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Physics 3311
Unit 5, Part II Notes
Gravitation
_________________________
STUDENT NAME
_________________________
TEACHER’S NAME
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Geocentrism –
Heliocentrism -
KEPLER’S THREE LAW’S OF PLANETARY MOTION
1st – The orbits of the planets are ellipses, with the sun at one focal point.
o The eccentricity of an ellipse ranges from 0 to 1 and describes the shape of the ellipse.
The greater the eccentricity (closer to 1) the more “oblong” the ellipse appears.
o Ellipses are not circles! The eccentricity of an ellipse does not say how large the ellipse
is, rather how un-circular the shape may be.
Celestial
Object
Mercury
Venus
Earth
Earth’s Moon
Mars
Jupiter
Orbital
Period
88 days
224.7 days
365.26 days
27.3 days
687 days
11.9 years
Eccentricity
of Orbit
0.206
0.007
0.017
0.055
0.093
0.048
Celestial
Object
Saturn
Uranus
Neptune
Pluto
Halley’s Comet
Comet Hyakutake
Orbital
Period
29.5 years
84.0 years
164.8 years
248.0 years
76.1 years
~ 70,000 yrs
Eccentricity
of Orbit
0.054
0.047
0.009
0.249
0.967
0.9998
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2nd – The line joining the sun and a planet (its radius vector) sweeps over equal areas in
equal times.
 What does this mean about the speed of a planet when it is closest/farthest from the
sun?
3rd – The squares of the periods of the planets’ motions around the sun are proportional
to the cubes of the average distance of each planet from the Sun.
o This ratio (r3/T2) is the same for all the planets, where T is the time for the planet to
complete one orbit about the sun (its orbital period) and c is the semi-major axis of
the planets orbit.
𝑟𝐴3 𝑟𝐵3
2 = 2
𝑇𝐴
𝑇𝐵
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Kepler’s Laws of Planetary Motion Questions:
1. Jupiter is 5.2 times farther from the Sun than Earth is. Find Jupiter’s orbital period in
Earth years.
2. Neptune requires 164 years to circle the Sun. Find Neptune’s orbital radius as a
multiple of Earth’s radius.
3. If a small planet, D, were located 8.0 times as far from the Sun as Earth is, how many
years would it take the planet to orbit the sun?
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NEWTON’S ANALYSIS OF PLANETARY MOTION
(1642 – 1727) Isaac Newton
 In 1687 Newton published his Philosophiæ Naturalis Principia Mathematica,
(Mathematical Principles of Natural Philosophy) which outlined his laws of motion
and became the foundation of classical mechanics.
o Newton discovered that the same force governing the motion of objects on
Earth was the same force governing the motion of the planets around the Sun
in our solar system.
 Newton proposed the following proportionality between the force, F and the distance
between the centers of the planet and the center of the sun, d.
𝐹∝
1
𝑑2
Change Result
2d
F/4
3d
F/9
d/2
4F
d/3
9F
 He also proposed that the Force is proportional to the product of the masses
𝐹 ∝ 𝑚1 𝑚2
Change Result
2m1m2
2F
3m1m2
3F
2m13m2
6F
½ m1m2
½F
He went through a derivation and came up with a law that he assumed that his law
should work for any two objects not just planets. Newton’s Law of Universal Gravitation
for any two objects is stated as follows…
𝐹𝑔 =
𝐺𝑚1 𝑚2
𝑑2
Defined, this law states that the force of gravity is proportional to the product of the
masses and inversely proportional to the square of the distances between them.
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 The force of gravity is a purely attractive force. It is always trying to bring things
together! Gravity obeys Newton’s 3rd Law. It is a mutual force shared between two
objects.
 The force of gravity is universal in that it extends FOREVER. However the farther
away you are from an object, the weaker and weaker the force of attraction
becomes.
 We are going to assume that all objects we are
dealing with are uniform in nature. All this means is
that the force of gravity acts on an object at the
center of the object. We are constantly being pulled
towards the center of the Earth.
 Gravity is always trying to pull two objects together, and acts in a straight line
between the centers of the two objects.
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THE CAVENDISH EXPERIMENT (1798)
OR
The torsion balance and the measurement of the gravitational
constant, “G”.
Scale
Light Beam
Mirr
or
Torsion Balance
Principle of Operation
1. Two small masses are placed at
the end of a light rod.
2. The force required to rotate the
fiber from its untwisted position (a),
is measured.
3. The force required to twist the
fiber is calibrated as a function of
Ftwist   measured
4. The light beam and scale are
5. When the mass M, is brought
near to mass m it applies a
gravitational force onto the fiber. To
check for symmetry it is placed in
positions AA and BB.
6. Knowing that the force causing
the twist was caused by gravity
Cavendish now used The Law of
Universal Gravitation. FTwist  FGravity  G
mM
d2
7. Cavendish was now able to isolate G, and calculated it as…
11 N m2
kg 2
G  6.67 x10
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Example 1: Using the data listed below, calculate the force of gravity between the Earth
and the Moon.
Earth Mass = 5.97x1024 kg
Moon Mass = 7.34x1022 kg
Moon orbital distance = 3.85x109 m
Example 2: Calculate the force of gravity on a spacecraft 12,800,000 m above the Earth’s
surface if its mass is 1400 kg. The mass of earth is 5.97 x 10 24 kg and the radius is
6.38x106m.
Example 3: Two balls have their centers 2.0 m apart. One ball has a mass of 8.0 kg. The
other has a mass of 6.0 kg. What is the gravitational force between them?
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SATELLITE MOTION / ORBIT
 Imagine placing a large cannon on top of Mount Everest.
 If we were to load the cannon with enough gun powder and fire it HORIZONTALLY to
the surface of the Earth below, we may hit a town far away.
 If we were to load the cannon with more gun powder and fire it HORIZONTALLY to the
surface of the Earth below, we may even hit a different country far away.
 Let’s say we were to load the cannon with just enough gun powder and fire it
HORIZONTALLY to the surface of the Earth below, we may even hit a country halfway
around the world.
 Well, let’s say we were to load the cannon with perfect amount of gun powder and
fire it HORIZONTALLY to the surface of the Earth below, so that it just keeps falling for
ever as the Earth curves away from its fall. If this cannon ball does not hit the back of
the cannon which it was fired from, then this cannon ball will continuously go around
the world; falling towards the Earth… FOREVER!!!!!
This is exactly what is happening to our moon with Earth and all the planets with our Sun.
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Orbiting Velocity and Escape Velocity
Example: satellite in orbit around the
Earth!
RECALL: from uniform circular motion…
𝐹𝑐 =
𝑚𝑣 2
𝑟
and universal gravitation…
𝐹𝑔 =
For an object in a circular (or very nearly circular) orbit…
FC  FG
2
𝑚𝑠𝑎𝑡𝑒𝑙𝑙𝑖𝑡𝑒 𝑣𝑜𝑟𝑏𝑖𝑡
𝐺𝑀𝐸𝑎𝑟𝑡ℎ 𝑚𝑠𝑎𝑡𝑒𝑙𝑙𝑖𝑡𝑒
=
𝑟
𝑟2
𝑣𝑜𝑟𝑏𝑖𝑡 = √
𝐺𝑀
𝑟
𝐺𝑀
𝑟
𝑣𝑒𝑠𝑐𝑎𝑝𝑒 > √
Figure 1https://xkcd.com/681/
𝐺𝑚1 𝑚2
𝑟2
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Example: Satellite Motion
A 5000.0 kg satellite is moving in a stable circular orbit at altitude of
12,800,000 m above the earth's surface.
Rearth= 6.38 x 106 m
Mearth= 5.98 x 1024 kg
a.
Please calculate the orbiting velocity of the satellite.
b.
What is its period (time to make one orbit) in hours.
Example 2
What’s the escape velocity of the astronauts for the Apollo 11 mission to the
Moon? The mass of the moon is 7.34x1022 kg, and it has a radius of 1.74x106
meters.
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Determination of the Acceleration Due to Gravity (g)
 The acceleration due to gravity would be a purely constant value at the surface of a
perfectly spherical planet or celestial object, no matter your location. However, the
Earth’s landscape is hardly constant with so many mountains and valleys on the
surface…. The acceleration due to gravity (g) is different at the top of Mount Everest
vs the bottom of Dead Sea.
Analyzing the gravitational attraction between a person and the center of the Earth….
𝐹𝑔 = 𝐹𝑔
𝑚𝑝𝑒𝑟𝑠𝑜𝑛 𝑔 = 𝐺
𝑚𝑝𝑒𝑟𝑠𝑜𝑛 𝑀𝐸𝑎𝑟𝑡ℎ
𝑑2
𝑀
𝑔=𝐺 2
𝑑
Example 1: Calculate the weight of a 50 kg person at:
A. The top of Mt. Everest (elevation of about 8800 meters above Earth’s surface)
B. The bottom of the Dead Sea (elevation of about 410 meters below Earth’s surface)
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Example 2: The Moon has a mass equal to 1/81 that of the Earth and a radius of 1.74x106
meters. Calculate the weight of 45 kg astronaut at the surface.
3. Assume that a satellite orbits Earth 225,000 m above its surface. Given that the mass of Earth is
5.97x1024 kg and the radius of Earth is 6.38x106 m, what are the satellite’s orbital speed and period?
4. Suppose that the satellite in number 2 is moved to an orbit that is 24,000 m larger in radius that it’s
previous orbit. What would its new speed be?
5. Given that Mercury has a radius of 2.44 x 106 m and a mass of 3.3x1023kg, what is the speed and
period of a satellite that orbits 260,000 m above Mercury’s surface?
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Review Problems for Test
1. NASA places a 100.0 kg satellite in a circular orbit just above the surface of the Earth. The mass of
the Earth is 5.98 x 1024 kg, and its radius is 6,371,000 m.
a. How much gravitation force does the Earth exert on the satellite? (983N)
b. What is the satellite’s orbital speed? (7912 m/s)
c. What is the satellites orbital period? (5059 s)
2. A 100 kg satellite is placed in orbit about the Earth at an altitude of 6.63 times the Earth’s average
radius.
a. How much gravitational force does the Earth exert on this satellite? (16.45N)
b. What is the satellite’s orbital speed? (2846 m/s)
c. What is the satellite’s orbital period? (30 hours)
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Data: Mass of Mars 6.42 x 10 23 kg
Mass of the Sun 1.991 x 10 30 kg
Mars’s distance from the Sun 2.279 x 10 11 m
3. Answer each of the questions below about the planet Mars.
a) Find the velocity with which Mars moves around the Sun.
b) How long in days does it take Mars to make one revolution about the Sun?
c) What is the Force of gravity experienced by Mars from the Sun?
4. You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they
respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it
isn’t so by calculating how much weaker gravity is 300,000 m above the Earth’s surface.