I'm dying to know what this 1 km value means! If you consider tossing something up from the surface and waiting 2760 s until it comes back down, it should only travel around 100 m upwards based on the expected gravity. Let's assume that we have constant gravitational acceleration to get some ballpark numbers:
distance = (acceleration)(time^2)
It takes an equal amount of time to go up as it does to come back down, ie. t = 1380 s, and surface gravity should be about 0.0001 m/s^2 (where g = mu/(r^2), mu = 667 m^3/s^2, r = 2.5 km), so that all leads to a vertical distance of about 190 m. In fact, gravity changes significantly over that distance, and when you include varying gravity you end up with a distance of about 52 m (since lower gravity means the bounce lasts longer for the same distance -- or a shorter distance over the same time).
If we consider the intermediary collision at 2760 s to be inconsequential regarding the bounce height, then you swap out t = 1380 s for t = 3343 s (half the duration of the first bounce), resulting in a distance of 1100 m based on constant gravity or 314 m based on actual varying gravity.
These numbers are only rough calculations, but it takes large changes in the assumed values to get to a height of 1 km (eg. you'd need 3x the expected gravity). I wonder if the 1 km altitude is referenced to some ellipsoid or datum radius, or if it has to do with the surface relief.
Regardless, if ESA isn't doing their math right then we wouldn't be talking about this, so there must be something I'm missing!