integraldisplay r r dr parenrightBigg d \u03a9 1 2 G \u03c1 r 2 r 2 d \u03a9 E3 The net

# Integraldisplay r r dr parenrightbigg d ω 1 2 g ρ r

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integraldisplay r ′′ 0 r dr parenrightBigg d Ω = 1 2 G ρ ( r 2 + r ′′ 2 ) d Ω . (E.3) The net potential at P is obtained by integrating over all solid angle, and dividing the result by two to adjust for double counting. This yields Ψ = 1 4 G ρ contintegraldisplay ( r 2 + r ′′ 2 ) d Ω . (E.4) From Figure E.1, the position vector of point Q , relative to the origin, O , is x = x + r n , (E.5) where x = ( x 1 , x 2 , x 3 ) is the position vector of point P , and n a unit vector pointing from P to Q . Likewise, the position vector of point R is x ′′ = x + r ′′ n . (E.6) However, Q and R both lie on the body’s outer boundary. It follows, from (E.1), that r and r ′′ are the two roots of summationdisplay i = 1 , 3 parenleftBigg x i + r n i a i parenrightBigg 2 = 1 , (E.7) which reduces to the quadratic A r 2 + B r + C = 0 , (E.8) where A = summationdisplay i = 1 , 3 n 2 i a 2 i , (E.9) B = 2 summationdisplay i = 1 , 3 x i n i a 2 i , (E.10) C = summationdisplay i = 1 , 3 x 2 i a 2 i 1 . (E.11)
280 FLUID MECHANICS P O x r Q R d Ω Figure E.1: Calculation of ellipsoidal gravitational potential. Now, according to standard polynomial equation theory, r + r ′′ = B / A , and r r ′′ = C / A . Thus, r 2 + r ′′ 2 = ( r + r ′′ ) 2 2 r r ′′ = B 2 A 2 2 C A , (E.12) and (E.4) becomes Ψ = 1 2 G ρ contintegraldisplay 2 parenleftBig i = 1 , 3 x i n i / a 2 i parenrightBig 2 parenleftBig i = 1 , 3 n 2 i / a 2 i parenrightBig 2 + 1 i = 1 , 3 x 2 i / a 2 i i = 1 , 3 n 2 i / a 2 i d Ω . (E.13) The above expression can also be written Ψ = 1 2 G ρ contintegraldisplay 2 i , j = 1 , 3 x i x j n i n j / ( a 2 i a 2 j ) parenleftBig i = 1 , 3 n 2 i / a 2 i parenrightBig 2 + 1 i = 1 , 3 x 2 i / a 2 i i = 1 , 3 n 2 i / a 2 i d Ω . (E.14) However, the cross terms ( i.e. , i nequal j ) integrate to zero by symmetry, and we are left with Ψ = 1 2 G ρ contintegraldisplay 2 i = 1 , 3 x 2 i n 2 i / a 4 i parenleftBig i = 1 , 3 n 2 i / a 2 i parenrightBig 2 + 1 i = 1 , 3 x 2 i / a 2 i i = 1 , 3 n 2 i / a 2 i d Ω . (E.15) Let J = contintegraldisplay d Ω i = 1 , 3 n 2 i / a 2 i . (E.16) It follows that 1 a i J a i = contintegraldisplay 2 n 2 i / a 4 i parenleftBig i = 1 , 3 n 2 i / a 2 i parenrightBig 2 d Ω . (E.17) Thus, (E.15) can be written Ψ = 1 2 G ρ J summationdisplay i = 1 , 3 A i x 2 i , (E.18) where A i = J a 2 i 1 a i J a i . (E.19)
Ellipsoidal Potential Theory 281 At this stage, it is convenient to adopt the spherical angular coordinates, θ and φ (see Section C.4), in terms of which n = (sin θ cos φ, sin θ sin φ, cos θ ) , (E.20) and d Ω = sin θ d θ d φ . We find, from (E.16), that J = 8 integraldisplay π/ 2 0 sin θ d θ integraldisplay π/ 2 0 sin 2 θ cos 2 φ a 2 1 + sin 2 θ sin 2 φ a 2 2 + cos 2 θ a 2 3 1 d φ. (E.21) Let t = tan φ . It follows that J = 8 integraldisplay π/ 2 0

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