It would, similar to how the mass of each object does have an effect, even if negligible. But the question is if the radius of the bowling ball vs feather has a greater effect than the mass of the bowling ball vs the feather.
You can adjust the value r in the universal gravitational equation by the radius of the bowling ball and compare the extremes (both plus and minus the radius) and the middle point to see the tidal effects.
If the feather starts at the middle height of the bowling ball, the tidal effects would help the bowling ball. If it starts at the lowest point of the bowling ball, the tidal effects would hinder the bowling ball.
But the magnitude of that effect depends on the distance from the center of the other mass.
I think the main thing would be the ratio of the small mass vs big mass compared to the ratio of the small radius vs the big radius.
Though, thinking of it more, since the bowling ball is a sphere (ignoring finger holes), the greater pull on the close side would be balanced by the lesser pull on the far side (assuming the difference between those two forces isn’t greater than the force holding the ball together), so now I think it doesn’t matter (up to that structural force and with the assumption that the finger holes aren’t significant).
If they are falling into a small black hole, then it does become relevant because the bowling ball will get stringified more than the feather once the forces are extreme enough to break the structural bonds, but the math gets too complicated to wrap my mind around right now. If I had to guess, the bowling ball would start crossing the event horizon first, but the feather would finish crossing it first. And an outside observer would see even more stretched out images of both of them for a while after that, which would make actually measuring the sequence of events impossible.
And who knows what happens inside, maybe each would become a galaxy in a nested universe.
Sasha@lemmy.blahaj.zone 1 month ago
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 massed.
BB84@mander.xyz 1 month ago
For the bowling ball, Newton’s shell theorem applies, right?
Sasha@lemmy.blahaj.zone 1 month ago
Yeah it would fair point, I’ll be honest I haven’t touched Newtonian gravity in a long time now so is forgotten that was a thing.
There’s a similar phenomenon in general relativity, but it doesn’t apply when you’ve got multiple sources because it’s non-linear.
BB84@mander.xyz 1 month ago
So if I have a spherically symmetric object in GR I can write the Schwarzschild metric that does not depend on the radial mass distribution. But once I add a second spherically symmetric object, the metric now depends on the mass distribution of both objects?
Your point about linearity is that if GR was linear, I could’ve instead add two Schwarzschild metrics together to get a new metric that depends only on each object’s total mass?
But even in a situation with one source, does the shell theorem work in GR? Say I put a infinitely light spherical shell around a black hole. Would it follow the same geodesic as a point particle?