anonymous
  • anonymous
how can we estimate the gravitational force of attraction between two people sitting side by side on a bench? How does this force compare with gravitational force exerted on each of them by the earth(weight)?
Physics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
I suppose you could "estimate" it by taking the product of their two masses, and about 7*10^-5...? \[\Large \begin{array}{l} g = G\frac{{{m_1}{m_2}}}{{{r^2}}}\\ = 6.673 \times {10^{ - 11}} \cdot \frac{{{m_1}{m_2}}}{{{r^2}}}\\ {m_1} = {m_2} = 60{\rm{kg}}\\ r = 0.01{\rm{m}}\\ {r^2} = 0.000001{\rm{m}}\\ 6.673 \times {10^{ - 11}} \times \frac{{3600}}{{0.000001}}\\ = 6.673 \times {10^{ - 11}} \times 3600000000 = 0.240228\\ 3600 \cdot 7 \times {10^{ - 5}} = 0.252 \end{array}\]
anonymous
  • anonymous
But to be honest I think this is a pretty askew question =)
anonymous
  • anonymous
This is a common question in "physics 101". Teachers like to get across the idea that gravitational forces exist between all things having mass, albeit these forces are usually negligible.

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anonymous
  • anonymous
Oh yes, I was referring to the asking to "estimate" the attraction. Unless by "estimate" they mean actually work it out to the best of your abilities, since this is in itself an estimation at best.
JamesJ
  • JamesJ
(One small thing. In the formula fsm has written above, he starts with g = .... It should be the force, F = .... )
anonymous
  • anonymous
the answers i have to it is 10^-6N is there anyway to find that out?
anonymous
  • anonymous
You'll need to set the radius and mass of the two people. This data isn't given, so a precise answer cannot be found.
anonymous
  • anonymous
but how do we estimate this value?
JamesJ
  • JamesJ
But the distance used in the calculation above isn't perhaps the best approximation. The distance between their center of masses is likely to be closer to 1 m than 0.01 m.
anonymous
  • anonymous
and the weight is greater by a factor of 10^9 ? what value of masses should we estimate? 50-80kg?
JamesJ
  • JamesJ
Sure. 70 kg is as good as any other point estimate. As has been pointed out above, the purpose of the exercise isn't the precise number, but showing it is so much less than the gravitational force of the earth.
anonymous
  • anonymous
Haha yeah, everything about my example was a bit awful. Apologies.

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