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Spring 2005
Dr. Mike Fanelli
Solutions to Assigned Problems
CHAPTER 7
PROBLEM 7-2: You are asked to modify the Earth’s mass by making its density
uniform, then assess how two properties of Earth would change. Escape speed
and surface gravity depend on the mass and radius of a planet. You are asked
to change the average density of the Earth to 3000 kg/m 3 from its actual value,
5500 kg/m3, which will produce a less massive Earth. Do you expect the escape
speed and surface gravity to be smaller or larger? Take the mass of the
“revised” Earth as the density X the Earth’s volume. Then use the formulae in
chapter 2 for escape speed and the acceleration due to gravity at the Earth’s
surface.
ANSWER: Remembering that the volume of a sphere = 4/3 x  x R3
Total M = density x volume = (amount of mass per volume) x volume
= 3000 kg/m3 x 4/3 x  x ( 6.4 x 106 m )3
Note that the Earth’s radius is expressed here in meters, because the density is
expressed in meters, and the units must be the same, for the answer to make
sense. Also note that the units of cubic length cancel, since there is an m3 in the
denominator on the left, and an m3 in the numerator on the right. The result is the
total mass for an Earth with the stipulated density.
Running the numbers, M = 3.3 x 1024 kilograms, which is ~ 55% of the actual
value.
The force due to gravity on the Earth’s surface is (from page 52):
M x m
F(grav) = G x -------------- .
R 2
Since F = m x a (Newton’s second law of motion, relating force to
acceleration), a, the acceleration due to gravity is just equal to --M
a = G x ------ .
R 2
Taking the value of “G”, the constant of gravitation from the table in the back of
the text, using the value for the mass of the Earth found above, and the radius of
the Earth, gives,
a = 5.4 meters / second2, a value to be compared to the actual value of 9.8
meters / second2
 A less massive Earth produces a weaker gravitational “pull”, as expected.
The expression for escape velocity is given on page 55 of the text:
V
2
escape
=
2 x G x M
-----------------------,
R
Where V 2escape is the square of the escape speed. Note that this expression
depends on the mass and radius of the Earth, and represents the escape speed
from a radius corresponding to the Earth’s surface.
Plugging in the numbers gives Vescape = 8.3 km/sec, compared to 11 km/sec
for a “normal” Earth.
 A less massive Earth produces a smaller escape speed, as expected.
PROBLEM 7-6: This question explores the time it will take for a wave, in this
case a seismic (pressure) wave, to pass across a distance, the diameter of the
Earth.
ANSWER: The diameter of Earth is 2 x 6378 km = 12,756 km. At 5 km/sec, the
time it will take for a P-wave to cross the Earth is:
time = distance  speed = 12756 km  5 km/sec
= 2551 seconds or 42.4 minutes.
PROBLEM 7-8: Continents attached to tectonic plates move around the Earth.
How long does it take a plate/continent to move about ? Given a rate, 3 cm/yr,
what is the time required to traverse 6000 kilometers ?
ANSWER: As in question 7-6, this is a rate question. How long does it take a
moving object to traverse a specified distance ? We are interested in the
timescale for the phenomena of continental drift. First convert the distance, here
the width of the Atlantic Ocean, into centimeters, because the rate of motion is
expressed in centimeters per year.
6000 kilometers = 6 x 108 centimeters (1 km = 1000 meter & 1 meter = 100
cm) therefore,
travel time = distance  speed
= (6 x 108 cm )  3 cm / yr = 2 x 108 years
It will take about 200 million years, for a piece of crust to cross the width of the
Atlantic Ocean.