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.
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.
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
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.
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.
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
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.
“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.
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?
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.
It won’t cause the perception to differ because the difference is so small it’s impossible to measure
The middle of the bell curve only works in a vacuum, and the top of the bell curve is true with wind resistanceEdit: I misread the post
Even in a perfect vacuum the bowling ball still falls faster. See my comment sibling to yours.
Oh, interesting. That’s a cool fact
Also, I very much misread the post lol
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.
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).
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
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
@[email protected] and @[email protected] are correct
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.
