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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 socalled 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 zaxis 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 = ——2u)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' (915)
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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 Spring '08
 LAM
 mechanics

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