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255

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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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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[–] [email protected] 16 points 3 weeks ago (1 children)

Wouldn’t this be equally offset by the increase in inertia from their masses?

[–] [email protected] 35 points 3 weeks ago* (last edited 3 weeks ago) (2 children)

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.

[–] [email protected] 19 points 3 weeks ago* (last edited 3 weeks ago) (1 children)

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.

[–] [email protected] 8 points 3 weeks ago* (last edited 3 weeks ago)

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.

[–] [email protected] 7 points 3 weeks ago

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.