1. Section 6.5 Newton’s Law of Gravity
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2. Gravity Obeys an Inverse-Square Law
Gravity is a universal force that affects all objects in the universe.
Newton proposed that the force of gravity has the following properties:
The force is inversely proportional to the square of the distance between the objects.
The force is directly proportional to the product of the masses of the two objects.
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3. Gravity Obeys an Inverse-Square Law
Newton’s law of gravity is an inverse-square law. This formula only works for spherically symmetric objects.
G is Newton’s Gravitational constant.
G = 6.67 × N · m2 / kg2
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4. Conceptual Example 6.11 Varying gravitational force
The gravitational force between two giant lead spheres is N when the centers of the spheres are 20 m apart. What is the distance between their centers when the gravitational force between them is N?
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5. Conceptual Example 6.11 Varying gravitational force
Answer 5.0 m. reason Gravity is an inverse-square relationship; the force is related to the inverse square of the distance. The force increases by a factor of (0.160 N)/(0.010 N) = 16, so the distance must decrease by a factor of = 4. The distance is thus (20 m)/4 = 5.0 m. assess This type of ratio reasoning is a very good way to get a quick handle on the solution to a problem.
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6. Example 6.12 Gravitational force between two people
You are seated in your physics class next to another student m away. Estimate the magnitude of the gravitational force between you. Assume that you each have a mass of 65 kg.
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7. Example 6.12 Gravitational force between two people
You are seated in your physics class next to another student 0.60 m away. Estimate the magnitude of the gravitational force between you. Assume that you each have a mass of 65 kg. prepare We will model each of you as a particle. This is not a particularly good model, but it will do for making an estimate. We will take the 0.60 m as the distance between you.
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8. Example 6.12 Gravitational force between two people (cont.)
solve The gravitational force is given by Equation 6.15: assess The force is quite small, roughly the weight of one hair on your head. This seems reasonable; you don’t normally sense this attractive force!
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9. QuickCheck 6.16
The force of Planet Y on Planet X is ___ the magnitude of
One quarter
One half
The same as
Twice
Four times 2M M
Answer: C
Planet X
Planet Y
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10. QuickCheck 6.16
The force of Planet Y on Planet X is ___ the magnitude of
One quarter
One half
The same as
Twice
Four times 2M M
Newton’s third law
Planet X
Planet Y
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11. QuickCheck 6.17
The gravitational force between two asteroids is 1,000,000 N. What will the force be if the distance between the asteroids is doubled?
250,000 N
500,000 N
1,000,000 N
2,000,000 N
4,000,000 N
Answer: A
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12. QuickCheck 6.17
The gravitational force between two asteroids is 1,000,000 N. What will the force be if the distance between the asteroids is doubled?
250,000 N
500,000 N
1,000,000 N
2,000,000 N
4,000,000 N
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13. Gravity on Other Worlds
If you traveled to another planet, your mass would be the same but your weight would vary. The weight of a mass m on the moon is given by
Using Newton’s law of gravity (Eq. (6.15)) the weight is given by
Since these are two expressions for the same force, they are equal and
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14. Gravity on Other Worlds
If we use values for the mass and the radius of the moon, we compute gmoon = 1.62 m/s2.
A 70-kg astronaut wearing an 80-kg spacesuit would weigh more than 330 lb on the earth but only 54 lb on the moon.
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15. QuickCheck 6.18
Planet X has free-fall acceleration 8 m/s2 at the surface. Planet Y has twice the mass and twice the radius of planet X. On Planet Y
g = 2 m/s2
g = 4 m/s2
g = 8 m/s2
g = 16 m/s2
g = 32 m/s2
Answer: B
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16. QuickCheck 6.18
Planet X has free-fall acceleration 8 m/s2 at the surface. Planet Y has twice the mass and twice the radius of planet X. On Planet Y
g = 2 m/s2
g = 4 m/s2
g = 8 m/s2
g = 16 m/s2
g = 32 m/s2
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17. QuickCheck 6.22
A 60-kg person stands on each of the following planets. On which planet is his or her weight the greatest?
Answer: A
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18. QuickCheck 6.22
A 60-kg person stands on each of the following planets. On which planet is his or her weight the greatest? A
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19. Example 6.14 Finding the speed to orbit Deimos
Mars has two moons, each much smaller than the earth’s moon. The smaller of these two bodies, Deimos, isn’t quite spherical, but we can model it as a sphere of radius 6.3 km. Its mass is 1.8 × 1015 kg. At what speed would a projectile move in a very low orbit around Deimos?
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20. Example 6.14 (cont.)
solve The free-fall acceleration at the Deimos’s surface is Given this, we calculate the orbital speed: assess This is quite slow. With a good jump, you could easily launch yourself into an orbit around Deimos!
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21. Section 6.6 Gravity and Orbits
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22. Gravity and Orbits
Because it’s moving in a circular orbit, Newton’s second law gives\
Solving for v gives
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23. Gravity and Orbits
A satellite must have this specific speed in order to maintain a circular orbit of radius r about the larger mass M.
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24. Gravity and Orbits
For a planet orbiting the sun, the period T is the time to complete one full orbit. The relationship among speed, radius, and period is the same as for any circular motion:
v = 2πr/T
Combining this with the value of v for a circular orbit from Equation 6.21 gives
If we square both sides and rearrange, we find the period of a satellite:
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25. QuickCheck 6.20
Two satellites have circular orbits with the same radius. Which has a higher speed?
The one with more mass.
The one with less mass.
They have the same speed.
Answer: C
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26. QuickCheck 6.20
Two satellites have circular orbits with the same radius. Which has a higher speed?
The one with more mass.
The one with less mass.
They have the same speed.
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27. QuickCheck 6.21
Two identical satellites have different circular orbits. Which has a higher speed?
The one in the larger orbit
The one in the smaller orbit
They have the same speed.
Answer: B
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28. QuickCheck 6.21
Two identical satellites have different circular orbits. Which has a higher speed?
The one in the larger orbit
The one in the smaller orbit
They have the same speed.
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29. QuickCheck 6.23
A satellite orbits the earth. A Space Shuttle crew is sent to boost the satellite into a higher orbit. Which of these quantities (for the satellite) increases after it is boosted?
Speed
Angular speed
Period
Centripetal acceleration
Gravitational force of the earth
Answer: C
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30. QuickCheck 6.23
A satellite orbits the earth. A Space Shuttle crew is sent to boost the satellite into a higher orbit. Which of these quantities increases?
Speed
Angular speed
Period
Centripetal acceleration
Gravitational force of the earth
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31. Example 6.15 Locating a geostationary satellite
Communication satellites appear to “hover” over one point on the earth’s equator. A satellite that appears to remain stationary as the earth rotates is said to be in a geostationary orbit. What is the radius of the orbit of such a satellite?
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32. Example 6.15 Locating a geostationary satellite
prepare For the satellite to remain stationary with respect to the earth, the satellite’s orbital period must be 24 hours; in seconds this is T = 8.64 × 104 s. solve We solve for the radius of the orbit using the previous equations.
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33. Example 6.15 Locating a geostationary satellite (cont.)
assess This is a high orbit, and the radius is about 7 times the radius of the earth (6371 km). Compare to the radius of the International Space Station’s orbit, which is only about 5% larger than that of the earth, and the radius of the moon’s orbit, which is about 385,000 km – over 60 times the Earth’s radius.
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34. Gravity on a Grand Scale
No matter how far apart two objects may be, there is a gravitational attraction between them.
Galaxies are held together by gravity.
All of the stars in a galaxy are different distances from the galaxy’s center, and so orbit with different periods.
In fact, most of the matter which keeps galaxies together is dark matter – which we know almost nothing about!
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35. © 2015 Pearson Education, Inc.

36. Chapter 7 Rotational Motion
Chapter Goal: To understand the physics of rotating objects.
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37. Chapter 7 Preview Looking Ahead: Rotational Kinematics
The spinning roulette wheel isn’t going anywhere, but it is moving. This is rotational motion.
You’ll learn about angular velocity and other quantities we use to describe rotational motion.
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38. Chapter 7 Preview Looking Ahead: Torque
To start something moving, apply a force. To start something rotating, apply a torque, as the sailor is doing to the wheel.
You’ll see that torque depends on how hard you push and also on where you push. A push far from the axle gives a large torque.
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39. Chapter 7 Preview Looking Ahead: Rotational Dynamics
The girl pushes on the outside edge of the merry-go-round, gradually increasing its rotation rate.
You’ll learn a version of Newton’s second law for rotational motion and use it to solve problems.
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40. Chapter 7 Preview Looking Back: Circular Motion
In Chapter 6, you learned to describe circular motion in terms of period, frequency, velocity, and centripetal acceleration.
In this chapter, you’ll learn to use angular velocity, angular acceleration, and other quantities that describe rotational motion.
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41. Section 7.1 Describing Circular and Rotational Motion
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42. Describing Circular and Rotational Motion
Rotational motion is the motion of objects that spin about an axis.
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43. Angular Position
We use the angle θ from the positive x-axis to describe the particle’s location.
Angle θ is the angular position of the particle.
θ is positive when measured counterclockwise from the positive x-axis.
An angle measured clockwise from the positive x-axis has a negative value.
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44. Angular Position
We measure angle θ in the angular unit of radians, not degrees.
The radian is abbreviated “rad.”
The arc length, s, is the distance that the particle has traveled along its circular path.
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45. Angular Position
We define the particle’s angle θ in terms of arc length and radius of the circle:
One revolution (rev) is when a particle travels all the way around the circle.
The angle of the full circle is
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46. Angular Displacement and Angular Velocity
Recall: for linear motion, a particle with a larger velocity undergoes a greater displacement
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47. Angular Displacement and Angular Velocity
For uniform circular motion, a particle with a larger angular velocity will undergo a greater angular displacement Δθ.
Angular velocity is the angular displacement through which the particle moves each second.
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48. Angular Displacement and Angular Velocity
The angular velocity ω = Δθ/Δt is constant for a particle moving with uniform circular motion.
Since the motion is uniform, this represents both instantaneous and average angular velocity – they are the same in this case.
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49. Example 7.1 Comparing angular velocities
Find the angular velocities of the two particles in the figure below.
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50. Example 7.1 Comparing angular velocities
prepare For uniform circular motion, we can use any angular displacement ∆θ, as long as we use the corresponding time interval ∆t. For each particle, we’ll choose the angular displacement corresponding to the motion from t = 0 s to t = 5 s.
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51. Example 7.1 Comparing angular velocities (cont.)
solve The particle on the left travels one-quarter of a full circle during the 5 s time interval. We learned earlier that a full circle corresponds to an angle of 2π rad, so the angular displacement for this particle is ∆θ = (2π rad)/4 = π/2 rad. Thus its angular velocity is
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52. Example 7.1 Comparing angular velocities (cont.)
The particle on the right travels halfway around the circle, or π rad, in the 5 s interval. Its angular velocity is
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53. Example 7.1 Comparing angular velocities (cont.)
assess The speed of the second particle is double that of the first, as it should be. We should also check the scale of the answers. The angular velocity of the particle on the right is rad/s, meaning that the particle travels through an angle of rad each second. Because 1 rad ≈ 60°, rad is roughly 35°, which looks about right.
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54. Angular Displacement and Angular Velocity
The linear displacement during a time interval is
Similarly, the angular displacement for uniform circular motion is
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55. Angular-Position and Angular-Velocity Graphs
We construct angular position-versus-time graphs using the change in angular position for each second.
Angular velocity-versus-time graphs can be created by finding the slope of the corresponding angular position- versus-time graph.
This is mathematically the same as (linear) position-vs.- time and velocity-vs.-time graphs.
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56. Angular Displacement and Angular Velocity
Angular speed is the absolute value of the angular velocity.
The angular speed is related to the period T:
Frequency (in rev/s) f = 1/T:
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57. QuickCheck 7.1
A ball rolls around a circular track with an angular velocity of 4 rad/s. What is the period of the motion? s
1 s
2 s
Answer: A
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58. QuickCheck 7.1
A ball rolls around a circular track with an angular velocity of 4 rad/s. What is the period of the motion? s
1 s
2 s
Answer: A
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59. Example 7.3 Rotations in a car engine
The crankshaft in your car engine is turning at 3000 rpm. What is the shaft’s angular speed?
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60. Example 7.3 Rotations in a car engine
The crankshaft in your car engine is turning at 3000 rpm. What is the shaft’s angular speed?
prepare We’ll need to convert rpm to rev/s and then use the equation to get the angular speed.
solve We convert rpm to rev/s by
Thus the crankshaft’s angular speed is
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