How Does Rosetta Maintain Its Position Relative to the Comet?

[docroc]

News stories make it sound like Rosetta is orbiting the comet. But presumably the comet's gravity is negligible, which means that orbiting it would require continuous acceleration (and therefore continuous use of energy) in order for Rosetta's motion to conform to a circle/ellipse, rather than simply moving in a straight line as it would naturally do without any input of energy.

Is Rosetta in fact simply moving alongside the comet in a straight line? Or am I missing something?

(Afterthought 5 mins later) Is Rosetta matching the comet's orbit around the sun?

 

[Mr.CROWLER]

I'm not sure if you heard yet, but it already landed.

 

[marcus]

News stories make it sound like Rosetta is orbiting the comet. But presumably the comet's gravity is negligible, which means that orbiting it would require continuous acceleration (and therefore continuous use of energy) in order for Rosetta's motion to conform to a circle/ellipse, rather than simply moving in a straight line as it would naturally do without any input of energy.

Is Rosetta in fact simply moving alongside the comet in a straight line? Or am I missing something?

(Afterthought 5 mins later) Is Rosetta matching the comet's orbit around the sun?

Rosetta is in orbit around the comet. It has changed its orbit from time to time by firing thrusters. But a lot of the time it just loops around and around.

I estimate orbit speeds to be on the order of 10 cm per second, but when closer in can be. say 20 cm per second, depending on the orbit radius, which the controllers have varied in order to study the surface and release the lander and communicate with the lander etc etc.

Just as a sample calculation how far can you travel in one Earth day (86,400 seconds) at 20 cm/s?
17 kilometers
So mightn't Rosetta sometimes complete an orbit in 2 Earth days?.

Maybe now she has retired to a farther out orbit because the work of the moment has been done, the lander is inactive so doesn't have to be communicated with, and so on. I don't know anything about the current status.

 

[lpetrich]

The comet's gravity is low enough so that Rosetta can easily travel in an unbound path without needing much delta-V.

A surface satellite on the comet would travel at something like 0.7 m/s, though the comet's highly irregular shape makes it difficult for there to be such a satellite.

 

[docroc]

I'm not sure if you heard yet, but it already landed.

The probe landed, but Rosetta (the mother ship) is up there watching.

 

[docroc]

The comet's gravity is low enough so that Rosetta can easily travel in an unbound path without needing much delta-V.

A surface satellite on the comet would travel at something like 0.7 m/s, though the comet's highly irregular shape makes it difficult for there to be such a satellite.

I'm still confused - maybe my question wasn't clear about what I don't get.

With a satellite orbiting earth, I understand that the Earth's gravitation keeps the satellite in orbit rather than flying off in a tangent to the orbit (?) and this requires a velocity of around 25k miles/hr for the satellite. So, with almost no gravitational field how does the comet do this for Rosetta? Is it just a question of matching the velocity of Rosetta so that it is appropriate to balance the low gravitation of the comet, ie is there always a velocity that can be used to "match" to any sized object around which you want to orbit?

 

[Nugatory]

Is it just a question of matching the velocity of Rosetta so that it is appropriate to balance the low gravitation of the comet, ie is there always a velocity that can be used to "match" to any sized object around which you want to orbit?

Yes. The radius of a circular orbit, the orbital speed, and the gravitational force are related by $F=mv^2/r$ so no matter how small $F$ is, there is always some $v$ that will allow an orbit at any given radius. Elliptical orbits are a bit more complicated, but it's the same general idea.

 

[Doug Huffman]

Yes. 67P masses 10^13 kg with gravity about 10^-3 m s^-2 or 10^-4 g Earth's gravity.

 

[docroc]

Yes. The radius of a circular orbit, the orbital speed, and the gravitational force are related by $F=mv^2/r$ so no matter how small $F$ is, there is always some $v$ that will allow an orbit at any given radius. Elliptical orbits are a bit more complicated, but it's the same general idea.

Thanks for your clear answer,
Dave