Does Earth/Moon Model Show Cavendish Is Wrong?

Discussion in 'Astronomy, Exobiology, & Cosmology' started by common_sense_seeker, Sep 26, 2008.

  1. common_sense_seeker Bicho Voador & Bicho Sugador Valued Senior Member

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    A scaled down version would have the Earth about the size of your eye and the Moon would be the size of a pea held at arms length. Even if they were made of everyday magnets there still wouldn't be enough attracive force to maintain an orbit. Let alone if they were made of rock or metal! The accepted Cavendish results for the gravitational constant are clearly wrong. Modern satellite technology is based on the measured surface gravity of the Earth, not the incorrectly calculated gravitational constant.
     
    Last edited: Sep 26, 2008
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  3. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    Explain exactly what this experiment with the magnets would involve.
     
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  5. common_sense_seeker Bicho Voador & Bicho Sugador Valued Senior Member

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    A scaled down model of the Earth/Moon system silly. Too difficult a concept for you?
     
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  7. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    Are you going to explain or not Mr. Know it all?
     
  8. Read-Only Valued Senior Member

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    That's absurd! Show PROOF that the Cavendish calculations were incorrect!!!

    Can you???? Or do you even begin to understand the math involved? I really, really doubt the latter. You've shown NO math ability at all in ANY of your posts here. In fact, all you've done the whole time is blow hot air!!!!:bugeye:
     
  9. Asguard Kiss my dark side Valued Senior Member

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    read only is right, its stupid to think that you could skale down the earth to that extent and still expect to see an atractive force. ESPECIALLY when they are both sitting on the earth
     
  10. common_sense_seeker Bicho Voador & Bicho Sugador Valued Senior Member

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    It requires an imagination, of course. I tend to take that for granted. How can you not understand what I'm saying? If there was a ball of steel on a table the size of your eye, and another smaller ball of steel the size of a pea at a distance apart of around 0.5m, this would represent the Earth and the Moon. The gravitational attraction between them based on the gravitational constant big G, is not enough to allow you to roll the 'pea' around the bigger steel ball (the 'eye') in an orbit!

    Try it out for real if you need to. I don't need to. Nor do I need to do any mathematical calculations! The people in www.thenakedscientists.com are smarter than you lot!
     
  11. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    Because you can't roll it at a constant and incredibly low speed on Earth.
     
  12. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    Okay, I decided to humour your suggestion of testing this scaled down model.

    I estimated the mass of the eye sized Earth to be about 0.5 kg (even if this is a gross underestimate the final outcome will still be unrealistic).

    To work out the orbital speed of the moon we use this formula, with your 0.5m figure, and my 0.5 kg figure...

    \(Speed=\sqrt{\frac {\mu}{r}}\approx8.2\times10^{-6}ms^{-1}\)

    I bet you not once thought about how incredibly small the speed would have to be for it to orbit did you?

    In fact I bet you didn't even think about the speed it would have to be travelling at all did you?
     
  13. Read-Only Valued Senior Member

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    What's actually important here is that you are a fruitcake that couldn't calculate his way out of a paper bag - and ALL of your postings are nothing but idiotic nonsense!!!
     
  14. D H Some other guy Valued Senior Member

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    Your scaled-down model is what is silly, silly. Do you have any common sense whatsoever? Mass scales with the cube of distance, and gravitational force therefore scales with the fourth power of distance. In scaling distance by a factor of 2*10[sup]-9[/sup], you have scaled mass by a factor of 8*10[sup]-27[/sup] and gravitational force by a factor of 16*10[sup]-36[/sup]. In other words, what you are doing is completely invalid.
     
  15. common_sense_seeker Bicho Voador & Bicho Sugador Valued Senior Member

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    Not true. I have considered it. Roll the 'pea' as slow as you like.

    Newton's law of universal gravitation is just that, a universal one. What I have done is not invalid. You just don't like it.
     
  16. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    Do you really expect the pea to be able to maintain the incredibly small speed that would be needed to orbit the eye?
     
  17. D H Some other guy Valued Senior Member

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    Your user name certainly is correct -- in the sense that you haven't a lick of common sense and are desperately seeking some.

    What you have done is to scale all linear dimensions by a factor of about 2*10[sup]-9[/sup] to make the Earth the size of an eyeball. With such a scaling,
    • The scaled Earth has a diameter of 25.48 millimeters and a mass of 47.7936 grams. The human eye has a diameter of about 24 millimeters.
    • The scaled Moon has a diameter of 6.9484 millimeters and a mass of 588.8 milligrams.
    • The scaled Earth and Moon are separated by 76.8798 centimeters.

    So, what is the gravitational force between the scaled Moon and the scaled Earth?

    \(F = \frac{GM_eM_m}{r^2} = 3.1771\times10^{-15} \, \text{newtons}\)

    What you have done is to assume that the force will also scale with distance, and this is what is completely invalid. Interestingly enough, acceleration does scale with distance because

    \(a_{\text{scaled}} = \frac{G\,M_{\text{scaled}}}{r_{\text{scaled}}^2} = \frac{G\,M\,sf^3}{r\,sf^2} = \frac{G\,M}{r^2}\,sf = a\,sf\)

    To demonstrate, the acceleration of the scaled Moon and Earth toward each other is

    \(a = \frac{G(M_e+M_m)}{r^2} = 5.4624\times10^{-11} \, \text{meters}/\text{sec}^2\)

    Using \(r\omega^2 = a\) for uniform circular motion and \(T=2\pi/\omega\), the period of the scaled Earth's and scaled Moon's orbit around one another is

    \(T = 2\pi \sqrt{\frac a r} = 27.28\,\text{days}\)

    or one sidereal month!
     
  18. Janus58 Valued Senior Member

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    IOW, if you had a smooth enough surface and downsized moon, and you could reduce friction enough, then yes, you could roll the minature "Moon" so that it "orbits" the minature "Earth" with a period close to one month.
     
  19. Steve100 O͓͍̯̬̯̙͈̟̥̳̩͒̆̿ͬ̑̀̓̿͋ͬ ̙̳ͅ ̫̪̳͔O Valued Senior Member

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    Gravity from all sorts of other things will most probably affect the moon too much as well.
     
  20. Janus58 Valued Senior Member

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    True, you'd have to isolate it quite a bit. As far as I can figure, you would have to watch it from around a 100 meter distance to prevent pulling the "moon" into orbit around you!(This would keep the radius of the Hill sphere to about twice the distance of the separation.)
     
  21. common_sense_seeker Bicho Voador & Bicho Sugador Valued Senior Member

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    The gravitational constant may well be correct, but this is only for baryonic matter. I'm convinced Cavendish's assumption of Newton's law of gravitation has led to an inaccurate calculation of the Earth's average density. I'm starting on the maths proof from now. With an assumption of DM at the center of the Earth, I'm sure the maths calculations will still work accordingly.
     
    Last edited: Sep 27, 2008

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