Page:Elementary Text-book of Physics (Anthony, 1897).djvu/72

58 major axis $$a$$ and the semi-minor axis $$b$$ of the hyperbola and ellipse are given by $$2a = \frac{m}{C},\, b = \frac{v^2_{0}p^2_{0}}{2C} = \frac{av^2_{0}p^2_{0}}{m}\cdot$$ Hence $$C = \frac{m}{2a},$$ and using this value of $$C$$ in the original equation, we get $$\frac{v^2}{2} = \frac{m}{r} \plusmn \frac{m}{2a},$$ the upper and lower signs holding for the hyperbola and ellipse respectively.

In case the particle is moving in an ellipse, its periodic time $$T$$, or the time in which it traverses the ellipse, may be found in terms of the elements of the ellipse and the constant $$m$$. The area of the ellipse is $$\pi ab,$$ and since the areas swept out in equal times by the radius vector drawn to the particle are equal, the rate at which the area is swept out is given by $$\frac{\pi ab}{T}\cdot$$ But $$\frac{v^2_{0}p^2_{0}}{2}$$ also represents this rate, so that $$\frac{\pi ab}{T} = \frac{v^2_{0}p^2_{0}}{2}\cdot$$ Substituting in this equation the value of $$b = v_{0}p_{0} \left(\frac{a}{m}\right)^{\tfrac{1}{2}},$$ we got (38) $$T = \frac{2 \pi a^{\tfrac{3}{2}}}{m^{\tfrac{1}{2}}},$$ or, $$m = \frac{4 \pi^2 a^3}{T^2}\cdot$$ If, therefore, different particles revolve in ellipses about a common centre of force in such a way that the squares of their periodic times are in the same ratio to the cubes of their semi-major axes, the constant $$m$$ is the same for all of them.

51. The Problem of Two Bodies.— The problem of two bodies may be reduced to the problem of the action of a central force. For, suppose two particles to attract each other with a force given by $$\frac{\mu m}{r^2},$$ where $$\mu$$ and $$m$$ are their masses and $$r$$ the distance between them. The acceleration of the particle $$m$$, relative to the centre of mass, which will remain fixed in position, is given by $$ma = \frac{\mu m}{r^2},$$ or by $$a = \frac{\mu}{r^2}\cdot$$ The acceleration of the mass $$\mu$$ relative to the centre of mass is similarly $$\frac{m}{r^2}\cdot$$ If now an acceleration equal to $$\frac{\mu}{r^2}$$ and opposite to it in direction be impressed on both particles, the particle $$m$$ will remain fixed, and the particle $$\mu$$ will move relatively to it