By E Buforn; Carmen Pro; Agustin Udias Vallina

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**Extra info for Solved problems in geophysics**

**Example text**

20 a3 M 4 ¼ M1 ¼ pð5r À rÞ 3 2 2 and for the inverse of the distance 1/q we use the approximation a 2 1 À Á 1 1 a ¼ 1 þ cos y þ 3cos2 y À 1 q r 2r 2r 2 ! Then, the total gravitational potential is V ¼ Á GM 3 a 1 a2 À þ cos y þ 3cos2 y À 1 r 2 4r 16 r ! The total potential U is the sum of the gravitational potential V plus the potential of rotation F, where 1 F ¼ r2 o2 sin2 y 2 and using the coefﬁcient m = o2a3/GM = 1/8, we have U¼ ! Á r 3 m 2 GM 3 a 1 a2 À þ cos y þ 3cos2 y À 1 þ sin y r 2 4r 16 r a 2 At the North Pole, y = 0 and r ¼ a, and the value of the potential is !

R 3 m 2 GM 3 a 1 a2 À þ cos y þ 3cos2 y À 1 þ sin y r 2 4r 16 r a 2 At the North Pole, y = 0 and r ¼ a, and the value of the potential is ! GM 3 1 1 15GM Up ¼ þ þ ¼ a 2 4 8 8a Earth’s gravity ﬁeld and potential 37 The form of the equipotential surface is found by putting U ¼ Up: Á r 3 m 2 15GM GM 3 a 1 a2 À ¼ þ cos y þ 3cos2 y À 1 þ sin y 8a r 2 4r 16 r a 2 ! Putting r ¼ a inside the square brackets and solving for r we ﬁnd ! 4 1 1 r ¼ a 1 þ cos y þ cos2 y 5 6 12 (b) The deviation of the vertical with respect to the radial direction is given by the angle i: tan i ¼ gy gr To ﬁnd this value we have to calculate the two components of gravity gr ¼ !

28 (a) The gravity anomaly produced by an inﬁnite horizontal cylinder buried at depth d, with centre at x ¼ 0 (Fig.