QUOTE (Art Martin @ Oct 8 2014, 06:45 PM)
Could someone give us an idea just how big across these boulders are and, given the gravity that's been calculated, what would they weigh?
A qualitative model calculation:
Take a boulder the shape of a cube with an edge length a little less than 14 pixels. Assume a density of 0.4 kg / liter.
The volume would be (20 m)³ = 8000 m³. The mass would be 8000 m³ x 400 kg / m³ = 3.2e6 kg ( = 3200 metric tons).
With a mass of 1e13 kg for the comet, the surface gravity at 2 km distance of the barycenter (neglecting rotation) is GM/r² = (6.672e-11 Nm²/kg² x 1e13 kg) / (2000 m)² = 1.668e-4 N / kg, hence 3.2e6 kg x 1.668e-4 N / kg = 533.76 N. That's about the weight of 54.4 kg on Earth.
The actual weight of a boulder depends on its size, shape, density, distance from the barycenter, distance from the rotation axis, details of the field of gravity of the comet. But the numbers should provide an idea of the order of magnitude.
Edit: Here a scale (right of the center):
Click to view attachment