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.
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Brian Cox shows ball and feathers falling together in vacuum: https://youtu.be/E43-CfukEgs
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.
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
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.
Obviously the bowling ball because it’s more MASSIVE.
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.
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.
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.
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.
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.
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.
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?
Stupid question, bowling balls don’t fit through the vacuum’s hose.
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 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.
