Results of the Contraction of a Solid Globe. 11 



The term to be integrated on the left-hand side becomes 



2r( € "■ 





between the limits. 



Let us take a for the unit of length. Then e~ r2 may be 

 neglected, and the equation may be written 



_l_ {(r _^ + l_| (r _- l .)» fl .} =B £ r e-^. . . (A) 



The value of x given by this equation will be the depth of 

 the level of no strain below the surface in terms of 2 y//et, or 

 a, as the unit of length. In order that the left-hand side may 

 be positive, we must have 



^r-x)\v<(r- X ? + l, 

 ° r $(r- *><! + • 



The last term is very small. We shall therefore get a close 

 inferior limit for x if 



which is satisfied by 



r . V^~6 



It is evident that the lower sign must be used. 



Putting a at 402832 feet, which corresponds to V=7000°F. 

 and the present epoch, we find log r = 1*7150379 and 

 r = 51'8845a. This gives as a limits = - 0287, and the large 

 factor in the term in (r— x) 3 x warns us that x must not be 

 otherwise than very small. 



The values of the definite integral *-*dx are tabulated 



for values at the interval of # 01*. 



And 



J e~* 2 dx= i ' e- x2 dx— j V* 2 dx 



" i;- 



= 0-8862269254- e~ x2 dx. 



We must therefore try values of x proceeding by intervals 

 of 0-01. 



Let x = - 02, and call the left-hand side of the equation (A) 

 p, and the right-hand side q. 



* Oppolzer, Lehrhuch z'tir Bahnbestimmung, vol. ii. tab. x. 



