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Physics 1050 – Fall 2003
Dr. Mike Fanelli
Solutions to Assigned Problems
Problems are taken from Chaisson and McMillan, fourth edition.
CHAPTER 7
PROBLEM 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/m3 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 in meters, because the density is given in meters,
and the units must be the same. 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:
M x m
F(grav) = G x -------------- (pg 52).
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 2escape =
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 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 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 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 plate to cross the width of the Atlantic.