Slides that go with Giancoli Chapter 5 section 5.6 to 5.9 on Universal Gravitation, Satellites, Weightlessness, and Kepler's Laws.

1. Newton’s Law of
Universal Gravitation
GIANCOLI - CHAPTER 5 – SECTION 5.6 – 5.9
MR. SHARICK

2. Section 5.6 – Newton’s Law of Universal
Gravitation
The geocentric model had now been proven
incorrect, and the heliocentric model of the
universe was very widely accepted.
Sir Isaac Newton, at the same time he developed
the three laws of motion, was also examining the
motions of the planets and the moon.
He was also trying to determine what the cause of
gravity was (Where did this force come from?)
He determined that gravity must act at a distance,
as opposed to most forces which are contact forces.
I wonder what causeth the apple to
falleth from the tree to the ground?

3. Section 5.6 – Deriving Newton’s
Universal Law
Newton found that the moons centripetal acceleration was 1/3600 of g.
He also realized that the moon was about 384,000km away from the Earth, which was about 60 times
the radius of the earth (6380 km).
Newton proposed that the proportionality between the gravitational force and the distance between the
two centers of the two objects was as follows:
𝐹 ∝
1
𝑑2

4. Section 5.6 – Deriving Newton’s
Universal Law
2
2
2 3
2 2
2 3
2 2
2
, 2
(get the total acceleration since sin )
(using r = sin )
(since c = d)
4
1. ( sin )
4
2. sin
4 1
3.
4
4. = P
P S
a a
da d
T
c
a
T d
c
a
T d
M
F
d












 
  
 
 
  
 
 
  
 
3
2
(using )
c
F ma
T
 
 
 

5. Section 5.6 – Deriving Newton’s
Universal Law
3
2
2 3
2
Using Kepler's 3rd law of planetary motion Newton knew that
c
was a constant for all planets orbiting the sun. Therefore, to
simplify Equ. # 4 Newton defined a new variable...
4
5. K
T
c
T

 3
2
20
m
, 2
, , 2
which he calculated as 1.33 10
Subsituting Equ # 5 into Equ. # 4 we get
6.
Using his own 3rd Law of motion Newton realized that
7.
s
P
P S
P
S P P S
M
F K
d
M
F F K
d


 

6. HERE IT COMES!!! ANOTHER STROKE OF GENIUS FROM NEWTON
Newton sees symmetry between these forces. If FS,P is proportional to the
mass of planet, then the FP,S must be proportional to the mass of the sun,
and therefore sets
where he called G the Universal Gravitational Constant.
With one last substitution Newton finds…
Section 5.6 – Deriving Newton’s
Universal Law
sK GM
, , 2
s P
S P P S
M M
F F G
d
 

7. Section 5.6 – Deriving Newton’s
Universal Law
Realize Newton still did not know G or MS
But, he did know GMS =
Newton 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…
The direction of the force is along the line joining the two objects and is
always an attractive force.
1 2
2
m m
F G
d

3
20
2
m
1.33 10
s


8. Section 5.6 – Cavendish Experiment
(1798)
•Used to measure the universal constant “G” performed by British
scientist Henry Cavendish
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 the angle,
twist measuredF 

9. Section 5.6 – Cavendish Experiment
(1798)
4. The light beam and scale are used to magnify theta.
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.
7. Cavendish was now able to isolate G, and calculated it as…
2Twist Gravity
mM
F F G
d
 
2
2
11
6.67 10 N m
kg
G x  


10. Example Problem – Using Newton’s
Universal Law of Gravitation
Assume the earth is moving in a circular orbit around the sun. Using the
following data calculate the speed of the earth in its orbit in miles/hr.
Mean Radius of Orbit = 1.5 x 1011 m
Mass of Sun = 1.99 x 1030 kg
1 mile = 1.61 km
G = 6.67 x 10-11
2
2
N m
kg

12. Section 5.6 – Example 5.11 – Can you
attract another person gravitationally?
A 50 kg person and a 75 kg person are sitting on a bench so that
their centers are about 50 cm apart. Estimate the magnitude of the
gravitational force each exerts on the other.

13. Section 5.6 – Example 5.12 – Spacecraft
at 2re
What is the force of gravity acting on a 2000 kg spacecraft when it
orbits two Earth radii from the Earth’s center (re = 6380 km)? The
mass of the Earth is me=5.98x1024kg.

14. Section 5.6 - Finding ‘g’ from Newton’s
Law of Universal Gravitation
Show that g = 9.81 m/s^2 using Newton’s Law of Universal
Gravitation, mass of the earth me=5.98x1024kg, and the radius of the
earth is 6380 km.

15. Section 5.6 – Example Problem - Gravity from
the moon
What is the gravitational force on a 50 kg person due to the mass of
the moon? The mass of the moon is 7.36 x 1022 kg and the distance
to the moon is 3.82 x 108m.

16. Section 5.7 – Gravity Near the Earth’s
Surface
Which ninja will experience a larger
acceleration from the earth?
How does this differ from g=9.81 m/s^2
that we calculate at sea level?

17. Section 5.7 – Gravity Near the Earth’s
Surface
The force from gravity a person experiences is his weight, which is:
Newton says that the attractive force between two objects is given by:
Gravity is the attractive force between the Earth and you!
1 2
2
gF mg
Gm m
F
r



18. Section 5.7 – Gravity Near the Earth’s
Surface
So making object 1 you, and object 2 the earth:
Cavendish was the first person to measure G, and used this to find the
mass of the Earth!
How could you use this equation to find ‘g’ on another planet or moon?
2
2
you earth
you
earth
earth
earth
Gm m
m g
r
Gm
g
r



19. Section 5.7 – Example 5.14 – Gravity on
Everest
Estimate the effective value of g on top of Mt. Everest, 8848m (29,028ft)
above the Earth’s surface.
What are some things that lower the precision of this calculated value?

20. Section 5.7 – Problem 5.32
What is the distance from the Earth’s center to a point outside the
Earth where the gravitational acceleration is 1/10 of its value at the
earth’s surface?

21. Section 5.7 – Losing weight
How would the weight of a 100 kg person change as he increases in
altitude? Find his weight at (a) sea level, (b) 5000 meters, (c) 10,000
meters, and (d) 20,000 meters.

22. Section 5.8 – Satellites and
“Weightlessness”

23. Section 5.8 – Newton’s cannonball
What causes a satellite to stay in orbit, rather than falling to the
Earth?
To answer this, let’s consider a cannon mounted on the top of Mt.
Everest.

24. Section 5.8 – Newton’s cannonball
What causes a satellite to stay in orbit, rather than falling to the
Earth?
To answer this, let’s consider a cannon mounted on the top of Mt.
Everest.

25. Section 5.8 – Newton’s cannonball
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.

26. Section 5.8 – Newton’s cannonball
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.

27. Section 5.8 – Newton’s cannonball
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.

28. Section 5.8 – Newton’s cannonball
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!!!!!

29. Section 5.8 – Newton’s cannonball
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!!!!!

30. Section 5.8 – Escape Velocity
And if it’s fired at a speed that is higher than its escape velocity, the
following happens!

31. Section 5.8 – Calculating Escape Velocity
2
c c
mv
F ma
r
 
RECALL:

32. Section 5.8 – Calculating Escape Velocity
1 2
2G
m m
F G
r

2
c c
mv
F ma
r
 
RECALL:
And

33. Section 5.8 – Calculating Escape Velocity
1 2
2G
m m
F G
r

2
c c
mv
F ma
r
 
C GF F
RECALL:
And

34. Section 5.8 – Calculating Escape Velocity
1 2
2G
m m
F G
r

2
c c
mv
F ma
r
 
C GF F
2
2
sat Earth satm v M m
G
r r

RECALL:
And

35. Section 5.8 – Calculating Escape Velocity
1 2
2G
m m
F G
r

2
c c
mv
F ma
r
 
C GF F
2
2
sat Earth satm v M m
G
r r

Earth
orbiting
M
v G
r

RECALL:
And

36. Section 5.8 – Calculating Escape Velocity
1 2
2G
m m
F G
r

2
c c
mv
F ma
r
 
C GF F
2
2
sat Earth satm v M m
G
r r

Earth
orbiting
M
v G
r

Earth
escape
M
v G
r

RECALL:
And

37. Section 5.8 – Example 5.15 –
Geosynchronous satellite
A geosynchronous satellite is one that stays above the same point on
the equator of the Earth. Such satellites are used for such purposes
as cable TV transmission, weather forecasting, and as
communication relays. Determine (a) the height above the Earth’s
surface such a satellite must orbit and (b) such a satellite’s speed.

38. Section 5.8 – Orbiting satellite
A 5000.0 kg satellite is moving in a stable circular orbit at altitude of
12,800 km above the earth's surface.
Rearth= 6.38 x 106 m Mearth= 5.98 x 1024 kg
G= 6.67 x 10-11 N*m2/kg2
a. Calculate the orbiting velocity of the satellite.
b. What is its period (time to make one orbit) in hours.

39. Section 5.8 – “Weightlessness”
Objects like satellites in orbit are said to experience apparent
“weightlessness”.
Let’s first consider a simpler case, a person standing in an elevator.

40. Section 5.8 – “Weightlessness”
If a = 0, then it is as simple as standing on
solid ground.
W = mg
Does this change if the elevator is moving
with constant speed?

41. Section 5.8 – “Weightlessness”
If the elevator is accelerating up at a rate of
a=1/2 g, then we have
ma = W – mg
W = ma + mg = (1/2)mg + mg = (3/2)mg

42. Section 5.8 – “Weightlessness”
If the elevator cable breaks, you have more
problems than just worrying about your
weight!
…like, dying when the elevator crashes into
the ground.

43. Section 5.8 – “Weightlessness”
But, let’s look at the physics anyway!
The overall acceleration is a = g, so
ma = mg - W
W = mg - ma
W = 0; thus “weightless!”

44. Section 5.8 – “Weightlessness”

45. These two situations are nearly the
same!

46. Section 5.8 – Finding the acceleration of
a “weightless” astronut
This man is plummeting toward the Earth
with an acceleration very close to 9.8 m/s^2.
If the satellite is orbiting 36,000 km above
the surface of the Earth, what is the value of
“g” this astronut experiences?

47. Section 5.8 – Problem 5.41
A cylindrical spaceship is rotated such that its occupants are to
experience simulated gravity of ½ g. Assume the spaceship’s
diameter is 32m. Find the time needed for one revolution (period).

48. Section 5.8 – Problem 5.46
A ferris wheel, 24.0m in diameter rotates once every 12.5s. What is
the person’s apparent weight (mass = 80kg) at the top and at the
bottom, compared to the person’s weight at rest?

49. Section 5.9 – Kepler’s Laws
1st Law- The orbits of the planets are ellipses, with the sun at one
focal point.

50. Section 5.9 – Kepler’s Laws
2nd Law- The line joining the sun and a planet (its radius vector)
sweeps over equal areas in equal times.
The time it takes to go from 1 to 2 (moving faster) is the same as it
takes to go from 3 to 4 (moving slower)

51. Section 5.9 – Kepler’s Laws
Also called the Law of Equal Areas

52. Section 5.9 – Kepler’s Laws
3rd - The squares of the periods of the planets’ motions are
proportional to the cubes of the semimajor axes of their elliptical
paths; that is c3/T2 is the same for all the planets, where T is the time
for the planet to complete one orbit about the sun and c is the
semimajor axis.
3 3
2 2
A B
A B
c c
T T


53. Deriving Kepler’s 3rd Law
2
1 1 1
2
1 1
1
1
1
2
1 1 1
2 2
1 1
Setting forces equal:
We know:
2
Substituting gives:
4
S
S
Gm M m v
r r
r
v
T
Gm M m r
r T





2 2
1
3
1
2 2
1 2
3 3
1 2
Rearrange to get:
4
This is the constant the Kepler derived!
Do this for a 2nd planet and then you can write:
S
T
r GM
T T
r r




54. Section 5.9 – Using Kepler’s 3rd Law
Mars’ period was first noted by Kepler to be about 687 days (Earth
days), which is 1.88 years. Determine the distance of Mars from the
Sun using the Earth as a reference (distance of the earth from the
sun is 1.50 x 1011m).

55. Section 5.9 – Using Kepler’s 3rd Law
How long does it take the planet Saturn to make one revolution
about the Sun if its orbital radius is 9.5 times larger than the earth?

56. Section 5.9 – Using Kepler’s 3rd Law
Determine the mass of the Sun given the Earth’s distance from the
Sun as r = 1.50 x 1011m.
2 2
1
3
1
4
S
T
r GM



57. Section 5.9 – Problem 5.56
The Sun rotates about the center of the Milky Way Galaxy at a
distance of about 30,000 light years from the center (1 light year =
9.5x1015m). If it take about 200 million years to make one rotation,
estimate the mass of our galaxy. Assume that the mass is
concentrated mostly in a central uniform sphere.
If all the stars had the mass of our Sun (about 2 x 1030kg) how many
stars would there be in our galaxy?

58. Section 5.9 – Kepler’s Laws
What happens next?
•Galileo used his telescopes to discover the four moons of Jupiter and times their
periods of orbit.
•He found that for all four moons Kepler’s ratio of c3/T2 was constant.
•This convinced most scientists that Kelper’s laws were not an accidental fit, and
Copernicus’ heliocentric universe model must be correct.
•The Catholic Church was not convinced and Galileo was arrested for his
teachings and excommunicated from the church. He has since been re-instated.
-Kepler and Galileo had worked to show how the planets behaved, and how the
motions of the moons of Jupiter and the planets of our sun obeyed the same
rules. It was Newton’s turn to explain what causes planets to obey the same
rules.