this post was submitted on 03 Nov 2024
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ur dada so buff he falls significantly faster than g (mander.xyz)
submitted 3 weeks ago* (last edited 3 weeks ago) by [email protected] to c/[email protected]
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Tap for spoilerThe 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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[–] [email protected] 24 points 3 weeks ago* (last edited 3 weeks ago) (2 children)

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.

[–] [email protected] 4 points 3 weeks ago

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.

[–] [email protected] -1 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

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.

[–] [email protected] 2 points 3 weeks ago

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.