1. Newton’s Universal Law of Gravitation
This cartoon mixes 2 legends:
1. The legend of Newton,
the apple & gravity
which led to
The Universal Law
of Gravitation.
2. The legend of William
Tell & the apple.

2. Newton's Universal Law of Gravitation!
It was very SIGNIFICANT & PROFOUND in the 1600's when Sir Isaac Newton first wrote
Newton's Universal
Law of Gravitation!
This was done at the age of about 30. It was this, more than any of his other achievements, which caused him to be well-known in the world scientific community of the late 1600's.

3. The orbits of the planets around
Newton's Universal Law
of Gravitation
Newton used this law, along with Newton's 2nd Law (his 2nd Law!) plus Calculus, which he also (co-) invented, to PROVE that
The orbits of the planets around
the sun must be ellipses.
For simplicity, we assume in the following that these orbits are circular.

4. THE COURSE THEME OF NEWTON'S LAWS OF MOTION
The topic of Gravitation fits
THE COURSE THEME OF
NEWTON'S LAWS OF MOTION
Newton used his Gravitation Law & his 2nd Law in his analysis of planetary motion. His prediction that planetary orbits are elliptical is in excellent agreement with Kepler's analysis of observational data & with Kepler's empirical laws of planetary motion.

5. Universal Law of Gravitation,
When Newton first wrote the
Universal Law of Gravitation,
it was the first time, anyone had EVER
written a theoretical expression (physics in
math form) & used it to PREDICT
something in agreement with observations!
For this reason, Newton's formulation of his Universal Gravitation Law is considered to be the BEGINNING OF THEORETICAL PHYSICS.

6. The Universal Law of Gravitation
Also gave Newton his major “claim to fame”. After this, he was considered to be a “major leader” in science & math among his peers. In modern times, this, plus the many other things he did, have led to the consensus that Sir Isaac Newton was the
GREATEST SCIENTIST
WHO EVER LIVED

7. Newton’s Universal Law of Gravitation The Gravitational Force
This is an EXPERIMENTAL LAW describing the gravitational force of attraction between 2 objects.
Newton’s reasoning:
The Gravitational Force
of Attraction
between 2 large objects (Earth - Moon, etc.) is the SAME force as the attraction of objects to the Earth.

8. Newton’s Universal Law of Gravitation
The Gravitational Force of Attraction
between 2 large objects (Earth - Moon, etc.) is the SAME force as the attraction of objects to the Earth.
Apple Story:
This is likely not a true historical account, but the reasoning discussed there is correct. This story is probably legend rather than fact.

9. this force is what keeps the Moon
Newton’s Question:
If the force of gravity is being exerted on objects
on Earth, what is the origin of that force?
Newton realized that
this force must come
from the Earth.
He further realized that
this force is what keeps the Moon
in its orbit.
Figure 6-1. Caption: Anywhere on Earth, whether in Alaska, Australia, or Peru, the force of gravity acts downward toward the center of the Earth.

10. Newton’s 3rd Law! Newton’s 3rd Law Pair
The gravitational force on you is half of a
Newton’s 3rd Law Pair
Earth exerts a downward force on you, & you exert an
upward force on Earth. When there is such a large
difference in the 2 masses, the reaction force (force you
exert on the Earth) is undetectable, but for 2 objects with
masses closer in size to each other, it can be significant.
Figure 6-2. Caption: The gravitational force one object exerts on a second object is directed toward the first object, and is equal and opposite to the force exerted by the second object on the first.
This must be
true from
Newton’s
3rd Law!

11. Newton’s Reasoning The Force of Attraction between 2
Always a Newton’s
3rd Law Pair!
Figure 6-2. Caption: The gravitational force one object exerts on a second object is directed toward the first object, and is equal and opposite to the force exerted by the second object on the first.
Newton’s Reasoning
The Force of Attraction between 2
small masses is the same as the force
between Earth & Moon, Earth & Sun, etc.

12. proportional to the product of the the square of the distance
By observing planetary orbits & analyzing other’s
observations, Newton concluded that the
gravitational force decreases as the inverse
square of the distance r between the 2 masses.
Newton’s Universal Law of Gravitation:
“Every particle in the Universe attracts every other particle in the Universe with a force that is
proportional to the product of the
masses & inversely proportional to
the square of the distance
between them”

13. Newton’s Universal Law of Gravitation:
“Every particle in the Universe attracts every other particle in the Universe with a force that is
proportional to the product of their masses &
inversely proportional to the square of the
distance between them:
F12 = -F21  [(m1m2)/r2]
F12 must = – F21 from
Newton’s 3rd Law!
The direction of this force is
along the line joining the 2 masses.

14. The FORCE between 2 small masses has the same origin (Gravity) as the FORCE between the Earth & the Moon, the Earth & the Sun, etc. 
From Newton’s 3rd Law!

15. Newton’s Universal Gravitation Law
This force is written as:
G  a constant
G  Universal Gravitational
Constant
G is measured & is the same for ALL objects. G must be small!

16. G  Universal Gravitational Constant
G is measured & is the same for ALL objects. G must be small!
Measurement of G in the
lab is tedious & sensitive because it is so small.
First done by Cavendish
in 1789.
Cavendish
Measurement
Apparatus

17. G  Universal Gravitational Constant
Modern version of the Cavendish experiment: Two small masses are fixed at ends of a light horizontal rod.
Two larger masses are placed near the smaller ones. The angle of rotation is measured. Use Newton’s 2nd Law to get the vector force between the masses. Relate to the angle of rotation & then extract G.
Cavendish
Measurement
Apparatus

18. strictly valid only for:
G = Universal Gravitational Constant
Measurements find, in SI Units:
The force given above is
strictly valid only for:
Very small (point) masses m1 & m2
Uniform spheres
For other objects: Need integral calculus!

19. The Universal Law of Gravitation is an example of an inverse square law
The force magnitude varies as the inverse square of the separation of the particles
The law can be expressed in vector form.
The negative sign means it’s an attractive force! Aren’t we glad it’s not repulsive?

20. Comments F12  Force exerted by particle 1 on particle 2
Its also true that F21 = - F12
This tells us that the forces form a Newton’s 3rd
Law action-reaction pair, as expected.
The negative sign in the vector equation tells us that
particle 2 is attracted toward particle 1

21. More Comments Gravity is a “field force” that always exists between
2 masses, regardless of the medium between them. The gravitational force decreases rapidly as the distance between the 2 masses increases. (This is an obvious consequence of the inverse square law)

22. Solar System Data (needed for calculations)
See also Table 5.1, page 147 or our book
Answer: Assume the distance between the two people is about ½ m; then the gravitational force between them is about 10-6 N.

23. Solar System Data

24. Gravitational Force Between 2 People
Example
Gravitational Force Between 2 People
A m1 = 50-kg person & m2 = 70-kg person
are sitting on a bench close to each other.
Estimate the magnitude of the gravitational
force each exerts on the other.
Answer: Assume the distance between the two people is about ½ m; then the gravitational force between them is about 10-6 N.

25. Gravitational Force Between 2 People
Example
Gravitational Force Between 2 People
A m1 = 50-kg person & m2 = 70-kg person
are sitting on a bench close to each other.
Estimate the magnitude of the gravitational
force each exerts on the other.
Solution: Suppose r = 1.0 m.
The force is then: F = (Gm1m2)/(r2)
= (6.67  10-11)(50)(70)/(1)2
F  2.33  10-7 N
Too small to be noticed!
Answer: Assume the distance between the two people is about ½ m; then the gravitational force between them is about 10-6 N.

26. Example: Spacecraft at 2rE
A spacecraft of mass
m = 2000 kg at an altitude of twice the Earth radius.
Numbers from tables:
Earth Radius: rE = 6320 km
Mass: ME = 5.98  1024 kg
FG = G(mME/r2)
Earth,
Mass ME m
At r = 2rE
FG = G[mME/(2rE)2] = 4900 N

27. Example: Force on the Moon
Find the net force on the
Moon due to the gravitational
attraction of both the Earth &
the Sun, assuming they are at
right angles to each other.
ME= 5.99  1024kg, MM= 7.35 1022kg
MS = 1.99  1030 kg
rME = 3.85  108 m, rMS = 1.5  1011 m
F = FME + FMS
(A vector sum!)

28. F = FME + FMS = 4.77  1020 N tan(θ) = 1.99/4.34  θ = 24.6º
(A vector sum!)
FME = G [(MMME)/(rME)2]
= 1.99  1020 N
FMS = G [(MMMS)/(rMS)2]
= 4.34  1020 N
F = [ (FME)2 + (FMS)2](½)
=  1020 N
tan(θ) = 1.99/4.34
 θ = 24.6º

29. Gravitational Force Due to a Mass Distribution
It can be shown with integral calculus that: The gravitational force exerted by a spherically symmetric mass distribution of uniform density on a particle outside the distribution is the same as if the entire mass of the distribution were concentrated at the center.
So, assuming that the Earth is such a sphere, the gravitational force exerted by the Earth on a particle of mass m on or near the Earth’s surface is
FG = G[(mME)/r2]
ME  Earth Mass, rE  Earth Radius

30. ME  Earth Mass, rE  Earth Radius
FG = G[(mME)/r2]
ME  Earth Mass, rE  Earth Radius
Similarly, to treat the gravitational force due to large spherical shaped objects, we can show with calculus, that:
1. If a (point) particle is outside a thin spherical shell, the gravitational force on the particle is the same as if all the mass of the sphere were at center of the shell.
2. If a (point) particle is inside a thin spherical shell, the gravitational force on the particle is zero. So, we can model a sphere as a series of thin shells. For a mass outside any large spherically symmetric mass, the gravitational force acts as though all the mass of the sphere is at the sphere’s center.

31. Vector Form of the Universal
Gravitation Law
In vector form:
The figure at the right gives the
directions of the displacement &
force vectors.
If there are many particles, the
total force is the vector sum of
the individual forces:
Figure 6-6. Caption: The displacement vector r21 points from particle of mass m2 to particle of mass m1. The unit vector shown, r21, is in the same direction as r21 but is defined as having length one.

32. Example: Billiards (Pool)
3 billiard (pool) balls, masses m1 = m2 = m3 = 0.3 kg on a table as in the figure. Triangle sides:
a = 0.4 m, b = 0.3 m, c = 0.5 m.
Calculate the magnitude & direction of the total gravitational force F on m1 due to m2 & m3.
Note: Gravitational force is a vector, so we have to add the vectors F21 & F31 to get the vector F (using the vector addition methods of earlier). F = F21 + F31

33. Example: Billiards (Pool)
3 billiard (pool) balls, masses m1 = m2 = m3 = 0.3 kg on a table as in the figure. Triangle sides:
a = 0.4 m, b = 0.3 m, c = 0.5 m.
Calculate the magnitude & direction of the total gravitational force F on m1 due to m2 & m3.
Note: Gravitational force is a vector, so we have to add the vectors F21 & F31 to get the vector F (using the vector addition methods of earlier). F = F21 + F31
Using components:
Fx = F21x + F31x =  N
Fy = F21y + F31y = 3.75  N
So, F = [(Fx)2 + (Fy)2]½ = 3.75  N
tanθ = 0.562, θ = 29.3º

34. Gravity Near the Earth’s Surface
The Gravitational
Acceleration g
and
Constant G

35. G vs. g
Obviously, it’s very important to distinguish between G and g!
They are clearly very different physical quantities!
G  Universal Gravitational Constant
It’s the same everywhere in the Universe
G =  N∙m2/kg2
ALWAYS at every location anywhere
g  The Acceleration Due to Gravity
g = 9.80 m/s2 (approximately!) on the Earth’s surface. g varies with location

36. g in terms of G FG = G[(mmE)/(rE)2]
Consider a mass m on Earth’s surface:
mE = mass of Earth (say, known)
rE = radius of Earth (known)
m = mass of the object (known)
Assume that the Earth is a uniform, perfect sphere.
The weight of m is FG = mg
The Gravitational force on m is
FG = G[(mmE)/(rE)2]
Setting these equal gives:
All quantities on the right are measured! mE
g = 9.8 m/s2

37. mE  6  1024 kg “Weigh” the Earth! Newton’s Gravitation Law
Using the same process, we can
“Weigh” the Earth!
(Determine it’s mass).
On Earth’s surface, equate the
usual weight of mass m to the
Newton’s Gravitation Law
form for the gravitational force: m mE
All quantities on the
right are measured!
Knowing g = 9.8 m/s2 & the radius of the Earth rE,
the mass of the Earth mE can be calculated:
mE  6  1024 kg

38. Effective Acceleration Due to Gravity
Consider the acceleration due to gravity at a distance r from Earth’s center.
Write the gravitational force as:
FG = G[(mME)/r2]  mg
( “Effective Weight” of m)
g  Effective acceleration
due to gravity.
SO : g = G(ME)/r2 ME

39. Altitude Dependence of g
If an object is some distance h
above the Earth’s surface, r
becomes RE + h. Again, set the
gravitational force equal to mg:
G[(mME)/r2]  mg
This gives: ME 
This shows that g decreases with increasing altitude
As r ® , the weight of the object approaches zero
Example, g on Mt. Everest

40. Altitude Dependence of g
Lubbock, TX:
Altitude: h  3300 ft  1100 m
 g  m/s2
Mt. Everest:
Altitude: h  8.8 km
 g  m/s2

41. Example: Effect of Earth’s Rotation on g
Assume that Earth is a perfect sphere.
Determine how its rotation affects
the value of g at the equator
compared to its value at the poles.
Let towards the center of the
Earth be positive y.
Figure 6-9.
Answer: The centripetal acceleration subtracts from g at the equator. First calculate the rotational speed at the equator: v = x 102 m/s. Then Δg = v2/r = m/s2, a difference of 0.3%.
Newton’s 2nd Law is:
∑F = maR = mg – W
so W = mg – maR = mg – m(v2/r)

42. Example: Effect of Earth’s Rotation on g
Newton’s 2nd Law:
∑F = maR = mg – W
so W = mg – maR
W = mg – m(v2/r)
At the pole, v = 0, so W – mg = 0 & W = mg
At the equator, the centripetal acceleration is
aR = [(v2)/(rE)]
So, the effective weight of a mass m there is:
W = mg – maR
= mg - m[(v2)/(rE)] = mg
Figure 6-9.
Answer: The centripetal acceleration subtracts from g at the equator. First calculate the rotational speed at the equator: v = x 102 m/s. Then Δg = v2/r = m/s2, a difference of 0.3%.

43. Example: Effect of Earth’s Rotation on g
Effective weight of mass m at equator:
W = mg – maR =
mg - m[(v2)/(rE)] = mg
Period = T = 1 day = 8.64  104 s
So the speed v there is
v = (2πrE)/T = 4.64  102 m/s
Figure 6-9.
Answer: The centripetal acceleration subtracts from g at the equator. First calculate the rotational speed at the equator: v = x 102 m/s. Then Δg = v2/r = m/s2, a difference of 0.3%.
The centripetal acceleration at the equator is
aC = [(v2)/(rE)] = m/s2
Effective gravitational acceleration at equator:
g = g - [(v2)/(rE)] = = m/s2

44. “Weighing” the Sun!
We’ve “weighed” the Earth, now lets “Weigh” the Sun!! (Determine it’s mass).
Assume: Earth & Sun are perfect uniform spheres & Earth orbit is a perfect circle (actually its an ellipse).
Note: For Earth: Mass ME = 5.99  1024kg
The orbit period is T = 1 yr  3 107 s
The orbit radius r = 1.5  1011 m
The orbit velocity v = (2πr/T), = v  3 104 m/s

45. ∑F = FG = MEa = MEac = ME(v2)/r
“Weighing” the Sun!
Note: For Earth: Mass ME = 5.99  1024kg
The orbit period is T = 1 yr  3 107 s
The orbit radius r = 1.5  1011 m
The orbit velocity v = (2πr/T), = v  3 104 m/s
The Gravitational Force between the Earth & the Sun:
FG = G[(MSME)/r2]
Circular orbit  Centripetal acceleration
Newton’s 2nd Law for the Earth gives:
∑F = FG = MEa = MEac = ME(v2)/r
OR: G[(MSME)/r2] = ME(v2)/r.
If the Sun mass is unknown, solve for it:
MS = (v2r)/G  2  1030 kg  3.3  105 ME

46. Gravitation and the Moon’s Orbit
The Moon follows an approximately circular orbit around the Earth
There is a force required for this motion
Gravity supplies the Centripetal Force

47. The Moon’s Motion We’ve assumed that the Moon orbits a “fixed” Earth
This is a good approximation
It ignores Earth’s motion around the Sun
The Earth and Moon actually both orbit their center of mass
We can think of the Earth as orbiting the Moon
The circle of the Earth’s motion is very small compared to the Moon’s orbit