Newton’s Universal Law
of Gravitation
This cartoon mixes 2 legends:
1. The legend of
Newton, the apple &
gravity which led to
Newton’s Universal
Law of Gravitation.
2. The legend of William
Tell & the apple.
• 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 wellknown in the world scientific community of the late 1600's.
• He 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.
• 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.
• 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.
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
Newton’s Law of Universal Gravitation
• 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.
• 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.
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.
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.

Must be true from
Newton’s 3rd Law!
The Force of Attraction between 2 small masses is the
same as the force between Earth & Moon, Earth & Sun, etc.
• By observing planetary orbits, Newton concluded that
the gravitational force decreases as the inverse of the
square of the distance r between the 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 their masses & inversely proportional to
the square of the distance between them:
F12 = -F21  [(m1m2)/r2]
This Must be true from
Newton’s 3rd Law!
• The direction of this force is
along the line joining the 2 masses.
• 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!
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!
• Measurement of G in the lab is tedious &
sensitive because it is so small.
– First done by Cavendish in 1789.
• A modern version of the Cavendish
experiment: Two small masses are fixed at
ends of a light horizontal rod. Two larger
masses were placed near the smaller ones.
• The angle of rotation is measured.
• Use Newton’s 2nd Law to get the vector
Cavendish Measurement
force between the masses. Relate to the
Apparatus
angle of rotation & then extract G.
G = Universal Gravitational Constant
• Measurements find, in SI Units:
• The force given above is
strictly valid only for:
– Very small masses m1 & m2 (point masses)
– Uniform spheres
• For other objects: Need integral calculus!
• 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?
Comments
• F12  Force exerted by particle 1 on particle 2
• F21  Force exerted by particle 2 on particle 1
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 above vector equation
tells us that particle 2 is attracted toward particle 1
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
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
-7
F  2.33  10 N
• Too small to be noticed!
Example: Spacecraft at 2rE
• A spacecraft at an altitude of twice the Earth radius.
• Numbers from tables:
Earth,
m
• Earth Radius: rE = 6320 km
Mass M
Earth Mass: ME = 5.98 
1024
kg
E
FG = G(mME/r2)
• At the surface (r = rE)
FG = weight = mg =
G[mME/(rE)2]
• At r = 2rE
FG = G[mME/(2rE)2] = (¼)mg = 4900 N
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!)
F = FME + FMS
(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)
= 4.77  1020 N
tan(θ) = 1.99/4.34
 θ = 24.6º
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
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:
•
l
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.
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:
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
• So,
• Using components:
Fx = F21x + F31x = 0 + 6.67  10-11
Fy = F21y + F31y = 3.75  10-11 N + 0
F = [(Fx)2 + (Fy)2]½ = 3.75  10-11 N
tanθ = 0.562, θ = 29.3º
Gravity Near the Earth’s Surface
• The Gravitational
Acceleration g
and
• The Gravitational
Constant G
G vs. g
• Obviously, it’s very important to distinguish
between G and g!
– They are obviously very different physical quantities!
G  Universal Gravitational Constant
– It is the same everywhere in the Universe
G = 6.673  10-11 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
g in terms of G
• Consider an object on Earth’s surface:
mE = mass of Earth (say, known)
rE = radius of Earth (known)
m = mass of the object (known)
m
mE
• 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!
g = 9.8 m/s2
• Using the same process, we can
m
“Weigh” the Earth!
(Determine it’s mass).
• On Earth’s surface, equate the
usual weight of mass m to the
Newton’s Gravitation Law
mE
form for the gravitational force:
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
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
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]
This gives:
g 
 mg
ME
GME
 RE  h 
2
• This shows that g decreases with increasing altitude
• As r , the weight of the object approaches zero
Example, g on Mt. Everest
g
Altitude Dependence of g
GME
 RE  h 
2
Lubbock, TX:
Altitude: h  3300 ft  1100 m
 g  9.798 m/s2
Mt. Everest:
Altitude: h  8.8 km
 g  9.77 m/s2
Example: Effect of Earth’s Rotation on g
• Assume that Earth is perfect sphere.
• Determine how its rotation affects the
value of g at equator compared to its
value at poles. In general,
Newton’s 2nd Law is:
∑F = maR = W – mg, so W = mg – maR
• At the pole, no acceleration, aR = 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
• Also, T = 1 day = 8.64  104 s, so the speed there is
v = (2πrE)/T = 4.64  102 m/s,
• So, the effective gravitational acceleration at the equator is
g = g - [(v2)/(rE)] = 0.037 m/s2
“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
“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
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
The Moon’s Motion
• We’ve assumed that the Moon orbits a
“fixed” Earth
– It 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
Solar System Data