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  1. Peter Renz

    Interesting. The period of a pendulum at the surface of the earth gives us the mass of the earth. For convenience we set the universal gravitational constant to 1 by our choice of units. The sidereal year gives the average angular velocity of the earth around the Sun, d theta/dt. The centripetal acceleration of the Earth is then R(d theta/dt)^2, where R is the distance from the Earth to the Sun. This acceleration times the mass m of the earth is equal to the gravitational force the Sun exerts on the Earth: mM/ R^2, where M is the mass of the Sun. After some algebra we have M = 1 m(R^3) (( d theta /dt)^2). (Remark. The units are straightened out by the units in the universal constant of gravitation, which is seen here as the leading factor of 1.) Here we know m and d theta / dt, so if we knew M we could find R.

    George Clark had students in Junior Lab at MIT determine the distance from the Earth to the Sun by determining the sun’s equatorial rate of angular rotation by observing the motion of Sun spots. Then they used a spectrometer to measure the shift of spectral line across an equatorial slice of the solar disc. This Doppler shift measurement gives the velocity of approach and recession at the edge of the solar disk, from which the radius of the sun can be found, and from that and the angular diameter of the Sun you get the distance to the Sun.

    Peter Renz, August 15, 2012

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  2. rolandflyoland boucher

    Dear Mr Renz

    There is no reason to introduce the mass of the Earth or the Universal Gravitational Constant into these calculations— The insight that the orbital pendulum ACTS AS IF IT WERE ORBITING THE CENTER OF THE EARTH is very powerful.

    If Robert Hooke had had this insight or if Newton had accept his proof, the controversy about the inverse square law of gravitation would have been over in his lifetime. The proof is simple if the pendulum gives the same value for the distance to the moon as direct geometrical methods
    then the argument would be over. Either Hooke did not have the insight or h Newton did not believe him.

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  3. Mike Zorn

    I wondered about the difference in period between your two pendulums (large ball, small one). It’s possible that air resistance plays a part. If you did the experiment in a vacuum, the periods might come out the same – but that’s not easy.

    Or if you used a grandfather clock type pendulum (disk) the results might be closer together.

    “… the square of the orbital period would now be equal to the cube of the distance ”

    That’s Kepler.

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    • Roland A. Boucher

      Mike

      The equations are quite straightforward. making the string heavier relative to the ball speeds up the pendulum because the center of mass of the string is only 1/2 its length of the from the pivot point. The effect of air resistance would be very small especially at the small total angle of swing.

      Ps what did you think of the pint and the Magna Carta

      Roland

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