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Gravity Chapter 8 Homework answers
1. Given the value of little-g at the equator is 9.78031 m/s^2 , what is the value of gravity at the North
Pole or South Pole?
Gravity varies with latitude according to this International Gravity Formula (IGF) as
2 2 3 6
g ( ) θ γ 1 (1 γ 2 sin θ γ 3 sin 2 ),θ γ 1 9.78031, γ 2 5.3024 e , γ 3 5.900 e.
− − = + + = = =
The latitude at the North Pole is Ѳ=+90° and the latitude at the South Pole is Ѳ=-90°.
Therefore, the predicted gravity at the North Pole is:
2 2 2 2 1 2 3 1 2 3 1 2
3 2
5 2 5 2
( 90 ) (1 sin (90 ) sin (2 * 90 ) (1 1 0 ) (1 )
g
m g e s
m mGal g e mgals mgals s m s
° ° °
° −
° −
eQ: State why the gravity may or may not be different at the north and South Pole according to the IGF?
eQ: State the number of significant figure for each of the five numbers in the IGF?
eQ: What MKS units does the terms
γ ,γ ,γ 1 2 3 have?
eQ: What units does Ѳ have and what is the difference between radian and degree units?
eQ: What two gravitational effects does the IGF account for?
2. A horizontal sill that extends well outside the survey area has a thickness of 30 m and density of 0.
Mg/m 3 in excess of the rocks it intrudes. Estimate the maximum depth at which it would be
detectable using a gravimeter that can measure to 0.1 mGal.
The equation for a flat infinite sheet is:
∆ g ( ∆ ρ , ) t = 2 π G ∆ρ t mGals
where G is gravitational constant, ∆ρ is density contrast, and t is the thickness of the infinite layer. First, we note that the infinite sheet equation
does not vary with the depth of the sheet. Hence, the answer is that the depth of the infinite sheet is
irrelevant to whether we can detect the gravitational anomaly with a gravimeter accurate to 0.1 mGal.
Second, let us calculate the gravitational pull of the infinite sheet mass anomaly. 3 8 6 2 2 3
m Mg g t G t m e m s mGals Mg s m
− − ∆ ∆ = ∆ = = = .
Therefore, our gravimeter with a 0.1 mGal accuracy could detect this anomaly. But, the important point is
that the gravity field for an infinite sheet does not change anywhere, it just makes the absolute level of the
gravity field 0.63 mGal greater EVERYWHERE! Therefore, the gravity anomaly from the infinite sheet
cannot be detected.
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3. An extensive dolerite sill was intruded at the interface between horizontal sandstones. Sketch the
gravity profiles expected if the sill and beds shave been displaced by: (a) A steeply dipping normal
fault. (b) A shallow thrust fault. (c) A strike-slip fault. (d) Repeat (c) when the beds dip at about 40'.
4. Calculate how much gravity changes, and whether it is an increase or decrease, on going one km
north from the following starting latitudes: (a) equator. (b) 45° N. (c) 45° S. What elevation changes
in air would give the same change in g?
The simplified latitudinal gravity equation for small (<20 km) poleward movements is approximated as:
∆ g lat ( λ) = 0.812 sin(2 λ) mGal / km − polewards
.
An important point to understand is that polewards in northern hemisphere is movement toward the
North Pole and polewards in southern hemisphere is movement towards the South Pole. If you are in either
hemisphere and move towards the equator that is anti-poleward motion.
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2 2
M (^) e m g r G r acceleration r s
Evaluation this equation at the Earth’s mean radius (r=6371 km) gives:
3 11 24 2 2 2 3 2 2
M (^) e m m m g r G r e e kg r s kg s e m s
− = = =
.
So, if you drop a ball, and ignore air friction, the velocity of the ball increases by 9.8 m/s every second.
Of course, if you know the gravitational acceleration, you can calculate your mass, but ONLY to the accuracy
that you know the gravitational field of the planet (which varies by hundreds of mGal).
However, a mass-balance can overcome the ‘weighing problem’ by NOT measuring the weight of an object,
but directly measuring its mass. This is done by ‘balancing’ an object’s mass against known masses placed
upon the opposite balance-pan. A mass-balance can perform this miraculous feat of measuring mass, not
weight, because the gravitational acceleration that pulls on the object’s mass in one pan and the known
masses placed in the other pan CANCEL OUT! Note: we are assuming that the gravity field does NOT vary
between the two pans of the mass-balance.
So weight is just the gravitational force that any mass ‘feels’ and your weight depends on where you are
with respect to other large masses. More quantitatively, the weight of an object depends on the inverse
squared distance between two objects center of masses.
So, now that I know what weight is, then what is mass?
Mass is a measure of an objects inertia. And, inertia is a property of all mass (matter) which makes a mass
either remain at rest (zero velocity) or remain at a constant velocity UNLESS acted upon by an unbalanced
net force. The fact that mass has inertia was only understood in 1594 by Galileo and before that all the
physicists (called natural philosophers then) had it wrong! They thought that a ball thrown in the air had
to have a force that continued to push on it to keep it moving: e.g., Aristotle said the air around the ball was
moving with the ball and that this force applied to the ball, was what keep the ball moving…WRONG! As a
corollary to this wrong-headed notion of motion, Aristotle also predicted that the more massive a ball, the
quicker it would fall. This is what Galileo finally tested with his experiment that dropped balls from the
Tower of Pisa and proved Aristotle to be wrong.
If you really want to know what mass is, the particle physics model QCD-lite predicts that 99% of mass is
the three quarks that fly-around each other to make the protons and neutrons that make the atomic
nucleus. Since relativity requires that E=mc^2 , mass is just spatially localized (condensed) energy. Hence,
most of the mass of the quarks that make the protons and neutrons is actually angular momentum energy
that keeps the quarks ALWAYS bound together and swirling around each other at a scale of 10-15^ m.
Another definition of mass is simply the number of atoms present in a given volume. That is called the
molar mass. We can calculate this approximately because the atomic masses are well-know: just add up the
mass of the protons and neutrons that make the nucleus (the electrons weigh nothing compared to the
nucleons).
But, most importantly, remember that mass is a measure of inertia; and, inertia means that you need to
apply an unbalanced net force to an object to change its state of motion, whether that motion be at rest or
moving along a straight line at constant velocity.
So, what then is a force?
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A force is ALWAYS and interaction between two (or more) objects that changes their motions. Force is not
conserved like energy or momentum. During the interaction (think of two billiard balls colliding), force is
ALWAYS an action-reaction pair. That is, the force of object 1 ON object two is equal and opposite (in a
vector sense) to the force of object 2 ON object 1. This MUST be true, otherwise the momentum (mass *
velocity) of the two balls is NOT conserved (i.e., conserved means the momentum remains constant). This is
Newton’s third law of motion published in 1687.
Physicist’s now know that there are only four fundamental forces: gravity, electro-magnetic, and the two
forces that are only significant in the nucleus of atoms (the strong and weak forces). From the gravity and electro-magnetic fundamental forces we can derive the concept of pressure and stress in a solid/liquid/gas
which are called ‘contact forces’. For example, when two balls collide, the fundamental force at work is the
electro-magnetic repulsion between the electron clouds that surround the atomic nuclei at the two balls
surfaces. BUT, we can ignore this fundamental force view and just called the force between the electrons a
contact force which is the pressure/stress that is impressed during the two balls collision. Force at distance
also occurs for gravity and electro-magnetism. In essence, a force-field does extend through both matter
(the earth) and empty space allowing objects to exchange force without ever touching each other! This
concept of forces without contact gave the physicist’s real fits for a long time and was not accepted until the
1890’s.
eQ: if an object with mass m 1 =1 kg with a finite speed, collides with a very heavy mass m 2 =1e^12 kg at rest,
what happens? (Use Newton’s second law).
eQ: So, if my body mass has the force of the earth’s gravity pulling me down, why I am not accelerating?
You are not accelerating downwards because the floor is pushing back against gravity’s force to make the
net Force zero. Of course, when you jump in the air, and find that it is the gravitational field of the earth that
decelerates your motion on the way up and accelerates your motion on the way down.
eQ: When I drop a ball at the Earth surface, the ball clearly falls (accelerates) towards the earth’s center of
mass, but the Earth does not ‘seem’ to move towards (accelerate) the ball. Newton’s gravitational equation says the Force of the earth on the ball is equal and opposite to the force of the ball on the earth.
eQ: Prove mathematically using Newton’s gravity law (Eq. 8.2) and Newton’s 2’nd law ( F = m*a) that two
balls of different mass fall at that same rate (ignore air friction).
eQ: Assume you are in a place where gravity is zero, how could you measure an objects mass? (Hint: use
Newton’s second law).
eQ: Is the inertial mass used in Newton’s second law the same as the gravitational mass used in Newton’s
gravitational equation?
eQ: What is the value of gravity at the center of the earth and why?
6. A spherical cavity of radius 8 m has its centre 15 m below the surface. If the cavity is full of water
and is in rocks of density 2.4 Mg/m 3 , what is the maximum size of the gravitational anomaly?
Note, we are NOT calculating the absolute value of the force interaction between the Earth’s mass and the
mass anomaly associated with the water in the buried spherical cavity. We are just calculating the variation in the gravity field associated with replacing the rock in the spherical cavity with water. Thus, we
can derive a simple approximation for the gravitational variation (Eqn. 8.5)
2
m g G d
δ
where ∆m is the change in mass from the reference state (all rock) and d is the distance
between the center of mass of the water filled cavity and the place where the gravity measurements is
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There are two gravitational effects (changes) to add together: the free air effect due to moving closer to the
earth’s center of mass and the Bouguer effect of the mass above the gravimeter whose mass would pull
upwards. The tricky part of this problem is getting the sign correct and remembering that we are asking for
what the gravitational ‘effect’ is not what the gravitational ‘correction’ is.
The magnitude of the free air effect is provided in Question 8 (309 mGal round off). The sign of this effect
will be positive (increase gravity).
The Bouguer gravity effect is given by the Bouguer formula:
3 ∆ g (^) b ( ∆ ρ , ) t = 2 π G ∆ρ t mGal = 2 π * G * (2.3 − 0.0) *1 e = 96 mGal
Because the mass is above the gravimeter which is 1 km down a well, the mass pulls the gravimeter-mass
upwards, opposite to the earth’s gravity pull which is down. Therefore, the Bouguer anomaly should be
negative.
Adding the two effects, the total change in gravity measured by the down-well gravimeter is +309 + (-96)
mGal which equals a +213 mGal INCREASE in the measured gravity.
10. A person having carried out a microgravity survey to locate a lost shaft (filled with air), creates a
gravity profile that shows a small gravitational dip in it (i.e., the measured gravity goes down or gets
less and the anomaly has a negative sign). The surveyor notes that the dip in the gravity coincides
with a dip (depression) in the otherwise level ground surface and thus she says she has located the
mine shaft. But, then you find out that she has not corrected her data for topography (both free air
and Bouguer effects). Discuss whether applying the topography corrections might result in the gravity dip disappearing.
The combined free air and Bouguer gravity equation 8.11 shows that free-air gravitational effects
(movement above or below one’s datum) will dominate the equation and determine its sign. Thus, using
the flat surface as a datum, the Bouguer gravity correction will have a negative value because the
topographic dip is below the flat surface datum. This could make the measured dip in the gravity field that
is coincident with the topographic dip get bigger.
Said another way, when one corrects for the fact that the gravity measured in the topographic dip will be
higher because the measurement point is closer to the earth’s center of mass, then the free-air correction (0.386 mGal/m) will be subtracted from the measured data. This will make the Bouguer anomaly even
more negative.
11. The mean radius of the Earth is 6371 km. On taking a gravimeter 1 km above the earth’s surface in a
balloon, you would expect the value of little-g to decrease by how much?
Form a ratio of the gravity at r (^) e (6371 km) and re+1 and then turn the ratio into a percent by multiplying by
2 2 1 2 (^1 2 22 2 2 2 ) 1 2 2 1 2 1
M (^) e M (^) e g r M (^) e Me r g r G g r G ratio G G r r g r r r r
To convert the ratio to percent change, multiple by 100. Thus, the gravity at 1 km height is 99.97% of its
values measured at the surface. So, the gravity has decrease by 0.03% (100.0 – 99.97)%.
12. The International Gravity formula describes gravity:
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(i) Only at the sea surface. Wrong. The IGF describes gravity only with respect to latitude and does not
care whether one is over sea or land.
(ii) On a surface simplified to be a sphere approximating the earth. Wrong. The IGF uses the best fitting
ellipsoid for the earth’s figure.
(iii) To ONLY allow for the equatorial bulge. Wrong. It allows for the bulge and centrifugal forces.
(iv) To allow for the centrifugal forces associated with the earth’s rotation. True.
(v) To allow for both the earth’s rotation and the equatorial bulge. Exactly what the IGF accounts for.
13. An ancient burial chamber is to be found using a microgravity survey. The chamber is about 4 m
across and is covered by about 3 m of material of density 2 Mg/m 3. Estimate: (a) the maximum
magnitude of the anomaly; (b) a suitable grid spacing assuming the total error in measurement is
0.1 mGal.
Assume the burial chamber is a sphere of radius 2 m with the sphere’s center at its burial depth (3 m) plus
the sphere’s radius (2 m) to give a spherical center depth of 5 m. Use the gravity formula (Eqn. 8.5) that
gives the maximum gravity value when measured directly over the center of a sphere.
3 2 2 2
7 2
m V g G G G r d G d d
g e m s mGal Gal
−
The relation between the depth of the center of a spherical mass anomaly and the gravitational half-width
is: d = 1.3 * half-width (m). Therefore, the half-width is equal to d/1.3 which for d=5 m is about 4 meters.
Thus, if one desires to measure the gravity anomaly associated with the burial chamber, one should sample at least every meter so that the gravity anomaly is not missed.
- Describe how you would carry out a microgravity survey to determine the lateral position of dense
mine workings (density of 2.6 Mg/m 3 ) that is buried beneath 20 m of rocks of density 2.1 Mg/m^3.
Assume the gravimeter is accurate to 5 μGal. Estimate the accuracy in height and position required
for these errors to be less than the gravimeters accuracy.
Little information is given, which is intentional and a real-world situation. All we know is that there is
some unknown volume of high density mine workings below 20 m. But, we do know that the mine working density is 2.6 Mg/m 3 and the overlying rock’s density is 2.1 Mg/m 3. Thus, we know that we are seeking to
find a positive gravity anomaly.
Inspection of the depth rules (Fig. 8.19 page 121) shows the gravity anomaly half-width relations for
different mass anomaly geometries: sphere, cylinder, dipping sheet, irregular body. Most useful for this
problem is figure (d) for an irregular body where the relation between the depth to the top of the mass
anomaly and the peak and slope of the anomaly are:
max
max
g d dg dx
δ ≤ where d is maximum depth to top surface (20 m), δgmax is the peak height (in
mGal) of the gravity anomaly, and (dg/dx)max is the maximum slope of the gravity anomaly (mgals/m).
Substituting d=20 m into the equation and rearranging gives:
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2 2 3 2 21 8 3 2
e
g r km r m s e m M kg e Mg G e m Mg s
−
Therefore, the mean earths density is 21 3 21
e Mg m e
ρ = =
which is about twice the density value of granite. This means the deep earth must be much denser than the
surface rocks with granite-like densities.
16. Explain why portions (portions means mass) weighed on the high Tibetan plateau (>4 km mean
elevation) would be larger (greater mass) than portions weighted at sea level, if the portions are
weighted using a spring balance? But, the portions would NOT have to be greater on the Tibetan
plateau if a mass balance is used.
The gravity on the Tibetan plateau will be less with respect to sea-level because the free-air gravity
decrease with altitude is bigger than the Bouguer gravity field effect due to the extra mass associated with
the topography (see equation 8.11). Thus, to get the same weight of ‘portions’ measured on the Plateau and
at sea level, the ‘portions’ (mass) weighed on the Plateau would have to be larger than the ‘portions’ (mass)
weighed at sea level.
eQ: What is the difference between an object’s mass and weight (again)?
17. The radius of the planet Mars is 3394 km and the mass is 0.108 (10.8%) of the mass of the earth.
Calculate the value of little-g at the surface of Mars in MKS units.
The mass of Mars is thus 0.108 * Mass-earth which equals 0.108* 5.97e^21 Mg which equals 6.45e 20 Mg.
Therefore, the gravitational acceleration at Mars surface is:
20 8 2 3 2 2
M (^) e e m g r G e r e s
− ^
eQ: So if you exert an upward force with your leg muscles to jump into the Mars air, why would the height
of your jump be different with respect to the earth?
18. Compared to the Earth, the mass of the Moon is about 1/80 and its radius is a quarter. How does the
surface gravity ratio compare between the Earth and the Moon?
m m
e e
M R
and M R
.
So, the calculated gravity ratio is
(^2 2 )
2 2
2
m
m m m e
e e e m
e
M
G
g r M r no units g M M r G r
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Convert to percentages by multiplying by 100 gives the Moon’s gravity to be 20% of the value of the earth’s.
A laborious way to answer this question would be to calculate the Moon and Earth gravity separately and
then ratio the values.
