it would not in and of itself exert any gravitational force. The most it could do would be to collect a smallish amount of dust in its own little Lagrangian point(s).

So, the greatest gravitic attraction to a mass on the surface of such an asteroid would be the vector towards the largest and closest attracting mass. Since a process of pulling material off of each body would be a dynamic process, I'd guess that whenever you got to a point where the other body exerted more pull than the body on which a given item (rock, dust particle or even astronaut) is sitting, all of the loose stuff would settle into new, stable configurations. Thus the rock pile masses that seem to connect several large pieces of Itokawa.

We're talking really small masses, here, when it comes to actual gravity field effects. We don't know exactly how much force holds the various pieces of Itokawa together, for example. The whole thing is only the size of a smallish hill. Could an astronaut, using just his/her muscles, actually shift one of the large pieces of Itokawa such that all of the stable configurations become newly dynamic? Or would it take a lot more energy than that to accomplish any serious movement of the big pieces against each other?

All I know for certain is that, until we get out and actually start doing it, we won't know just how non-intuitive all of this might end up being. But I bet it'll take a lot of getting used to...

-the other Doug


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“The trouble ain't that there is too many fools, but that the lightning ain't distributed right.” -Mark Twain
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fredk
post Feb 25 2007, 06:32 PM
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As others have said, "down" isn't in general towards the centre of mass, but instead it's towards the sum of the individual gravitational force vectors from all pieces of the body. Since those individual forces decrease like one over the square of the distance to the piece, "down" is heavily weighted towards the closer parts of the body. (Rotation complicates this, of course.)

But you're absolutely right that on a (non-rotating) squashed-at-the-poles smooth asteroid, a ball would roll from the equator to the poles. In practice this will only be an issue for small enough (few 100 km or less) bodies. For these bodes the gravitational force at mid-latitudes (which won't be normal to the surface) won't be strong enough to overcome the structural forces and redistribute mass and collapse the body into a sphere (recall the pull-itself-into-a-sphere criterion for defining planets!).

I could definitely imagine weird stuff near small irregular bodies!
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David
post Feb 25 2007, 10:14 PM
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I tried to figure out what would happen on the simplest variant of the "dumbell asteroid" I could think of: two spheres, of equal size and density, rotating around each other and in contact at a single point on the equator of both of them. That produces two centers of gravity to work with, and I suppose if you were halfway decent at physics you could figure out the exact direction and strength of the combined force of gravity from the two spheres.

I am not halfway decent at physics, but I think that the general picture is like this: if you start on the equator of one of the spheres (Sphere A) at the exact opposite side from the other sphere, "down" will be straight down, in the direction of the center of gravity of the sphere you're standing on. The gravity will be somewhat greater than it would be if you were on the sphere alone due to the additional pull of the second sphere.

Proceed along the equator in any direction, and you will seem to be walking down a very gentle slope, because the pull of Sphere B is now coming from a different direction than that of Sphere A, and you are also getting closer, so it's getting stronger. "Down" wouldn't be at the center of gravity of the two spheres, but would rather be somewhere on a line between the common center and the center of gravity of Sphere A.

After you pass the halfway point, the slope will start to get steeper much faster. However, as the pull of the two objects is now coming from different directions, you will start to get lighter, as the forces to some extent cancel each other out.

As you approach the point of contact of the two objects, you will be "falling" nearly straight down, but you would also be extremely light, due to Sphere A and Sphere B pulling on you with nearly equal strength from opposite directions. If you separated the spheres slightly, and arrived at the midpoint between them, you'd be floating weightless at the common center of gravity. So stuff would fall toward that point, but it wouldn't fall very hard, because the combined forces would cancel each other out, lessening as you got closer. So dust and rocks might end up floating to the midpoint, but they would also be very easily dislodged. Of course, if they began to accumulate, that would in itself change the gravitational characteristics of the asteroid. The thicker the point of connection between the two objects, the greater the local gravity (due to there being some matter directly underfoot, as opposed to the big masses on both sides).

What I don't quite see is, as the valley fills up with debris and eventually creates an ellipsoidal asteroid, at what point do the two separate centers of gravity of the two objects coalesce into a single center? I suppose if the rubble were significantly lower density than the original objects, that might not happen at all?
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helvick
post Feb 25 2007, 10:41 PM
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David,

One of the problems with this is that the effect is highly dependent on the track taken. If you move parallel to the central axis of your dumbbell shaped object but remain outside the envelope described by a cylinder covering both ends then local gravity will peak when you are directly above the midpoint between the two end spheres (which is also the center of gravity of the whole system obviously).

Once you follow the surfaces of the spheres into the mid zone then the effect of reducing local gravity occurs but the same could be said of a hypothetical voyage to the centre of any spherical object.

For those with the ability to build graphical gee whiz apps there's an opportunity to make an interesting interactive educational "toy" to demonstrate these effects and allow people to play with rolling marbles down some very weird hills. smile.gif
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