this post was submitted on 03 Nov 2024
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Tap for spoilerThe 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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[–] superkret@feddit.org 64 points 3 weeks ago (1 children)

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

[–] Nomecks@lemmy.ca 23 points 3 weeks ago

Ur mom could suck it through

[–] Shard@lemmy.world 50 points 3 weeks 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 46 points 3 weeks ago (1 children)

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

[–] SzethFriendOfNimi@lemmy.world 17 points 3 weeks ago (1 children)

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

[–] BB84@mander.xyz 35 points 3 weeks ago* (last edited 3 weeks 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 19 points 3 weeks ago* (last edited 3 weeks 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.

[–] WhatAmLemmy@lemmy.world 8 points 3 weeks ago* (last edited 3 weeks ago)

Considering the mass of the ~~earth~~ (?) moon, I wouldn't be surprised if it'd be nearly impossible to capture a difference between a feather or bowling ball. You might have to release them at 100m or 1000m above the surface, but then maybe the moons miniscule atmosphere or density variances will have more of an effect.

[–] Robust_Mirror@aussie.zone 7 points 3 weeks ago

But if you're dropping them at the same time right next to each other, the earth is so large they would functionally be one object and pull the earth at the same combined acceleration.

[–] reliv3@lemmy.world 35 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

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

  1. 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.
  2. 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 24 points 3 weeks ago* (last edited 3 weeks 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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[–] MrsDoyle@sh.itjust.works 31 points 3 weeks ago (2 children)

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

[–] Xavienth@lemmygrad.ml 15 points 3 weeks ago* (last edited 3 weeks ago)

The difference in relative acceleration implied by the meme is on the order of tens of yoctometres (10⁻²³ m) per second per second.

It's a difference so small that it would be overshadowed by the fact that you're holding one object femtometres (10⁻¹⁵ m) higher or lower than the other in the gravitational field.

Additional sources of error to consider at this scale might be the heat radiation from the surroundings providing radiation pressure on the object, the sloshing of Earth's core causing time-dependent variations in the gravitational field, the location-dependent variations in the Earth's gravitational field, and the difference in centrifugal (yes, centrifugal in this reference frame) force due to latitude differences of one micrometre, and also due to natural variations in the rate of Earth's rotation over time.

I love it when scientists who know something to be true in theory get to see practical experiments like this. The jubilation on thier faces.

[–] Kolanaki@yiffit.net 26 points 3 weeks ago (13 children)

But what weighs more:

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

[–] KoboldCoterie@pawb.social 63 points 3 weeks 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...

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[–] CatZoomies@lemmy.world 26 points 3 weeks 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 weeks ago

This person sciences good

[–] pumpkinseedoil@mander.xyz 24 points 3 weeks ago (3 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 weeks 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 weeks 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 weeks 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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[–] red@lemmy.zip 10 points 3 weeks ago (18 children)

the fact that you got upvoted, you clearly said force on both objects is gM and the feather or ball will move with g BUT earth will move with gM/m1 which is more in case of ball, and no its not acceleration between mases, its the force experiencec by both mases so, fg=m1.a

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[–] roscoe@lemmy.dbzer0.com 23 points 3 weeks ago* (last edited 3 weeks 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.

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[–] fubarx@lemmy.ml 21 points 3 weeks ago (1 children)

Depends on the color of the feather and the ball.

There's a simple explanation.

[–] tetris11@lemmy.ml 8 points 3 weeks ago (1 children)

Exactly, red has way more up-quarks than blue

[–] fubarx@lemmy.ml 7 points 3 weeks ago (2 children)

Because light-blue weighs less than blue.

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[–] Sasha@lemmy.blahaj.zone 18 points 3 weeks ago* (last edited 3 weeks 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 9 points 3 weeks ago (9 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?

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[–] Slovene@feddit.nl 14 points 3 weeks ago* (last edited 3 weeks ago)

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

[–] TriflingToad@sh.itjust.works 14 points 3 weeks ago (1 children)

Reading that spoiler, I hate scientists sometimes.

[–] chatokun@lemmy.dbzer0.com 14 points 3 weeks 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 3 weeks 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

[–] BmeBenji@lemm.ee 12 points 3 weeks ago (4 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?”

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[–] jerkface@lemmy.ca 11 points 3 weeks ago (4 children)

Obviously the bowling ball because it's more MASSIVE.

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[–] BB84@mander.xyz 7 points 3 weeks ago* (last edited 3 weeks ago) (3 children)

Here’s a problem for y’all: how heavy does an object have to be to fall 10% faster than g? Just give an approximate answer.

[–] WolfLink@sh.itjust.works 8 points 3 weeks ago* (last edited 3 weeks ago)

10% of the earths mass

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[–] RumorsOfLove@lemmy.dbzer0.com 7 points 3 weeks ago (1 children)

the feather falling toward the earth will also be attracted to the bowling ball (which is on the earth)

doesnt offset, because the feather-ball attraction is not as large as the earth-ball. just wanted to say

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[–] DavidGarcia@feddit.nl 7 points 3 weeks ago

uhmmm ackchshickzually, it's the space-time that's falling

[–] noisefree@lemmy.world 6 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

This may be a stupid question, but: assuming an object (the bowling ball) is created from materials found on Earth and that it remains within the gravity well of Earth from material procurement stage to the point where it is dropped, wouldn't the acceleration of the Earth towards the object be kind of a null considering the whole timeline of events? I mean, I get the distinction of higher mass objects technically causing the Earth to accelerate towards them faster if we're talking a feather vs a bowling ball that both originated somewhere else before encountering Earth's gravity well in a vacuum, it just seems kind of weird to consider Earth's acceleration towards objects that are originating and staying within its gravity well?

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[–] NateNate60@lemmy.world 6 points 3 weeks ago (5 children)

So obviously I ended up in the middle of this bell curve. How would that cause the perception of the ball's acceleration to differ?

[–] BB84@mander.xyz 22 points 3 weeks ago (3 children)

When the earth pulls on an object with some F newtons of force, the object is also pulling on the earth with the same force. It’s just that the earth is so massive that its acceleration F/m will be tiny. Tiny is not zero though, so the earth is still accelerating toward the object. The heavier the object, the faster earth accelerates toward it.

Both the bowling ball and the feather accelerates toward earth at the same g=9.81m/s^2, but the earth accelerates toward the bowling ball faster than it does toward the feather.

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[–] trxxruraxvr@lemmy.world 6 points 3 weeks ago

It won't cause the perception to differ because the difference is so small it's impossible to measure

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