The document discusses Newton's realization that the force of gravity on Earth must come from Earth itself and must also be what keeps the Moon in orbit. It then explains Newton's third law and how it applies to gravitational forces between objects of different masses. The document also defines the universal law of gravitation, including how the gravitational force between two objects is proportional to their masses and inversely proportional to the square of the distance between them. It further discusses how to calculate gravitational forces and acceleration due to gravity at different locations and distances from Earth.

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2. Newton’s Question: If the force of gravity is being exerted
on objects on Earth, what is the origin of that force?
Newton’s realization was that
this force must come from
the Earth.
He further realized that this
force must be what keeps
the Moon in its orbit.

3. 
Must be true from
Newton’s 3rd Law
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.
The gravitational force one body exerts
on a 2nd body , is directed toward the
first body, and is equal and opposite to
the force exerted by the second body on
the first

4. Every particle of matter in the
universe attracts every other particle with a
force that is directly proportional to the product
of the masses of the particles and inversely
proportional to the square of the distance
between them.
F12 = -F21  [(m1m2)/r2]
Direction of this force:  Along the line joining
the 2 masses

5.  G = the Universal Gravitational constant
 Measurements in SI Units:
 The force given above is strictly valid only for:
› Very small masses m1 & m2 (point asses)
› Uniform spheres
 For other objects: Need integral calculus!

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

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

8. 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

9. • Earth Radius: rE = 6320 km
Earth Mass: ME = 5.98  1024 kg
FG = G(mME/r2)
Mass of the Space craft m
• At surface r = rE
FG = weight
or mg = G[mME/(rE)2]
• At r = 2rE
FG = G[mME/(2rE)2]
or (¼)mg = 4900 N
• A spacecraft at an altitude of twice the Earth radius

10. 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

11. F = FME + FMS
(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]
= 4.77  1020 N
tan(θ) = 1.99/4.34
 θ = 24.6º

12. Gravity Near Earth’s Surface
Gravitational
Acceleration g
and
Gravitational
Constant G

13.  Obviously, it’s very important to distinguish
between G and g
 They are obviously very different physical
quantities
 G  The Universal Gravitational Constant
› It is the same everywhere in the Universe
G = 6.673  10-11 N∙m2/kg2
Always same on every location
 g  The Acceleration due to Gravity
g = 9.80 m/s2 (approx) on Earth’s surface
g varies with location
G vs. g

14. Consider an object on Earth’s surface:
mE = mass of the Earth
rE = radius of the Earth
m = mass of object
Let us the Earth is a uniform, perfect
sphere.
The weight of m: FG = mg
The Gravitational force on m:
FG = G[(mmE)/(rE)2]
Setting these equal gives:
g in terms of G m
mE
g = 9.8 m/s2All quantities on the right are measured!

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

16. Acceleration due to gravity at a
distance r from Earth’s center.
Write gravitational force as:
FG = G[(mME)/r2]  mg
(effective weight)
g  the effective acceleration
due to gravity.
SO : g = G (ME)/r2
ME

17.  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[(m ME)/r2]  mg
This gives:
 This shows that g decreases with increasing altitude
 As r , the weight of the object approaches zero
 
2
E
E
GM
g
R h


ME

18. Altitude Dependence of g

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