Page:Elementary Text-book of Physics (Anthony, 1897).djvu/70

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56

ELEMENTARY PHYSICS.

[§ 50

is in general an ellipse, with its centre at the centre of force. For, the force acting along the radius vector may be resolved into two components along the two axes, which will be proportional to the displacement of the particle from the axes. Each of these components will cause proportional accelerations along the axes, and these accelerations will be those of a point having simple harmonic motions parallel to the axes. Since the constants which enter into the measure of the components of the force, and therefore into the measure of the accelerations produced by them, are the same for each component, the periods of these component simple harmonic motions will be the same. The motion of the particle will therefore be the resultant of two simple harmonic motions of equal periods at right angles to each other, and its path is therefore (§ 21) an ellipse, with its centre at the centre of force.

50. Central Force Proportional to the Inverse Square of the Radius Vector.— If the central force vary inversely with the square of the distance of the particle from the centre, the path described by the particle is in general a conic section, with the centre of force at one of its foci. To prove this we will use a theorem that will be demonstrated in § 55. It will there be shown that if a particle of mass $\mu$ be moved from an infinite distance under the action of a central force equal to ${\frac {m}{r^{2}}}$ , where $m$ is a constant and $r$ the distance between the particle and the centre, the potential energy which it will lose by moving to a point distant $r$ from the centre is given by ${\frac {\mu m}{r}}\cdot$ Thus, if $\mu P$ represent the potential energy of the particle at an infinite distance, $\mu P-{\frac {\mu m}{r}}$ will represent its potential energy at the distance $r$ . The sum of its potential and kinetic energies is constant: and hence $\mu P-{\frac {\mu m}{r}}+{\frac {\mu v^{2}}{2}}=\mu A$ , a constant, or

{\frac {v^{2}}{2}}-{\frac {m}{r}}=A-P=C,

(34)

a constant. $C$ may be greater or less than zero, or equal to zero.