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anonymous

  • 4 years ago

Newton's first law of gravitation by integration.

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  1. anonymous
    • 4 years ago
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    My question is: can you get the same result by integrating the force on all points of an object (planet) by, say, the sun, as if you treated just its centre of mass?

  2. anonymous
    • 4 years ago
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    Firstly, consider the sun. If we draw a circle around the it, we can say that the same force will act on 1kg anywhere in the circle.|dw:1328041988435:dw|

  3. anonymous
    • 4 years ago
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    Now, consider a planet orbiting the sun|dw:1328042007572:dw|

  4. anonymous
    • 4 years ago
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    All points on the arc will feel the same force. Now... |dw:1328042035156:dw|

  5. anonymous
    • 4 years ago
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    |dw:1328042090769:dw| Okay, the area of the arc will be: \[\pi2q*\theta/360\] z and r are constants. q=sqrt{y^2+p^2} x=sqrt{z^2-y^2} y=sqrt{z^2-x^2} p=r-x t=q-p theta=arctan(y/p) What would you do now to integrate the arc with respect to d(x-t) (as the force is the same as on the arc at x-t, not x)?

  6. JamesJ
    • 4 years ago
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    This can be dealt with, but not as you're proceeding. The best way to deal with this is to reformulate classical gravity using what's known as Gauss' law of gravitation. Then what that formulation shows is that we can consider a mass at a height (radial distance) from the earth d, to be attracted to a point mass at the center of the earth. We do this all of the time without thinking about why exactly this works. The Gaussian formulation of gravity makes that explicit. http://en.wikipedia.org/wiki/Gauss'_law_for_gravity

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