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Rosetta, the Comet Hunter
Roland Piquepaille writes: "After being delayed for about a year because of a failure of the Ariane-5 rocket, the Rosetta spacecraft is scheduled to be launched on February 26. Rosetta is a special spacecraft, including an orbiter and a lander. And it will take up to 2014 before landing on Comet 67P/Churyumov-Gerasimenko -- with the help of a harpoon. Then, as says the European Space Agency (ESA), Rosetta will help to solve planetary mysteries. This news release looks at the goals of Rosetta's mission and explains why it will take more than ten years to reach the comet. But here the 'funny' part of the story: the landing. 'In November 2014, the lander will be ejected from the spacecraft from a height which could be as low as one kilometre. Touchdown will be at walking speed, about one metre per second. Immediately after touchdown, the lander will fire a harpoon into the ground to avoid bouncing off the surface back into space, since the comet's extremely weak gravity alone would not hold onto the lander.' This overview contains more details and includes illustrations of the Rosetta's spacecraft and its landing on the comet."
From Rosetta's webpage: The relative speeds of the spacecraft and comet will gradually be reduced, slowing to 2 metres per second after about 90 days. If it moves slowly enough, the comet's weak gravity can hold it in.
It's all about escape velocity. The mass needed to keep a person on an object or in an orbit comes down to the speed the person can obtain by its own force. (Jumping or pushing or something.) Since an object like this is evacuated there is little to slow things down so should the get a little push in a direction, it will have a relatively large impact.
(And no, I don't care to do the math.)
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Fg = G*m1*m2/d^2
with m1 your mass, m2 the rock's mass, G being 6.67e-11 for our universe and d being the distance between you and the rock.
So there is ALWAYS gravity, but when you hit an asteroid at 1m/s, your momentums (m1*v1) and the asteroid's momentum (m2*v2) adjust, and propel you and the asteroid in opposite directions because momentum, like energy and forces, is conserved.. and since m2 >> m1, this results in a bouncing off situation (there's a formula for it, but I can't be bothered to break out the notes from first year physics).. The gravitation force between you and the asteroid now has to be enough to counteract this bouncing-off-one-another for you to stay on it.
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As far as we know, every object "pulls" on every other object. However, if you're moving away from an object at a sufficient speed (escape velocity) then its pull will never reverse your motion.
For a small object, escape velocity can be quite small. Take a spherical comet 1 mile in diameter. This is about 1/4000th that of Earth. Suppose it has the same density as our planet (surely an overestimate). Then its gravity would be about 6.4e10 times weaker.
More importantly, as you stand on the surface, your potential energy would be 1.6e7 times smaller than it is on Earth's surface. To achieve escape velocity (at the surface), a spacecraft's kinetic energy must be larger than its potential energy. So escape velocity on the comet is 4e3 times larger than on Earth. Escape velocity on earth is about 11.18 km/s, so on the comet it is only about 2.8 m/s.
So if you try to make a soft landing on the comet, and with your initial bounce you are moving away from the comet at 2.8 m/s (i.e. 6.3 miles per hour), then you will need to make some corrective measure, or else you will just fly away from the comet and never return.
As mentioned, you have to be moving slower than the escape velocity to be in orbit around something. The formula is v = sqrt(2GM/r). G is 6.67x10^-11 m^3/s^2kg everywhere.
For Earth, M is 6x10^24 kg, and the highest relevent velocity as at the surface, so r = 6x10^6 m. That's 11.2 km/s. Very fast. Which is why it's hard just to get into orbit.
Now for the comet. If it's 4 km across, r = 2000 m. I can't find a value for the mass, but based on the common description of comets as dirty snowballs let's guess the density is about that of water, or 1000 kg/m^3. The volume of a sphere is 4/3 r^3 so our guess for M is 3.35x10^13 kg.
That makes the escape velocity for 67P/Churyumov-Gerasimenko at 1.5 m/s which pretty much the same brisk walking-speed which which the lander is expected to hit the comet, especially if our guess at the density is high. Thus, the lander could easily bounce off, and a person could with some effort jump off, fast enough that the comet's gravity wouldn't bring them back. On the other hand, an rocky asteroid (denser) the size of Manhattan (bigger) would probably be hard to get away from under your own power. This comet is right on the edge.