Download - ECTED RECTED PAGE PAGE PROOFS
c04Gravitation 107 24 May 2016 2:12 PM
Understanding gravitational forces has allowed us to put satellites in orbit around the Earth.
REMEMBER
Before beginning this chapter, you should be able to: ■ model forces as vectors acting at the point of application (with magnitude and direction), labelling these forces using the convention ‘force on A by B’ or Fon A by B
■ model the force due to gravity, Fg, as the force of gravity acting at the centre of mass of a body
■ apply Newton’s three laws of motion to a body on which
forces act: aFmnet= , Fon A by B = −Fon B by A
■ analyse uniform circular motion in a horizontal plane ■ resolve vectors into components ■ apply the energy conservation model to energy transfers and transformations.
KEY IDEAS
After completing this chapter, you should be able to: ■ apply Newton’s Law of Universal Gravitation to the motion of planets and satellites
■ describe gravitation using a � eld model ■ describe the gravitational � eld around a point mass in terms of its direction and shape
■ calculate the strength of the gravitational � eld at a point a distance, r, from a point mass
■ analyse the motion of planets and satellites by modelling their orbits as uniform circular orbital motion
■ describe potential energy changes of an object moving in the gravitational � eld of a point mass
■ analyse energy transformations as objects change position in a changing gravitational � eld, using area under a force–distance graph and area under a � eld–distance graph multiplied by mass
■ apply the concepts of force due to gravity, Fg, and normal reaction force, FN, to satellites in orbit.
4 Gravitation
CHAPTER
UNCORRECTED apply Newton’s Law of Universal Gravitation to the motion
UNCORRECTED apply Newton’s Law of Universal Gravitation to the motion
UNCORRECTED
UNCORRECTED
UNCORRECTED PAGE the gravitational � eld of a point mass
PAGE the gravitational � eld of a point massanalyse energy transformations as objects change
PAGE analyse energy transformations as objects change position in a changing gravitational � eld, using area under
PAGE position in a changing gravitational � eld, using area under a force–distance graph and area under a � eld–distance
PAGE a force–distance graph and area under a � eld–distance
PAGE graph multiplied by mass
PAGE graph multiplied by massapply the concepts of force due to gravity,
PAGE apply the concepts of force due to gravity, reaction force,
PAGE reaction force,
PROOFSdescribe gravitation using a � eld model
PROOFSdescribe gravitation using a � eld modeldescribe the gravitational � eld around a point mass in
PROOFSdescribe the gravitational � eld around a point mass in
calculate the strength of the gravitational � eld at a point a
PROOFScalculate the strength of the gravitational � eld at a point a
, from a point mass
PROOFS, from a point mass
analyse the motion of planets and satellites by modelling
PROOFSanalyse the motion of planets and satellites by modelling their orbits as uniform circular orbital motion
PROOFStheir orbits as uniform circular orbital motiondescribe potential energy changes of an object moving in PROOFS
describe potential energy changes of an object moving in the gravitational � eld of a point massPROOFS
the gravitational � eld of a point massanalyse energy transformations as objects change PROOFS
analyse energy transformations as objects change
UNIT 3108
c04Gravitation 108 24 May 2016 2:12 PM
Explaining the solar systemIsaac Newton (1642–1726) published his Law of Universal Gravitation in 1687. � is law provided a mathematical and physical explanation for several impor-tant observations about the movement of planets in the solar system that had been made over the previous two centuries.
In 1542, Nicolas Copernicus (1473–1543) published ‘On the Revolution of the Heavenly Orbs’, outlining an explanation for the observations of planetary motion with the Sun at the centre. In his explanation, the planets moved in circular orbits about the Sun. Copernicus’s model became increasingly pre-ferred over the geocentric model of Ptolemy because it made astronomical and astrological calculations easier. � e publication had a signi� cant scienti� c, social and political impact during the latter part of the 16th century.
Galileo Galilei (1564–1642) was a strong advocate for the view that the Coper-nican model was more than ‘a set of mathematical contrivances, merely to pro-vide a correct basis for calculation’ and instead represented physical reality. (� is had also been Copernicus’s view, but he could not express this in print.) Galileo thought that astronomy could now ask questions about the structure, fabric and operation of the heavens, but as with so many of his scienti� c inter-ests, Galileo did not pursue these questions further.
Johannes Kepler (1571–1630) decided his purpose in life was to reveal the fundamental coherence of a planetary system with the sun as its centre. In 1600–1601 he was working an assistant to Tycho Brahe (1546–1601), a Danish astronomer who had been compiling very precise measurements of the planets’ positions for over twenty years. Brahe’s data was so accurate that they
are still valid today. Without the aid of a telescope, he was able to measure angles to an accuracy of half a minute of arc (for example 23°34′ ± 0.5′).
Kepler was seeking to � nd patterns and relationships between motion of the various planets. He used the data to calculate the positions of the planets as they would be observed by someone outside the solar system, rather than from the revolving platform of the Earth. Initially he was looking for circular orbits, but Brahe’s precise data did not � t such orbits. Eventually he tried other shapes, until in 1604 he formulated what is known as Kepler’s First Law:
Each planet moves, not in a circle, but in an ellipse, with the sun, o� centre, at a focus.
An ellipse is like a stretched circle. � e shape can be drawn by placing two pins on the page several cm apart, with a loose piece of string tied between the pins. If a pencil is placed against the string to keep it tight and then the pencil is moved around the page, the drawn shape is an ellipse with a focus at each of the pins. � e closer the two foci, the more like a circle the ellipse becomes.
Evidence of elliptical orbits� e equinoxes are the two days in the year when the sun is directly above the equator and the durations of night and day are equal. � ey occur when the line drawn from the Sun to the Earth is at right angles to the Earth’s orbit. Because the Earth’s orbit is an ellipse, when the Sun is o� centre at one of the two foci, these two points are not directly opposite each other. � is means the time for the Earth to go from the March equinox to the September equinox is longer by a few days than the time to go from the September equinox to the March equinox.
Drawing an ellipse
Sun
UNCORRECTED arc (for example 23
UNCORRECTED arc (for example 23Kepler was seeking to � nd patterns and relationships
UNCORRECTED Kepler was seeking to � nd patterns and relationships between motion of the various planets. He used the data
UNCORRECTED between motion of the various planets. He used the data
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED PAGE ests, Galileo did not pursue these questions further.
PAGE ests, Galileo did not pursue these questions further.Johannes Kepler (1571–1630) decided his purpose in life was to reveal the
PAGE Johannes Kepler (1571–1630) decided his purpose in life was to reveal the fundamental coherence of a planetary system with the sun as its centre. In
PAGE fundamental coherence of a planetary system with the sun as its centre. In 1600–1601 he was working an assistant to Tycho Brahe (1546–1601), a Danish
PAGE 1600–1601 he was working an assistant to Tycho Brahe (1546–1601), a Danish astronomer who had been compiling very precise measurements of the
PAGE astronomer who had been compiling very precise measurements of the planets’ positions for over twenty years. Brahe’s data was so accurate that they
PAGE planets’ positions for over twenty years. Brahe’s data was so accurate that they
are still valid today. Without the aid of a telescope, he was
PAGE are still valid today. Without the aid of a telescope, he was able to measure angles to an accuracy of half a minute of PAGE able to measure angles to an accuracy of half a minute of arc (for example 23PAGE
arc (for example 23
PROOFScircular orbits about the Sun. Copernicus’s model became increasingly pre-
PROOFScircular orbits about the Sun. Copernicus’s model became increasingly pre-ferred over the geocentric model of Ptolemy because it made astronomical
PROOFSferred over the geocentric model of Ptolemy because it made astronomical and astrological calculations easier. � e publication had a signi� cant scienti� c,
PROOFSand astrological calculations easier. � e publication had a signi� cant scienti� c, social and political impact during the latter part of the 16th century.
PROOFSsocial and political impact during the latter part of the 16th century.Galileo Galilei (1564–1642) was a strong advocate for the view that the Coper-
PROOFSGalileo Galilei (1564–1642) was a strong advocate for the view that the Coper-
nican model was more than ‘a set of mathematical contrivances, merely to pro-
PROOFSnican model was more than ‘a set of mathematical contrivances, merely to pro-vide a correct basis for calculation’ and instead represented physical reality.
PROOFSvide a correct basis for calculation’ and instead represented physical reality. (� is had also been Copernicus’s view, but he could not express this in print.)
PROOFS(� is had also been Copernicus’s view, but he could not express this in print.) Galileo thought that astronomy could now ask questions about the structure,
PROOFSGalileo thought that astronomy could now ask questions about the structure, fabric and operation of the heavens, but as with so many of his scienti� c inter-PROOFS
fabric and operation of the heavens, but as with so many of his scienti� c inter-ests, Galileo did not pursue these questions further.PROOFS
ests, Galileo did not pursue these questions further.Johannes Kepler (1571–1630) decided his purpose in life was to reveal the PROOFS
Johannes Kepler (1571–1630) decided his purpose in life was to reveal the
109CHAPTER 4 Gravitation
c04Gravitation 109 24 May 2016 2:12 PM
Sun
Septemberequinox
Marchequinox
� ere was much speculation in Newton’s time that gravitational attraction might vary inversely with the square of the distance, but it was Newton who was able to show mathematically, using a geometric proof, that the Earth’s elliptical orbit means that the inverse square law applies to the attraction between the Sun and the Earth.
Kepler’s Second LawKepler also looked at the speed of the planets in their orbits. His analysis of the data showed that speeds of the planets were not constant. � e planets were slower when they were further away from the sun and faster when closer. He also found that their angular speed, the number of degrees a line from the sun to planet sweeps through every day, was not constant. Both results reinforced his � rst law. However, he did � nd in 1609 that the planets sweep out equal areas with time.
Kepler’s Second Law: � e linear speed and angular speed of a planet are not constant, but the areal speed of each planet is constant. � at is, a line joining the sun to a planet sweeps out equal areas in equal times.
Sun
PlanetOne month
One month
Newton was able to show mathematically that a constant areal velocity meant that the force acting on a planet must always act along the line joining the planet to the Sun.
Kepler’s Third LawKepler was keen to � nd a mathematical relationship between the period of a planet’s orbit around the sun and its average radius that gave the same result for each planet. He tried numerous possibilities and eventually in 1619 he found a relationship that � tted the data.
Kepler’s � ird Law: For all planets, the cube of the average radius is
proportional to the square of the orbital period; that is, R
T
3
2 is a constant for all
planets going around the sun.
Location of the two equinoxes in the Earth’s orbit
Kepler’s Second Law
Look up the dates of the March and September equinoxes and determine the two times between them.
UNCORRECTED his � rst law. However, he did � nd in 1609 that the planets sweep out equal
UNCORRECTED his � rst law. However, he did � nd in 1609 that the planets sweep out equal
Kepler’s Second Law:
UNCORRECTED Kepler’s Second Law:
UNCORRECTED constant, but the areal speed of each planet is constant.
UNCORRECTED constant, but the areal speed of each planet is constant.the sun to a planet sweeps out equal areas in equal times.
UNCORRECTED the sun to a planet sweeps out equal areas in equal times.
UNCORRECTED
UNCORRECTED
Kepler’s Second Law
UNCORRECTED
Kepler’s Second Law
UNCORRECTED PAGE
Kepler also looked at the speed of the planets in their orbits. His analysis of the
PAGE Kepler also looked at the speed of the planets in their orbits. His analysis of the data showed that speeds of the planets were not constant. � e planets were
PAGE data showed that speeds of the planets were not constant. � e planets were slower when they were further away from the sun and faster when closer. He
PAGE slower when they were further away from the sun and faster when closer. He also found that their angular speed, the number of degrees a line from the sun
PAGE also found that their angular speed, the number of degrees a line from the sun to planet sweeps through every day, was not constant. Both results reinforced PAGE to planet sweeps through every day, was not constant. Both results reinforced his � rst law. However, he did � nd in 1609 that the planets sweep out equal PAGE
his � rst law. However, he did � nd in 1609 that the planets sweep out equal
PROOFS
PROOFS� ere was much speculation in Newton’s time that gravitational attraction
PROOFS� ere was much speculation in Newton’s time that gravitational attraction
might vary inversely with the square of the distance, but it was Newton who
PROOFSmight vary inversely with the square of the distance, but it was Newton who was able to show mathematically, using a geometric proof, that the Earth’s
PROOFSwas able to show mathematically, using a geometric proof, that the Earth’s elliptical orbit means that the inverse square law applies to the attraction
PROOFSelliptical orbit means that the inverse square law applies to the attraction
UNIT 3110
c04Gravitation 110 24 May 2016 2:12 PM
Kepler was also able to show that the relationship held for the orbits of the moons of Jupiter.
Kepler had constructed as detailed a description of the solar system as was possible without a mechanism to explain the motion of the planets, although he did understand gravity as a reciprocal attraction. Kepler wrote, “Gravity is the mutual tendency between bodies towards unity or contact (of which the magnetic force also is), so that the Earth draws a stone much more than the stone draws the Earth . . . ”
TABLE 4.1 The solar system: some useful data
Body Mass (kg)Radius of body (m)
Mean radius of orbit (m)
Period of revolution (s)
Sun 1.98 × 1030 6.95 × 108 Not applicable Not applicable
Earth• Moon
5.98 × 1024
7.35 × 10226.37 × 106
1.74 × 1061.50 × 1011
3.84 × 1083.16 × 106
2.36 × 107
Mercury 3.30 × 1023 2.44 × 106 5.79 × 1010 7.60 × 106
Venus 4.87 × 1024 6.05 × 106 1.08 × 1011 1.94 × 107
Mars 6.42 × 1023 3.40 × 106 2.28 × 1011 5.94 × 107
Jupiter 1.90 × 1026 7.15 × 107 7.78 × 1011 3.74 × 108
Saturn 5.69 × 1026 6.00 × 107 1.43 × 1012 9.30 × 108
Uranus 8.66 × 1025 2.61 × 107 2.87 × 1012 2.65 × 109
Neptune 1.03 × 1025 2.43 × 107 4.50 × 1012 5.20 × 109
Pluto* 1 × 1022 1 × 106 5.90 × 1012 7.82 × 109
*Pluto is no longer classi�ed as a planet. Scientists have recently hypothesised that a ninth planet may exist, but it has not yet been directly observed.
Revision question 4.1
Use the data in table 4.1 to calculate the value of RT
3
2 for each of the planets in
the solar system and therefore con�rm Kepler’s �ird Law.
Newton’s Law of Universal GravitationNewton combined his deductions from Kepler’s Laws with his own Laws of Motion to develop an expression for a law of universal gravitation.
From Kepler’s �rst law, Newton had determined that the force on a planet was inversely proportional to the square of the distance.
∝FR
1on planet by Sun 2
Using his second law of motion, Fnet = ma, Newton reasoned that the force Fon planet by Sun depended on the mass of the planet. By using his third law of motion, Fon planet by Sun = −Fon Sun by planet, he reasoned that the force Fon Sun by planet depended on the mass of the sun.
Combining these two statements produces:
Gravitational force between the sun and the planet ∝ ×R
MassMass
.Sunplanet2
In general,
=Fm m
R
G 1 22
where G is the universal gravitational constant and m1 and m2 are the masses of any two objects.
Unit 3 Newton’s Law of Universal GravitationConcept summary and practice questions
AOS 1
Topic 2
Concept 1
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED planet may exist, but it has not yet been directly observed.
UNCORRECTED planet may exist, but it has not yet been directly observed.
Revision question 4.1
UNCORRECTED Revision question 4.1
Use the data in table 4.1 to calculate the value of
UNCORRECTED Use the data in table 4.1 to calculate the value of
the solar system and therefore con�rm Kepler’s �ird Law.
UNCORRECTED the solar system and therefore con�rm Kepler’s �ird Law.
UNCORRECTED
N
UNCORRECTED
Newton’s Law of Universal Gravitation
UNCORRECTED
ewton’s Law of Universal GravitationNewton combined his deductions from Kepler’s Laws with his own Laws of
UNCORRECTED
Newton combined his deductions from Kepler’s Laws with his own Laws of
UNCORRECTED
UNCORRECTED
L
UNCORRECTED
Law
UNCORRECTED
aw of Universal
UNCORRECTED
of UniversalGravitation
UNCORRECTED
GravitationConcept summary
UNCORRECTED
Concept summary and practice
UNCORRECTED
and practice questions
UNCORRECTED
questions
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE 10
PAGE 107
PAGE 7
×
PAGE × 10
PAGE 107
PAGE 7
2.61
PAGE 2.61 ×
PAGE × 10
PAGE 10
PAGE 7
PAGE 7
2.43
PAGE 2.43 ×
PAGE × 10
PAGE 107
PAGE 7
1
PAGE 1 ×
PAGE × 10
PAGE 10
*Pluto is no longer classi�ed as a planet. Scientists have recently hypothesised that a ninth PAGE *Pluto is no longer classi�ed as a planet. Scientists have recently hypothesised that a ninth planet may exist, but it has not yet been directly observed.PAGE
planet may exist, but it has not yet been directly observed.
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFSPeriod of
PROOFSPeriod of revolution (s)
PROOFSrevolution (s)
Not applicable Not applicable
PROOFSNot applicable Not applicableNot applicable Not applicable
PROOFSNot applicable Not applicable
10
PROOFS 1011
PROOFS11
×
PROOFS× 10
PROOFS 108
PROOFS8
5.79
PROOFS5.79 ×
PROOFS× 10
PROOFS 1010
PROOFS10
1.08
PROOFS1.08 ×
PROOFS× 10
PROOFS 1011
PROOFS11
2.28 PROOFS
2.28 ×PROOFS
× 10PROOFS
10
7.78 PROOFS
7.78
111CHAPTER 4 Gravitation
c04Gravitation 111 24 May 2016 2:12 PM
�e value of G could not be determined at the time because the mass of the Earth was not known. It took another 130 years before Henry Cavendish was able to measure the gravitational attraction between two known masses and calculate the value of G.
�e value of G is 6.674 × 10−11 N m2 kg−2. Alternatively, replacing newtons with kg m s−2, G = 6.674 × 10−11 m3 kg−1 s−2.
�e value of G is very small, which indicates that gravitation is quite a weak force. A large quantity of mass is need to produce a gravitational e�ect that is easily noticeable.
Sample problem 4.1
Calculate the force due to gravity of:(a) Earth on a 70 kg person standing on the equator(b) a 70 kg person standing on the equator on Earth.
(a) mEarth = 5.98 × 1024 kg, mperson = 70 kg, radiusEarth = 6.38 × 106 m,
G = 6.67 × 10−11 N m2 kg−2
=
=× × × ×
×=
− −
Fm m
r
G
6.67 10 N m kg 5.98 10 kg 70 kg
(6.38 10 m)
686 N towards the centre of Earth
Earth person2
11 2 2 24
6 2
(b) Newton’s �ird Law of Motion states that if one object exerts a force on another object, then the other object exerts an equal and opposite force on the �rst object. In this situation, if Earth is exerting a force of 686 N down-wards on the person, then the person is exerting a 686 N force upwards on Earth! �e same result could be calculated with the formula used in part (a).
Revision question 4.2
Use the data in table 4.1 to calculate the force due to gravity by:(a) the Earth on the Moon(b) the Moon on Earth.
The falling appleNewton published his law of universal gravitation in 1687, in his famous book titled Philosophiae Naturalis Principia Mathematica. In this book he included an anecdote about observing an apple falling from a tree. �ere is no record of such an event in his earlier papers, so the story may just have served an explanatory purpose. Nevertheless, the story is instructive.
Newton said he observed that the falling apple had been pulled from the tree by the attractive force of the Earth, which had acted across the space between the Earth’s surface and the apple, that is, ‘action at a distance’. He speculated that the e�ect of the Earth’s pull might reach higher into the atmosphere, possibly beyond the atmosphere to the moon. He had previously applied his Laws of Motion to circular motion and developed expressions for the inward acceleration,
av
R
2
= and = πa
R
T
4 2
2.
�e obvious questions that arise from this are: 1. How does the acceleration of the apple compare to that of the moon?2. How are these two values related to their respective distances from the
centre of the Earth?
Solution:
UNCORRECTED
UNCORRECTED
UNCORRECTED another object, then the other object exerts an equal and opposite force on
UNCORRECTED another object, then the other object exerts an equal and opposite force on the �rst object. In this situation, if Earth is exerting a force of 686
UNCORRECTED the �rst object. In this situation, if Earth is exerting a force of 686wards on the person, then the person is exerting a 686
UNCORRECTED wards on the person, then the person is exerting a 686Earth! �e same result could be calculated with the formula used in part (a).
UNCORRECTED Earth! �e same result could be calculated with the formula used in part (a).
Revision question 4.2
UNCORRECTED Revision question 4.2
UNCORRECTED
Use the data in table 4.1 to calculate the force due to gravity by:
UNCORRECTED
Use the data in table 4.1 to calculate the force due to gravity by:(a)
UNCORRECTED
(a) th
UNCORRECTED
the Earth on the Moon
UNCORRECTED
e Earth on the Moon(b)
UNCORRECTED
(b) th
UNCORRECTED
th
UNCORRECTED PAGE
PAGE × ×
PAGE × × 5.98 10
PAGE 5.98 10× ×5.98 10× ×
PAGE × ×5.98 10× ×(6.38 10
PAGE (6.38 10×(6.38 10×
PAGE ×(6.38 10× m)
PAGE m)
th
PAGE the c
PAGE e centr
PAGE entre o
PAGE e of E
PAGE f E
6 2
PAGE 6 2m)6 2m)
PAGE m)6 2m)
ewton’s �ird Law of Motion states that if one object exerts a force on PAGE ewton’s �ird Law of Motion states that if one object exerts a force on
another object, then the other object exerts an equal and opposite force on PAGE
another object, then the other object exerts an equal and opposite force on the �rst object. In this situation, if Earth is exerting a force of 686PAGE
the �rst object. In this situation, if Earth is exerting a force of 686
PROOFS
PROOFS
PROOFSg person standing on the equator on Earth.
PROOFSg person standing on the equator on Earth.
g, radius
PROOFSg, radiusEarth
PROOFSEarth =
PROOFS= 6.38
PROOFS 6.38
UNIT 3112
c04Gravitation 112 24 May 2016 2:12 PM
TABLE 4.2 The relationship between the Earth, an apple and the Moon
Body Apple Moon
Distance to the centre of the Earth (m) 6.38 × 106 3.84 × 108
Period of orbit (s) 2.36 × 106
Acceleration towards the centre of the Earth (m s−2) 9.8 m s−2 2.72 × 10−3
With the data from the table, we can make the following calculations:
= ××
=
distance of the Moon from the centre of the Earth
distance of the apple from the centre of the Earth
3.84 10
6.38 10
60.1
8
6
=×
=−
acceleration of the apple towards the centre of the Earth
acceleration of the Moon towards the centre of the Earth
9.8
2.72 103603
3
�e value of the second ratio, 3603, is very close to 60.12.�e ratio of the accelerations is the square of the ratio of the distances, but
note that the Moon is in the numerator for the �rst ratio, while it is in the denominator for the second ratio. Newton used this calculation to show that the gravitational force is inversely proportional to the square of the separation of the two masses.
Newton’s expression for the centripetal acceleration, aR
T
4 2
2
π= , was used to
con�rm Kepler’s �ird Law, that R
T
3
2 is a constant for all planets or satellites
orbiting a central body.
π
=
× =
F F
mR
T
M m
R
4 G
net g
planet
2
2Sun planet
2
Cancelling mplanet and rearranging gives
π=
R
T
MG
4
3
2Sun2
which depends only on the mass of the Sun and thus has the same value for all planets orbiting the Sun.
Similarly, the Moon and all other satellites orbiting the Earth will have the
same value for R
T
3
2, though in this case the value will equal
MG
4Earth
2π.
Sample problem 4.2
Calculate the value of R
T
3
2 for the Moon using the data in table 4.1 and use that
value to calculate the mass of the Earth.
Radius of Moon’s orbit, R = 3.84 × 108 m; period, T = 2.36 × 106 s; G = 6.67 × 10−11 Nm2 kg−2; mass of Earth, MEarth = ?
=××
= × −
R
T
(3.84 10 m)
(2.36 10 s)
1.02 10 m s
3
2
8 3
6 2
13 3 2
Solution:
UNCORRECTED con�rm Kepler’s �ird Law, that
UNCORRECTED con�rm Kepler’s �ird Law, that orbiting a central body.
UNCORRECTED orbiting a central body.
π
UNCORRECTED π
UNCORRECTED × =
UNCORRECTED × =× =
UNCORRECTED × =
F F
UNCORRECTED F F=F F=
UNCORRECTED =F F=
R
UNCORRECTED R
T
UNCORRECTED T
4
UNCORRECTED 4 G
UNCORRECTED G
ne
UNCORRECTED neF FneF F
UNCORRECTED F FneF Ft g
UNCORRECTED t gF Ft gF F
UNCORRECTED F Ft gF F=F F=t g=F F=
UNCORRECTED =F F=t g=F F=
anet
UNCORRECTED anet
2
UNCORRECTED 2
2
UNCORRECTED 2
× =2
× =
UNCORRECTED × =
2× =
Cancelling
UNCORRECTED Cancelling
UNCORRECTED
R
UNCORRECTED
R
T
UNCORRECTED
T
3
UNCORRECTED
3
2
UNCORRECTED
2
PAGE �e ratio of the accelerations is the square of the ratio of the distances, but
PAGE �e ratio of the accelerations is the square of the ratio of the distances, but note that the Moon is in the numerator for the �rst ratio, while it is in the
PAGE note that the Moon is in the numerator for the �rst ratio, while it is in the denominator for the second ratio. Newton used this calculation to show that
PAGE denominator for the second ratio. Newton used this calculation to show that the gravitational force is inversely proportional to the square of the separation
PAGE the gravitational force is inversely proportional to the square of the separation
Newton’s expression for the centripetal acceleration,
PAGE Newton’s expression for the centripetal acceleration,
con�rm Kepler’s �ird Law, that PAGE con�rm Kepler’s �ird Law, that
PROOFS
PROOFS6.38 10
PROOFS6.38 10×6.38 10×
PROOFS×6.38 10×60.1
PROOFS60.1
6
PROOFS6
PROOFS
PROOFS=
PROOFS=he Eart
PROOFShe Earth
PROOFShhe Earthhe Eart
PROOFShe Earthhe Eart
th
PROOFSthe E
PROOFSe Earth
PROOFSarth
�e value of the second ratio, 3603, is very close to 60.1PROOFS
�e value of the second ratio, 3603, is very close to 60.1�e ratio of the accelerations is the square of the ratio of the distances, but PROOFS
�e ratio of the accelerations is the square of the ratio of the distances, but
113CHAPTER 4 Gravitation
c04Gravitation 113 24 May 2016 2:12 PM
Using π
=R
T
MG
4,
3
2Earth
2
MR
T
4
G(3.84 10 m) 4
(2.36 10 s) 6.67 10 N m kg
6.02 10 kg
Earth
3
2
2
8 3 2
6 2 11 2 2
24
π
π
= ×
=× ×
× × ×
= ×
− −
Revision question 4.3
Use the values in table 4.1 to calculate the value of R
T
3
2 for the planets, and use
the average value to calculate the mass of the Sun.
Newton’s Law of Universal Gravitation can also be used to calculate the average speed of planets around the Sun. �is involves using the other expression for the centripetal acceleration.
=
× =
F F
mv
R
M m
R
G
net g
planet
2Sun planet
2
Cancelling mplanet and rearranging gives
=
=
vM
R
vM
R
G
G.
2 Sun
Sun
Sample problem 4.3
Calculate the average speed of the Earth around the Sun using the values in table 4.1.
Radius of Earth’s orbit, R = 1.50 × 1011 m; mass of Sun, MSun = 1.98 × 1030 kg; G = 6.67 × 10−11 Nm2 kg−2; speed, v = ?
Using =
= × × ××
= ×
− −
−
vM
R
v
G,
6.67 10 N m kg 1.98 10 kg
1.50 10 m
2.97 10 m s .
Sun
11 2 2 30
11
4 1
Revision question 4.4
Use the values in table 4.1 to calculate the average speed of the other planets around the Sun. Graph the average speed as a function of average orbital radius. Does the graph �t your expectation? Are there any outliers in the data? If so, suggest an explanation.
Graphing the gravitational force�e gravitational force is an attractive force, whereas the force between electric charges can be either attractive or repulsive.
Solution:
UNCORRECTED
UNCORRECTED
UNCORRECTED R
UNCORRECTED R
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED .
UNCORRECTED .n
UNCORRECTED n
Sample problem 4.3
UNCORRECTED Sample problem 4.3
Calculate the average speed of the Earth around the Sun using the values in
UNCORRECTED Calculate the average speed of the Earth around the Sun using the values in table 4.1.
UNCORRECTED
table 4.1.
Radius of Earth’s orbit,
UNCORRECTED
Radius of Earth’s orbit, G
UNCORRECTED
G =
UNCORRECTED
=Solution:
UNCORRECTED
Solution:
PAGE and rearranging gives
PAGE and rearranging gives
PROOFS
PROOFS
PROOFS for the planets, and use
PROOFS for the planets, and use
PROOFSNewton’s Law of Universal Gravitation can also be used to calculate the average
PROOFSNewton’s Law of Universal Gravitation can also be used to calculate the average speed of planets around the Sun. �is involves using the other expression for
PROOFS
speed of planets around the Sun. �is involves using the other expression for
UNIT 3114
c04Gravitation 114 24 May 2016 2:12 PM
The Earth exerts gravitational force on the Moon.
Distance vector Force vector
Earth
Moon
For the force the Earth exerts on the Moon, there is a distance vector from the centre of the Earth to the centre of the Moon, whereas the force vector points in the opposite direction, back to the Earth. For this reason, the gravi-tational force equation should really have a negative sign and the force should be graphed under the distance axis. � us, more correctly,
= −Fm m
R
G.g
1 22
This diagram shows how the Earth’s gravitational force varies with distance.
RE
Fg = −GMmRE
2
R
g
0
Earth
� e straight blue line in the graph shows how the gravitational force by the Earth on you would decrease if you were to drill down to the centre of the earth. Newton calculated that if you were inside a hollow sphere, the gravi-tational force from the mass in the shell would cancel out, no matter where you were inside the sphere. � is means that if you were inside the Earth, only the mass in the inner sphere between you and the centre of the Earth would exert a gravitational force on you. � is force will get smaller the closer to the centre you go, and at the centre of the Earth the gravitational force will be zero.
Gravitational � eldsNewton’s Law of Universal Gravitation describes the force between two masses. However, the solar system has many masses, each attracting each other. � e sun, the heaviest object in the solar system, determines the orbits of all the other masses, but each planet can cause minor variations in the orbital paths of the other planets. Precise calculation of the path of a planet or comet
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED PAGE R
PAGE RE
PAGE E
0 PAGE 0 PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE PROOFSFor the force the Earth exerts on the Moon, there is a distance vector from
PROOFSFor the force the Earth exerts on the Moon, there is a distance vector from the centre of the Earth to the centre of the Moon, whereas the force vector
PROOFSthe centre of the Earth to the centre of the Moon, whereas the force vector points in the opposite direction, back to the Earth. For this reason, the gravi-
PROOFSpoints in the opposite direction, back to the Earth. For this reason, the gravi-tational force equation should really have a negative sign and the force should
PROOFStational force equation should really have a negative sign and the force should be graphed under the distance axis. � us, more correctly,
PROOFSbe graphed under the distance axis. � us, more correctly,
115CHAPTER 4 Gravitation
c04Gravitation 115 24 May 2016 2:12 PM
becomes a complicated exercise with many gravitational forces needing to be considered.
Physicists after Newton realised it was easier to determine for each point in space the total force that would be experienced by a unit mass, that is, 1 kilo-gram, at that point. � is idea slowly developed and in 1849 Michael Faraday, in explaining the interactions between electric charges and between magnets, formalised the concept, calling it a ‘� eld’.
A � eld is more precisely de� ned as a physical quantity that has a value at each point in space. For example, a weather map showing the pressure across Australia could be described as a diagram of a pressure � eld. � is is an example of a scalar � eld. In contrast, gravitational, electric and magnetic � elds are vector � elds; they give a value to the strength of the � eld at each point in space, and also a direction for that � eld at that point. For example, the arrows in the diagram of the Earth’s gravitational � eld show the direction of the � eld, and the density of the lines (how close together the lines are) indicates the strength of the � eld.
Diagram of the Earth’s gravitational � eld
lines of equal �eld strength
A value for the strength of the gravitational � eld around a mass M can be determined from the value of the force on a unit mass in the � eld. If the mass m2 in Newton’s Universal Law of Gravitation equation is assigned a value of 1 kg, then the force expression will give the strength of the gravi-tational � eld.
Gravitational � eld strength, = −gM
R
G2
� e unit of gravitational � eld strength is Newtons per kilogram, N kg−1.
Unit 3 Classical � eld modelsConcept summary and practice questions
AOS 1
Topic 1
Concept 1
Unit 3 Characteristics of � eldsConcept summary and practice questions
AOS 1
Topic 1
Concept 2
Unit 3 Gravitational � eld strengthConcept summary and practice questions
AOS 1
Topic 2
Concept 2
UNCORRECTED
UNCORRECTED
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UNCORRECTED
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UNCORRECTED
UNCORRECTED
UNCORRECTED
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UNCORRECTED
UNCORRECTED PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE PROOFS
across Australia could be described as a diagram of a pressure � eld. � is is an
PROOFSacross Australia could be described as a diagram of a pressure � eld. � is is an example of a scalar � eld. In contrast, gravitational, electric and magnetic � elds
PROOFSexample of a scalar � eld. In contrast, gravitational, electric and magnetic � elds are vector � elds; they give a value to the strength of the � eld at each point in
PROOFSare vector � elds; they give a value to the strength of the � eld at each point in space, and also a direction for that � eld at that point. For example, the arrows
PROOFSspace, and also a direction for that � eld at that point. For example, the arrows in the diagram of the Earth’s gravitational � eld show the direction of the � eld,
PROOFSin the diagram of the Earth’s gravitational � eld show the direction of the � eld, and the density of the lines (how close together the lines are) indicates the
PROOFSand the density of the lines (how close together the lines are) indicates the
PROOFS
PROOFSlines of equal �eld strength
PROOFSlines of equal �eld strength
UNIT 3116
c04Gravitation 116 24 May 2016 2:12 PM
� e strength of the gravitational � eld at the Earth’s surface can be calculated using the values for the mass and radius of the Earth from table 4.1:
Gravitational � eld strength, = − × × ××
= −
− −
−
g6.67 10 N m kg 5.98 10
(6.38 10 )
9.80 N kg .
11 2 2 24
6 2
1
� is is the acceleration due to gravity at the Earth’s surface.
Graph of the magnitude of the strength of the Earth’s gravitational � eld
R (m)REGra
vita
tiona
l �el
d s
tren
gth
(N k
g−1 )
2RE 3RE
9.8
9.84
9.89
Revision question 4.5
(a) Use the data in table 4.1 to calculate the gravitational � eld strength on the
surface of the Moon. Show that it is about 16
of the Earth’s gravitational � eld at its surface.
(b) Determine which planet has the largest gravitational � eld strength at its surface. Table 4.1 is also available as a spreadsheet in your eBookPLUS.
At the time Newton developed his Law of Universal Gravitation, he knew it did not provide an explanation for how gravity works, that is, how ‘action at a distance’ was achieved.
It is inconceivable . . . that Gravity should be innate, inherent and essential to Matter, so that one body may act upon another at a distance thro’ a Vacuum, without the Mediation of any thing else . . . is to me so great an Absurdity that I believe no Man who has in philosophical Matters a competent Faculty of thinking can ever fall into it. Gravity must be caused by an Agent acting constantly according to certain laws; but whether this Agent be material or immaterial, I have left to the Consider-ation of my readers. Newton, 1692
� e concept of a � eld now provides an explanation for action at a distance.
Kinetic energy and potential energy in a gravitational � eldConsider the following scenarios.1. On 15 February 2013, an asteroid approached the Earth, gaining speed in
the Earth’s gravitational � eld. By the time it reached the atmosphere, it was travelling at a speed of 19 km s−1. With a mass of about 1.2 × 107 kg, its kinetic energy was about 2.2 × 1015 J. It exploded about 30 km above Chelyabinsk in Russia.
Unit 3 Gravitational potential energyConcept summary and practice questions
AOS 1
Topic 2
Concept 5
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED Revision question 4.5
UNCORRECTED Revision question 4.5
Use the data in table 4.1 to calculate the gravitational � eld strength on the
UNCORRECTED Use the data in table 4.1 to calculate the gravitational � eld strength on the
surface of the Moon. Show that it is about
UNCORRECTED surface of the Moon. Show that it is about
at its surface.
UNCORRECTED at its surface.
Determine which planet has the largest gravitational � eld strength at its
UNCORRECTED Determine which planet has the largest gravitational � eld strength at its
surface. Table 4.1 is also available as a spreadsheet in your eBookPLUS.
UNCORRECTED surface. Table 4.1 is also available as a spreadsheet in your eBookPLUS.
UNCORRECTED
At the time Newton developed his Law of Universal Gravitation, he knew it
UNCORRECTED
At the time Newton developed his Law of Universal Gravitation, he knew it did not provide an explanation for how gravity works, that is, how ‘action at a
UNCORRECTED
did not provide an explanation for how gravity works, that is, how ‘action at a
PAGE
PAGE
PAGE
PAGE Graph of the magnitude of the strength of the Earth’s
PAGE Graph of the magnitude of the strength of the Earth’s gravitational � eld
PAGE gravitational � eld
PAGE
PAGE E
PAGE E 3
PAGE 3R
PAGE RE
PAGE E
PAGE
Revision question 4.5PAGE
Revision question 4.5
PROOFS
PROOFS
117CHAPTER 4 Gravitation
c04Gravitation 117 24 May 2016 2:12 PM
2. Edmund Halley used Newton’s law of gravitation to calculate the e� ect of Jupiter and Saturn on the orbits of comets. He concluded that comet sightings in 1531 and 1607 were sightings of the same comet and that it should appear again in 1758. We now call it Halley’s Comet.
Halley’s Comet orbits the sun every 75.3 years in a very stretched path. � e closest it gets to the Sun is about 0.6 times the radius of the Earth’s orbit, while its furthest distance is 35 times the radius. At its closest approach its kinetic energy is 1.6 × 1023 J with a speed of 38 km s−1, but at its furthest it has only 4.5 × 1019 J, travelling at 0.64 km s−1.
3. Ignoring the initial air resistance, a rock thrown at 11 km s−1 would eventu-ally escape the e� ect of the Earth’s gravitational pull, slowing down to zero only at an in� nite distance away.
In each of these scenarios there are changes in speed and height, and thus in kinetic energy and gravitational potential energy. How do we describe these changes using Newton’s law of gravitation?
� e change in gravitational potential energy can be obtained from the area under a force–distance graph. Because gravitation is an attractive force, the force–distance graph is below the distance axis and the area under the graph has a negative value.
Change in potential energy needs a reference point or zero point. � e Earth’s surface is an obvious reference point for objects on or near the Earth; we can assume a constant value of 9.8 N kg−1 for the strength of the gravitational � eld at the Earth’s surface. But out in space, with the gravitational force getting weaker with distance, the preferred reference point is at in� nity where the gravitational force is zero. � is means that the gravitational potential energy of a mass at a distance R from the Earth is the area under the graph from the distance R out to in� nity.
Graph of the strength of the Earth’s gravitational force. The gravitational potential energy of a mass at distance R from the Earth is equal to the shaded area.
r (m)
F g (N
)
R
Let’s look at situation 1, the Chelyabinsk asteroid. When the asteroid is some distance from Earth, its kinetic energy is relatively small. As it falls towards the Earth, its kinetic energy increases, its gravitational potential energy becomes more negative, and the total energy remains the same throughout. As the asteroid falls from A to B in the � gures on the following page, the orange shaded area in the third graph is the gain in kinetic energy. � at is,
change in GPE = change in KE.
UNCORRECTED out to in� nity.
UNCORRECTED out to in� nity. PAGE Change in potential energy needs a reference point or zero point. � e Earth’s
PAGE Change in potential energy needs a reference point or zero point. � e Earth’s surface is an obvious reference point for objects on or near the Earth; we can
PAGE surface is an obvious reference point for objects on or near the Earth; we can assume a constant value of 9.8 N kg
PAGE assume a constant value of 9.8 N kg−
PAGE −1
PAGE 1 for the strength of the gravitational � eld
PAGE for the strength of the gravitational � eld at the Earth’s surface. But out in space, with the gravitational force getting
PAGE at the Earth’s surface. But out in space, with the gravitational force getting weaker with distance, the preferred reference point is at in� nity where the
PAGE weaker with distance, the preferred reference point is at in� nity where the gravitational force is zero. � is means that the gravitational potential energy
PAGE gravitational force is zero. � is means that the gravitational potential energy of a mass at a distance PAGE of a mass at a distance R PAGE
R from the Earth is the area under the graph from the PAGE from the Earth is the area under the graph from the
out to in� nity. PAGE out to in� nity.
PROOFS would eventu-
PROOFS would eventu-
ally escape the e� ect of the Earth’s gravitational pull, slowing down to zero
PROOFSally escape the e� ect of the Earth’s gravitational pull, slowing down to zero
In each of these scenarios there are changes in speed and height, and thus
PROOFSIn each of these scenarios there are changes in speed and height, and thus in kinetic energy and gravitational potential energy. How do we describe these
PROOFSin kinetic energy and gravitational potential energy. How do we describe these
� e change in gravitational potential energy can be obtained from the area
PROOFS� e change in gravitational potential energy can be obtained from the area
under a force–distance graph. Because gravitation is an attractive force, the
PROOFSunder a force–distance graph. Because gravitation is an attractive force, the force–distance graph is below the distance axis and the area under the graph
PROOFSforce–distance graph is below the distance axis and the area under the graph
Change in potential energy needs a reference point or zero point. � e Earth’s PROOFS
Change in potential energy needs a reference point or zero point. � e Earth’s surface is an obvious reference point for objects on or near the Earth; we can PROOFS
surface is an obvious reference point for objects on or near the Earth; we can
UNIT 3118
c04Gravitation 118 24 May 2016 2:12 PM
r (m)
F g (N
)
A r (m)
F g (N
)
B
Change in PE
r (m)
F g (N
)
B A
+
At A At B
=
Ene
rgy
+
−GPE GPE
KE
Total
The total energy at B equals the total energy at A.
Sample problem 4.4
A mass of 10 kg falls to the surface of the Earth from an altitude equal to two Earth radii. What is the gain in kinetic energy?
� ere are three methods, two of which give an approximate answer. � e accu-racy of each depends on the care you take.
Method 1
Use this method if the graph has a relatively coarse grid.• Divide the area up into simple geometric shapes such as rectangles and
triangles.• Calculate the area of each shape using graph-based units.
Solution:
UNCORRECTED
Ene
rgy
UNCORRECTED
Ene
rgy
+
UNCORRECTED +
UNCORRECTED PAGE Change in PE
PAGE Change in PE
PAGE PROOFS
PROOFS
Change in PE PROOFS
Change in PE
r
PROOFSr (m)
PROOFS(m)
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
119CHAPTER 4 Gravitation
c04Gravitation 119 24 May 2016 2:12 PM
• Total the areas.• Convert the total area to SI units for energy.
Take care in deciding the height of the rectangles or triangles so that their areas (approximating the area under the curve) will produce more representative results.
0 1 2 3 4 5 6 7 8 9 10
−20
−40
Gra
vita
tiona
l for
ce o
n 10
kg
mas
s b
y E
arth
(N)
Distance from the centre of the Earth (in Earth radii)
−60
−80
−100
−120
4
1
2
3
= ×== ×=
= × ×
=
Area1 (blue) 40 0.520 energy units
Area 2 (purple) 10 1.515 energy units
Area 3 (orange)1
224 1.5
18 energy units
(Note: �e triangle with area 1
230 1.5× × would be larger than the orange area,
so the height of 30 was reduced to a level where the areas matched.)
= × ×
=
Area 4 (yellow)1
253 0.5
13.25 energy units
Total area = 20 + 15 + 18 + 13.25
= 66.25 energy units
= ×= × ×= ×
1 energy unit 1 N 1 Earth radius1 N 6.38 10 m6.38 10 J
6
6
�erefore, the kinetic energy gained = 66.25 × 6.38 × 106
= 4.23 × 108 J.
UNCORRECTED = ×
UNCORRECTED = ×=
UNCORRECTED =
blue
UNCORRECTED blue) 4
UNCORRECTED ) 4= ×) 4= ×
UNCORRECTED = ×) 4= ×0 0.5
UNCORRECTED 0 0.5= ×0 0.5= ×
UNCORRECTED = ×0 0.5= ×
20
UNCORRECTED 20 energy
UNCORRECTED energy
2 (
UNCORRECTED 2 (purp
UNCORRECTED purple
UNCORRECTED le) 1
UNCORRECTED ) 1= ×) 1= ×
UNCORRECTED = ×) 1= ×
Area
UNCORRECTED Area 3 (
UNCORRECTED 3 (or
UNCORRECTED orange)
UNCORRECTED ange)
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
UNIT 3120
c04Gravitation 120 24 May 2016 2:12 PM
Method 2
Use this method when the graph has a relatively �ne grid.
0
−20
−40
Gra
vita
tiona
l for
ce o
n 10
kg
mas
s b
y E
arth
(N)
1 2 3 4 5 6 7 8 9 10
Distance from the centre of the Earth (in Earth radii)
−60
−80
−100
−120
• Count the number of small squares between the graph and the zero-value line or horizontal axis. Tick each one as you count it to avoid counting it twice. For partial squares, �nd two that add together to make one square and tick both.
• Calculate the area of one small square.• Multiply the area of one small square by the number of small squares.
Number of small squares = 80.5
= × ×
= × × ×
= ×
Area of one small square 4 N 0.2 1 Earth radius
4 N 0.2 6.38 10 m
5.1 10 J
6
6
�erefore, the gain in kinetic energy = 80.5 × 5.1 × 106 J = 4.11 × 108 J.
Method 2 can be very accurate, but it is laborious.
Method 3• Print out the graph.• Cut out the required shape.• Measure the mass of the shape with a top-loading balance.• Using the mass of a piece of the same paper with known dimensions, calcu-
late the area of the cut-out shape.• Use the scales on the axes of the graph to determine the value for the area
under the graph.
Revision question 4.6
(a) Use the graph of the gravitational force on the Chelyabinsk asteroid (shown on the next page) to show that in moving from an altitude of two Earth radii down to an altitude of one Earth radius, it gained 1.25 × 1014 joules of kinetic energy.
(b) Use the graph to �nd, to the nearest whole number, approximately how much kinetic energy was gained from falling from an altitude of one Earth radius to the Earth’s surface. Compare this value with your answer to (a) above.
UNCORRECTED alculate the area of one small square.
UNCORRECTED alculate the area of one small square.ultiply the area of one small square by the number of small squares.
UNCORRECTED ultiply the area of one small square by the number of small squares.
Number of small squares
UNCORRECTED Number of small squares
= ×
UNCORRECTED = ×
= ×
UNCORRECTED = ×
of on
UNCORRECTED of one s
UNCORRECTED e small
UNCORRECTED mall
4 N
UNCORRECTED 4 N= ×4 N= ×
UNCORRECTED = ×4 N= × 0.2 6.38 10 m
UNCORRECTED 0.2 6.38 10 m× ×0.2 6.38 10 m× ×
UNCORRECTED × ×0.2 6.38 10 m× ×
5.1 10 J
UNCORRECTED 5.1 10 J= ×5.1 10 J= ×
UNCORRECTED = ×5.1 10 J= × 6
UNCORRECTED 6
�erefore, the gain in kinetic energy
UNCORRECTED
�erefore, the gain in kinetic energy
M
UNCORRECTED
Method 2 can be very accurate, but it is laborious.
UNCORRECTED
ethod 2 can be very accurate, but it is laborious.
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE
PAGE ount the number of small squares between the graph and the zero-value
PAGE ount the number of small squares between the graph and the zero-value
line or horizontal axis. Tick each one as you count it to avoid counting it
PAGE line or horizontal axis. Tick each one as you count it to avoid counting it twice. For partial squares, �nd two that add together to make one square
PAGE twice. For partial squares, �nd two that add together to make one square
alculate the area of one small square.PAGE
alculate the area of one small square.ultiply the area of one small square by the number of small squares.PAGE
ultiply the area of one small square by the number of small squares.
PROOFS
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PROOFS
PROOFS
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PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
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PROOFS
PROOFS
PROOFS
PROOFS
PROOFS
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121CHAPTER 4 Gravitation
c04Gravitation 121 24 May 2016 2:12 PM
Gravitational force exerted on the Chelyabinsk asteroid by the Earth
0
−20
−40
Gra
vita
tiona
l for
ce (1
06 N
)
−60
−80
−100
−120
1 2 3 4 5 6 7 8 9 10
Distance from the centre of the Earth (in Earth radii)
Using the area under a � eld graph� e graphs for the 10 kg mass and the Chelyabinsk asteroid have di� erent values on the force axis. To � nd the changes in energy of a rock escaping the Earth, a di� erent graph would be needed, because the mass of the rock is dif-ferent, and thus the gravitational force on it is di� erent. It would be simpler if we could use the same graph for di� erent objects regardless of their mass.
� e graph that can be used for this purpose is a graph of the gravitational � eld against distance. � e gravitational force on a mass at a point in space is just the value of the gravitational � eld at the point times its mass. Similarly, the change
in energy for an object that moves from one point to another can obtained by multiplying the area under the graph of the gravi-tational � eld against distance by its mass.
� e unit for gravitational � eld is Newtons per kilogram. � e unit for the area under a graph of gravitational � eld against distance is (Newtons per kilo-gram) × metre, hence Newton metre per kilogram or simply Joule per kilogram. � e change in energy can be obtained from this area by multiplying by the mass of the object.
� is method was also used in Chapter 2 on page 57 with the gravitational � eld close to the Earth’s surface where the � eld strength is usually constant.
Graph of the strength of the Earth’s gravitational � eld
0
−2
−4
Gra
vita
tiona
l �el
d s
tren
gth
(N k
g−1
or m
s−
2 )
−6
−8
−10
−12
1 2 3 4 5 6 7 8 9 10
Distance from the centre of the Earth (in Earth radii)
UNCORRECTED values on the force axis. To � nd the changes in energy of a rock escaping the
UNCORRECTED values on the force axis. To � nd the changes in energy of a rock escaping the Earth, a di� erent graph would be needed, because the mass of the rock is dif-
UNCORRECTED Earth, a di� erent graph would be needed, because the mass of the rock is dif-ferent, and thus the gravitational force on it is di� erent. It would be simpler if
UNCORRECTED ferent, and thus the gravitational force on it is di� erent. It would be simpler if we could use the same graph for di� erent objects regardless of their mass.
UNCORRECTED we could use the same graph for di� erent objects regardless of their mass.
� e graph that can be used for this purpose is a graph of the gravitational � eld
UNCORRECTED � e graph that can be used for this purpose is a graph of the gravitational � eld
against distance. � e gravitational force on a mass at a point in space is just the
UNCORRECTED against distance. � e gravitational force on a mass at a point in space is just the value of the gravitational � eld at the point times its mass. Similarly, the change
UNCORRECTED value of the gravitational � eld at the point times its mass. Similarly, the change
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
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Distance from the centre of the Earth (in Earth radii)
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PAGE
PAGE
PAGE Gravitational force exerted on the Chelyabinsk asteroid by the Earth
PAGE Gravitational force exerted on the Chelyabinsk asteroid by the Earth
PAGE
PAGE
PAGE Using the area under a � eld graph
PAGE Using the area under a � eld graph� e graphs for the 10 kg mass and the Chelyabinsk asteroid have di� erent PAGE � e graphs for the 10 kg mass and the Chelyabinsk asteroid have di� erent values on the force axis. To � nd the changes in energy of a rock escaping the PAGE
values on the force axis. To � nd the changes in energy of a rock escaping the Earth, a di� erent graph would be needed, because the mass of the rock is dif-PAGE
Earth, a di� erent graph would be needed, because the mass of the rock is dif-
PROOFS
PROOFS
PROOFS
Gravitational force exerted on the Chelyabinsk asteroid by the EarthPROOFS
Gravitational force exerted on the Chelyabinsk asteroid by the EarthPROOFS
PROOFS
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UNIT 3122
c04Gravitation 122 24 May 2016 2:12 PM
Revision question 4.7
If a rock of mass 1 kg was thrown upwards from the Earth’s surface with suf-�cient kinetic energy to escape the Earth’s gravitational �eld, the amount of kin-etic energy required would be the area under the graph out to in�nity. Using a distance of 10 Earth radii as an approximation for in�nity, show that the required initial speed is about 11 km s−1.
Astronauts and satellites in orbitAs we saw earlier, Newton used his law of Universal Gravitation to show that
Kepler’s �ird Law, π
=R
T
MG
4
3
2 2, applies to all satellites going around the same
central mass. In the context of the Earth, this means that R
T
3
2 is the same for
every single arti�cial satellite, regardless of the orientation of its orbit, as well as the Moon itself. Because we know the period and the radius of the Moon’s orbit, we can use the method of ratios to calculate the characteristics of any other satellite:
or
=
=
R
T
R
T
R
R
T
T
M3
M2
sat3
sat2
M
sat
3M
sat
2
.
�e bene�t of this method is that because you are working with ratios, you don’t need to use metres and seconds for your data. Earth radii and days can be used, making for simpler calculations.
�e orbit of the Moon is slightly elliptical, but the average radius of the Moon’s orbit is about 384 000 km or about 60 Earth radii.
�e period of the Moon in relation to the stars is called the sidereal period and has been measured at 27.321 582 days (or approximately 2.36 × 106 sec-onds). For our purposes we can use 27.3 days. �e period of the Moon in relation to the Sun, that is the time between full moons, is 29.5 days; this is longer than the sidereal period because in that time the Earth has moved fur-ther around the Sun.
Geostationary satellitesArti�cial satellites are used for communication and exploration. Some transmit telephone and television signals around the world, some photograph cloud patterns to help weather forecasters, some are �tted with scienti�c equipment that enables them to collect data about X-ray sources in outer space, whereas others spy on our neighbours! �e motion of an arti�cial satellite depends on what it is designed to do. �ose satellites that are required to rotate so that they stay constantly above one place on Earth’s surface are called geostationary satellites and they are said to be in geostationary orbit. In order to stay in pos-ition, a geostationary satellite must have the same period as the place it is above. �erefore, geostationary satellites have a period of 24 hours or 1 day.
Sample problem 4.5
What is the radius of the orbit of a geostationary satellite as a multiple of the Earth’s radius and also in metres?
Unit 3 Satellites and gravitational forceConcept summary and practice questions
AOS 1
Topic 2
Concept 3
Unit 3 Satellite equations of motionConcept summary and practice questions
AOS 1
Topic 2
Concept 4
A satellite in geostationary orbit is stationary relative to a point directly below it on Earth’s surface. A geostationary orbit has the same period as the rotation of Earth.
UNCORRECTED �e bene�t of this method is that because you are working with ratios, you
UNCORRECTED �e bene�t of this method is that because you are working with ratios, you don’t need to use metres and seconds for your data. Earth radii and days can
UNCORRECTED don’t need to use metres and seconds for your data. Earth radii and days can
UNCORRECTED be used, making for simpler calculations.
UNCORRECTED be used, making for simpler calculations.
�e orbit of the Moon is slightly elliptical, but the average radius of the
UNCORRECTED �e orbit of the Moon is slightly elliptical, but the average radius of the
Moon’s orbit is about 384
UNCORRECTED Moon’s orbit is about 384
�e period of the Moon in relation to the stars is called the sidereal period
UNCORRECTED �e period of the Moon in relation to the stars is called the sidereal period
and has been measured at 27.321
UNCORRECTED and has been measured at 27.321onds). For our purposes we can use 27.3 days. �e period of the Moon in
UNCORRECTED
onds). For our purposes we can use 27.3 days. �e period of the Moon in relation to the Sun, that is the time between full moons, is 29.5 days; this is
UNCORRECTED
relation to the Sun, that is the time between full moons, is 29.5 days; this is longer than the sidereal period because in that time the Earth has moved fur
UNCORRECTED
longer than the sidereal period because in that time the Earth has moved fur
UNCORRECTED
ther around the Sun.
UNCORRECTED
ther around the Sun.
UNCORRECTED PAGE
�e bene�t of this method is that because you are working with ratios, you PAGE
�e bene�t of this method is that because you are working with ratios, you
PROOFSAstronauts and satellites in orbit
PROOFSAstronauts and satellites in orbitAs we saw earlier, Newton used his law of Universal Gravitation to show that
PROOFSAs we saw earlier, Newton used his law of Universal Gravitation to show that
, applies to all satellites going around the same
PROOFS, applies to all satellites going around the same
central mass. In the context of the Earth, this means that
PROOFScentral mass. In the context of the Earth, this means that
PROOFSR
PROOFSR
T
PROOFST
3
PROOFS3
2
PROOFS2
is the same for
PROOFS is the same for
every single arti�cial satellite, regardless of the orientation of its orbit, as well
PROOFSevery single arti�cial satellite, regardless of the orientation of its orbit, as well as the Moon itself. Because we know the period and the radius of the Moon’s
PROOFSas the Moon itself. Because we know the period and the radius of the Moon’s orbit, we can use the method of ratios to calculate the characteristics of any PROOFS
orbit, we can use the method of ratios to calculate the characteristics of any
123CHAPTER 4 Gravitation
c04Gravitation 123 24 May 2016 2:12 PM
RM = 60 × RE, TM = 27.3 days, Tsat = 1 day, Rsat = ?
Rearranging
=
R
R
T
TM
sat
3M
sat
2
to �nd Rsat:
RT
TR
R
R
R R
1
27.3(60 )
289.8
6.62
sat3 sat
M
2
M3
2
E3
E3
sat E
=
×
=
×
==
In metres,
R 6.62 6.38 10 m
= 4.22 10 m
sat6
7
= × ×
×
Revision question 4.8
(a) Use a small coin to draw a circle in the middle of a blank page to represent the equator of the Earth.
(b) Using the answer from sample problem 4.5, put a dot on the page where you estimate a geostationary satellite would be.
(c) Draw two lines from the dot to the circle representing the Earth, to touch the circle at a tangent. �e part of the circle facing the satellite between these two lines represents how much of the Earth’s surface could receive signals from the satellite.
(d) Use your diagram to determine the minimum number of geostationary satellites required to cover all of the Earth’s equator.
(e) Which parts of the Earth could not receive signals from any of these geosta-tionary satellites?
Revision question 4.9
Global Positioning System (GPS) satellites are used for navigation. �e Navstar 66 satellite, launched in 2011, has an orbital radius of about 20 100 km. What is the period of its orbit expressed in days?
AS A MATTER OF FACT
Why are geostationary satellites always above the equator? Why isn’t there a geostationary satellite directly above central Australia?
Newton showed that the motion of a large object can be analysed as if all of its mass was concentrated at a single point, called its centre of mass. For a symmetrical object such as the Earth, the centre of mass is located at its geometric centre. �us, all satellites around Earth are in orbit about the centre of the Earth.
If a satellite was to be directly above central Australia for 24 hours each day, the centre of its orbit would be at the centre of the matching circle of latitude, some distance away from the centre of the Earth.
If instead a satellite was to be only momentarily directly above central Australia, given the centre of its orbit is the centre of the Earth, the orbit would take it into the sky above the northern hemisphere for half the time.
�is is why geostationary communication satellites orbit around the equator. A satellite dish has to be angled at the latitude of that point on the Earth to point towards one of these satellites.
Solution:
UNCORRECTED se your diagram to determine the minimum number of geostationary
UNCORRECTED se your diagram to determine the minimum number of geostationary satellites required to cover all of the Earth’s equator.
UNCORRECTED satellites required to cover all of the Earth’s equator.
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED
UNCORRECTED arts of the Earth could not receive signals from any of these geosta
UNCORRECTED arts of the Earth could not receive signals from any of these geosta
tionary satellites?
UNCORRECTED tionary satellites?
UNCORRECTED Revision question 4.9
UNCORRECTED Revision question 4.9
Global Positioning System (GPS) satellites are used for navigation. �e Navstar
UNCORRECTED
Global Positioning System (GPS) satellites are used for navigation. �e Navstar 66 satellite, launched in 2011, has an orbital radius of about 20
UNCORRECTED
66 satellite, launched in 2011, has an orbital radius of about 20the period of its orbit expressed in days?
UNCORRECTED
the period of its orbit expressed in days?
UNCORRECTED PAGE se a small coin to draw a circle in the middle of a blank page to represent
PAGE se a small coin to draw a circle in the middle of a blank page to represent
sing the answer from sample problem 4.5, put a dot on the page where you
PAGE sing the answer from sample problem 4.5, put a dot on the page where you estimate a geostationary satellite would be.
PAGE estimate a geostationary satellite would be.
aw two lines from the dot to the circle representing the Earth, to touch the
PAGE aw two lines from the dot to the circle representing the Earth, to touch the
circle at a tangent. �e part of the circle facing the satellite between these
PAGE circle at a tangent. �e part of the circle facing the satellite between these two lines represents how much of the Earth’s surface could receive signals
PAGE two lines represents how much of the Earth’s surface could receive signals
se your diagram to determine the minimum number of geostationary PAGE
se your diagram to determine the minimum number of geostationary satellites required to cover all of the Earth’s equator.PAGE
satellites required to cover all of the Earth’s equator.
PROOFS
PROOFS
PROOFS
se a small coin to draw a circle in the middle of a blank page to represent PROOFS
se a small coin to draw a circle in the middle of a blank page to represent
UNIT 3124
c04Gravitation 124 24 May 2016 2:12 PM
‘Floating’ in a spacecraft
An astronaut inside the International Space Station
� e appearance of an astronaut � oating around inside a spacecraft suggests that there is no force acting on them, leading some people to mistakenly think that there is no gravity in space. In fact, both the astronaut and the spacecraft are in a circular orbit about the Earth.
However, you also know that if an object is moving in a curved path, changing its direction, there must be an acceleration. If the path is circular the acceleration is directed towards the centre of that path.
� e astronaut and the space craft are in the same gravitational � eld. � ey are at the same distance from the centre of the Earth. � ey are travelling at the same speed, taking the same time to orbit the earth. � erefore, their centrip-etal accelerations provided by the gravitational � eld are the same.
For the spacecraft: For the astronaut:
π
π
= ×
=
M m
Rm
R
TM
R
R
T
G 4
G 4
E s2 s
2
2
E2
2
2
π
π
= ×
=
M m
Rm
R
TM
R
R
T
G 4
G 4
E a2 a
2
2
E2
2
2
� ere is no need for a normal reaction force by the spacecraft on the astro-naut to explain the astronaut’s motion. � e astronaut inside the spacecraft circles the Earth as if the spacecraft was not there. Indeed, if the astronaut is outside the spacecraft doing a space walk, the astronaut’s speed and accel-eration around the Earth will be unchanged as they ‘� oat’ beside the space-craft. Once back inside, their speed and acceleration are still unchanged, and this time they are ‘� oating’ inside the spacecraft.
If the astronaut steps onto a set of bathroom scales, they will give a reading of zero. As shown in the second photo opposite, an astronaut running on a treadmill needs stretched springs attached to his waist to pull him down to the treadmill.
Unit 3 Normal reaction forces in uniform orbitsConcept summary and practice questions
AOS 1
Topic 2
Concept 6
UNCORRECTED � e appearance of an astronaut � oating around inside a spacecraft suggests
UNCORRECTED � e appearance of an astronaut � oating around inside a spacecraft suggests that there is no force acting on them, leading some people to mistakenly think
UNCORRECTED that there is no force acting on them, leading some people to mistakenly think that there is no gravity in space. In fact, both the astronaut and the spacecraft
UNCORRECTED that there is no gravity in space. In fact, both the astronaut and the spacecraft are in a circular orbit about the Earth.
UNCORRECTED are in a circular orbit about the Earth.
However, you also know that if an object is moving in a curved path,
UNCORRECTED However, you also know that if an object is moving in a curved path,
changing its direction, there must be an acceleration. If the path is circular the
UNCORRECTED changing its direction, there must be an acceleration. If the path is circular the acceleration is directed towards the centre of that path.
UNCORRECTED acceleration is directed towards the centre of that path.
UNCORRECTED � e astronaut and the space craft are in the same gravitational � eld. � ey
UNCORRECTED � e astronaut and the space craft are in the same gravitational � eld. � ey
are at the same distance from the centre of the Earth. � ey are travelling at the
UNCORRECTED
are at the same distance from the centre of the Earth. � ey are travelling at the same speed, taking the same time to orbit the earth. � erefore, their centrip-
UNCORRECTED
same speed, taking the same time to orbit the earth. � erefore, their centrip-etal accelerations provided by the gravitational � eld are the same.
UNCORRECTED
etal accelerations provided by the gravitational � eld are the same.
PAGE
PAGE
PAGE An astronaut inside the International Space Station
PAGE An astronaut inside the International Space Station
PAGE
PAGE � e appearance of an astronaut � oating around inside a spacecraft suggests PAGE � e appearance of an astronaut � oating around inside a spacecraft suggests
that there is no force acting on them, leading some people to mistakenly think PAGE
that there is no force acting on them, leading some people to mistakenly think
PROOFS
125CHAPTER 4 Gravitation
c04Gravitation 125 24 May 2016 2:12 PM
This astronaut is � oating inside the International Space Station. Both the astronaut and the station are in orbit around the Earth.
The cloth-covered stretched springs are pulling the astronaut down so he can exercise on the treadmill.
UNCORRECTED
UNCORRECTED PAGE
PAGE PROOFS
PROOFS
UNIT 3126
c04Gravitation 126 24 May 2016 2:12 PM
Chapter reviewUnit 3
Fields and interactions Gravitational �elds and forces
Sit Topic test
AOS 1
Topics 1 & 2
Summary ■ �e gravitational �eld strength g at a distance r from a
body of mass M is given by the formula g = MR
G2 where
G is the gravitational constant. �e force of gravity F
on an object of mass m at a distance r from the same
body is therefore given by F = MmR
G2 . �is equation is
referred to as Newton’s Law of Universal Gravitation.
■ For a given planetary or satellite system, RT
3
2 is con-
stant. �e value of the constant is equal to πMG
4 2 where
M is the mass of the central body. ■ Gravitational attraction can be explained using a
�eld model. ■ �e �eld model enables the gravitational �eld around
a point mass to be described in terms of its direction and shape, and also its strength.
■ �e �eld model also enables descriptions of the changes in potential energy of an object moving in the gravitational �eld of a point mass.
■ Changes in potential energy can be calculated from the area under a force–distance graph and from the area under a �eld–distance graph when multiplied by mass.
■ Satellites in orbit and their occupants (who are also in orbit) experience no reaction force.
QuestionsIn answering the questions on the following pages, assume, where relevant, that the magnitude of the gravitational �eld at Earth’s surface is 9.8 N kg−1. Additional data required for questions relating to bodies in the solar system can be found in table 4.1 on page 110.
Modelling the motion of satellites 1. A gravitational �eld strength detector is released
into the atmosphere and reports back a reading of 9.70 N kg−1.(a) If the detector has a mass of 10 kg, what is the
force of gravity acting on it?(b) If the detector is to remain stationary at this
height, what upwards force must be exerted on the detector?
(c) How far is the detector from the centre of Earth? 2. Use the information provided in table 4.1 on page
110 to calculate (i) the gravitational �eld strength and (ii) the weight of a 70 kg person at the surface of the following bodies of the solar system:(a) Earth (c) Venus(b) Mars (d) Pluto.
3. A space probe orbits a distance of 5.0 × 105 m from the centre of an undiscovered planet. It experiences a gravitational �eld strength of 4.3 N kg−1. What is the mass of the planet?
4. Calculate the force of attraction between Earth and the Sun.
5. If the Earth expanded to twice its radius without any change in its mass, what would happen to your weight?
6. By how much would your reading on bathroom scales change with the Moon on the opposite side of the Earth to you, compared with being above you?
7. Determine the value of the ratio F
Fon Moon by Sun
on Moon by Earth.
Assume the Moon is the same distance from the Sun as the Earth is.
8. How many Earth radii from the centre of the Earth must an object be for the gravitational force by the Earth on the object to equal the gravitational force that would be exerted by the Moon on the object if the object was on the Moon’s surface?
9. A space station orbits at a height of 355 km above Earth and completes one orbit every 92 min.(a) What is the centripetal acceleration of the
space station?(b) What gravitational �eld strength does the
space station experience?(c) Your answers to (a) and (b) above should be
the same. (i) Explain why. (ii) Explain any discrepancy in your answers.
(d) If the mass of the space station is 1200 tonnes, what is its weight?
(e) �e mass of an astronaut and the special spacesuit he wears when outside the space station is 270 kg. If he is a distance of 10 m from the centre of mass of the space station, what is the force of attraction between the astronaut and the space station?
10. What is the centripetal acceleration of a person standing on Earth’s equator due to Earth’s rotation about its axis? (Radius of Earth is 6.38 × 106 m.) Would the centripetal acceleration be greater or less for a person standing in Victoria? Justify your answer.
11. In the future, it is predicted that space stations may rotate to simulate the gravitational �eld of Earth and therefore make life more normal for the occupants. Draw a diagram of such a space station. Include on your diagram:
UNCORRECTED Changes in potential energy can be calculated from the
UNCORRECTED Changes in potential energy can be calculated from the area under a force–distance graph and from the area
UNCORRECTED area under a force–distance graph and from the area under a �eld–distance graph when multiplied by mass.
UNCORRECTED under a �eld–distance graph when multiplied by mass.Satellites in orbit and their occupants (who are also
UNCORRECTED Satellites in orbit and their occupants (who are also in orbit) experience no reaction force.
UNCORRECTED in orbit) experience no reaction force.
In answering the questions on the following pages,
UNCORRECTED
In answering the questions on the following pages, assume, where relevant, that the magnitude of the
UNCORRECTED
assume, where relevant, that the magnitude of the gravitational �eld at Earth’s surface is 9.8
UNCORRECTED
gravitational �eld at Earth’s surface is 9.8Additional data required for questions relating to bodies
UNCORRECTED
Additional data required for questions relating to bodies in the solar system can be found in table 4.1 on page 110.
UNCORRECTED
in the solar system can be found in table 4.1 on page 110.
Modelling the motion of satellites
UNCORRECTED
Modelling the motion of satellitesavitational �eld strength detector is released
UNCORRECTED
avitational �eld strength detector is released into the atmosphere and reports back a reading of
UNCORRECTED
into the atmosphere and reports back a reading of N
UNCORRECTED
N
UNCORRECTED
kg
UNCORRECTED
kg−
UNCORRECTED
−1
UNCORRECTED
1.
UNCORRECTED
.If the detUNCORRECTED
If the detector has a mass of 10UNCORRECTED
ector has a mass of 10force of gravity acting on it?UNCORRECTED
force of gravity acting on it?If the detUNCORRECTED
If the det
9.
UNCORRECTED 9.PAGE sume the Moon is the same distance from the
PAGE sume the Moon is the same distance from the Sun as the Earth is.
PAGE Sun as the Earth is.
PAGE ow many Earth radii from the centre of the
PAGE ow many Earth radii from the centre of the
Earth must an object be for the gravitational
PAGE Earth must an object be for the gravitational force by the Earth on the object to equal the
PAGE force by the Earth on the object to equal the gravitational force that would be exerted by the
PAGE gravitational force that would be exerted by the Moon on the object if the object was on the PAGE Moon on the object if the object was on the Moon’s surface?PAGE
Moon’s surface?
PROOFSalculate the force of attraction between Earth
PROOFSalculate the force of attraction between Earth
th expanded to twice its radius without
PROOFSth expanded to twice its radius without any change in its mass, what would happen to
PROOFSany change in its mass, what would happen to
y how much would your reading on bathroom
PROOFSy how much would your reading on bathroom
scales change with the Moon on the opposite side
PROOFSscales change with the Moon on the opposite side of the Earth to you, compared with being above
PROOFSof the Earth to you, compared with being above
etermine the value of the ratio PROOFS
etermine the value of the ratio
sume the Moon is the same distance from the PROOFS
sume the Moon is the same distance from the
127CHAPTER 4 Gravitation
c04Gravitation 127 24 May 2016 2:12 PM
■ the axis of rotation ■ the distance of the occupants from the axis ■ arrows indicating which direction the occupants
would consider as ‘down’. (Remember to consider the frame of reference of the occupants!) Make an estimate of the period of rotation your space station would need to simulate Earth’s gravitational �eld.
12. Neutron stars are thought to rotate at about 1 revolution every second. What is the minimum mass for the neutron star so that a mass on the star’s surface is in the same situation as a satellite in orbit, that is, the strength of the gravitational �eld equals the centripetal acceleration at the surface?
13. �e Sun orbits the centre of our galaxy, the Milky Way, at a distance of 2.2 × 1020 m from the centre with a period of 2.5 × 108 years. �e mass of all the stars inside the Sun’s orbit can be considered as being concentrated at the centre of the galaxy. �e mass of the Sun is 2.0 × 1030 kg. If all the stars have the same mass as the Sun, how many stars are in the Milky Way?
14. �e asteroid 243 Ida was discovered in 1884. �e Galileo spacecraft, on its way to Jupiter, visited the asteroid in 1993. Search online for images of the �yby. �e asteroid was the �rst to be found to have a natural satellite, that is, its own moon, now called Dactyl. Dactyl orbits Ida at a radius of 100 km and with a period of 27 hours. What is the mass of the asteroid?
Motion of the planets15. What force holds the solar system together?
Explain how this results in the planets moving in roughly circular orbits.
16. Venus and Saturn both orbit the Sun. Using only information about the Sun and the periods of the two planets, calculate the value of the ratio:
distance of Saturn from the Sun
distance of Venus from the Sun.
17. A spacecraft leaves Earth to travel to the Moon. How far from the centre of the Earth is the spacecraft when it experiences a net force of zero?
Use the data in table 4.1 to determine where that point is, and draw a scale diagram to show its location.
18. A satellite is in a circular orbit around the Earth with a radius equal to half of the radius of the Moon’s orbit. What is the satellite’s period expressed as a fraction of the Moon’s period about the Earth?
Satellites of the Earth 19. A geostationary satellite remains above the
same position on Earth’s surface. Once in orbit, the only force acting on the satellite is that of gravity towards the centre of Earth. Why doesn’t the satellite fall straight back down to Earth?
20. A new geostationary satellite is to be launched. At what height above the centre of Earth must the satellite orbit?
21. Can a geostationary satellite remain above Melbourne? Why or why not?
22. Explain why the area under a gravitational force–distance graph gives the energy needed to launch a satellite, but the area under a gravitational �eld strength–distance graph gives the energy per kilogram needed to launch a satellite.
23. A space shuttle, orbiting Earth once every 93 mins at a height of 400 km above the surface, deploys a new 800 kg satellite that is to orbit a further 200 km away from Earth.(a) Use the following graph to estimate the
work needed to deploy the satellite from the shuttle.
(b) Use the mass and radius of Earth to assist you in determining the period of the new satellite.
(c) Show how the period of the new satellite can be determined without knowledge of the mass of Earth.
(d) If the new satellite was redesigned so that its mass was halved, how would your answers to (a) and (b) change?
2
2
4
6
8
10
4 6 8 10Distance above the
surface of the Earth (× 105 m)
Gra
vita
tiona
l �el
d s
tren
gth
g (N
kg–1
)
24. A disabled satellite of mass 2400 kg is in orbit around Earth at a height of 2000 km above sea level. It falls to a height of 800 km before its built-in rocket system can be activated to stop the fall continuing.
UNCORRECTED to have a natural satellite, that is, its own moon,
UNCORRECTED to have a natural satellite, that is, its own moon, now called Dactyl. Dactyl orbits Ida at a radius of
UNCORRECTED now called Dactyl. Dactyl orbits Ida at a radius of 100 km and with a period of 27 hours. What is the
UNCORRECTED 100 km and with a period of 27 hours. What is the
hat force holds the solar system together?
UNCORRECTED
hat force holds the solar system together? Explain how this results in the planets moving in
UNCORRECTED
Explain how this results in the planets moving in
enus and Saturn both orbit the Sun. Using
UNCORRECTED
enus and Saturn both orbit the Sun. Using only information about the Sun and the periods
UNCORRECTED
only information about the Sun and the periods of the two planets, calculate the value of the
UNCORRECTED
of the two planets, calculate the value of the
UNCORRECTED
ance
UNCORRECTED
ance of
UNCORRECTED
of Saturn
UNCORRECTED
Saturn fr
UNCORRECTED
fr
ance
UNCORRECTED
ance of Ve
UNCORRECTED
of Venu
UNCORRECTED
nu
A s UNCORRECTED
A spacecraft leaves Earth to travel to the Moon. UNCORRECTED
pacecraft leaves Earth to travel to the Moon. How far from the centre of the Earth is the UNCORRECTED
How far from the centre of the Earth is the
PAGE k
PAGE k
PAGE g satellite that is to orbit a further 200
PAGE g satellite that is to orbit a further 200way from Earth.
PAGE way from Earth.U
PAGE Use the following graph to estimate the
PAGE se the following graph to estimate the
work needed to deploy the satellite from the
PAGE work needed to deploy the satellite from the shuttle.
PAGE shuttle.
(b)
PAGE (b) U
PAGE Use the mass and radius of Earth to assist
PAGE se the mass and radius of Earth to assist
you in determining the period of the new PAGE you in determining the period of the new
PROOFSAt what height above the centre of Earth must the
PROOFSAt what height above the centre of Earth must the
an a geostationary satellite remain above
PROOFSan a geostationary satellite remain above
xplain why the area under a gravitational force–
PROOFSxplain why the area under a gravitational force–
distance graph gives the energy needed to launch
PROOFSdistance graph gives the energy needed to launch a satellite, but the area under a gravitational �eld
PROOFSa satellite, but the area under a gravitational �eld strength–distance graph gives the energy
PROOFSstrength–distance graph gives the energy
needed to launch a satellite.
PROOFS needed to launch a satellite.
pace shuttle, orbiting Earth once every 93 mins
PROOFS
pace shuttle, orbiting Earth once every 93 mins at a height of 400PROOFS
at a height of 400 PROOFS
km aPROOFS
km above the surface, deploys a PROOFS
bove the surface, deploys a PROOFS
g satellite that is to orbit a further 200PROOFS
g satellite that is to orbit a further 200
UNIT 3128
c04Gravitation 128 24 May 2016 2:12 PM
Distance from the Earth’s centre (× 106 m)
Gra
vita
tiona
l �el
d s
tren
gth
g (N
kg–1
)
7.06.00
5
10
8.0 9.0
(a) Calculate the gravitational force on the satellite while it is in its initial orbit.
(b) Calculate the loss of gravitational potential energy of the satellite during its fall.
(c) If the speed of the satellite during its initial orbit is 6900 m s−1, what is its speed when the rocket system is activated?
25. In a space shuttle that is in orbit around Earth at an altitude of 360 km, what is the magnitude of:(a) the gravitational �eld strength(b) the reaction force by the shuttle on a 70-kg
astronaut(c) the gravitational force by the Earth on this
astronaut? 26. Why does the gravitational force do no work on a
satellite in orbit?
UNCORRECTED PAGE P
ROOFSavitational force by the Earth on this
PROOFSavitational force by the Earth on this
hy does the gravitational force do no work on a
PROOFShy does the gravitational force do no work on a