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.
Comment on your mom falls significantly faster than g
BB84@mander.xyz 1 month agoRe 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 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.
reliv3@lemmy.world 1 month ago
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 affect 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 they would measure the acceleration of the Earth to also 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…
BB84@mander.xyz 1 month 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.