| Ok, here's an extremely rough back-of-the-envelope calculation. As you'll see, these numbers can be out by orders of magnitude, and it doesn't greatly change the conclusion. Oumuamua interstellar asteroid.
230x35x35m, ~= 280000 m^3 Density assumption: 2 x water. => mass is ~500,000 metric tonnes. Spotted only after passing the Sun.
Assume we'd spot such objects only if they came within the orbit of
mercury so are well illuminated.
Assume one such object every 10 years (we've not been searching very
long with automated telescopes), and we spot all of them. Mean mercury orbit radius ~ 60,000,000 km Area of mercury's orbit: 1.1 x 10^16 km^2 Mercury's orbital area x path length in 10 years = volume swept by one visible object
in 10 years. Asteroid velocity ~100,000 km/h Path length in 10 years = 100,000 x 10 x 24x365.
Swept volume ~ 10 x 10^25 km^3 Distance to Alpha Centauri: 4.37 light years = 4.37 x 9.5 x 10^12 km = 4.15 x 10^13km Sol's "cube of influence" ~= 7 x 10^40 km^3 Cube of influence / swept volume = rough estimate of number of asteroids in cube of influence.
Number of asteroids: 7 x 10^14 Mass of asteroids: 3.5 x 10^20 tonnes.
Mass of sun: 2 x 10^27 tonnes. Conclusion: dark interstellar asteroids like Oumuamua are a tiny fraction of the visible mass of the galaxy. |