Introducción a la Geofísica
2010-01
TAREA 3
1) FoG. Using the data in Table 1.1 calculate the gravitational acceleration on the
surface of the Moon as a percentage of that on the surface of the Earth
GE
R2
Gravitation Acceleration of the Earth:
aE = −
Gravitation Acceleration of the Moon:
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aM = −
Gm
RL 2
Ratio
2
aM −Gm RL 2  m  R 
=
=   
aE
−GE R 2  E  RL 
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Inserting values from Table 1.1 for the mass E (5.974 x 1024 kg) and radius R (6371
km) of the Earth and the mass m (0.0735 x 1024 kg, or 0.0123 E) and radius RL (1738
km) of the Moon gives
 6371 2
aM
= 0.0123
 = 0.165
 1738 
aE
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Gravitational acceleration on the Moon is 16.5% of that on the surface of the Earth.
The mean value of gravity on the Earth is 9.81 m s-2, on the Moon it is 1.62 m s-2.
2) FoG An Olympic high-jump champion jumps a record height of 2.45 m on the
Earth. How high could this champion jump on the Moon?
The general relationship between initial velocity (u), final velocity (v), constant
acceleration (a) and distance (s) is v2 = u2 + 2as.
In the case of a high-jumper, the final velocity is 0, the distance is the height (h)
jumped, and the acceleration is –g, so u2 = 2gh.
The gravity (gL) and height jumped (hL) on the Moon are different from gravity (gE)
and height jumped (hE) on Earth, but u is the same, being determined by the high
jumper’s ability. Thus,
u 2 = 2gE hE = 2gL hL
hL =
gE
9.81
hE =
2.45 = 14.8 m
gL
1.62
3) FoG A communications satellite is to be placed in a geostationary orbit.
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(a) What must the period and orientation of the orbit be?
(b) What is the radius of the orbit?
(c) If a radio signal is sent to the satellite from a transmitter at latitude 45°N,
what is the shortest time taken for its reflection to reach the Earth?
(a) A geostationary orbit is one for which the period of rotation of the satellite
about the Earth is equal to the rotation of the Earth about its own axis. This
keeps the satellite “stationary” above a given location. The orbit must be in the
plane of the equator.
(b) The radius of the satellite’s orbit is found as in the previous exercise.
Satellite rotation period, T = 1 day = 86.400 seg
Rotation speed, ω s = 2π = 7.2722 ×10−5 rad s-1
T
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2
Centripetal acceleration at distance rs, ac = ω s rs
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GE
Gravitational acceleration at distance rs, aG = 2
rs
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GE
2
Equating the accelerations, ω s rs = 2
rs
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2
2
R
GE  R 
rs3 = 2   = aG  
R ωs 
ωs 
€ 6 2
 6.371×10
3
rs = 9.81
= 7.5297 ×10 22
−5 
 7.272 ×10 
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The radius of the stationary orbit is 42,226 km (aprox. 42,200 km).
(c) The quickest reflection travels along the shortest path to the satellite from the
point on the Earth’s surface at 45°N where the transmitter and receiver are
located (point P in the diagram); the satellite is above the equator (point S in
the diagram).
The side d is given by
 1 
2
2
2
d 2 = R 2 + rs − 2Rrs cos(45º ) = (6371) + ( 42226) − 2(6371)( 42226) 
 2
Distance from station to satellite, d = 37,989 km
Speed of light, c = 299,792 km s–1
The two-way travel time of the signal is 2(d/c) = 0.253 s
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4) FoG By differentiating the normal gravity formula given by Eq. (2.56)
develop an expression for the change of gravity with latitude. Calculate the
gravity change in milligals per kilometer of northward displacement at latitude
45°.
gn = ge (1+ β1 sin 2 λ + β 2 sin 2 2 λ )
where
ge = 9.780327m /s2
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β1 = 5.30244 ×10 –3
β 2 = –5.8 ×10 –6.
The normal gravity formula is
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gN = ge (1+ β1 sin 2 λ + β 2 sin 2 2 λ )
where ge = 9.780327 m s−2 ;β1 = 5.30244 ×10−3 ;β1 = 5.8 ×10−6
Noting that 1 mgal = 10-5 m s-2 we can write ge = 978,032.7 mgal.
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Differentiating with respect to λ:
dgN
d
d
= gE β1
sin 2 λ) + gE β 2
(
(sin2 2λ)
dλ
dλ
dλ
d
(sin2 λ) = 2sin λ cos λ = sin2λ
dλ
d
(sin2 2λ) = 4 sin2λ cos2λ
dλ
dgN
= gE β1 sin2 λ + 4gE β 2 sin2 λ cos2 λ
dλ
dgN
= gE sin2 λ(β1 + 4 β 2 cos2 λ)
dλ
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This expression gives the change of gravity in m s-2 with latitude, where the
latitude is in radians. To convert to kilometers of north-south displacement (s)
it is necessary to use the relationship s =(Rλ), where R is the Earth’s radius.
dgN dgN gE
=
= sin2 λ (β1 + 4 β 2 cos2 λ )
ds Rdλ R
inserting numerical values for the parameters in the equation
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dgN 978032.7
=
(5.30244 ×10−3 ) sin2λ(1+ 4.375 ×10−3 cos2λ)
ds
6371
= 0.81399sin2 λ
The cos 2λ term in the equation is much smaller than the first term. At latitude
45°N, as in this exercise, sin2 λ = 1 and cos2 λ = 0, and the last term in the
equation is zero.
 dgN 
978033
=


(5.30244 ×10−3 ) sin2λ = 0.814
 ds  45º N
6371
Thus, at 45°N gravity increases northwards at a rate of 0.814 mgal/km.
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5) FoG The following gravity measurements were made on a traverse across a
rock formation. Use the combined elevation correction to compute the
apparent density of the rock (Hint: Free air + Bouguer Plate).
Elevation (m)
100
150
235
300
385
430
Gravity (mgal)
-39.2
-49.5
-65.6
-78.1
-95.0
-104.2
The measured gravity anomaly contains (1) a variation due to the elevation above the
reference ellipsoid and the corresponding changing thickness of the “Bouguer plate”
beneath the traverse, and (2) a contribution from deep sources. Ignoring finer details
such as tidal and topographic corrections, the Bouguer anomaly ΔgB is
ΔgB = gm + Δge − gN
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where gm is the measured gravity, given in the table, and gN is the theoretical value on
the reference ellipsoid. Δge is the combined elevation correction for elevation h :
Δge = (0.3086 − 0.0419 ρ ×10−3 ) h
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The mean rock density ρ is unknown. Using the Nettleton method (Section 2.5.5.4)
trial values of density are inserted in the equation and added to the observed gravity
values to give an elevation-corrected gravity at each measurement station. The
optimum density is found when the corrected anomaly shows the least correlation
with the elevation profile.
The table above gives the gravity anomaly at each station corrected with the
combined elevation correction using densities of 2500, 2700 and 2900 kg m-3.
A plot of these data (below) shows that using ρ = 2500 kg m-3 the anomaly is parallel
to the elevation profile; this indicates that the chosen density is too low. Using ρ =
2900 kg m-3 the data are over-corrected; with increasing eleavation the corrected
anomaly becomes ever more negative. With ρ = 2700 kg m-3 there is minimum
correlation with the elevation.
A better method is to draw a graph in which the gravity anomaly, corrected only for
the free-air effect (last column in the above table), is plotted against the station
elevation.
The points fall on a straight line whose equation may be found using linear regression.
In this case it is
Δ g = 0.1129h - 19.73
The first term corresponds to the Bouguer correction; the second term is the
background anomaly, equal to –19.7 mgal. Thus the coefficient 0.1129 is equal to
0.0419 ρ x 10-3. This gives the optimum density, ρ = 2695 kg m-3 (aprox 2700 kg m-3)
as in the first method.