The comet 67P has a
mass of 3.14E12 kg
Today the comet is 186,444,271 km from the Sun
Where is Rosetta?
Using
F=GMm/R^2, the Sun's gravity on Rosetta is equal to 67P's gravity on Rosetta at 700m from the
center of Rosetta on 6 August 2014, which means that Rosetta will never really be completely within 67P's field. (At Perihelion on 13 Aug 2015, 67P's gravity field will be as strong as the Sun's only 250m from the centre)
However, now that Rosetta is in the same orbit as 67P we can mostly disregard the Sun's gravity and the elliptical path that Rosetta and 67P now share as of today. (Earth's pull on Rosetta is at least a million times weaker than the Sun's pull - so forget any influence from the Earth's mass.)
The "orbits" at 100km are called
hyperbolic because Rosetta is not trapped in 67P's gravity well since the gravity is so weak and because Rosetta is still moving FAST at 1 m/s. But this hyperbola is so weak it is effectively a straight line.
Rosetta will turn 60 degrees after every 100 km of a hyperbolic path to make a triangular "orbit". This triangular path cannot be called an orbit because it is not a
conic section, nor is the comet at a focal point of the conic section
Kepler's First Law.
These "straight"/"hyperbolic" paths of 100km and 50km are deliberately done for two reasons:
-to calculate exactly the gravity field of the comet, because it is clearly not a uniform sphere. They will likely use radar&cameras to continuously measure the precise distance to the comet
-to keep in front of the comet to avoid its coma and tail.
After these maneuvers, Rosetta will go into a 30 km "orbit", so that the task of mapping 80% of the surface all happens from the same distance. This orbit is not natural and will be powered because a natural 30km orbit of 67P takes 26 days.
Here's how to calculate the natural circular orbits for 67P (it won't be circular, because of the crazy shape, but close enough).
Kepler's 3 Law gives us
T^2=4pi^2/GM*r^3. 4pi^2/GM=0.19 for this comet. G=
6.67×1011 N(m/kg)2
if r=30km=3e4m, the natural orbit would have a period of T=2.3e6 seconds=26.11 days
If r=2.5km, the natural orbit would have a period of T=15 hours
If r= 5km, the natural orbit would have a period of T=1.77 days
If r= 100km, the natural orbit would have a period of 159 days
So I could imagine that when Rosetta gets within 5km it is mostly using the natural orbit and hence saving fuel.