If this force were to be transmitted by a forest of weightless pillars each a square foot in cross-section, with a tension of 30 tons to the square inch throughout, there would have to be 5 million million of them. _Arithmetical Calculation of the Pull of the Sun on the Earth._ The mass of the earth is 6 × 10²¹ tons. The intensity of solar gravity at the sun's surface is 25 times ordinary terrestrial gravity. At the earth's distance, which is nearly 200 solar radii, solar gravity will be reduced in the ratio of 1:200 squared. Hence the force exerted by the sun on the earth is (25 × 6 × 10²¹)/(200)² tons weight. That is to say, it is approximately equal to the weight of 37 × 10¹⁷ ordinary tons upon the earth's surface. Now steel may readily be found which can stand a load of 37 tons to every square inch of cross-section. The cross-section of a bar of such steel, competent to transmit the sun's pull to the earth, would therefore have to be 10¹⁷ square inches, or say 700 × 10¹² square feet. And this is equivalent to a million million round rods or pillars each 30 feet in diameter. Hence the statement in the text (page 26) is well within the mark. _The Pull of the Earth on the Sun._ The pull of the earth on the sun is, of course, equal and opposite to the pull of the sun on the earth, which has just been calculated; but it furnishes another mode of arriving at the result, and may be regarded as involving simpler data--i.e. data more generally known. All we need say is the following:-- The mass of the Sun is 316,000 times that of the Earth. The mean distance of the sun is, say, 23,000 earth radii. Hence the weight or pull of the sun by the earth is 316000/(23000)² × 6 × 10²¹ tons weight. In other words, it is approximately equal to the ordinary commercial weight of 36 × 10¹⁷ tons, as already calculated. _The Centripetal Force acting on the Earth._ Yet another method of calculating the sun's pull is to express it in terms of the centrifugal force of the earth; namely, its mass, multiplied by the square of its angular velocity, multiplied by the radius of its orbit;--that is to say, F = M (2φ/T)² r where T is the length of a year. The process of evaluating this is instructive, owing to the manipulation of units which it involves:-- F = 6 x 10²¹ tons x (4φ² x 92 x 10⁶ miles)/(365¼ days)² which of course is a mass multiplied by an acceleration. The acceleration is-- (40 x 92 x 10⁶)/133300 x (24)² miles per hour per hour = (3680 x 10⁶ x 5280)/133300 x 576 x (3600)² feet per sec. per sec. = (115 x 5280)/133300 x 576 x 12·96 feet per sec. per sec. = g/1640 Hence the Force of attraction is that which, applied to the earth's mass, produces in it an acceleration equal to the 1/1640th part of what ordinary terrestrial gravity can produce in falling bodies; or F = 6 × 10²¹ tons × g/1640 = 6/1640 x 10²¹ tons weight; which is the ordinary weight of 37 × 10¹⁷ tons, as before.