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Week 12: Chapter 13
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Week 12: Chapter 13
Due: 11:59pm on Sunday, April 19, 2015
To understand how points are awarded, read the Grading Policy for this assignment.
± Gravity on Another Planet
Description: ± Includes Math Remediation. Using a simple pendulum, calculate the gravitational acceleration on
another planet.
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 53.0cm . The explorer
finds that the pendulum completes 104 full swing cycles in a time of 145s .
Part A
What is the magnitude of the gravitational acceleration on this planet?
Express your answer in meters per second per second.
Hint 1. How to approach the problem
Calculate the period of the pendulum, and use this to calculate the magnitude of the gravitational acceleration
on the planet.
Hint 2. Calculate the period
Calculate the period T of the pendulum.
Express your answer in seconds.
ANSWER:
T
= = 1.39 s Hint 3. Equation for the period
The period of a simple pendulum is given by the equation T
−
−−−−−−
= 2π√ L/g
planet
, where L is the length of the
pendulum and g planet is the magnitude of the gravitational acceleration on the planet.
ANSWER:
g planet
= = 10.8 m/s 2 Exercise 13.4
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Description: Two uniform spheres, each with mass M and radius R, touch one another. (a) What is the magnitude
of their gravitational force of attraction?
Two uniform spheres, each with mass M and radius R, touch one another.
Part A
What is the magnitude of their gravitational force of attraction?
Express your answer in terms of the variables M , R, and appropriate constants.
ANSWER:
Exercise 13.6
Description: Each mass is ## kg. (a) Find the magnitude of the net gravitational force on mass A due to masses B
and C in the figure (a). (b) Find the direction of the net gravitational force on mass A due to masses B and C in the
figure (a). (c) Find the...
Each mass is 1.00kg .
Part A
Find the magnitude of the net gravitational force on mass A due to masses B and C in the figure (a).
ANSWER:
F
= = 6.94×10−9 N Part B
Find the direction of the net gravitational force on mass A due to masses B and C in the figure (a).
ANSWER:
To the left
To the right
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Part C
Find the magnitude of the net gravitational force on mass A due to masses B and C in the figure (b).
ANSWER:
F
= = 6.25×10−9 N Part D
Find the direction of the net gravitational force on mass A due to masses B and C in the figure (b).
ANSWER:
To the left
To the right
Exercise 13.11
Description: (a) At what distance above the surface of the earth is the acceleration due to the earth's gravity g_1 if
the acceleration due to gravity at the surface has magnitude 9.80 m/s^2?
Part A
At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.940m/s 2 if the
acceleration due to gravity at the surface has magnitude 9.80 m/s 2 ?
ANSWER:
= 1.42×107 m Exercise 13.18
Description: Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft
(mass 629 kg) was 2.87 * 10^6 (km) from the earth and traveling at 1.20 * 10^4 (km)/h relative to the earth. (a) At
this time, what was the...
Ten days after it was launched toward Mars in December 1998, the Mars Climate Orbiter spacecraft (mass 629 kg) was 6
4
2.87 × 10
km from the earth and traveling at 1.20 × 10
km/h relative to the earth.
Part A
At this time, what was the spacecraft's kinetic energy relative to the earth?
ANSWER:
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K
= 3.49×109 J Part B
What was the potential energy of the earth­spacecraft system?
ANSWER:
U
= −8.73×107 J Exercise 13.19
Description: (a) For a satellite to be in a circular orbit h above the surface of the earth, what orbital speed must it be
given? (b) What is the period of the orbit (in hours)?
Part A
For a satellite to be in a circular orbit 840km above the surface of the earth, what orbital speed must it be given?
ANSWER:
v orbital
= = 7430 m/s Part B
What is the period of the orbit (in hours)?
ANSWER:
T
= = 1.70 hours Exercise 13.22
Description: The International Space Station makes 15.65 revolutions per day in its orbit around the earth. (a)
Assuming a circular orbit, how high is this satellite above the surface of the earth?
The International Space Station makes 15.65 revolutions per day in its orbit around the earth.
Part A
Assuming a circular orbit, how high is this satellite above the surface of the earth?
ANSWER:
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h
= 370 km Problem 13.53
Description: (a) What is the escape speed from an asteroid of diameter D with a density of rho?
Part A
What is the escape speed from an asteroid of diameter 265km with a density of 2300kg/m3 ?
ANSWER:
v
= = 150 m/s Problem 13.59
Description: An astronaut, whose mission is to go where no one has gone before, lands on a spherical planet in a
distant galaxy. As she stands on the surface of the planet, she releases a small rock from rest and finds that it
takes the rock t to fall 1.90 m. (a) ...
An astronaut, whose mission is to go where no one has gone before, lands on a spherical planet in a distant galaxy. As
she stands on the surface of the planet, she releases a small rock from rest and finds that it takes the rock 0.560s to fall
1.90 m.
Part A
If the radius of the planet is 8.50×107m , what is the mass of the planet?
Express your answer to three significant figures and include the appropriate units.
ANSWER:
mp
= = 1.31×1027
Problem 13.62
Description: The 0.100­kg sphere is released from rest at the position shown in the sketch, with its center 0.400 m
from the center of the 5.00­kg mass. Assume that the only forces on the 0.100­kg sphere are the gravitational
forces exerted by the other two...
The 0.100­kg sphere is released from rest at the position shown in the sketch, with its center 0.400 m from the center of
the 5.00­kg mass. Assume that the only forces on the 0.100­kg sphere are the gravitational forces exerted by the other
two spheres and that the 5.00­kg and 10.0­kg spheres are held in place at their initial positions.
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Part A
What is the speed of the 0.100­kg sphere when it has moved 0.200 m to the left from its initial position?
Express your answer to three significant figures and include the appropriate units.
ANSWER:
v
−5
= 3.33×10
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