PHOs Approaching Closer Than Geostationary Ring

[Ben the Space Brit]

According to Spaceweather.com, the very small PHO 2017 BH30 will pass closer to Earth than the geostationary ring on 1/30/2017 (UT). I know that the 'big sky, small rock, small spacecraft' rule applies here. However, how reasonably assured can we be that the rock will not collide with any major object in the ring, even decommissioned ones?

[Bynaus]

First of all, it should be tested whether the track of the PHO even passes through the ring of geosynchronous satellites (the geostationary orbit). If it approaches to a closer distance than the geosynchronous orbit, but without also crossing the geostationary orbit (possible in 3D), the chance of a collision is simply zero. The chance of a collision would be largest in the case of the PHO approaching within the ecliptic equatorial plane, where the PHO has two chances of striking a satellite (on the way in to closest approach, and on the way out again). For any other geometry, there is only a single possible collision point. The estimate below is based on that latter scenario. Of course, given that the position of all the satellites is precisely known, in any specific case it can be determined with very high certainty whether a collision will occur or not. The estimate is thus more general in the sense that it gives us the overall probability that such a collision might happen at some point in the future, for all encounters with PHOs which certainly cross the geostationary orbit (in 3D).

Given about 600 geosynchronous satellites, an assumed spherical radius R of 4 m per satellite, the volume of geosynchronous space filled by satellites is 600 * (4/3*Pi*R^3) / (Pi * R^2 * 2 * Pi * R_G), with R_G the radius of the geosynchronous orbit (35 786 km). This gives 1.4 * 10e-5. So this is the chance that a given volume of geosynchronous space (e.g., the point where the orbit of the PHO crosses the geostationary orbit) will be filled with a satellite at a given time (e.g., the moment the PHO crosses the geostationary orbit). If we assume the PHO is point-like, this is the chance we are looking for (odds of 1 in 70 000). In fact, chances of a collision are a bit higher since the size of the PHO is non-zero, but that effect is small. But remember also that this is only the chance for PHOs which definetly cross the actual geostationary orbit, not all of those which approach to a closer distance (which is a much, much larger number).

Edit: replace "ecliptic" by "equatorial"

[Lar]

4m seems low radius to me, wouldn't there be some non negligible gravity tractor effects from a near miss that near? Also some solar panels extend out farther but I guess not in all directions making 4m a good approximation...

but 1:70,000 ? that's scary.

[Bynaus]

4m seems low radius to me, wouldn't there be some non negligible gravity tractor effects from a near miss that near? Also some solar panels extend out farther but I guess not in all directions making 4m a good approximation...

but 1:70,000 ? that's scary.

I think gravity is negligible here. The hill sphere of a small PHO is most likely inside it, so the sun (or the Earth's) gravity dominates in every situation where no collision occurs.

It might sound scary, but remember, it is 1:70 000 for PHOs which cross the ecliptic somewhere within 4 m of the 35 786 km circle, and zero for all other PHOs. I don't know how many PHO satisfying that first condition we have had since we started observations, but it cannot be a very high number (in fact, I would expect that number to be less than 1). We would need an approximation of that number though for a realistic estimate of the collision rate between PHOs and geostationary satellites.

[kevin-rf]

Considering it's estimated size is 8m, the size of a small house, negligible might be overstating it's gravitational effect on anything.

I was going to say it's chances of running a muck in GSO was very low until I looked at it's inclination. At 1.8 deg it's coming in quite flat.

[Bynaus]

But this is 1.8° relative to the ecliptic - the Earth's equatorial plane is tilted by 23° relative to the ecliptic. So unless the encounter happens at one of the two equinoxes, an asteroid having 0° inclination relative to the ecliptic will have a 23° inclination relative to the equatorial plane, where all the GEO sats are. So again, there is - when approximating the path travelled by the PHO close to Earth with a straight line - only one point where the collision might happen.