CMChap9 - Chapter 9 The Shape of the Earth We will use the...

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Unformatted text preview: Chapter 9 The Shape of the Earth We will use the potential theory as developed in the last chapter to estimate the values of two parameters that characterize the shape and mass distribution of the earth. To a good approximation the earth has the shape of an oblate spheroid flattened slightly at the poles (see Fig. 8.5). The departure from perfect spheric- ity is described by the so-called oblateness: CL—C 5 , (9.1) CI, where a is the equatorial radius and c is the distance from the center of the earth to either pole. For the earth, a — 0 ~ 21.5 Km. Using a ~ 6.37 x 103 Km, one obtains 5 ~ 1/300. Assume for the moment that the mass distribution of the earth is uniform (which is not the case), its quadrupole moment Q is given by (8.34). It can be rewritten in terms of the oblateness e as follows: 2 5 _ 2 __ M(a'~’ c2): 2M(a+c)(a C)a~ M26“ c)“: :IVICLEE. 5 a 5 a (9.2) But the mass density p of the earth is not constant; in fact it increases towards the center. To account for this fact, we introduce a parameter A in place of 5 in the expression for Q and write Q: 4 Q : —gMa2/\, A < 5. (9.3) On recalling (8.16) and (8.33), we then have the following expression for the gravitational potential due to the earth up to the quadrupole term [with respect to a CM (center—of—mass) frame Whose z-axis points from the center of the earth to the north pole]: _% GIVIaQA 2 -1 . .4 r + 5T3 (Secs 6’ ) (9) (15(7‘) = 39 40 CHAPTER 9. THE SHAPE OF THE EARTH We will work with a body—fixed CM frame, that is, one that is rotating with the earth. With respect to such a frame, every part of the earth is in equilibrium under the combined gravitational and centrifugal forces. In particular the surface of the earth must be an equipotential surface under these combined forces. The centrifugal potential (be can be written as: 1 1 (:50 = ——2-u)2(3;2 + 3/2) = —§w2r2 sing 6’ , (9.5) where w is the angular speed of rotation of the earth about its axis. Using (9.4) for the gravitational potential (25 and (9.5) for the centrifugal potential (be, and equating the values of the total potential (1) + (256 at the north pole and the equator, we have GJl/f 2GJl/[a2/\ GZVI GMA 1 2 2 c + 503 —_ (1 5a 2w a I (9.6) Use (9.1) and the fact that for the earth, the oblateness 6 << 1, to show that 1 1 — ~ — 1 . 9.7 C a< +5) < > Using the above equation, (9.6) reduces to GJVIE — + a 5a 3m” = gm)? . (9.8) Define the quantity : GM 90 — (99) a2 which is an approximate value for the acceleration due to gravity anywhere on the earth’s surface. Then (9.8) can be written as 2 can 90 25 — 2A 2 (9.10) We will obtain an independent equation relating e and /\ in order to solve for these parameters. The gravitational field g('r) can be obtained from the gravitational potential ¢(r) by g = —V¢>. Thus, on using (9.4) and (3.37) (applied to spherical coordinates), we have _ a¢ _ G'M 30Ma2A 2 gr — 8T — 72 + 5T4 (3 cos 6 — 1) , (9.11) 1 8gb _ 6GJl/Ia2/\ cos 0 sin (9 99: T86 — 5T4 (9.12) 41 Note that 99 = 0 at both poles (6 = 0, 7r) and the equator (6 = 7r/2). We then have G114 6GJl/[a2/\ GM GGJl/IA lgpolel: 2 “TN 2 (1+26)— 2 c 50 a 5a (9 13) ~G‘M(1+25—g)— (ins—9) ' — a2 5 _ go 5 a GM' 3G.Mx\ 3A 19qu = a2 + 5&2 ~ go(1 + 3), (9.14) where 1 91,018] and I gag] are the magnitudes of the gravitational field 9 at the poles and the equator, respectively. From (9.5), the centrifugal force field 98 at the poles vanishes, while at the equator it is in the outward radial direction and has a magnitude given by 1.9%. Exercise 9.2 Verify the above statement by calculating the gradient of (be. We then obtain the following expression for the difference in acceleration (due to both gravitational and centrifugal forces) at a pole and the equator: A9 E lgpole l — lgeq l + w2a' (9-15) Using (9.10), (9.13) and (9.14) in the above equation we have E = 45 — 3A. (9.16) 90 One can then use the approximate empirical values: Ag : 5.2 cm/sec2 , wga = 3.4 cm/sec2 , go 2 980 cm/sec2 , (9.17) to solve the simultaneous equations (9.10) and (9.16) and obtain the following estimates for a and A: a = 3.37 x 10—3 , /\ = 2.72 x 10—3 . (9.18) Confirm the above numerical results by solving (9.10) and (9.16). 42 CHAPTER 9. THE SHAPE OF THE EARTH ...
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CMChap9 - Chapter 9 The Shape of the Earth We will use the...

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