Last summer, while visiting the California wine country with my wife, I awoke from a dream about an ancient Astronomer who used a long pendulum to prove that the Earth was spinning around once each day and not the stars. He tasked a young assistant to give the pendulum a gentle push now and then to keep it swinging. In my dream his assistant gave the pendulum a giant push at the top of its swing but at 90 degrees to the normal motion. In my dream the pendulum continued to swing, but now it was in a perfectly circular orbit around an imaginary vertical line from the pivot point. I was startled, could it be that the pendulum acted as though it was circling the center of the Earth? I awoke and made myself a note to check out the physics of this insight in the morning. This is what I found.

The period of a pendulum in an orbital motion (conical swing) is the same as if it were swinging in the normal back and forth motion. I knew that a one meter pendulum swung with a period of about 2 seconds and that the length of a pendulum is proportional to the square of its period.

The radius of the Earth is about 6366000 times the length of our one meter pendulum, so the period of this giant pendulum would be twice the square root of 6366000 or 5046 seconds. This period is 84 minutes six seconds which is very near the period of a low altitude satellite.

I knew the time it took the Moon to orbit the Earth in relation to the fixed stars was about 27 1/3 days. This is 2,361,312 seconds, or 468 times as long as our giant pendulum. If gravity were constant, regardless of the distance from the surface of the Earth, the Moon would orbit at the square of 468 or 218,983 Earth radii and it would be much larger the the Earth.

But we know the Earth has only so much gravity and it is spread equally all over its surface, or over any larger sphere centered on the Earth. The strength of gravity at the distance to the Moon in this case would be reduced by the square of the distance. A little algebra can quickly show that the ** square **of the orbital period would now be equal to the

**of the distance from the center of the Earth. Our simple pendulum predicts that the Moon orbits the Earth at 60.27 radii or 383,769 km.**

*cube***The actual distance is 384,400 km.**

The Moon subtends a visual angle of 29.3 arc minutes or about 0.4883 degrees. Its diameter would be 0.2568 Earth radii. T**he actual value is 0.2259 Earth radii.**

The calculation of the distance to the Moon worked out quite well so lets try to find the distance to the Sun. The Sun orbits the Earth in 365.25 days. This is 13.364 times the period of the Moon. Its orbit would be 5.6315 times as large or 339.411 Earth radii.

The Sun subtends an angle of 32 arc minutes or 0.5333 degrees. Its diameter would be 1.5797 earth diameters and its weight and volume would be 3.9418 times that of the Earth. If true, the Earth would circle the larger Sun with an orbital period of only 184 days.

We could try to correct the period to one year by increasing the orbital distance by 158.740 percent to 3,429,068 km. But then the Sun’s size would increase another 1.5874 times and the orbital period would remain 184 days..

Continuing to make these corrections has no effect on the period but pushes the sun farther and further away. The only conclusion we can reasonably make is that the Sun has a much lower density than the earth. This makes sense, the Sun is very hot and heat makes things expand.

It is obvious that we cannot determine the distance to the Sun by this method, nevertheless we have found that the density of the Sun is about 1/4 that of Earth and that it is quite far away. *Today we know the Earth orbits the Sun at a distance of 149,000,000 km and that the density of the Sun is 0.2554 that of the earth.*

This accuracy of these calculations based on observations of a simple pendulum is amazing. I hope you have enjoyed reading about my discovery of this property of the simple pendulum.

References

1. http://home.online.no/~sigurdhu/Grid_1deg.htm Earth according to WGS84

The Circumference of the Earth through the Poles

2. http://en.wikipedia.org/wiki/Gravity_of_Earth (WGS84 defines Earth)

The Gravity of the earth

3. http://en.wikipedia.org/wiki/ The_moon

4. PHYSICS Principles and Applications Margenau,Watson,& Montgomery

Mcgraw-Hill 1949 Physical Constants Page 731 The Simple Pendulum

Page 128, The Physical Pendulum page 179, The Conical Pendulum page 102

5. Advanced Mathematics in Physics and Engineering Arthur Bronwell

Mcgraw-Hill 1953 The Pendulum Section 7.10 page 137

## 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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