I actually thought the answer might be never, but a quick back of the envelope calculation suggests you can do this by dropping a ~1kg bowling ball from a height of 10-11m. (Above the surface of the earth ofc)
This is an extremely rough calculation, I’m basically just looking at how big a bunch of numbers are and pushing all that through some approximate formulae. I could easily be off by a few orders of magnitude and frankly I didn’t take care to check I was even doing any of it correctly.
10-11m seems wrong, and it probably is. But that’s still 1,000,000,000,000,000,000,000,000 times further than the earth moves in this situation. Which hey, fun What If style fact for you: that’s about the same ratio of 1kg to the mass of the Earth at ~1024kg.
That makes perfect sense because the approximations I made are linear in mass, so the distance ratio should be given by the mass ratio.
Possibly?
A bowling ball is more dense than a feather (I assume) and that’s probably going to matter more than just the size. Things get messy when you start considering the actual mass distributions, and honestly the easiest way to do any calculations like that is to just break each object up into tiny point like masses that are all rigidly connected, and then calculate all the forces between all of those points on a computer.
I full expect it just won’t matter as much as the difference in masses.