[X] My Assistant

[David]

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?

[AndyG]

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 Chryseïs, 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

[JRehling]

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.