this post was submitted on 03 Nov 2024

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267

ur dada so buff he falls significantly faster than g (mander.xyz)

submitted 3 days ago* (last edited 3 days ago) by BB84@mander.xyz to c/science_memes@mander.xyz

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The bowling ball isn’t falling to the earth faster. The higher perceived acceleration is due to the earth falling toward the bowling ball.

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[–] Sasha@lemmy.blahaj.zone 16 points 2 days ago* (last edited 2 days ago) (2 children)

If anyone's wondering, I used to be a physicist and gravity was essentially my area of study, OP is right assuming an ideal system, and some of the counter arguments I've seen here are bizarre.

If this wasn't true, then gravity would be a constant acceleration all the time and everything would take the same amount of time to fall towards everything else (assuming constant starting distance).

You can introduce all the technicalities you want about how negligible the difference is between a bowling ball and a feather, and while you'd be right (well actually still wrong, this is an idealised case after all, you can still do the calculation and prove it to be true) you'd be missing the more interesting fact that OP has decided to share with you.

If you do the maths correctly, you should get a=G(m+M)/r^2 for the acceleration between the two, if m is the mass of the bowling ball or feather, you can see why increasing it would result in a larger acceleration. From there it's just a little integration to get the flight time. For the argument where the effect of the bowling ball/feather is negligible, that's apparent by making the approximation m+M≈M, but it is an approximation.

I could probably go ahead and work out what the corrections are under GR but I don't want to and they'd be pretty damn tiny.

[–] Simulation6@sopuli.xyz 8 points 2 days ago (2 children)

Physics books always say to assume the objects are points in doing calculations. Does the fact that the ball is thicker then the feather make a difference?

[–] Sasha@lemmy.blahaj.zone 4 points 2 days ago* (last edited 2 days ago) (1 children)

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.

[–] BB84@mander.xyz 2 points 1 day ago (1 children)

For the bowling ball, Newton’s shell theorem applies, right?

[–] Sasha@lemmy.blahaj.zone 2 points 1 day ago* (last edited 1 day ago) (1 children)

Yeah it would fair point, I'll be honest I haven't touched Newtonian gravity in a long time now so I'd forgotten that was a thing. You'd still need to do a finite element calculation for the feather though.

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 2 points 14 hours ago* (last edited 13 hours ago) (1 children)

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 position and total mass?

Anyway, assuming we are in a situation with only one source, will the shell theorem still work in GR? Say I put a infinitely light spherical shell close to a black hole. Would it follow the same trajectory as a point particle?

[–] Sasha@lemmy.blahaj.zone 1 points 12 hours ago* (last edited 10 hours ago) (1 children)

Yeah, once you add in a second mass to a Schwarzschild spacetime you'll have a new spacetime that can't be written as a "sum" of two Schwarzschild spacetimes, depending on the specifics there could be ways to simplify it but I doubt by much.

If GR was linear, then yeah the sum of two solutions would be another solution just like it is in electromagnetism.

I'm actually not 100% certain how you'd treat a shell, but I don't think it'll necessarily follow the same geodesic as a point like test particle. You'll have tidal forces to deal with and my intuition tells me that will give a different result, though it could be a negligible difference depending on the scenario.

Most of my work in just GR was looking at null geodesics so I don't really have the experience to answer that question conclusively. All that said, from what I recall it's at least a fair approximation when the gravitational field is approximately uniform, like at some large distance from a star. The corrections to the precession of Mercury's orbit were calculated with Mercury treated as a point like particle iirc.

Close to a black hole, almost definitely not. That's a very curved spacetime and things are going to get difficult, even light can stop following null geodesics because the curvature can be too big compared to the wavelength.

Edit: One small point, the Schwarzschild solution only applies on the exterior of the spherical mass, internally it's going to be given by the interior Schwarzschild metric.

[–] Sasha@lemmy.blahaj.zone 1 points 12 hours ago* (last edited 12 hours ago)

On that first point, calculating spacetime metrics is such a horrible task most of the time that I avoided it at all costs. When I was working with novel spacetimes I was literally just writing down metrics and calculating certain features of the mass distribution from that.

For example I wrote down this way to have a solid disk of rotating spacetime by modifying the Alcubierre warp drive metric, and you can then calculate the radial mass distribution. I did that calculation to show that such a spacetime requires negative mass to exist.

[–] Buddahriffic@lemmy.world 2 points 1 day ago

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.

[–] barsoap@lemm.ee 2 points 1 day ago* (last edited 1 day ago)

Quick intuition boost for the non-believers: What do things look like if you're standing on the surface of the bowling ball? Are feather and earth falling towards you at the same speed, or is there a difference?

[–] jerkface@lemmy.ca 10 points 2 days ago (4 children)

Obviously the bowling ball because it's more MASSIVE.

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[–] superkret@feddit.org 60 points 3 days ago (1 children)

Stupid question, bowling balls don't fit through the vacuum's hose.

[–] Nomecks@lemmy.ca 23 points 2 days ago

Ur mom could suck it through

[–] roscoe@lemmy.dbzer0.com 22 points 2 days ago* (last edited 2 days ago) (1 children)

This would make a good "What if?" for XKCD. In a frictionless vacuum with two spheres the mass of the earth and a bowling ball how far away do they need to start before the force acting on the earth sized mass contributes 1 Planck length to their closure before they come together? And the same question for a sphere with the mass of a feather.

[–] Sasha@lemmy.blahaj.zone 3 points 2 days ago* (last edited 2 days ago)

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^-11^m. (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^-11^m 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 ~10^24^kg.

That makes perfect sense because the approximations I made are linear in mass, so the distance ratio should be given by the mass ratio.

[–] Shard@lemmy.world 48 points 3 days ago (1 children)

So will the bowling ball gravitationally attract the earth to itself there by reach the earth an infinitesimally small amount?

[–] BB84@mander.xyz 44 points 3 days ago (1 children)

Yes, the earth accelerates toward the ball faster than it does toward the feather.

[–] SzethFriendOfNimi@lemmy.world 16 points 3 days ago (1 children)

Wouldn’t this be equally offset by the increase in inertia from their masses?

[–] BB84@mander.xyz 33 points 3 days ago* (last edited 3 days ago) (2 children)

If your bowling ball is twice as massive, the force between it and earth will be twice as strong. But the ball’s mass will also be twice as large, so the ball’s acceleration will remain the same. This is why g=9.81m/s^2 is the same for every object on earth.

But the earth’s acceleration would not remain the same. The force doubles, but the mass of earth remains constant, so the acceleration of earth doubles.

[–] Venator@lemmy.nz 17 points 3 days ago* (last edited 3 days ago) (1 children)

I wonder how many frames per... picosecond you'd need to capture that on camera... And what zoom level you'd need to see it.

I think the roughness of the surface of the bowling ball would have a bigger impact on the time, in that the surface might be closer at some points if it were to rotate while falling.

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[–] MrsDoyle@sh.itjust.works 30 points 3 days ago (1 children)

Brian Cox shows ball and feathers falling together in vacuum: https://youtu.be/E43-CfukEgs

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[–] CatZoomies@lemmy.world 26 points 3 days ago (1 children)

There’s too many words in this meme that’s making me dizzy from all your fancy science leechcraft, wizard.

I reject your reality and substitute my own: the feather falls faster. It’s more streamlined than the bowling ball, and thus it slips through the vacuum much faster and does hit the ground and stay on the ground, I think. The ball will bounce at least once, maybe even three times. On each bounce, parts of it probably break off, which change the weight. Thankfully those broken pieces won’t hurt anyone because they’re sucked up by the vacuum. Thus, rendering your dungeon wizard spells ineffective against me.

[–] Masta_Chief@lemmy.world 9 points 3 days ago

This person sciences good

[–] TriflingToad@sh.itjust.works 14 points 2 days ago (1 children)

Reading that spoiler, I hate scientists sometimes.

[–] chatokun@lemmy.dbzer0.com 14 points 2 days ago (1 children)

For some reason on my client, it can't remove the spoiler (gives a network error). I'm assuming it says that since the ball has more mass, it has a higher attraction rate of its own gravity to Earth's, so does fall faster in a vacuum but so miniscule it would be hard to measure?

[–] TriflingToad@sh.itjust.works 12 points 2 days ago

"The bowling ball isn't falling to the earth faster. The higher perceived acceleration is due to the earth falling toward the bowling ball." is what the spoiler says

[–] fubarx@lemmy.ml 20 points 3 days ago (1 children)

Depends on the color of the feather and the ball.

There's a simple explanation.

[–] tetris11@lemmy.ml 8 points 2 days ago (1 children)

Exactly, red has way more up-quarks than blue

[–] fubarx@lemmy.ml 7 points 2 days ago (2 children)

Because light-blue weighs less than blue.

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[–] reliv3@lemmy.world 34 points 3 days ago* (last edited 3 days ago) (1 children)

This argument is deeply flawed when applying classical Newtonian physics. You have two issues:

Acceleration of a system is caused by a sum of forces or a net force, not individual forces. To claim that the Earth accelerates differently due to two different forces is an incorrect application of Newton's second law. If you drop a bowling and feather in a vacuum, then both the feather and the bowling ball will be pulling on the Earth simultaneously. The Earth's acceleration would be the same towards both the bowling ball and the feather, because we would consider both the force of the feather on the Earth and the force of the bowling ball on the Earth when calculating the acceleration of the Earth.
You present this notion that two different systems can accelerate at 9.81 m/s/s towards Earth according to an observer standing on the surface of Earth; but when you place an observer on either surface of the two systems, Earth is accelerating at a different rate. This is classically impossible. If two systems are accelerating at 9.81 m/s/s towards Earth, then Earth must be accelerating 9.81 m/s/s towards both systems too.

[–] BB84@mander.xyz 22 points 3 days ago* (last edited 3 days ago) (3 children)

Re your first point: I was imagining doing the two experiments separately. But even if you do them at the same time, as long as you don’t put the two objects right on top of each other, the earth’s acceleration would still be slanted toward the ball, making the ball hit the ground very very slightly sooner.

Re your second point: The object would be accelerating in the direction of earth. The 9.81m/s/s is with respect to an inertial reference frame (say the center of mass frame). The earth is also accelerating in the direction of the object at some acceleration with respect to the inertial reference frame.

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[–] BmeBenji@lemm.ee 12 points 2 days ago (1 children)

“In our limited language that tries to describe reality and does so very poorly, how would you describe this situation that would literally never happen?”

[–] Fleur_@lemm.ee 5 points 2 days ago (1 children)

I'm pretty sure bowling balls and feathers fall all the time

[–] zqps@sh.itjust.works 6 points 2 days ago (2 children)

I think they mean the vacuum part.

To which I'd add that we had astronauts perform this experimentally on the surface of the moon.

[–] Fleur_@lemm.ee 2 points 2 days ago

True fair enough, but since I'm here, being an internet clown, I might as well double down...

Obviously heavy and light objects never experience gravitational attraction in a vacuum throughout the vastness of the universe. Clearly F = G(m1m2)/R^2 only applies to objects in earths atmosphere.

[–] blind3rdeye@lemm.ee 2 points 2 days ago

Also, I've seen a video of an experiment done in a vacuum chamber. (Although they kind of botched the point of the video by showing lots of slow-mo and junk like that.)

[–] Kolanaki@yiffit.net 26 points 3 days ago (11 children)

But what weighs more:

A ton of bowling balls or a ton of feathers? 🤔

[–] KoboldCoterie@pawb.social 64 points 3 days ago (1 children)

When you carry a ton of feathers, you also have to carry the weight of what you did to those poor birds...

[–] OriginalUsername7@lemmy.world 3 points 2 days ago

What about all the bowling cattle you had to castrate for those balls?

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[–] pumpkinseedoil@mander.xyz 25 points 3 days ago (21 children)

Why your spoiler is wrong:

The gravitational force between two objects is G(m1 m2)/r²

G = ~6.67 • 10^-11 Nm²/kg²

m1 = Mass of the earth = ~5.972 • 10^24 kg

m2 = Mass of the second object, I'll use M to refer to this from now on

r = ~6378 • 10^3 m

Fg = 6.67 • 10^-11^ Nm²/kg² • 5.972 • 10^24^ kg • M / (6378 • 10^3 m)² = ~9.81 • M N/kg = 9.81 • M m kg / s² / kg = 9.81 • M m/s² = g • M

Since this is the acceleration that works between both masses, it already includes the mass of an iron ball having a stronger gravitational field than that of a feather.

So yes, they are, in fact, taking the same time to fall.

[–] NoneOfUrBusiness@fedia.io 13 points 3 days ago (1 children)

Uh... That's not how that works. The distance between two objects changes with acceleration a1-a2 where object 1 moves with acceleration a1 and object 2 a2 (numbers interchangeable). In the bowling ball's case a2 is the same but a1 is bigger in the negative direction so the result is that the bowling ball falls faster.

[–] pumpkinseedoil@mander.xyz 10 points 3 days ago (4 children)

Calculate the force between the earth and the bowling ball. It'll be G • (m(earth) • m(bowling ball)) / (r = distance between both mass centers)²

Simplify. You're getting g • m(bowling ball).

Now do the same for the feather. Again, the result is g • m(feather).

Both times you end up with an acceleration of g. If you want to put it that way: The force between the earth and the bowling ball is m(bowling ball)/m(feather) times as high as the force between the earth and the feather, but the second mass also is m(bowling ball)/m(feather) times as high, resulting in the same acceleration g.

Higher force on same mass results in stronger acceleration. Same force on higher mass results in lower acceleration. Higher force on equally higher mass results on equally high acceleration.

I just asked my professor this exact thing (if the ball would get to the earth sooner because it accelerates the earth towards it) like two weeks ago and my previous message + this message was his explanation.

PS: If you're looking at this from outside, the ball travels less distance before touching the ground (since the ground is slightly nearer due to pulling the earth more towards it), but also accelerates slower while accelerating the earth faster towards it. The feather gets accelerated faster towards the earth and travels a longer distance before touching the ground but doesn't accelerate the earth as fast towards it.

But because we're not outside, we only care about the total acceleration (of the earth towards the object and the object towards the earth), and that's g. We don't notice if (fictional numbers) the earth travels 1m and the object travels 1m or if the earth stays in place and the object travels 2m, what matters for us is how long it takes an object 2m away from the earth to be 0m away from the earth.

[–] NoneOfUrBusiness@fedia.io 11 points 3 days ago (2 children)

So let's just look at that again. The bowling ball's (mass m1) acceleration is GM/R². The feather's is also GM/R². They have the exact same acceleration, which is g. I'm not sure where you're getting that the bowling bowl accelerates slower. Meanwhile in the bowling ball's case the Earth's acceleration is higher, as you already said. This results in less free fall time overall.

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[–] Sasha@lemmy.blahaj.zone 2 points 2 days ago

This is not correct, the force on the objects is the same sure, but the accelerations aren't so you can't calculate them both in one go like this.

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[–] Slovene@feddit.nl 13 points 3 days ago* (last edited 3 days ago)

But ... Steel is heavier than feathers ...

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