Tap for spoiler

The bowling ball isn’t falling to the earth faster. The higher perceived acceleration is due to the earth falling toward the bowling ball.

  • originalfrozenbanana@lemm.eeEnglish
    2·
    2 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.eeEnglish
      4·
      2 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.