Wednesday, June 25, 2014

Newton's Indian rope trick

how long would a vertical rope need to be so that it is in dynamical equilibrium in the Earth's gravitational field with the centrifugal force?

2 comments:

  1. Let $L$ denote the rope's length, $\mu$ its mass density, and $M$ the mass, $R$ the radius, $\omega$ the angular velocity of the Earth. The $z$ axis points away from the Earth. The two forces are then the gravitational force $F_G$ pointing inwards,

    $$F_G = - \int_{R}^{R+L} \frac{G M \mu}{z^2} dz$$,

    and the centrifugal force $F_C$ pointing outwards,

    $$F_C = \int_R^{R+L} \mu \omega^2 z dz$$.

    The condition $F_G = - F_C$ gives the length $L$.

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  2. thank you very much for your answer. I hope I didn't make a mistake: solving the above integrals leads to a quadratic equation in $x=L/R$ of the form
    \begin{equation}
    x^2+3x+2-2\frac{GM}{\omega^2R^3} = 0
    \end{equation}
    the term $2GM/\omega^2/R^3$ is dimensionless and has the value $\simeq 105$ for the Earth, which yields as a solution $x=8.6$, meaning that $L\simeq52e3$ kilometers (a bit more than the Earth's circumference).

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