anonymous
  • anonymous
Newton's second law: I'm I right that a bowling ball and a tennis ball fall at the same rate because of the force of gravity and the mass of the two objects does not matter?
MIT 8.01 Physics I Classical Mechanics, Fall 1999
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
From the universal law of gravitation\[F _{g} = ma = GM _{earth} m/r^{2} \]Notice that the mass of the object, m, cancels leaving\[a = GM _{earth} /r^{2} =g\]
anonymous
  • anonymous
You are right - but keep in mind that you are neglecting air-drag (which is often the case and not). On the moon there is no air-drag - therefore a feather and hammer touch the ground at the same time if the fall from the same height - proof -> http://en.wikipedia.org/wiki/File:Apollo_15_feather_and_hammer_drop.ogg - pretty cool huh? ;) seeing is believing
Zivko
  • Zivko
What about the other object (Earth,Moon...)? Don't they accelarate towards the 'falling body' as well? In that case, mass of the 'falling object' does matter.

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anonymous
  • anonymous
The forces felt by each object is the same, namely Fg from above, but of course the accelerations felt are very different... a = F/m. Masses are different so the accelerations are different but the forces are the same. The mass of whatever body you are calculating the acceleration of is the one that doesn't matter in the calculation (cancels out)
Zivko
  • Zivko
Yes, but doesn't the Earth accelerates more towards the body with higher mass? Of course, then we would have to drop one body at a time and I doubt that we could measure the difference. But, exactly speaking, Earth will meet with higher mass object sooner.
anonymous
  • anonymous
the point is that it doesn't accelerate at a rate which depends it ITS mass
anonymous
  • anonymous
this was Galileo's theory no Newton.

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