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.
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
They are on opposite sides of the earth
Isn’t “heavier” only used when describing weight and not mass?
It’s almost analogous. A more massive object experiences a larger force caused by gravity, so assuming the gravity field stays the same, a larger mass is heavier.
You’re right that it’s technically incorrect, especially when talking about something like moving the Earth with gravity.
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
You can stuff your misogynist fatshaming where it would hurt the most too wtf is going on with lemmy lately?
fixed it sorry
Oh wow, awesome! Was bracing for a whole host of anti-woke commentary but you genuinely seem to care about respecting others so I should point out that you only fixed the misogynist part and not the fatshaming part.
Okay how about now
yeeeeee thank you
lol. What had you said?
the original title was “your mom falls significantly faster than g”
No, it isn’t. Because earth wouldn’t fall towards the ball. Why?
Go to your frige right now and try to push it with one finger. It doesn’t move does it? You may say “That’s because of static friction!” And you would be correct. The force of static friction. Because the object moves in the direction of vector sum of all forces.
Tap for spoiler
(In the example with fridge the static friction force cancels all other forces up to certain value and after that - motion)
And adding microscopic attraction force towards the ball absolutely doesn’t change the full vector sum of forces, that are applied to Earth constantly (which is probably pointed towards the sun).
It does change the vector though. There’s no (well barely any) friction in space. In fact pushing the fridge with one finger rather than pushing the fridge will move and rotate the earth at incomprehensibly small velocity for a short time.
It does change the vector, yes, but to such a small amount that it does not become pointed at a ball.
You can substitute static friction with a gravitational pull of the sun, and the result is going to be the same as in fridge example
Static friction causes that result because it matches the force it’s resisting. Gravity doesn’t do that, so while the total vector will still be pointing away from the ball, it will point away from it with a slightly smaller magnitude.
Yes. Therefore Earth would not “fall towards the ball” and therefore it isn’t falling faster
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
Tho, with a cloud of gas you can’t get as close to the center of mass without passing a bunch of it
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
Does this imply that if I am standing on an object moving at a constant speed in a straight line, and I am lifting and dropping a sufficiently massive object such that I’m causing the object in standing on to accelerate towards the object I’m dropping, that eventually I’ll slow or stop the object I’m standing on?
Nope. The argument only works if you conjured the bowling ball and feather out of
thin airvacuum. https://lemmy.world/comment/13237315 discusses what happens when the objects were lifted off earth.Drat. Thanks 😂
For the sake of simplicity, let’s say you have negligible mass, while the two masses, m1 and m2, have equal masses and sizes. Everything is moving at some velocity in a vacuum.
When the two masses are touching, the Centre of Gravity is midway between their Centres of Mass, which in this scenario would mean it is where they touch.
When you pick up m2, an equal and opposite force would push m1 away. Because they both have equal mass, both would end up the same distance away from the CoG. If you lifted m2 on your head, the CoG would be right at the middle of your height.
For as long as you’re holding m2, your body is resisting the force of attraction due to gravity between m1 and m2. When you drop m2, both it and m1 accelerate towards the CoG. When they meet, the energy you put into lifting m2 would be converted into heat in the collision. From an outside observer, while you were doing all that, the CoG was moving in a perfectly straight line with no change in velocity.
Now, if you instead threw m2 away from m1 faster than its escape velocity, then that would change the velocity. If m1 and m2 weren’t equal in mass and size, the CoG would still be moving in a straight line, but the distance m1 and m2 moves away from the CoG would be proportional to their masses.
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?
I didn’t think about that! If the object was taken from earth then indeed the total acceleration between it and earth would be G M_total / r^2, regardless of the mass of the object.
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.
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.
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.
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?
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.
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.
@[email protected] and @[email protected] are correct
One tenth the mass of the earth?
10% of the earths mass
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.
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.
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.
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.
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.
