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Asteroid gravity question, Which way is down? |
Feb 23 2007, 07:23 AM
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![]() Member ![]() ![]() ![]() Group: Members Posts: 809 Joined: 11-March 04 Member No.: 56 |
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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Feb 23 2007, 07:58 PM
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![]() Senior Member ![]() ![]() ![]() ![]() Group: Members Posts: 3419 Joined: 9-February 04 From: Minneapolis, MN, USA Member No.: 15 |
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 -------------------- “The trouble ain't that there is too many fools, but that the lightning ain't distributed right.” -Mark Twain
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David Asteroid gravity question Feb 23 2007, 07:23 AM
tasp Keeping in mind any impacts on this object will te... Feb 23 2007, 07:36 AM
AndyG QUOTE (David @ Feb 23 2007, 07:23 AM) On ... Feb 23 2007, 09:53 AM
JRehling Imagine that a thread attached the Sun and the Ear... Feb 23 2007, 03:38 PM
stevesliva QUOTE (JRehling @ Feb 23 2007, 10:38 AM) ... Feb 23 2007, 06:49 PM
David QUOTE (JRehling @ Feb 23 2007, 03:38 PM) ... Feb 23 2007, 07:37 PM
ngunn Just an anecdote. I was once on a gravity survey i... Feb 23 2007, 03:29 PM
Bill Harris Look at the gravity and slope maps produced by/dur... Feb 23 2007, 04:12 PM
centsworth_II QUOTE (Bill Harris @ Feb 23 2007, 11:12 A... Feb 23 2007, 06:18 PM
fredk As others have said, "down" isn't in... Feb 25 2007, 06:32 PM
David I tried to figure out what would happen on the sim... Feb 25 2007, 10:14 PM
helvick David,
One of the problems with this is that the ... Feb 25 2007, 10:41 PM![]() ![]() |
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