A Computation of the Distance of the Sun from the Earth. By S. Horsley, LL. B. Rector of Saint Mary, Newington, in Surry, F. R. S.

XVIII. A Computation of the Distance of the Sun from the Earth. By S. Horsley, LL.B. Rector of Saint Mary, Newington, in Surry, F.R.S. Received February 12, 1767. Read March 26, 1767. I offer the following computation, rather as a verification than an amendment of Dr. Stewart's. The method, in which I have pursued, is different from what is used by that great and able geometrician, in his treatise on the distance of the sun, but founded entirely on the theorems established in that and the preceding tracts of the same author. Let TA be a given line. Take Am, so that TA may be to Am, as the moon's accelerating attraction to the earth, to the sun's mean disturbance of that attraction. Take AG quintuple of Am. Take AP, such that twice Am may be to AP, as TG to TA. Now it is proved in the twenty-fifth proposition of Dr. Stewart's fourth tract, that the cube of TA is to the cube of TP, in the duplicate proportion of the periodic month to the anomalistic month. Therefore the proportion of $TA^3$ to $TP^3$, and consequently that of TA to TP, is given; and by division, that of TA to AP is given. Therefore TA being given, AP is given. Now $TG : TA = 2Am : AP$. That is, $Aa^2$ TA - 5 Am : TA = 2 Am : AP. Therefore TA × 2 Am = \frac{TA \times AP}{TA} - 5 Am \times AP. That is \(2 Am + \frac{5 Am \times AP}{TA}\), or \(\frac{2 TA + 5 AP \times Am}{TA} = AP\). That is, \[Am = \frac{TA \times AP}{2 TA + 5 AP}.\] 2. Let TA represent the moon's mean distance from the earth. Take TV, such that TA may be to TV, in the duplicate proportion of the periodic month to the sidereal year. Take TG, such that VT may be to TG in the proportion of the moon's accelerating attraction to the earth, to the sun's mean disturbance of that attraction. Take TE, such that TE may be to TA, as TA to TG. Take EO, such that the rectangle EOA may be equal to \(3 TE \times TA\). Upon the centre T, with the interval TA, describe a circle. Draw OX perpendicular to AB, meeting the circle in X. Take AD = AT. The proportion of TA to TV being given, and TA being given, TV is given. But the proportion of TV to TG is given. Therefore TG is given, and the proportion of TG to TA is given. TG : TA = TA : TE. Therefore the proportion of TA to TE is given. Therefore TE is given. Therefore \(3 TE \times TA\) is given. Therefore EO × OA is given. And EA (= TE - TA) is given. Therefore AO is given. But AB (= 2 AT) is given. Therefore OB is given. Therefore AO × OB is given. AO × OB = OX² (by the circle). Therefore OX², and consequently OX is given. But DB (= 3 AT) is given. Therefore the proportion of DB to OX is given. And the proportion of DB to O$x$, is that of the mean distance of the sun, to the mean distance of the moon. This is in brief the method of my computation. The computation is as follows: The periodic month is to the anomalistic month, as 57600 to 58091. Therefore (in Fig. 1.) $TA^3 : TP^3 = 57600^2 : 58091^2 = 3317760000 : 3374564281$ $$\frac{3374564281}{3317760000} = 1,0171212748963155c864197530c864197530, \text{ &c.}$$ Hence, by extracting the cube root, I find $TA : TP = 1 : 1,005674827053$. Therefore put $TA = 1$. Then $TP = 1,005674827053$; and $AP = 0,005674827053$. Hence $$\frac{TA \times AP}{2TA + 5AP} = 0,002797722 = Am.$$ (See Fig. 2). The square of the periodic month is to the square of the sidereal year, as 1 to 178,725. Therefore $TA : TV = 1 : 178,725$. But $TV : TG = 1 : 0,002797722$. Therefore $TA : TG = 1 : 178,725 \times 0,002797722 = 1 : 0,50002286445$. $TA : TG = TE : TA$. Therefore $TE : TA = 1 : 0,50002286445$. Therefore put $TE = 1$. Then $TA = 0,50002286445$ And $EA = 0,49997713555$ And $3TE \times TA = 1,50006859335 = EO.A$. Hence $AO (\sqrt{TE \times TA + \frac{EA^2}{4}}) = 1,00003658292$. But $AB = 2TA = 1,00004572890$ Therefore $OB = 0,00000914598$ Therefore $AO \times OB = 0,0000091463145866546616$ Therefore $\sqrt{AO \times OB} = 0,0032024287 = O$x But $DB = 3TA = 1,500068593$ Hence $DB : O$x = 496,0073 : 1. These computations have been made with no small rigor. I was sensible that, to obtain an accurate conclusion, it was necessary to determine AO with extreme precision; and for that purpose I submitted to the laborious task of computing the foregoing numbers to the 11th or 12th decimal place, by the common operations of arithmetic. In the result I differ from Dr. Stewart, by much less than \(\frac{1}{3000}\)th part of the whole distance, that is, by less than 5 semi-diameters of the earth; a very contemptible difference in so nice a calculation. That great mathematician indeed seems to have flattered himself, that he had determined the sun's distance within \(\frac{1}{3000}\) of the truth. I suspect that when he affirmed this, he did not consider that to attain so great an accuracy in the conclusion, the line \(E_7\) in his method (vide Stewart on the sun's distance, Fig. 10.), or AO in mine, should be determined strictly to the 11th or 12th decimal place. And after the utmost rigor of computation, I am afraid any pretensions to such extreme nicety in the result will be but ill-founded. For it is very likely that these computations represent the sun's distance less than it really is: because the whole progression of the moon's apogee (which is the basis of the calculation) is ascribed to the sun's disturbance of the moon's gravitation to the earth. Whereas part of it must be due to the disturbances of the planets. What part is due to them we cannot tell, and therefore cannot allow for it. But in giving the whole to the sun we certainly overrate his disturbing force, and by that means must obtain too small a distance. It It is most likely indeed, that the motion of the apogee produced by the disturbing forces of the planets bears but a very small, perhaps insensible, proportion to the whole. But those who are masters of Dr. Stewart's theorems will easily perceive that an insensible error in the proportion of the moon's gravity, and the sun's disturbance, may produce a very sensible error in the proportion of the mean distances. And therefore the real distance is probably greater by two or three semidiameters of the earth than these computations make it. This, however, is much too nice a point for the approaching transit, or, perhaps, for any method of observation, to determine. The highest expectations of astronomers will be answered, if they can come within 50 or 60 semidiameters of the earth. It is to be hoped, that every civilized nation of the universe will give due attention to that interesting phenomenon, which we, the present possessors of these sublunary regions, shall behold no more; and that proper persons will be sent in due time, and duly equipped, to the most advantageous stations. If the decisions of observation in so nice a point should be found to agree with the previous conclusions of theory, the disciples of Newton will have no small reason to exult in a new attestation of nature, to the truth of their great master's doctrine. But it is much to be wished, that they, who shall be deputed to prosecute this curious search, in distant and sequestered parts, may divest themselves of all prejudice; that they may have nothing at heart, but, that which the world will expect from them, the advancement of real science; that they may be diligent in their observations, and faithful in their reports; and not sacrifice the repose of their own minds, or the interests of philosophy, to the credit of an admired hypothesis, the memory of a friend, or the jealousies of rival nations. If the moon's mean distance from the earth be 60 $\frac{1}{3}$ semidiameters of the earth, the sun's mean distance is 30008,4416 semidiameters of the earth. The sun's semidiameter is to the semidiameter of the earth, as 139,876 to 1. The globe of the sun is to the globe of the earth, as 2736718,8 to 1; and the sun's horizontal parallax is $6''52''',415$. February 8. To satisfy myself more fully of the accuracy of my work, I this day re-computed the whole, from the determination of EA, in Dr. Stewart's approximating method. I found the proportion of DB to $v_s$ (see Dr. Stewart on the distance of the sun, Fig. 10.), that of 496,00579 to 1; and the proportion of DB to $tv$, that of 496,00805 to 1. The mean of these two gives the proportion of DB to $ox$, nearly that of 496,0069 to 1. Which differs from the result of my former computation by less than $\frac{1}{1248000}$ of the whole; and the method of the former computation is undoubtedly the most accurate. Supplement Supplement to the foregoing Paper. Read June 19, 1767. In deducing the distance of the sun in semidiameters of the earth, and his horizontal parallax, from the proportion above concluded between the sun's mean distance and that of the moon; I have supposed the latter to be $60\frac{1}{4}$ semidiameters of the earth, as it is reckoned by Sir Isaac Newton. According to the hypothesis which seems to be now generally received, that the density of the moon is very nearly equal to that of the earth, (the French reckon it rather less), the moon's mean distance should be little more than $60,23207$, that is, not quite $60\frac{1}{4}$ semidiameters of the earth. But from some computations that I have formed with great care; I have reason to think, that Sir Isaac Newton's determination is much nearer to the truth; that the density of the moon is actually greater than that of the earth, in the proportion of 6 to 5 nearly; and that the moon's mean distance amounts to $60,441$ semidiameters of the earth; which differs from the distance assigned by Sir Isaac Newton, by less than $\frac{1}{1000}$ of the whole. S. Horfley.