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Universal Gravitation
Ck12 Science
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Printed: April 26, 2014
AUTHOR
Ck12 Science
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Chapter 1. Universal Gravitation
C HAPTER
1
Universal Gravitation
C HAPTER O UTLINE
1.1
Kepler’s Laws of Planetary Motion
1.2
Universal Law of Gravity
1.3
Orbital Motion
1.4
References
We know that gravity is the force that keeps us from falling away from the surface of the earth. The force of gravity
reaches a significant distance, though as the distance increases, the force decreases by the square of the distance.
However, Earth’s mass is so large, and its gravity so strong, that the force reaches beyond the surface and atmosphere
into space. The Gemini 4 flight of June 1965 was NASA’s second manned space flight, and circled the Earth 66 times
in four days. Astronaut Edward H. White II was the first American to take a space walk, held in place orbiting the
planet because of the force of gravity.
1
1.1. Kepler’s Laws of Planetary Motion
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1.1 Kepler’s Laws of Planetary Motion
• Describe Kepler’s three laws.
• Calculate satellite periods given average radius and vice versa.
Though a drawing, not an accurate portrayal of the solar system, the elliptical appearance of the orbits is correct.
The elliptical orbits around the sun are not limited to the planets; comets, asteroids, and other orbiting objects also
follow elliptical paths.
Kepler’s Laws
Fifty years before Newton proposed his three laws of motion and his law of universal gravitation, Johannes Kepler
(1571 –1630) published a number of astronomical papers with detailed descriptions of the motions of the planets. Included in those papers were the findings that we now refer to as Kepler’s Laws of Planetary Motion. These
are summarized below.
Kepler’s First Law: The path of each planet around the sun is an ellipse with the sun at one focus.
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Chapter 1. Universal Gravitation
Though it seems at first glance that this law is incorrect (the sun appears to be in the center of our orbit), remember
that a perfect circle is an ellipse with the foci in the same place. Since the Earth’s orbit is nearly circular, the sun
appears to stay in the center.
Kepler’s Second Law: As a planet moves in its orbit, a line from the sun to the planet sweeps out equal areas in
equal times.
The image above illustrates this relationship. Though the green wedges may appear significantly different in area,
Kepler’s Second Law states that the areas are equal if the planet travels along the perimeter of the segments in equal
periods of time. From this, we can clearly see that the planet moves with greater speed when it is near the sun and
slower when it is far away.
Kepler’s Third Law: The ratio of the squares of the periods of any two planets revolving around the sun is equal to
the ratio of the cubes of their average distance from the sun.
T1
T2
2
=
r1
r2
3
This is the only one of Kepler’s three laws that deals with more than one planet at a time.
3
1.1. Kepler’s Laws of Planetary Motion
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This equation can be reworked to reveal that the ratio between the period and the radius of the planet’s orbit is always
the same:
(T1 )2
(r1 )3
=
(T2 )2
(r2 )3
In truth, it has been calculated that this ratio holds for all the planets in our solar system, in addition to moons and
other satellites.
Example Problem: The planet Venus has a mean distance from the sun of 108.2 ×106 km and a period of 0.615
years. The planet Mars has an average mean distance from the sun of 227.9 ×106 km and a period of 1.88 years. Do
these planets follow Kepler’s third law?
Solution: The average mean distance of Venus divided by the average mean distance of Mars = 0.475. The period
of Venus divided by the period of Mars = 0.327.
The square of the period ratio is 0.107 and the cube of the mean distance ratio is 0.107. It is clear that these two
planets follow Kepler’s third law.
Summary
• Kepler’s First Law: The path of each planet around the sun is an ellipse with the sun at one focus.
• Kepler’s Second Law: Each planet moves such that an imaginary line drawn from the sun to the planet sweeps
out equal areas in equal periods of time.
• Kepler’s Third Law: The ratio of the squares of the periods of any two planets revolving around the sun is
equal to the ratio of the cubes of their average distance from the sun.
Practice
The following is a video about Kepler’s three laws. Use this resource to answer the questions that follow.
http://www.youtube.com/watch?v=6TGCPXhMLtU
MEDIA
Click image to the left for more content.
1. What is the shape of a planetary orbit?
2. How are the areas swept out by the line able to be equal, when the line is much longer at some times than
others?
3. What is the T in Kepler’s Third Law? What is the r?
Review
1. The average mean distance of the earth from the sun is 149.6 ×106 km and the period of the earth is 1.0
year. The average mean distance of Saturn from the sun is 1427 ×106 km. Using Kepler’s third law, calculate
the period of Saturn.
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Chapter 1. Universal Gravitation
2. Which of the following is one of Kepler’s Laws of Planetary Motion?
(a)
(b)
(c)
(d)
Planets move in elliptical orbits with the Sun at one focus.
Gravitational force between two objects decreases as the distance squared.
An object in motion remains in motion.
Inner planets orbit in a different direction that outer ones.
3. If a planet’s orbital speed is 20 km/s when it’s at its average distance from the sun, which is most likely orbital
speed when it is nearest the sun?
(a)
(b)
(c)
(d)
10 km/s
15 km/s
20 km/s
25 km/s
• Kepler’s First Law: The path of each planet around the sun is an ellipse with the sun at one focus.
• Kepler’s Second Law: Each planet moves such that an imaginary line drawn from the sun to the planet
sweeps out equal areas in equal periods of time.
• Kepler’s Third Law: The ratio of the squares of the periods of any two planets revolving around the sun is
equal to the ratio of the cubes of their average distance from the sun.
5
1.2. Universal Law of Gravity
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1.2 Universal Law of Gravity
• Describe and give the formula for Newton’s Universal Law of Gravity.
• Using Newton’s Law of Gravity and the equations of uniform circular motion, solve problems involving the
force of gravity.
In 1798, Henry Cavendish designed and created an apparatus and experiment to determine the density of the planet
and the value of the gravitational constant, G. His apparatus involved a light, rigid rod about 2-feet long with two
small lead spheres attached to the ends. The rod was suspended by a thin wire. When the rod rotated, the twisting of
the wire pushed backwards to restore the rod to the original position.
Force of Gravity
In the mid-1600’s, Newton wrote that the sight of a falling apple made him think of the problem of the motion of the
planets. He recognized that the apple fell straight down because the earth attracted it and thought this same force of
attraction might apply to the moon. It further occured to him that motion of the planets might be controlled by the
gravity of the sun. He eventually proposed the universal law of gravitational attraction as
F =G
6
m1 m2
d2
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Chapter 1. Universal Gravitation
where m1 and m2 are the masses being attracted, d is the distance between the centers of the masses, G is the universal
gravitational constant, and F is the force of attraction. The formula for gravitational attraction applies equally to two
rocks resting near each other on the earth and to the planets and the sun. The value for the universal gravitational
constant, G, was determined by Henry Cavendish (using the apparatus described in the introduction) to be 6.67 ×
10−11 N·m2 /kg2 .
The moon is being pulled toward the earth and the earth toward the moon with the same force but in the opposite
direction. The force of attraction between the two bodies produces a greater acceleration of the moon than the earth
because the moon has smaller mass. Even though the moon is constantly falling toward the earth, it never gets
any closer. This is because the velocity of the moon is perpendicular to the radius of the earth (as shown in the
image above) and therefore the moon is moving away from the earth. The distance the moon moves away from the
orbit line is exactly the same distance that the moon falls in the time period. This is true of all satellites and is the
reason objects remain in orbit. In the case of orbiting bodies, the centripetal force is the gravitational force, and they
undergo imperfect circular motion.
Example Problem: Since we know the force of gravity on a 1.00 kg ball resting on the surface of the earth is 9.80
N, and we know the radius of the earth is 6380 km, we can use the equation for gravitational force to calculate the
mass of the earth.
2
(9.80 m/s )(6.38×106 m)2
Fd 2
Solution: me = Gm
=
= 5.98 × 1024 kg
2
1
(6.67×10−11 N·m2 /kg )(1.00 kg)
Sample Problem: John and Jane step onto the dance floor about 20. m apart at the Junior Prom and they feel an
attraction to each other. If John’s mass is 70. kg and Jane’s mass is 50. kg, assume the attraction is gravity and
calculate its magnitude.
Solution: Fg =
Gm1 m2
d2
=
(6.67×10−11
N·m2 /kg2 )(70. kg)(50. kg)
= 1.2 × 10−8 N
(20. m)2
This is an extremely weak force; it is probably not the force of attraction they truly felt.
Summary
• Newton proposed the universal law of gravitational attraction as F = G md1 m2 2 .
• The universal gravitational constant, G, was determined by Cavendish to be 6.67 × 10−11 N·m2 /kg2 .
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1.2. Universal Law of Gravity
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Practice
Use this resource to answer the questions that follow.
https://www.youtube.com/watch?v=p_o4aY7xkXg
MEDIA
Click image to the left for more content.
1.
2.
3.
4.
What is gravity?
What caused the sun to form?
What is the relationship between the strength of a gravitational force and distance?
What is the relationship between the strength of a gravitational force and mass?
Review
1. The earth is attracted to the sun by the force of gravity. Why doesn’t the earth fall into the sun?
2. If the mass of the earth remained the same but the radius of the earth shrank to one-half its present distance,
what would happen to the force of gravity on an object that was resting on the surface of the earth?
3. Lifting an object on the moon requires one-sixth the force that would be required to lift the same object on
the earth because gravity on the moon is one-sixth that on earth. What about horizontal acceleration? If you
threw a rock with enough force to accelerate it at 1.0 m/s2 horizontally on the moon, how would the required
force compare to the force necessary to accelerate the rock in the same way on the earth?
4. The mass of the earth is 5.98 × 1024 kg and the mass of the moon is 7.35 × 1022 kg. If the distance between
the earth and the moon is 384,000 km, what is the gravitational force on the moon?
• gravity: A natural phenomenon by which physical bodies appear to attract each other with a force proportional
to their masses and inversely proportional to the distance separating them.
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Chapter 1. Universal Gravitation
1.3 Orbital Motion
• Learn to calculate how objects are put into orbit.
• Understand why objects in orbit are weightless.
• Using Newton’s Law of Gravity and the equations of uniform circular motion, solve problems involving orbital
velocity and period.
Orbital Motion
We commonly talk about satellites orbiting Earth. But what does that really mean? When a satellite, space shuttle,
or some other object is orbiting a planet, it maintains a circular orbit around the planet a constant distance off the
surface. Manmade satellites typically orbit between 200 and 400 miles. For example, the International Space Station
(ISS) orbits at 370 km, or 230 miles.
The ISS has an average velocity of 7.66 km/sec tangential to its orbit. An orbiting satellite is close enough to be
acted upon by Earth’s gravity. This force is constantly pulling the satellite in toward the center of the earth –it is
a centripetal force and causes a centripetal acceleration. At this height, however, Earth’s gravity is only about 8.7
m/s2 . As was discussed in Motion in Two Dimensions: Circular Motion, the velocity and centripetal acceleration
are perpendicular.
NASA scientists, in designing this and all other satellites, must carefully calculate the velocity necessary to keep the
satellite orbiting. To keep the satellite from falling back to Earth, the horizontal velocity must be large enough. The
satellite must travel far enough horizontally that it follows the curve of the planet, as shown below.
When the satellite is orbiting in this way, it is falling straight down towards Earth. Imagine standing in an elevator
when the bottom drops out from under you. The elevator, you, and anything you might have had with you all fall
straight down. If you had a backpack on your back, the weight of the books can no longer be felt because the books
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1.3. Orbital Motion
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are in free fall as well. A similar thing is happening in the space shuttle or orbiting satellite, where objects are
weightless.
For a full description of weight, look at Forces: Mass versus Weight. In sum, weight is the result of the force of
gravity being opposed by the normal force. As we just learned, objects in orbit are in freefall; they have nothing
exerting a normal force against them and thus no weight.
Sample Problem: Consider a satellite of the earth orbiting at 225 km above the surface of the earth. Keep in mind
that this is NOT the radius of the satellite’s orbit. The satellite’s orbit is measured from the center of the earth, so
its radius will be the radius of the earth, 6.37 × 106 m, plus the 225,000 m. The mass of the earth is 5.98 × 1024
kg. What is the velocity of the satellite?
Solution: Since the centripetal force keeping this satellite traveling in orbit is provided by the gravitational force of
the earth, we can set the formula for centripetal force equal to the formula for gravitational force.
Fc =
mv2
r
and Fg =
Gm1 m2
d2
r
GmE
=
so v =
r
s
(6.67 × 10−11 N · m2 /kg2 )(5.98 × 1024 kg)
= 7770 m/s
v=
6.60 × 106 m
m s v2
r
GmE ms
r2
Sample Problem: Set the equation for Newton’s law of gravity equal to the equation for centripetal force shown
below and derive an equation for the period of a planet around the earth.
Fg =
GmE ms
r2
and Fc =
4π2 msat r
T2
Solution:
s
GmE ms
r2
=
4π2 ms r
T2
and T = 2π
r3
GmE
Summary
• A satellite in a circular orbit accelerates toward the object it is orbiting at a rate equal to the acceleration of
gravity at its orbital radius.
• As in a falling elevator, objects in orbit are in constant freefall.
• Objects in free fall have no normal force acting upon them, and thus no weight.
Practice
Use the resources below to answer the following questions.
http://www.youtube.com/watch?v=N6ZJnIJRUCI
MEDIA
Click image to the left for more content.
The following website has a discussion of weightlessness. http://www.sparknotes.com/testprep/books/sat2/physic
s/chapter11section4.rhtml
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1.
2.
3.
4.
5.
Chapter 1. Universal Gravitation
What are some of the things the astronauts are learning to do/practicing in this video?
These astronauts are in a plane, flying over Earth. How do they simulate weightlessness?
How is a falling elevator similar to an orbiting satellite?
Why does your weight double when the elevator is rising with an acceleration of g?
Why doesn’t the pen fall when you drop it while weightless?
Review
1. Sally’s mass on the earth is 50. kg.
(a) What is her weight on the earth?
(b) What is her weight on the moon?
(c) What is her mass on the moon?
2. The radius of the planet Mercury is 2.43 × 106 m and its mass is 3.2 × 1023 kg.
(a) Find the speed of a satellite in orbit 265,000 m above the surface.
(b) Find the period of the satellite.
3. A geosynchronous orbit is an orbit in which the satellite remains over the same spot on the earth as the earth
turns. This is accomplished by matching the velocity of the satellite to the velocity of the turning earth. The
orbital radius of a geosynchronous satellite is 4.23 × 107 m (measured from the center of the earth).
(a) What is the speed of the satellite in orbit?
(b) What is its period?
Kepler described the motion of the planets back in the 17th century, but it wasn’t until Isaac Newton created his
Law of Universal Gravitation that we had a solid explanation of the why of planetary motion. This chapter explained
Kepler’s Laws of Planetary Motion, and explained that all objects, regardless of size or distance, exert a gravitational
pull on each other. Incorporating the lessons of circular motion from earlier, we explored satellite orbits and why
astronauts experience weightlessness in space.
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1.4. References
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1.4 References
1. Courtesy of NASA. http://spaceflight.nasa.gov/gallery/images/gemini/gemini4/html/s65-30431.html . Public
Domain
2. Image copyright Andrey Burmakin, 2013. http://www.shutterstock.com . Used under license from Shutterstock.com
3. CK-12 Foundation - Joy Sheng. . CC-BY-NC-SA 3.0
4. CK-12 Foundation - Joy Sheng. . CC-BY-NC-SA 3.0
5. Chris Burks (Wikimedia: Chetvorno). http://commons.wikimedia.org/wiki/File:Cavendish_Torsion_Balance
_Diagram.svg . Public Domain
6. CK-12 Foundation - Samantha Bacic. . CC-BY-NC-SA 3.0
7. Courtesy of NASA. http://www.nasa.gov/audience/forstudents/k-4/stories/what-is-the-iss-k4.html . Public
Domain
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