hAB from which we get gB= 979.877 84 Gal
The Helmert height of B is found as in station A:
HB¼ 678:077 m
Station C
From the known values of gravity and Helmert height we find the geopotential number (Equation 15.1)
CC¼ gHCþ 0:0424HC2¼ 661:574 gpu
To calculate the difference in height of C with respect to B we begin with the expression CC¼ CBþ gBþ gC
2
hCB from which
hCB¼2 Cð C CBÞ gBþ gC
¼ 2:937 m The dynamic height is found directly from the geopotential number:
HDC¼CC
g45 ¼ 674:642 m The complete table is:
Earth’s gravity field and potential
16. Suppose an Earth is formed by a sphere of radius a and density r, and within it there are two spheres of radius a/2 with centres located on the axis of rotation. The density of that of the northern hemisphere is 5r and of that of the southern hemi-sphere is r /5. The value of the rotation is such that m¼ 0.1. Determine:
(a) The potential U in the r3approximation.
(b) The values of grand gufor a point on the equator in the r2approximation.
Station
Gravity (Gal)
Height increment (m)
Geopotential number (gpu)
Dynamic height (m)
Helmert height (m)
A 979.88696 – 664.982 678.118 678.611
B 979.87784 0.541 664.452 677.577 678.077
C 979.88665 2.937 661.574 674.642 657.134
(c) The error made in (b) with respect to the exact solution.
(d) The deviation of the vertical from the radial at the equator.
(a) The gravitational potential is the sum of the potentials of the three spheres V ¼ V1þ V2þ V3¼GM
r þGM1 q þGM2
q0 ð16:1Þ
where r is the distance from a point P to the centre of the sphere of radius a and mass M, where M is given by
M ¼4 3pra3
q and q0are the distances to the centres of the two spheres in its interior in the northern and southern hemispheres which have differential masses M1and M2, respectively (Fig. 16).
The differential masses are those corresponding to the difference in density in each case with respect to the large sphere:
M1 ¼4
3pð5r rÞa3 8 ¼M
2 differential mass of the sphere in the northern hemisphere
M2 ¼4 differential mass of the sphere in the southern hemisphere
The distance q can be calculated using the cosine law q¼
Considering this expression, 1/q corresponds to one of the generating functions of the Legendre polynomials. Then 1/q, in the first-order approximation, is given by
1
If we substitute in Equation (16.1), the potentials for each sphere are given by
Then, the total gravity potential is the sum of the three gravitational potentials plus the potential of the centrifugal force due to the rotation:
U ¼ GM 1þ1
(b) Using this first-order approximation of the potential, the radial and tangential components of gravity at the equator, r = a and y = 90, putting m¼ 0.1, are
(c) To calculate exactly the value of gr at the equator we have to calculate the exact contribution of each of the three spheres plus the centrifugal force (m¼ 0.1):
gr¼ g1rþ g2r þ g3r mGM
sin a¼a=2
The error made in the first-order approximation with respect to the exact solution is
gr ¼ 1:19 þ 1:3ð ÞGM
(d) The deviation of the vertical with respect to the radial direction is given by the angle i which is determined from the gravity components grand gy. At the equator this angle is:
• Using the first order approximation tan i¼gy
gr
¼0:3
1:3) i ¼ 13:0
• Using the exact values
tan i¼0:16
1:19) i ¼ 7:6
17. A spherical planet is formed by a sphere of radius a and density r, and inside it a sphere of radius a/2 and density 5r centred at the midpoint of the radius of the northern hemisphere. There is no rotation.
(a) Determine J0, J1, and J2.
(b) What is the deviation of the vertical from the radial at the equator?
(a) The total gravitational potential is the sum of the potentials of the two spheres (Fig. 17) where g, is the attraction due to the potential V1and g2that due to the potential V2:
V ¼ V1þ V2¼GM r þGM0
q
where r and q are the distances from a point P to the centres of the large and small spheres, respectively.
As we did in Problem 16, for the small sphere of radius a/2 we take the differential where M is the mass of the sphere of radius a and density r.
For 1/q we take the first-order approximation of the Legendre polynomial, as we did in Problem 16:
Then, the expression for the gravitational potential V is:
V ¼GM
(b) We know that the potential can be expressed by an expansion in zonal spherical harmonics (Legendre polynomials) given in the first-order approxima-tion by
Comparing the two expressions we obtain,
5r q P
J0¼3
The components of gravity at the surface of the large sphere (r¼ a) are:
gr¼@V
At the equator the deviation of the vertical with respect to the radial direction is tan i¼gy
gr
¼ 8 45 i¼ 10:08
18. Suppose an Earth is formed by a sphere of radius a and density r, and within it there are two spheres of radius a/2 and density 2r with centres located on the axis of rotation in each hemisphere. If M is the mass of the sphere of radius a, calculate:
(a) The potential U(r,u) and the form of the equipotential surface passing through the Poles.
(b) The component grof gravity in the first-order approximation for points on the surface.
(c) Calculate grdirectly at the Pole and the equator, and compare with the first-order approximation.
(a) This problem is similar to Problem 16, but now the density of the two spheres is the same. The total gravity potential is the sum of the gravitational potentials of the three spheres (V, V1and V2) plus the potential due to the rotation F:
U¼ V þ V1þ V2þ F ð18:1Þ
and where M is the mass of the large sphere of radius a and M0the differential mass of each of the small spheres of radius a/2, r is the distance from a point P to the centre of the large sphere, and q1and q2the distances from P to the centres of the small spheres (Fig. 18). As in Problem 16 the differential mass is given by the difference in density between the large and the small spheres:
M0¼4
The inverse of the distance 1/q can be approximated by 1
The potential of the rotation can be written in terms of the coefficient m = a3o2/GM, F¼GM Substituting in Equation (18.1)
U¼GM
At the Poles, r¼a and y = 0, and the potential is
The form of the equipotential surface which passes through the Poles (r¼ a) is obtained from Equation (18.2)
Making the approximation r¼ a in the right-hand side:
r¼32
This is the equation of an ellipse with flattening a ¼ (32/42)(3/32m/2). Since there is symmetry with respect to the axis of rotation, the equipotential surface is an ellipsoid of revolution.
Depending on the value of m, we have the following cases, m
< a) prolate ellipsoid
(b) For the gravity at the Pole, in the first-order approximation, we take the derivative of the potential (18.2) and substitute y¼ 0 ) rp¼ a:
(c) The exact solution for the gravity at the pole is the sum of the attractions of the three spheres:
At the equator we take the derivative of the potential and substitute r = a and For the exact solution we write
gr¼ GM a2 2GM
8q2 cos aþ o2a From Fig. 18 the distance q is given by
q2¼a2 The approximated values are smaller than the exact solutions.
19. For the case of Problem 18, if GM = 4 103m3s2, a = 6 103km, and v = 7 105s1, calculate the values of J2, a, m, H, and b.
From the definition of m we obtain m¼a3o2
GM ¼216 1018 49 1010
4 1014 ¼ 2:6 103
The value of J2 is obtained by comparing the two expressions for the potential U (Problem 18):
The flattening is obtained from the relation a¼3
The gravity flattening is given by b¼gp ge
ge
¼1:555 1:175
1:175 ¼ 0:323
where we have used the values of gravity at the Pole and equator obtained in Problem 18, and in the latter we have substituted the value obtained for the coefficient m.
The dynamic ellipticity H is defined as the ratio of the moments of inertia with respect to the polar and equatorial radius (Fig. 19a):
H¼C A C
where A and C are the moments of inertia of a sphere respect to the polar and equatorial radi: (axes x1and x3). The moment of inertia of a sphere of radius R is
Isph¼2 5MR2
We have to add to the moment of inertia of the sphere of radius a the moments of inertia of the two internal spheres of radius a/2. For the C-axis (x3) we have
IC¼ Isph aþ 2Ispha2¼2
5Ma2þ 22 5
M 8
a2
4 ¼ 0:425Ma2
For the A-axis (x1) the moment of inertia of each of the small spheres is given by (Fig. 19b)
X3
X1 2r
2r r
a/2
a
Fig. 19a
I ¼ ICMþ Mh2
since in this case the A axis does not coincide with the centre of mass, where R ¼ a/2 and h¼ a/2:
Isph a=2¼2 5
M 8
a2 4 þM
8 a2
4 ¼ 0:044Ma2 IA¼ Isph aþ 2Isph a=2¼2
5Ma2þ 2 0:044Ma2¼ 0:488Ma2 Finally
H ¼C A
C ¼IC IA IC
¼0:425 0:488
0:425 ¼ 0:147
20. Suppose an Earth is formed by a sphere of radius a and density r, and within it there is a sphere of radius a/2 and density 5r centred at the midpoint of the northern-hemisphere polar radius. If m ¼ 1/8 and M is the mass of the sphere of radius a, determine:
(a) The form of the equipotential surface passing through the North Pole.
(b) For latitude 45º, the astronomical latitude and the deviation of the vertical from the radial.
(a) The gravitational potential is the sum of the potentials for the sphere of radius a and that of the sphere of radius a/2 (Fig. 20):
V ¼ V1þ V2¼GM r þGM1
q
As in the previous problems the potential of the small sphere is given in terms of differential mass M1:
a
R
h
b
Fig. 19b
M1 ¼4 and for the inverse of the distance 1/q we use the approximation
1
Then, the total gravitational potential is V ¼GM
The total potential U is the sum of the gravitational potential V plus the potential of rotation F, where
The form of the equipotential surface is found by putting U¼ Up:
Putting r¼ a inside the square brackets and solving for r we find
r¼ a4 5 1þ1
6cos yþ 1 12cos2y
(b) The deviation of the vertical with respect to the radial direction is given by the angle i:
tan i¼gy
gr
To find this value we have to calculate the two components of gravity gr¼@U
and for latitude 45º
gr¼ GM The astronomical latitude is
fa¼ 90 y i ¼ 36:0
21. If the internal sphere of Problem 20 is located on the equatorial radius at longitude zero, find expressions for the components of gravity: gr, gu, gl.
As in the previous problem the differential mass of the small sphere M1is (Fig. 21a):
M1 ¼M 2
The total potential U is the sum of the gravitational potentials V and V1, and the potential due to rotation F. According to Fig. 21b, using the relations of spherical triangles, if’ and l are the coordinates of the point where the potential is evaluated, then
cosc¼ cos 90º cos(90º’) þ sin 90º sin(90º’) cosl cosc¼ cos’ cosl
Using the expression for 1/q as in Problem 16, 1
q¼1 r 1þ a
2rcos cþ a 2r
21
23cos2c 1
The gravitational potential of the small sphere is given by V1 ¼GM
2r 1þ a
2rcos’ cos lþ a2
8r23cos2’cos2l 1
The total potential U is given by:
U¼GM r
3 2þ a
4rcos’ cos lþ a2
16r23cos2’cos2l 1
þ1
2r2o2cos2’
l = 0° A
l a/2
y
P B
q r
j
Fig. 21a
90° 90°–j
y
l B
P
A
Fig. 21b
Using the coefficient m and sin y = cos ’, we obtain
The three components of gravity are found by differentiating U with respect to r, y, and l:
gr¼@U
22. A planet is formed by a sphere of radius a and density r, with a spherical core of density 5r and radius a/2 centred on the axis of rotation in the northern hemisphere and tangential to the equator. The planet rotates with m ¼ 1/4. For the point at coordinates (45º N, 45º E), calculate:
(a) The astronomical latitude.
(b) The deviation of the vertical from the radial.
(c) The angular velocity of rotation that would be required for this deviation to be zero.
(a) The gravitational potential is the sum of the potentials of the two spheres (Fig. 22):
V ¼ V1þ V2¼GM
As in Problem 16, the inverse of the distance from a point P to the centre of the small sphere, 1/q, can be approximated by
1
As in previous problems we use the differential mass of the small sphere, M0¼4
Substituting in Equation (22.1) we obtain V ¼GM
The total potential U is the sum of V plus the potential due to rotation F:
U ¼ V þ F ¼GM The components of gravity grand gyare
gr¼@U For a point on the surface of the large sphere and coordinates 45º N, 45º E, we have that y = 45, r¼ a, and
m¼a3o2 GM ¼1
4 Putting these values in (22.2) and (22.3), we obtain
gr¼ 1:82GM
a2 and gy¼ 0:24GM a2
To find the astronomical latitude we first have to find the deviation of the vertical with respect to the radial:
tan i¼gy
gr
¼ 0:13 ; i ¼ 7:5
The astronomical latitude is, then, f =’ i = 45 7.5 = 37.5.
(b) The deviation of the vertical with respect to the radial direction, as already found, is i = 7.5. (c) If we want the deviation of the vertical to be null, i¼ 0, this implies gy= 0.
writing Equation (22.3) in terms of the coefficient m, where
Substituting in Equation (22.4) we obtain o¼
23. A planet consists of a very thin spherical shell of mass M and radius a, within which is a solid sphere of radius a/2 and mass M0 centred at the midpoint of the equatorial radius of the zero meridian. The planet rotates with angular velocity v about an axis normal to the equatorial plane. Calculate:
(a) The potential at points on the surface as a function of latitude and longitude.
(b) The components of the gravity vector.
(c) If M0¼ 10 M, what is the ratio between the tangential and radial components of gravity at the North Pole?
(a) The potential U is the sum of the gravitational potentials due to the spherical shell V1, and to the interior sphere V2, plus the potential due to the rotation of the planet F (Fig. 23):
U¼ V1þ V2þ F
where r is the distance from a point P on the surface of the planet to its centre, q is the distance from point P to the centre of the interior sphere, and’ the latitude of point P.
Using the cosine law,
q¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r2þa2
4 ar cos c r
where c is the angle between r and the equatorial radius, and its inverse can be approxi-mated by (Problem 16)
1 Using the relation for spherical triangles
cos a¼ cos b cos c þ sin b sin c cos A
putting b¼ 90º ’, c ¼90º, A ¼ l, and a ¼ c, where l is the longitude of P, then cos c = cos’ cos l
Substituting in (23.1), the potential due to the small sphere is V2¼ GM0
r 1þ a
2rcos’ cos lþ a2
8r23cos2’cos2l 1
The total potential U is U ¼GM
(b) The components of the gravity vector are obtained from Equation (23.3):
gr ¼@U
gl¼ 1 previous equations we obtain
gr¼ GM
The ratio between the radial and the tangential components of gravity at the North Pole is
gr gy
¼ 1:45
24. An Earth consists of a sphere of radius a and density r, within which there are two spheres of radius a/2 centred on the axis of rotation and tangent to each other. The density of that of the northern hemisphere is 4r and that of the southern hemisphere is r/4.
(a) Express the gravitational potential in terms of M (the mass of the large sphere) up to terms of 1/r3.
(b) What astronomical latitude corresponds to points on the equator (without rotation)?
(c) What error is made by using the 1/r3approximation in calculating the value of gr at the equator?
(a) The total gravitational potential V is the sum of the potentials of the sphere of radius a (V0) and of the two spheres of radius a/2 situated in the northern (V1) and southern (V2) hemispheres (Fig. 24):
V ¼ V0þ V1þ V2
As in previous problems the large sphere is considered to have uniform density r and the effect of the two interior spheres is calculated using their differential masses
V0¼GM
where r1 and r2 have been calculated as in Problem 16. Then, the total gravitational potential in the 1/r3approximation is
V ¼ GM 41
(b) The components of the gravity vector, taking into account that there is no rotation, are gr¼@V
gr¼ 1:175GM a2 gy¼ 0:243GM
a2
The astronomical latitude (’a) is the angle between the vertical and the equatorial plane. In our case at the equator this is given by the deviation of the vertical from the radial direction:
tan’a¼gy gr
¼ 0:207 Then
’a¼ 11:68N
(c) If we want to calculate the exact value of grat the equator, we calculate the exact attractions of each sphere and add them:
gr0¼ GM
a2 gr1¼ 3GM
8r21 cos a gr2¼ 3GM
32r22 cos a ð24:2Þ where r1 and r2 are the distances from the centre of each of the two interior spheres (Fig. 24):
r1¼ r2¼
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2þa2 4 r
¼a ffiffiffi p5
2
and a is the angle which r1and r2form with the equatorial plane:
sin a¼a=2 r1 ¼ 1
ffiffiffi5 p The radial component of gravity is given by
gTr ¼ g0r þ gr1þ gr2¼ 1:335GM a2 The error we make using the approximation is
gapprox gexact¼ 0:160GM
a2 ; that is; 16%:
25. An Earth consists of a sphere of radius a and density r within which there are two spheres of radius a/2 centred on the axis of rotation and tangent to each other.
The density of that of the northern hemisphere is 2r and that of the southern hemisphere is r/2.
(a) Express the potential V in terms of M (the mass of the large sphere), G, and r up to terms in 1/r3.
(b) According to the value of this potential V, which astronomical latitudes corres-pond to the geocentric latitudes 45º N and 45º S?
(c) What must the rotation period be for the astronomical and geocentric latitudes to coincide?
(d) What error is made by the 1/r3approximation in calculating the value of grat the equator? And at the North Pole?
(a) As in previous problems the effect of the interior spheres is given in terms of their differential masses (Fig. 25):
M1¼4
The distances q1and q2from the centre of each sphere to an arbitrary point P are found using the cosine law:
q1¼ r2þa2
Using the approximation for 1/q (Problem 16), the total gravitational potential V is the sum of the potentials of the three spheres:
V ¼GM
(b) The components of the gravity vector are given by gr¼@V
At geocentric latitude 45º N, y¼ 45º,
gr¼ 1:21GM a2 gy¼ 0:09GM
a2
The deviation of the vertical with respect to the radial direction i is given by tan i¼gy
gr ¼ 0:074 ) i45¼ 4:2
According to Fig. 25, the astronomical latitude’acan be determined from the deviation of the vertical i, Then the astronomical latitude is
’a¼ 45 2:5 ¼ 47:5 ¼ 47:5S
(c) If we want the astronomical and geocentric latitudes to coincide, then the deviation of the vertical must be null, i¼ 0º. This implies that gytotal must be zero. To do this by means of the rotation, we have to make the tangential component of the centrifugal force gyR be equal and of opposite sign to that of the gravitational potential gyV:
gytotal¼ gyVþ gyR¼ 0 ) gyV¼ gyR
The tangential component due to rotation is gRy ¼1 r
@F
@y where F¼1
2o2r2sin2y. Then
gyR¼ o2rcos y sin y
For a point on the surface at latitude 45º N, r = a and y = 45, so gyV¼ gyR) 0:09 GM =a2¼ o2 a=2 From here we can calculate the period of rotation
T ¼2p o ¼ 2p
ffiffiffiffiffiffiffiffiffi p0:18
ffiffiffiffiffiffiffiffi a3 GM r
For a point at latitude 45º S, r = a and y = 135, so
gyR¼ gVy ) 1
2o2a¼ 0:04GM
a2 ) T ¼ 2p ffiffiffiffiffiffiffiffiffi p0:08
ffiffiffiffiffiffiffiffi a3 GM r
(d) The value of the radial component of gravity at the equator, r¼ a, y = 90, by substitution in (25.1), is
gr ¼ 1:04GM a2
If we calculate the exact value by adding the contributions of the three spheres (Fig. 25)
gexactr ¼ grMþ g1r þ g2r
gMr ¼ GM a2 g1r ¼ g1cos b g2r ¼ g2cos b where
cos b¼ a The error in the approximation is:
gerrorr ¼ 0:96GM The error in the approximation is
gerrorr ¼ 1:47GM
26. A spherical Earth of radius a has a core of radius a/2 whose centre is displaced a/2 along the axis of rotation towards the North Pole. The core density is twice that of the mantle.
(a) What should the period of rotation of the Earth be for the direction of the plumb-line to coincide with the radius at a latitude of 45º S?
(b) What are the values of J0, J1, J2, and m?
(a) As in previous problems we calculate the gravitational potential by the sum of the potentials of the two spheres, using for the core the differential mass (Fig. 26):
V ¼ V1þ V2 ¼GM
As we saw in Problem 16, we use for 1/q the first-order approximation The total potential U is the sum of the gravitational potential V and the potential due to rotation
In order that the direction of the plumb-line coincides with the radial direction, the tangential component of gravity, gy, must be null:
gy¼1
For a point on the surface at latitude 45º S, the tangential component of gravity is, with r= a, y = 135,
Putting this component equal to zero, we find the value of the period of rotation T:
(b) The gravitational potential V of Equation (26.1) can be written as V ¼GM
We obtain the values of J1and J2by comparison with the equation V ¼GM
Since the total mass is (9/8)M, we obtain J0¼ 1 J1¼ 1
18 J2¼ 1
72
27. Within a spherical planet of radius a and density r there are two spherical cores of radius a/2 and density r0 with centres located on the axis of rotation at a/2 from the planet’s centre, one in the northern hemisphere and the other in the southern hemisphere.
(a) Neglecting rotation of the planet, calculate what the ratio r0/r should be for the gravity flattening to be 1/8.
(b) If the planet rotates so that m¼ 1/16, and the ratio of the densities is that found in part (a), calculate the astronomical latitude which corresponds to the geocentric latitude 45º N.
(a) Since there is no rotation the total potential U is the sum of the gravitational potentials of the three spheres (Fig. 27). As in previous problems we use the mass M of the planet with uniform density r and for the two cores the differential masses M0. For 1/q we use the approximation as in Problem 16:
M0¼4 The potential U is
U ¼GM
The radial components of gravity at the equator and the Pole are found by taking the derivative of the potential U:
gr¼@U
On the surface r¼ a, and at the equator y = 90and at the Pole y = 0, so The gravity flattening is given by
On the surface r¼ a, and at the equator y = 90and at the Pole y = 0, so The gravity flattening is given by