To find the mass m which, at the distance of 1 centim. from an equal mass, would attract it with a force of 1 dyne, we have 1 = Cm2; 86. To find the acceleration a produced at the distance of 7 centims. by the attraction of a mass of m grammes, where C has the value 6:48 × 10-8 as above. To find the dimensions of C we have C dimensions of a are LT-2. The dimensions of C are therefore L2M-LT-2; that is, L3M-1T-2. m 87. The equation a = C shows that when a = 1 and 1 C 12 1 = 1, m must equal ; that is to say, the mass which produces unit acceleration at the distance of 1 centimetre is 1.543 × 107 grammes. If this were taken as the unit of mass, the centimetre and second being retained as the units of length and time, the acceleration produced by the attraction of any mass at any distance would be simply the quotient of the mass by the square of the distance. It is thus theoretically possible to base a general system of units upon two fundamental units alone; one of the three fundamental units which we have hitherto employed being eliminated by means of the equation mass = acceleration × (distance)2, which gives for the dimensions of M the expression L3T-2. Such a system would be eminently convenient in astronomy, but could not be applied with accuracy to ordinary terrestrial purposes, because we can only roughly compare the earth's mass with the masses which we weigh in our balances. 88. The mass of the earth on this system is the product of the acceleration due to gravity at the earth's surface, and the square of the earth's radius. This product is 981 × (6.37 × 108)2 = 3·98 × 102o, and is independent of determinations of the earth's density. The new unit of force will be the force which, acting upon the new unit of mass, produces unit acceleration. It will therefore be equal to 1·543 × 107 dynes; and its dimensions will be mass acceleration = (acceleration)2 × (distance)2 89. If we adopt a new unit of length equal to l centims., and a new unit of time equal to t seconds, while we define the unit mass as that which produces unit acceleration at unit distance, the unit mass will be 73 t−2 × 1·543 × 107 grammes. If we make the wave-length of the line F in vacuo, 4.86 x 10-5, say, and t the period of vibration of the same ray, so that t is the velocity of light in vacuo, say, and the unit mass will be the product of this quantity into 1.543 × 107 grammes. This product is 6·75 × 1023 grammes. The mass of the earth in terms of this unit is 3.98 x 1020 (4.374 x 1016) 9100, = and is independent of determinations of the earth's density. 76 CHAPTER VII. VELOCITY OF SOUND. 90. THE propagation of sound through any medium is due to the elasticity of the medium; and the general formula for the velocity of propagation s is where D denotes the density of the medium, and E the coefficient of elasticity. 91. For air, or any gas, we are to understand by E the quotient that is to say, if P, P+p be the initial and final pressures, and V, V v the initial and final volumes, p and v being small in comparison with P and V, we have If the compression took place at constant temperature, we should have But in the propagation of sound, the compression is effected so rapidly that there is not time for any sensible part of the heat of compression to escape, and we have where y = 1.41 for dry air, oxygen, nitrogen, or hydrogen. P The value of D for dry air at t° Cent. (see p. 44) is (1 + .00366ť) × 7·838 × 108. Hence the velocity of sound through dry air is = 33240 √1+00366t; or approximately, for atmospheric temperatures, S = 33240+ 60t. 92. The following are the principal experimental determinations of the velocity of sound in air, reduced to 0° Moll, van Beek, and Kuytenbrouwer (1822), . 332.2 93. In the case of any liquid, E denotes the elasticity of volume.* * See foot note in following page. 330.7 330.7 |