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If everything in earth's atmosphere contributes to earth's overall weight, including air etc. Does an objects weight in free-fall effect this overall weight in anyway.

Just say the weight of earth was 10kg, and consisted of 8kg earth mass, 1kg air and 1kg as an object. If that 1kg was in free-fall, what effect will this have on the 10kg weight of the planet.

I understand that this can not be replicated in any other environment as as soon as the object stops exerting downward force on the object it's inside, it no longer contributes to the containing objects weight. But how does this work with our entire planet, which creates the gravitational pull that defines that very rule? Or is this much simpler than I am imagining?

LBPLC
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  • Related: http://physics.stackexchange.com/q/12756/ – Brandon Enright Jul 09 '14 at 21:40
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    Reminds me of a joke I heard: A trucker stopped before crossing a bridge and gave the trailer a strong whack with a tire iron, got back in and crossed the bridge. When asked why he did that, he said,"My cargo is exotic birds and I'm over weight. If half the birds are flying, I'm fine." – LDC3 Jul 10 '14 at 01:17
  • 3-body problem. –  Jul 11 '14 at 17:04

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First a counter-question as an idea for further thinking: How do you measure Earths weight? On Earth we weigh things using Earth's gravity.

Think of it that every little particle creates it's own gravitational field (that's actually true).

So every particle there is attracts every other in the entire universe (according to our current equations).

So if you measure Earth's weight you probably mean it's gravitational strength. The free falling thing is producing gravity as much as anything else, so if it's near Earth, it contributes to its gravity.

As DWin has noticed, whenever something falls, the Earth falls towards it. But usually, the falling stuff is so much smaller that the gravitational pull doesn't move the Earth very much.
In other words same force affects both Earth and the falling object according to 2nd Newtons law. When you think about it that way, you can't even tell who's falling when you've got 8kg and 10kg objects.

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Tomáš Zato
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  • I think you need to add the fact that the Earth would be falling toward the object as well and so the COM of the system would remain the same. Otherwise the apparent COM would be moving away from a test mass( if it were on the same side as the falling object.) – DWin Jul 09 '14 at 23:34
  • What does COM mean? – Tomáš Zato Jul 10 '14 at 06:13
  • Centre Of Mass? – John Rennie Jul 10 '14 at 11:00
  • You're saying it almost like it was commonly used abbreviation everyone should know... It's awkward 1 grader habit to point at people who don't know something. Particularly awkward on Q&A portal. – Tomáš Zato Jul 10 '14 at 18:48
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    @TomášZato You may be reading some negative implications into what seemed like a helpful suggestion. It was pretty common in my physics training. Sorry about not upvoting until now. I thought your answer provided a useful perspective. – DWin Jul 11 '14 at 00:17