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[-] pikmeir@lemmy.world 246 points 5 months ago

For anyone wondering it's because the bowling ball slightly pulls the earth faster toward itself. This amount is too small to possibly measure. But imagine if the bowling ball were the size of another Earth and it's easier to see why it happens.

[-] TheControlled@lemmy.world 63 points 5 months ago

Thanks for the non-jargon version

[-] nova_ad_vitum@lemmy.ca 48 points 5 months ago

This amount is too small to possibly measure

What the fuck did you say to me you little bitch? I'm going to go get $300 million in funding to create a device so complex and so sensitive that a butterfly sneezing 30 miles away will fuck it up and then I'm going to directly measure the the acceleration of the earth as a result of the mass of that bowling ball. You fucked up, kiddo.

  • Average metrologist, probably
[-] Tlaloc_Temporal@lemmy.ca 5 points 5 months ago

The issue isn't so much the sensitivity (although that is a significant issue), it's all the other crap going on. You'll probably be able to filter out the Mains Hum, but every anything moving in the same axis as the test will confount the data.

I'm thinking we might set up the instuments near counterweight energy storage or pumped hydro, and some on the exact opposite side of the planet, and try to measure the movement of the earth that way.

We can already see a change in the length of a day after big earthquakes and dam construction/destruction, but I don't think the acceleration has ever been measured directly.

[-] Tar_alcaran@sh.itjust.works 9 points 5 months ago

When you drop them at the same time, it doesn't matter though.

[-] bort@sopuli.xyz 14 points 5 months ago

because of two bodies can not occupy the same space, the feather and the ball will be in different position when you drop them. And therefor gravitation will pull the earth slightly more toward the ball and slightly less toward the feather.

[-] KazuchijouNo@lemy.lol 5 points 5 months ago

But being more massive means that due to inertia the ball will take just a tiny little wee bit longer to start moving no? So they end up falling at the same time.

Also, are these Newtonian mechanics? How do they compare to relativity at the "bowling ball and feather" scale?

Someone please correct me if I'm wrong. It's been a while since I read anything physics-related.

[-] MajorSauce@sh.itjust.works 17 points 5 months ago

The acceleration from gravity would be the same no matter the object mass (~9.8m/s²).

[-] KazuchijouNo@lemy.lol 3 points 5 months ago

Oh yes! I omitted that part, but what I meant to say is that mass and inertia balance each other, so that in the end the acceleration from gravity ends up the same for any object.

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[-] qjkxbmwvz@startrek.website 6 points 5 months ago

The above is just referring to the fact that the standard "feather vs. bowling ball" question assumes the earth/moon/ground is immovable. In that case, Newton says they fall the same.

The fact that the ground is not immovable is what's being referenced


in this picture, things don't "fall," they are each accelerated towards each other.

[-] brown567@sh.itjust.works 111 points 5 months ago

This is fascinating! Both of them accelerate toward the earth at the same rate, but because of the bowling ball's greater mass, the EARTH accelerates faster toward the bowling ball than it does toward the feather, so it's imperceptibly faster XD

[-] chicken@lemmy.dbzer0.com 26 points 5 months ago

But they are being dropped at the same time for dramatic effect, so the earth will also be accelerating towards the feather at bowling ball speeds because the feather is next to the bowling ball, therefore they still land at the same time.

[-] Dwomen@lemmy.dbzer0.com 32 points 5 months ago

That’s only be true if the feather was in the same position as the ball. Otherwise, the earth is moving ever so slightly more towards the ball.

[-] mexicancartel@lemmy.dbzer0.com 6 points 5 months ago

You are forgetting the sun. The earth turns in the direction of sun ever so slightly so if you align the feather next to the ball in the side of the sun, then probably the feather falls faster

[-] dumbass@leminal.space 5 points 5 months ago

So, what you're saying is, to do this experiment correctly, we have to stop the earth from moving. . I'm keen.

[-] clickyello@lemmy.world 4 points 5 months ago

🎵 I'll stop the world and melt with you 🎵

[-] lauha@lemmy.one 10 points 5 months ago

No, because the earth is accelerating towards the bowling ball and the feather is next to the bowling ball, the force vector is (ever so slightly) greater towards the bowling ball than the feather, thus the bowling ball drops faster

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[-] iz_ok@lemmynsfw.com 10 points 5 months ago

Can someone explain how the Earth accelerates towards an object? Is this just because objects with mass attract things?

[-] candybrie@lemmy.world 15 points 5 months ago

You got it.

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[-] Tudsamfa@lemmy.world 49 points 5 months ago
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[-] Granixo@feddit.cl 16 points 5 months ago

It's not even because it's heavier, it's because it's way more dense.

[-] shutz@lemmy.ca 61 points 5 months ago

It's not density, it's mass. A mass of 1kg compressed to the density of the Sun's core would pull the Earth with just as much force as a 1kg ball of styrofoam.

[-] rockerface@lemm.ee 21 points 5 months ago

And is the Sun was replaced with a black hole of the same mass, the Earth would just keep on rotating around it without issues, if slightly frozen

[-] chatokun@lemmy.dbzer0.com 6 points 5 months ago

Xkcd did a what if on a black hole moon (getting it to collapse into one may be impossible, but a black hole the mass of the moon is theoretically stable), and it has the same conclusion, except just slightly colder instead of slightly frozen. And by slightly, I mean almost imperceptible.

https://what-if.xkcd.com/129/

[-] rockerface@lemm.ee 8 points 5 months ago

I mean, the Sun also contributes slightly more heat to the Earth compared to the Moon. And by slightly, I mean the difference between cold and frozen

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[-] Chrobin@discuss.tchncs.de 3 points 5 months ago* (last edited 5 months ago)

Just to add some formality to this, the original commenter might want to look up the shell theorem for classical mechanics and Birkhoff's theorem for general relativity.

[-] Carrolade@lemmy.world 22 points 5 months ago* (last edited 5 months ago)

The guy on the right, if he be so wise in the ways of science, should be using the word "massive" instead of "heavier".

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[-] hddsx@lemmy.ca 4 points 5 months ago
[-] KillerTofu@lemmy.world 10 points 5 months ago* (last edited 5 months ago)

Heavy is a subjective term based on the force of gravity. You are heavier if we weigh you on the earth compared to if you are weighed on the moon.

Your mass in those two examples is unchanged. The amount of mass you have is finite and not subjective like weight.

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[-] JohnDClay@sh.itjust.works 12 points 5 months ago

Does the bowling ball ever so slightly increase the gravitational constant because of it's greater mass? Is that what the right guy is getting at?

[-] dream_weasel@sh.itjust.works 39 points 5 months ago* (last edited 5 months ago)

The gravitational constant G, no, the mutual gravitational force between the earth and the ball approximated as g, yes.

Edit: Since this is a little pedantic, G is used to calculate g.

[-] Faresh@lemmy.ml 7 points 5 months ago* (last edited 5 months ago)

But how would that make the bowling ball fall faster? F = G × m₁ × m₂ / r² and F = m₁ × a ⇒ a = F / m = G × m₂ / r², where m₁ is the mass of the ball and m₂ the mass of the planet. So the gravitational acceleration of a bowling ball is independent of its mass (assuming the planet has way more mass than a bowling ball).

[-] Zehzin@lemmy.world 18 points 5 months ago* (last edited 5 months ago)

I guess the bowling ball attracts the Earth towards it, shortening the distance so it hits the ground faster

[-] hddsx@lemmy.ca 5 points 5 months ago

No. F=GMm/d2. The mass of the earth doesn’t change so g=GM/d2 will not change

[-] Tar_alcaran@sh.itjust.works 8 points 5 months ago

Ah but the earth doesn't just attract the ball or feather. The bowling ball attracts the earth as well, and since it has more mass, it will pull the earth towards it faster than the feather.

But if you drop them at the same time, that's moot.

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[-] JohnDClay@sh.itjust.works 6 points 5 months ago

So why does the bowling ball fall faster in a vacuum? Does it appear faster locally because the heavier object makes local time slower than the lighter object compared to a distant observer? I'm trying to understand what the meme is getting at.

[-] Tja@programming.dev 6 points 5 months ago

That's the neat thing: it doesn't

[-] Drewfro66@lemmygrad.ml 4 points 5 months ago

The bowling ball also pulls the earth towards itself. This amount is imperceptibly small but still there

[-] hddsx@lemmy.ca 3 points 5 months ago

I’m trying to understand as well.

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[-] Heavybell@lemmy.world 11 points 5 months ago

There's a video of astronauts doing the heavy thing vs feather in vacuum experiment. I think it was a hammer rather than a bowling ball tho.

[-] mako@lemmy.today 9 points 5 months ago

I get that the heavier bowling ball affects the acceleration of the earth more than the lighter feather, but I don't see how that means it's falling faster as the meme is stating. The bowling ball would meet the earth first when dropped separately and from the same height because the earth is (imperceivably) accelerating toward it faster than it does the falling feather, but both the bowling ball and feather are falling at the same rate due to Earth's gravitational force.

Or am I missing something?

[-] Adalast@lemmy.world 13 points 5 months ago

One definition for a "rate of falling" would comfortably be "the time it takes the surfaces of two free gravitational separated by some distance to meet." With this in mind, the imperceptible but very real difference in the acceleration of the earth towards the bowling ball would become part of that equation, as it shortens the distance between the two from the other side.

Think of it like a head on collision of two vehicles. You can do the math as two bodies colliding with opposite velocity vectors, or you can arrive at the same mathematical result (at least for some calculations) by considering one of them to be stationary and the other to have the sum of the two speeds in the direction of its original velocity. "Two cars colliding head on at 60mph is the same as one car hitting a brick wall at 120mph." It is rough and doesn't work for all calculations, but the idea is the same.

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[-] iceonfire1@feddit.nl 9 points 5 months ago

I think the answer to this question changes based on your interpretation of 'falling faster'. I.e. whether that refers to the total time between the start and end of the fall or to the speed of the feather/ball to an outside observer.

[-] thisfro@slrpnk.net 7 points 5 months ago

Since we are in avaccum

That's where you're wrong kiddo

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this post was submitted on 28 May 2024
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