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submitted 4 weeks ago* (last edited 4 weeks ago) by BB84@mander.xyz to c/science_memes@mander.xyz

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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[-] noisefree@lemmy.world 6 points 4 weeks ago* (last edited 3 weeks ago)

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?

[-] BB84@mander.xyz 5 points 4 weeks ago

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.

[-] Muffi@programming.dev 3 points 4 weeks ago

Isn't "heavier" only used when describing weight and not mass?

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[-] ultrahamster64@lemmy.world 3 points 4 weeks ago

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).

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[-] marcos@lemmy.world 3 points 3 weeks ago

Hum... What is your measurement error?

[-] originalfrozenbanana@lemm.ee 2 points 4 weeks ago

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?

[-] BB84@mander.xyz 4 points 4 weeks ago

Nope. The argument only works if you conjured the bowling ball and feather out of ~~thin air~~ vacuum. https://lemmy.world/comment/13237315 discusses what happens when the objects were lifted off earth.

[-] originalfrozenbanana@lemm.ee 3 points 4 weeks ago

Drat. Thanks 😂

[-] sheepy@lemm.ee 4 points 4 weeks ago* (last edited 4 weeks ago)

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.

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this post was submitted on 03 Nov 2024
272 points (86.6% liked)

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