Tap for spoiler
The bowling ball isn’t falling to the earth faster. The higher perceived acceleration is due to the earth falling toward the bowling ball.
Stupid question, bowling balls don’t fit through the vacuum’s hose.
Ur mom could suck it through
So will the bowling ball gravitationally attract the earth to itself there by reach the earth an infinitesimally small amount?
Yes, the earth accelerates toward the ball faster than it does toward the feather.
Wouldn’t this be equally offset by the increase in inertia from their masses?
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.
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.
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.
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.
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.
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.
If the earth would be accelerating towards you, then g would be less than 9.81.
Think of free falling, where your experienced g would be 0.
Even if you imagine doing them separately, the acceleration of the Earth cannot be calculated based on just a singular force unless you assume nothing else is exerting a force on the Earth during the process of the fall. For a realistic model, this is a bad assumption. The Earth is a massive system which interacts with a lot of different systems. The one tiny force exerted on it by either the feather or bowling ball has no measurable effect on the motion of Earth. This is not just a mass issue, it’s the fact that Earth’s free body diagram would be full of Force Vectors and only one of them would either be the feather or bowling ball as they fall.
As for my second point, I understand your model and I am defining these references frames by talking about where an observer is located. An observer standing still on Earth would measure the acceleration of the feather or bowling ball to be 9.81 m/s/s. If we placed a camera on the feather or bowling ball during the fall, then it would also measure the acceleration of the Earth to be 9.81 m/s/s. There is no classical way that these two observers would disagree with each other in the magnitudes of the acceleration.
Think of a simpler example. A person driving a car towards someone standing at a stop sign. If the car is moving 20 mph towards the pedestrian, then in the perspective of the car’s driver, the pedestrian is moving 20 mph towards them. There is no classical way that these two speeds will be different.
Earth is in this case not an inertial reference frame. If you want to apply Newton’s second law you must go to an inertial reference frame. The 9.81m/s/s is relative to that frame, not to earth.
Brian Cox shows ball and feathers falling together in vacuum: https://youtu.be/E43-CfukEgs
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.
But what weighs more:
A ton of bowling balls or a ton of feathers? 🤔
When you carry a ton of feathers, you also have to carry the weight of what you did to those poor birds…
What about all the bowling cattle you had to castrate for those balls?
That’s a trick question. Feathers have lower density than bowling balls; a ton of feathers would have a larger volume compared to the same mass in bowling ball, thus the feathers are heavier
Why does a larger volume mean the feathers are heavier?
Because big
So a beach ball weights more than a bowling ball?
If they have the same mass, yes.
If they have the same mass they weigh the same until you blow up the beach ball and are weighing the beach ball + the air inside.
You know, I’m not strong willed enough to keep going, but this comment thread is starting to remind me of this post
Strong sharks are smooth energy.
nice chatgpt roleplay
Thanks, I
was trained bylearned from the bestReddit comments
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.
This person sciences good
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 • 1024 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.
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.
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.
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.
The acceleration relative to the earth is the same, relative to some point from another system the bowling ball accelerates very slightly slower but accelerates the earth very slightly more towards it. The total acceleration of these two bodies towards each other is g.
Yeah you’re making that statement but it’s not true. Their acceleration relative to an inertial reference frame is g. That’s what the law of universal gravitation says, I have no idea where you’re getting that stuff from.
You said the two objects accelerate at the same rate, but then in the PS you said the feather gets accelerated faster. What do you mean?
Are you saying the feather gets pulled on more because the mass of earth minus feather is greater than the mass of earth minus ball? You would be right. If you lift the feather, measure how long it takes to fall, then lift the ball and measure, you should get the same number. This meme was assuming you either let them fall side by side, or measure them separately but each time conjure the object out of thin air.
You said the two objects accelerate at the same rate, but then in the PS you said the feather gets accelerated faster. What do you mean?
Both accelerate at the same rate relative to the earth (the bowling ball accelerates slightly slower relative to some outside point, but it accelerates the earth slightly more towards it, resulting in the same relative acceleration to the earth as the feather)
Newton’s second law works in inertial frames. The acceleration of both objects would be the same in the inertial frame. But in the inertial frame, the earth would accelerate faster toward the object if the object was a bowling ball than if it was a feather.
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
BUT earth will move with gM/m1
No. Multiplication is associative, you can switch the masses around as you please, nowhere in the formula does it say “the greater mass” or “the smaller mass” you could just as well re-arrange the formula and come up with “earth moves with gm1/M”. Last but not least there’s only one force acting on both objects… and gM/m1 is neither a speed nor a force. G * 100kg / 20kg is 5G. Measured in Nm²/kg² which is the same we started with because the two kg cancel each other out.
They both fall towards their shared centre of gravity. It’s this “the earth revolves around the sun” thing again, no it doesn’t, they both revolve around their shared centre of gravity (which, yes, is within the sun but still makes it wobble). That centre is very far away from the ball and very close to the earth and both are moving at the same speed towards it (because acceleration doesn’t depend on mass), blip to the next frame of the simulation now the centre of gravity moved towards the ball, next frame still closer to the ball, that is the reason both reach it at the same time, not because one is faster than the other.
…or so it would be, if the shared centre of gravity of ball and earth wouldn’t lie within the earth so they don’t actually both reach it, the earth is in the way, the rest of the acceleration is turned into static friction: Because they both are still falling even when in contact. But really that complication only exists because they have volumes which is why I factored it out from the rest of the reasoning.
all that is only brain-rot statements with no technical meaning. lemme make this completly clear
mf= mass of feather mb= mass of ball me= mass of earth ae=accelaration of earth fg=force experienced by both
now in case of feather
force on earth is what? yes thats fg =G.mf.me/r^2
now thats the net force on earth, now what is newtons law? me.ae=G.mf.me/r^2
we get ae=G.mf/r^2
similarly in case of ball ae=G.mb/r^2
and accelaration of earth is clearly more in case of ball, and yes this is accelaration in non inertial frame study newtons laws of motion again if you didnt know, so your second paragraph is utter nonsense
instead of nonsense brainrot statements like 'Multiplication is associative, you can switch the masses around as you please, nowhere in the formula does it say “the greater mass” or “the smaller mass” you could just as well re-arrange the formula and come up with “earth moves with gm1/M” tell me where in equations you think i am wrong
It’s not nonsense when it makes people understand, buddy. And don’t get all “oh be technical” on me when you say things like “earth will move with <something with the same units as G>”. Something that’s definitely something, but not m/s.
inertial frame
I was talking about time-steps when I said frame. Hence “simulation”, and “one frame, then another, then another”, referencing successive moments in time.
yet another brain rot reply, man i am done,
““earth will move with <something with the same units as G>”. Something that’s definitely something, but not m/s” you idiot i was talking about accelwration, if you need units just put in dementions of all the variables, thats trivial stuff you dont understand nlm at all.
second para is another non technical nonesense
you idiot i was talking about accelwration,
Then why did you say “move” instead of “accelerate”. And the units don’t match acceleration, either. Best I can tell it’s some fraction of a term. If you want it to be an acceleration then you’re missing a squared distance, and if you want it to be acceleration, why are both mass terms in there.
For someone who throws around things like “that’s non-technical brainrot” damn is your prose fuzzy.
tell me how Gm/r^2 dosent match acceleration, the fact that i wasted my time on low iq person like you
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.
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.
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.
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.
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?
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.
For the bowling ball, Newton’s shell theorem applies, right?
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.
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?
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.
even light can stop following null geodesics because the curvature can be too big compared to the wavelength
Very interesting! How do you study something like this? Is it classical E&M in a curved space time, or do you need to do QED in curved space time?
Also, are there phenomena where this effect is significant? I’m assuming something like lensing is already captured very well by treating light as point particles?
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.
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.
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?
Depends on the color of the feather and the ball.
There’s a simple explanation.
Exactly, red has way more up-quarks than blue
Because light-blue weighs less than blue.
and Diet light-blue weighs less than light-blue
Please stop by the office and pick up your combo Nobel Prize in Physics and Chemistry.
Reading that spoiler, I hate scientists sometimes.
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?
“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
“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?”
I’m pretty sure bowling balls and feathers fall all the time
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.
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.)
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.
Obviously the bowling ball because it’s more MASSIVE.
Or dense?
It’s the mass that results in gravity, not the density. A giant cloud of gas will have the same gravitational effects as if it were compressed into its solid phase
Ehh, just thinking of possible air resistance and such. I don’t know which variables they like to include and which not. But I’m guessing this is not one for just treating gravity as a one way street cuz the earth is so big
Tho, with a cloud of gas you can’t get as close to the center of mass without passing a bunch of it
But … Steel is heavier than feathers …
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.
10% of the earths mass
@WolfLink@sh.itjust.works and @theturtlemoves@hexbear.net are correct
