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."

Re:Gravity?

[C17GMaster]

From Rosetta's webpage [esa.int]: 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.

Re:Anyone good with gravity?

[halftrack]

It's all about escape velocity. [wikipedia.org] 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.)

Re:Anyone good with gravity?

[kryptkpr]

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.

Re:Anyone good with gravity?

[Anonymous Coward]

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.

Ok, here's the math

[rillian]

As mentioned, you have to be moving slower than the escape velocity [wikipedia.org] 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 [seds.org] 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.

Re:Anyone good with gravity?

[Anonymous Coward]

Here's the way you'd go about calculating the orbital velocity. This is a Newtonian (in the very essence of the term) question, and can be solved using only forces. The key relation is that for a particle moving in a circular path, the force required to keep it moving in that path is

F = m v*v / r

Where m is the object's mass, v is its velocity, and r is its radius. Newton's law of gravitation states that the force exerted upon a mass by another mass is

F = G m M / r*r

G being the gravitational constant, m and M the objects respective masses, and r their seperation. We'll assume that M >> m, so it can be regarded as having no acceleration. We can then set these equal, and find

v * v/R = G M / R*R ...
v = SqrRoot(G M / R)

Looking at the article, we see that r = 2 km = 2x10^3 m. We can do a seat of the pants calculation of M by assuming that the comet is solid and made of water (1g/cm^3 = 1000 kg/m^3)

M = density*V = (1000) (4/3 pi r^3)
= 3*10^13 kg

So, finally, using G = 6.67*10^-11 and R=3 km (this is the orbital radius), we have v = 0.3 m/s, or about one mile per hour.

Re:Gravity?

[mindriot]

This is going to be a very difficult mission. I would love to have a job constructing the lander... I am simply amazed by the fact that we're able to hurl a piece of fragile technology at tiny objects in space that are far, far away (yes, considering how big space is, I would call Mars 'small' too) -- and they will actually get there in one piece and work.

I really hope they'll make it with this one. The German Max-Planck Institute for Aeronomy [linmpi.mpg.de] (soon to be called Institute for Solar System Research) is responsible for the lander. My mom works there, so from what she talked about I could tell how complicated the development of such a lander is.

Considering how long one of the computer scientists there has been working on the lander software, and what kinds of stress testing procedures the parts had to go through (some of them were done at Astrium in Munich [astrium-space.com] and my mom had the honor of personally delivering the components...), I have deep respect for the engineers who work on such projects - even more when they actually make it work (Spirit/Opportunity).

Also, I am glad that the Rosetta project got to keep going at all, considering that originally it was supposed to visit Wirtanen [linmpi.mpg.de] (German link, for English see maybe here [esa.int]), a whole different comet. That also means that for the new target, the lander's software and some components even had to be redesigned to suit the new comet's features. So, good luck to Rosetta - hitting this target would be one cool achievement.

Oh, also, this ESA page [esa.int] has some nice information about the mission as well.

Re:Ok, here's the math

[sirsex]

His formula is correct for any object outside the surface of the earth.

The effective pull of gravity decreases as you go below the surface, as the rock above you pulls UP on you. Gravity cancels out ot zero at the center of a spherical mass. We'll leave the diffy-q up to the reader

Re:Dangerous route

[juhaz]

The asteroid belt is really quite sparse, not at all what most people would expect after watching/reading bit too much scifi.

Very unlikely that it will even even get to see any of the rocks if it's not intentionally directed to fly by one of the big ones to get pretty pictures, much less get hit by them.