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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.

  • roscoe@lemmy.dbzer0.com
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    edit-2
    13 days ago

    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.

  • pumpkinseedoil@mander.xyz
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    13 days ago

    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.

    • red@lemmy.zip
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      13 days ago

      the fact that you got upvoted, you clearly said force on both objects is gM and the feather or ball will move with g BUT earth will move with gM/m1 which is more in case of ball, and no its not acceleration between mases, its the force experiencec by both mases so, fg=m1.a

  • mathemachristian [he/him]
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    13 days ago

    You can stuff your misogynist fatshaming where it would hurt the most too wtf is going on with lemmy lately?

  • fubarx@lemmy.ml
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    13 days ago

    Depends on the color of the feather and the ball.

    There's a simple explanation.

  • RumorsOfLove@lemmy.dbzer0.com
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    13 days ago

    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

  • BmeBenji@lemm.ee
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    13 days ago

    “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?”

  • BB84@mander.xyz
    hexagon
    ·
    edit-2
    13 days ago

    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.

  • originalfrozenbanana@lemm.ee
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    13 days 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?

    • sheepy@lemm.ee
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      edit-2
      13 days 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.

    • BB84@mander.xyz
      hexagon
      ·
      13 days 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.