Just to establish a reference point from the radar images, I made a crude estimate for the moons orbital radius (upper bound) and orbital period. Assuming QE2 is a sphere 2.7km in diameter with a density ≈ 3g/cm3, I get an orbital radius ≈ 22km and a period ≈ 5 days.neufer wrote:neufer wrote:The Earth pulls almost equally on QE2 and its satellite.mjimih wrote:
can someone direct me to an article that explains why the Earth won't try to steal the little satelite away from the asteroid when it flies by?
The L1 Lagrangian point: is about 800 kilometers above the surface of QE2 such that the satellite would have to orbit at least that high above QE2 for the Earth to have a chance to steal it away.Upon reconsideration: the tidal effects by the Earth on the satellite are much less than I originally thought.neufer wrote:geckzilla wrote:
What about just changing its orbit rather than stealing it?
- Maybe 400 km.
This is because tidal effects by the Earth are:Hence, any satellite of an Earth grazing asteroid must be many L1 distances away to be affected.
- 1) on average, perpendicular to the satellite velocity (and, hence, ineffective)
2) of a fleeting duration much shorter than the L1 satellite period (~4.5 years in this case).
The radar data actually describes a projected view the orbital plane wrt Earth, i.e. it is most likely inclined from our perspective. I assumed the orbital plane is not inclined, and the moon's motion was directly toward Earth. Therefore the true orbital velocity can only be larger than what I estimate, which in turn means a smaller radius and shorter period. As I said this is very rough estimate.
As suggested earlier for a moon rotation that's locked to the orbital period (Doppler data), then the orbital period ≈ 2 days. This shorter period is well in line with expectation and my estimate's uncertainty.