UNIT 6
Circular Motion and
Gravitation
1
ConcepTest 12.4 Averting Disaster
1) It’s in Earth’s gravitational field
The Moon does not
crash into Earth
because:
2) The net force on it is zero
3) It is beyond the main pull of Earth’s
gravity
4) It’s being pulled by the Sun as well as by
Earth
5) none of the above
ConcepTest 12.4 Averting Disaster
1) It’s in Earth’s gravitational field
The Moon does not
crash into Earth
because:
2) The net force on it is zero
3) It is beyond the main pull of Earth’s
gravity
4) It’s being pulled by the Sun as well as by
Earth
5) none of the above
The Moon does not crash into Earth because of its high
speed. If it stopped moving, it would, of course, fall
directly into Earth. With its high speed, the Moon
would fly off into space if it weren’t for gravity
providing the centripetal force.
Follow-up: What happens to a satellite orbiting Earth as it slows?
ConcepTest 12.5 In the Space Shuttle
1) They are so far from Earth that Earth’s gravity
doesn’t act any more.
Astronauts in the
space shuttle
float because:
2) Gravity’s force pulling them inward is cancelled
by the centripetal force pushing them outward.
3) While gravity is trying to pull them inward, they
are trying to continue on a straight-line path.
4) Their weight is reduced in space so the force of
gravity is much weaker.
ConcepTest 12.5 In the Space Shuttle
1) They are so far from Earth that Earth’s gravity
doesn’t act any more.
Astronauts in the
space shuttle
float because:
2) Gravity’s force pulling them inward is cancelled
by the centripetal force pushing them outward.
3) While gravity is trying to pull them inward, they
are trying to continue on a straight-line path.
4) Their weight is reduced in space so the force of
gravity is much weaker.
Astronauts in the space shuttle float because
they are in “free fall” around Earth, just like a
satellite or the Moon. Again, it is gravity that
provides the centripetal force that keeps them
in circular motion.
Follow-up: How weak is the value of g at an altitude of 300 km?
Thursday January 12th
CIRCULAR MOTION AND GRAVITATION
6
TODAY’S AGENDA
Thursday, January 12
 Kepler’s Laws of Planetary Motion
 Hw:
Practice D (All) p251
UPCOMING…
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


Fri: Torque and Rotational Equilibrium
Mon: MLK Holiday NO SCHOOL
Tue: Torque Lab
Wed: Problem Quiz #2
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Units of Chapter 7
Kinematics of Uniform Circular Motion
Dynamics of Uniform Circular Motion
Newton’s Law of Universal Gravitation
Kepler’s Laws and Newton’s Synthesis
Torque and Newton’s Laws of Motion
8
Kepler’s Laws and Newton's Synthesis
Kepler’s laws describe planetary motion.
• All planets move in elliptical orbits with the Sun at one of
the focal points.
• A line drawn from the Sun to any planet sweeps out equal
areas in equal time intervals.
• The square of the orbital period of any planet is
proportional to cube of the average distance from the Sun
to the planet.
Kepler’s Laws and Newton's Synthesis
• Based on observations made by Tycho Brahe
• Newton later demonstrated that these laws were
consequences of the gravitational force between any two
objects together with Newton’s laws of motion
Kepler’s 1st Law
• All planets move in
elliptical orbits with the
Sun at one focus.
– Any object bound to
another by an inverse
square law will move
in an elliptical path
– Second focus is empty
Kepler’s 2nd Law
• A line drawn from the
Sun to any planet will
sweep out equal areas
in equal times
– Area from A to B
and C to D are the
same
Kepler’s 3rd Law
• The square of the orbital period of any planet is proportional
to cube of the average distance from the Sun to the planet.
𝐓 2 = 𝐊 s 𝒓3
• For orbit around the Sun, K = KS = 2.97x10-19 s2/m3
• K is independent of the mass of the planet in orbit
K=
4𝜋2
𝐺𝐦
where m is the mass of the Sun being orbited.
Kepler’s Laws and Newton's Synthesis
Kepler’s laws can be derived from Newton’s laws. Irregularities
in planetary motion led to the discovery of Neptune, and
irregularities in stellar motion have led to the discovery of many
planets outside our Solar System.
Kepler’s 3rd Law
Kepler’s 3rd Law leads to an equation for the period of an
object in circular orbit. The speed of an object in a circular
orbit depends on the same factors:
𝑟3
𝑇 = 2𝜋
𝐺𝒎
𝑣=
𝒎
𝐺
𝑟
Note that m is the mass of the central object that is being
orbited. The mass of the planet or satellite that is in orbit
does not effect its speed or period.
The mean radius (r) is the distance between the centers of
the two bodies.
Kepler’s 3rd Law
𝟐
𝟒𝝅
𝑻𝟐 =
𝒓𝟑
𝑮𝒎
𝒎𝟏 𝒎𝟐
𝐅𝒈 = 𝐆
𝒓𝟐
𝒗𝟐
𝐅𝒈 = 𝐦
𝒓
𝑻=
𝟒𝝅𝟐 𝒓𝟑
𝑮𝒎
𝒎𝐬 𝒎
𝒗𝟐
𝐆 𝟐 =𝐦
𝒓
𝒓
𝒓𝟑
𝑻 = 𝟐𝝅
𝑮𝒎
𝒗=
𝒎
𝑮
𝒓
Kepler’s 3rd Law
Planetary Data
Period and Speed of an Orbiting Object Problem
Magellan was the first planetary spacecraft to be launched from a
space shuttle. During the spacecraft’s 5th orbit around Venus,
Magellan traveled at a mean altitude of 361 km. If the orbit had
been circular, what would Magellan’s period and speed have
been?
𝒓 = 361 km + 6050 km = 6.41 x 106 m
𝑟3
= 2𝜋
𝑇 = 2𝜋
𝐺𝒎
𝑣=
𝒎
𝐺
𝑟 =
6410000m 3
6.67 𝑥 10−11 4.87 𝑥 1024
6.67 𝑥
10−11
4.87 𝑥 1024
6.41 𝑥 106
5.66 x 𝟏𝟎𝟑 𝐬
𝟕. 𝟏𝟐
𝒙 𝟏𝟎𝟑
𝐦
𝐬
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Torque
To make an object start rotating, a force is needed; the position and
direction of the force matter as well.
The perpendicular distance from the axis of rotation to the line along
which the force acts is called the lever arm (r).
Torque
A longer lever arm is
very helpful in rotating
objects.
Torque
Here, the lever arm for FA is the distance from the knob to the
hinge; the lever arm for FD is zero; and the lever arm for FC is rC.
Torque
The torque is defined as:
𝝉 = 𝑭 𝐬𝐢𝐧 𝜽 𝒓 𝒐𝒓 𝝉 = 𝑭𝒓 𝐬𝐢𝐧 𝜽
Torque
Sign of Torque
Torque Problem
A basketball is being pushed by two players during a tip-off. One
player exerts an upward force of 15.0 N at a perpendicular
distance of 14.0 cm from the axis of rotation. The second player
applies a downward force of 11.0 N at a perpendicular distance of
7.00 cm from the axis of rotation. Find the net torque acting on
the ball about its center of mass.
t1 = F1d1 = (15 N)(–0.14 m) = –2.1 N•m
t2 = F2d2 = (–11 N)(0.070 m) = –0.77 N•m
tnet = t1 + t2 = –2.1 N•m – 0.77 N•m
tnet = –2.9 N•m
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END
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