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The Gravitational Pendulum.

As a further example, consider a simple pendulum at a point on the earth's surface 90° from the pole of its motion, so that the string is perpendicular to the direction of motion. When it vibrates in a plane at right angles to the motion the path of the bob is a circular arc and the period is $\scriptstyle {T=2\pi {\sqrt {\frac {m_{0}L}{G}}}}$ where $\scriptstyle {G}$ is the force with which the earth attracts the bob. When it vibrates in the plane of motion its path is the arc of an ellipse whose axes are $\scriptstyle {L}$ and $\scriptstyle {L{\sqrt {1-\beta ^{2}}};}$ for the same vertical height (that is for the same potential energy), the infinitesimal arc described will be in this case less than in the other in the ratio of $\scriptstyle {\sqrt {1-\beta ^{2}}}$ to unity. So that the period is $\scriptstyle {T=2\pi {\sqrt {\frac {m_{1}(1-\beta ^{2})L}{G}}}}$ giving the same ratio of masses as before.

Comparing, say, the first of these with the period which the pendulum would have if the earth were at rest, we have $\scriptstyle {{\sqrt {1-\beta ^{2}}}{\sqrt {\frac {m_{2}L}{G}}}={\sqrt {\frac {m_{0}L}{G_{0}}}}}$ and since $\scriptstyle {m_{2}=m_{0}{\frac {1}{\sqrt {1-\beta ^{2}}}}}$, $\scriptstyle {G={\sqrt {1-\beta ^{2}}}G_{0}}$ Thus the gravitational force between two bodies moving at right angles to the line joining them is the same function of the velocity as the electric force between two moving charges in a corresponding position.^[1]

If we imagine the pendulum suspended at the place on the earth which is foremost or rearmost in its motion, the length of the string will be $\scriptstyle {L{\sqrt {1-\beta ^{2}}}}$ and the period $\scriptstyle {T=2\pi {\sqrt {\frac {m_{0}L{\sqrt {1-\beta ^{2}}}}{G^{\prime }}}}}$ whence $\scriptstyle {G_{1}=(1-\beta ^{2})G_{0}}$ which again corresponds to the electrical case when the line joining the charges is parallel to the motion.^[2]

1. See p. 503.
2. See below p. 503.

Am. Jour. Sci.—Fourth Series, Vol. XXVI, No. 155.—November, 1908.

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