“A rocket moving out of a gravity well does not actually need to attain escape velocity to do so, but could achieve the same result at walking speed with a suitable mode of propulsion and sufficient fuel. Escape velocity only applies to ballistic trajectories.”

Apparently escape velocity applies only to ballistics? Also, as I’ve read in another article (actually the same article but on Slovenian Wikipedia ), spacecrafts don’t need to reach the actual speed as it’s specified as planet Earth’s escape velocity (in that same Wikipedia article); they first travel to lower orbit (with lower speed, obviously) and then accelerate further to escape velocity at this specific height (which is slightly less; the article says it’s about 10.9 km/s) so in the end that extra required speed is lower as the vessel is already accelerated to a speed of about 8 km/s.

I really hope I’ve translated it correctly (I did it as best as can).

The topography reminds me a little of the Martian moons, which is neat… in spite of the difference in size/mass, might they all be of similar compostion to have had their terrain affected in similar ways?

Anyway, even though it’s just a ‘chunk of rock’ in space, I find this asteroid amazing… heck, all the asteroids have been amazing, once seen up close! I can’t wait to see more details of Vesta, and am really looking forward to Ceres!

When our planet’s magnetic field flips again, North on the globe will still be North, even though the earth’s magnetic “North” pole will be in the Southern hemisphere.

@ Jamey:

It’s not actually true that your trajectory necessarily will bring you back to the same point. The amount of gravitational potential energy for an object that is at infinite distance from the source of gravity is not infinite — the integral of 1/r^2, from r0 > 0 to infinity, converges.

Ergo there is some amount of kinetic energy that you can apply to an object where it will never slow to a stop or turn around due the gravity of the body it originally was launched from. This is what “escape velocity” means.

The other and in my opinion more logical choice is to use the sense of the rotation to define north by sense of rotation (look down on the north pole and the planet appears to rotate anticlockwise), but this can be more difficult to determine observationally.

Because it’s spinning. At two points on Vesta’s surface, the axis of rotation intersects the surface, just like on Earth. As to which one is the North Pole and which one is the South Pole, I assume it’s determined by whether it’s spinning the same direction as the Earth or the opposite direction.

“Are poles defined by the magnetic poles, or is it simply by the axis of spin?”

I’m going to guess the second. I doubt Vesta has a magnetic field, though I suppose that just means it’s waiting there to surprise astronomers…

I mean in my layman’s opinion if the “chunk of rock” is VERY small (small as Vesta or even smaller, see below paragraph) there might not be enough gravitational force/pull to pull you back (to the “point of your last impulse”) and return you to the ground.

For example, what about if the asteroid in question would be only 10 miles in diameter (or only 1 mile etc.), would this rule that you’ve described still apply?

Well, I am not a physicist / astronomer either, but if I understand correctly it doesn’t matter how far away from the rock you get, you will still experience its gravitational influence. The only way in which you can ever not fall back to the rock is if some other massive body captures you.

So, if you go into orbit around the rock, the point of your last mpulse will be on that orbit, and since you pushed off from the rock to get that impulse, your orbit must intersect with the surface of the rock.

This must be where they build and test monster trucks.

Hmmm…you sure about that?

Escape velocity for Vesta, according to Wikipedia, is .35 km/sec – 350 meters/sec, or about 790 miles per hour. So, not jumping off of *that* one, though if there were an atmosphere, almost any jet capable of reaching supersonic speeds would work. And yet, there’d have to be more gravity to have that much atmosphere, so…

23 miles per hour nets you escape velocity from Eros. Still not quite running speed. Visiting Itokawa, about a quarter mile long, finds you with an escape velocity of half of a mile per hour, so even a normal walk ends with you in trouble. Phobos and Deimos are in the same size range as Eros, so you’re looking at about the same speeds there.

So yeah, in theory, on a really small rock, you’re going to get to jump to a very cold and lonely death, unless Tom Corbett happens to be handy.

I’m still thinking computer animations and wild claims at the bar later are a better solution.

Some of those craters really stand out.

I wonder what else we’ll find here?

I mean in my layman’s opinion if the “chunk of rock” is VERY small (small as Vesta or even smaller, see below paragraph) there might not be enough gravitational force/pull to pull you back (to the “point of your last impulse”) and return you to the ground.

For example, what about if the asteroid in question would be only 10 miles in diameter (or only 1 mile etc.), would this rule that you’ve described still apply?

But then again, I still have so much to learn…

And grooooves!!!

@3. Steve Says: “Can I get a Holy Haleakala brothers and sisters?”

You can indeed : HOLY HALEAKALA!

Guess we can now say that Vesta is no longer virgin territory as a new Dawn in our understanding of the brightest asteroid arrives!

The three craters that appear to be chained together on the left side of the picture; Are they from simultaneous impacts? If so did they strike at the same time?