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anonymous

  • 5 years ago

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)?

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  1. anonymous
    • 5 years ago
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    I suppose you could "estimate" it by taking the product of their two masses, and about 7*10^-5...? \[\Large \begin{array}{l} g = G\frac{{{m_1}{m_2}}}{{{r^2}}}\\ = 6.673 \times {10^{ - 11}} \cdot \frac{{{m_1}{m_2}}}{{{r^2}}}\\ {m_1} = {m_2} = 60{\rm{kg}}\\ r = 0.01{\rm{m}}\\ {r^2} = 0.000001{\rm{m}}\\ 6.673 \times {10^{ - 11}} \times \frac{{3600}}{{0.000001}}\\ = 6.673 \times {10^{ - 11}} \times 3600000000 = 0.240228\\ 3600 \cdot 7 \times {10^{ - 5}} = 0.252 \end{array}\]

  2. anonymous
    • 5 years ago
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    But to be honest I think this is a pretty askew question =)

  3. anonymous
    • 5 years ago
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    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.

  4. anonymous
    • 5 years ago
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    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.

  5. JamesJ
    • 5 years ago
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    (One small thing. In the formula fsm has written above, he starts with g = .... It should be the force, F = .... )

  6. anonymous
    • 5 years ago
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    the answers i have to it is 10^-6N is there anyway to find that out?

  7. anonymous
    • 5 years ago
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    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.

  8. anonymous
    • 5 years ago
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    but how do we estimate this value?

  9. JamesJ
    • 5 years ago
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    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.

  10. anonymous
    • 5 years ago
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    and the weight is greater by a factor of 10^9 ? what value of masses should we estimate? 50-80kg?

  11. JamesJ
    • 5 years ago
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    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.

  12. anonymous
    • 5 years ago
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    Haha yeah, everything about my example was a bit awful. Apologies.

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