Gravitation
Forces Causing Centripetal
Acceleration

Newton’s Second Law says that the
centripetal acceleration is accompanied
by a force
F = maC = mv2/r
 F stands for any force that keeps an object
following a circular path
 Tension in a string
 Gravity
 Force of friction

Problem Solving Strategy
Draw a free body diagram, showing and
labeling all the forces acting on the
object(s)
 Choose a coordinate system that has
one axis perpendicular to the circular
path and the other axis tangent to the
circular path

Problem Solving Strategy,
cont.
Find the net force toward the center of the
circular path (this is the force that causes
the centripetal acceleration)
 Solve as in Newton’s second law problems:
∑ F = mac (2 dimensions)

The directions will be radial and tangential
 The acceleration on the right side will be the
centripetal acceleration.
 Do not put centripetal force on the left side – it
is the right side of the equation already.

Newton’s Law of Universal
Gravitation - 1687

“Every particle in the Universe attracts
every other particle with a force that is
directly proportional to the product of
the masses and inversely proportional
to the square of the distance between
them.”
m1m 2
FG 2
r
r does not go to zero here
because the objects have some
radius of their own.
Law of Gravitation, cont.
G is the constant of
universal gravitational
 G = 6.673 x 10-11 N m² /kg²
 This is an example of an
inverse square law - force is
proportional to 1/r2

m1m 2
FG 2
r
Do the units of G make sense
to you? Could you derive
them?
Features of gravitation
1.
2.
m1m 2
FG 2
r
The gravitational force is a field force
that always exists between two
particles regardless of the medium that
separates them.
The force varies as one over the square
of the distance between the particles
and therefore decreases rapidly with
increasing separation.
Features of gravitation
3.
4.
m1m 2
FG 2
r
The force is proportional to the
product of the particles’ masses.
The gravitational force is actually very
weak, as shown by the size of G
(6.67 x 10-11 Nm2/kg2)
Gravitation Constant
Determined experimentally
by Henry Cavendish in 1798
using this torsion balance.
 The small spheres m are
attracted by large spheres
M and the rod rotates.
 The light beam and mirror
serve to amplify the motion.

Quick Quiz – which are false?
Scenario: A ball falls to the ground
 The force that the ball exerts on the Earth is
equal in magnitude to the force that Earth
exerts on the ball.
 The ball undergoes the same acceleration as
Earth.
 Earth is much more massive than the ball, so
Earth pulls much harder on the ball than the
ball pulls on the Earth. Therefore the ball falls
while Earth remains stationary.

Quick Quiz

Superman circles Earth at a radius of 2R
where R is Earth’s radius. He then
moves out to a radius of 4R. The
gravitational force on him at this second
orbit as compared to the first is a) the
same, b) 2x more, c) 4x more, d) ½ as
great, e) ¼ as great.
Example 1 – 2D problem
Mass of all balls= 0.3 kg
Find net force on cue ball 1
 F21= (6.67E-11)(.3)(.3)/(.4)2
= 3.75E-11 N
 F31= (6.67E-11)(.3)(.3)/(.3)2
= 6.67E-11 N
 Fnet = (F212 + F312)1/2
=7.65E-11 N
Can use ϴ= tan-1(Fy/Fx) to
find angle of Fnet

Example 2– Find total force on B
4 cm
6 cm
B
A
𝐹𝐴𝐵 =
𝐺𝑚𝐴 𝑚𝐵
𝑟2
𝐹𝐶𝐵 =
𝐺𝑚𝐵 𝑚𝐶
𝑟2
C
mA = 10 kg
mB = 8 kg
mC = 12 kg
=
6.67 𝑥 10−11 (10)(8)
.04 2
= 3.33 x 10-6 N
=
6.67 𝑥 10−11 (8)(12)
.062
= 1.78 x 10-6 N
Ftotal = FAB + FBC = -3.33 x10-6 + 1.78 x 10-6 =
1.55 x 10-6 N
Applications of Universal
Gravitation

Find the mass of the
earth

Use an example of an
object close to the
surface of the earth

r ~ RE, W = weight
ME m
W  mg  G
RE2
gRE2
ME 
G
Example 2





Find the mass of Earth.
Use an object like a baseball of mass m,
falling to Earth. The magnitude of the
gravitational force exerted by the Earth on the
ball = the weight of the ball.
mg = GMem /RE 2
Cancel m on both sides, solve for Me
Me = g RE2 / G = (9.8) (6.38E6)2/ 6.67E-11 =
5.98E24 kg is the mass of the Earth
Applications of
Universal Gravitation

We can rearrange the
equation used on the last
2 pages to find the
acceleration due to gravity
at various radius positions.
ME
gG 2
r

g will vary with altitude
(1000 km is about 600 miles)
Gravitational Potential Energy
This next section has to do with Energy.
We actually have our Energy unit as Unit
6, later in the semester.
 That means you will use some equations
here that we haven’t introduced earlier,
but trust me, the proof will come later in
unit 6.
 No matter which order you do Physics in,
there are always some before/after
entanglements, and this is one of them.

Energy equations





Here’s just a quick intro to some of the energy equations
we will be using
Potential energy PE = mgh where m=mass,
g=acceleration due to gravity and h = height above the
ground (or some reference plane)
Kinetic energy KE = ½ mv2 where m = mass,
v = velocity
Work
W = Force x distance Work is in Joules, the unit
of energy, and it is equal to the net applied force x the
distance that the object is moved.
Total energy E = mgh + ½ mv2 This is PE + KE
Gravitational Potential Energy



You may know that potential energy, PEg = mgh
This equation is actually only valid when the object is
near Earth’s surface where g is constant. For objects high
above Earth’s surface, like a satellite, you need to use
this equation, derived by taking
𝑃𝐸 = −𝑊 = −𝐹 𝑥 𝑑 = −
 𝑷𝑬


=
−𝑮𝑴𝑬 𝒎
𝒓
𝐺𝑀𝐸 𝑚
𝑟2
𝑥𝑟=
−𝐺𝑀𝐸 𝑚
𝑟
for planetary scale problems
Where did the negative sign come from?
As two planetary bodies get closer together (r gets
smaller), their PE decreases (so need the negative sign).
Gravitational
Potential Energy




−𝑮𝑴𝑬 𝒎
𝑷𝑬 =
𝒓
Here, PE is zero at infinite distance from Earth’s center,
because gravity goes to zero at infinite distance from Earth.
(Distant planets can’t feel the pull of the Earth.)
So if r gets smaller, PE decreases!
The negative sign indicates that the work done by an external
force in moving an object from infinity to a distance r away
from Earth’s center is negative, or the gravitational potential
energy decreases in such a process. (it doesn’t gain PE
because it gets pulled in w/o an external force)
Another way to say that is that the work is negative because
the external force has to hold the object back against the
attractive force of gravity.
Potential Energy
Let’s review:
• Looking at this graphic, can
you see how the potential
energy goes to zero when
the object is very far away
from Earth?
• Do you see how the PE
function is always
negative?
• Note that the value of PE at
the Earth’s surface is
PE = - GMEm / RE
Total Energy in Planetary Systems

If an isolated system consists of an object of mass m
moving with a speed v in the vicinity of a massive object M,
the total energy E of the system is the sum of kinetic and
gravitational potential energies:
𝟏
𝑮𝑴𝒎
𝟐
𝑬𝒕𝒐𝒕𝒂𝒍 = 𝒎𝒗 −
that’s just KE + PE
𝟐
𝒓
 This equation shows that E may be positive, negative or
zero, depending on the value of v. But for a bound
system, such as a planet and the Sun, the PE must be < 0
because we have chosen the convention that PE >> 0 as r
approaches infinity.
 Let’s try setting gravitational force = centripetal force for
this orbit.
Total Energy in Planetary
Systems (continued)

Setting gravitational force = centripetal force
𝑮𝑴𝒎
𝒎𝒗𝟐
 𝟐 =
𝒓
𝒓
 Multiplying both sides by r and dividing by 2 gives
𝟏
 𝒎𝒗𝟐
𝟐

=
𝑮𝑴𝒎
𝟐𝒓
Substituting this back into the energy equation on last
page
𝑮𝑴𝒎
𝑮𝑴𝒎
𝑮𝑴𝒎
𝑬=
−
=−
𝟐𝒓
𝒓
𝟐𝒓
 So total energy is negative and equal in magnitude to ½
the potential energy.
Escape Speed
The escape speed is the speed needed for an object to soar
off into space and not return.
 We assume that the escape speed is just large enough to
allow the object to reach infinity with a speed of zero.
When the object is an infinite distance from Earth its kinetic
energy is zero because vf = 0.
 The gravitational potential energy is also zero because our
zero level of PE was selected at r>>∞, so the sum is zero.

𝟏
 𝒎𝒗𝒆𝒔𝒄 𝟐
𝟐


−
𝑮𝑴𝒎
𝒓
=𝟎
so 𝒗𝒆𝒔𝒄 =
𝟐𝑮𝑴
𝒓
For the earth, vesc is about 11.2 km/s
Note, v is independent of the mass of the object (m) !
Newton’s Cannon applet and
PhET simulation


This applet allows you to see the
trajectory of a projectile as a function of
launch speed to explore escape speed.
http://waowen.screaming.net/revision/force&motion/ncananim.htm
This PhET simulation models the sun and
earth. You can change the masses and
watch what happens to the trajectory. You
can look it up oneline at PhET too.
 gravity-and-orbits_en.jar

Kepler’s Laws – 3 laws
1)
2)
3)
All planets move in elliptical orbits with
the Sun at one of the focal points.
A line drawn from the Sun to any
planet sweeps out equal areas in equal
time intervals.
The square of the orbital period of any
planet is proportional to cube of the
average distance from the Sun to the
planet.
Kepler’s Laws, cont.
Based on observations made
by Tyco Brahe
 Newton later demonstrated
that these laws were the
consequence of the
gravitational force between
any two objects together
with Newton’s laws of
motion

Kepler’s First Law



All planets move in
elliptical orbits with
the Sun at one focus.
Any object bound to
another by an inverse
square law will move
in an elliptical path
(F ~ 1/r2)
Second focus is empty
Kepler’s Second Law
A line drawn from the
Sun to any planet will
sweep out equal areas
in equal times.
 Area from A to B and
C to D are the same
 This is a result of conservation of angular
momentum (a topic for AP Physics).

Kepler’s Second Law
Kepler’s Second Law
• So did you notice that when
Earth gets closer to the Sun, it
has to go faster, to sweep out
that area?
• Did you notice that when Earth is
farther away from the Sun, it
goes slower?
• Why? Check out the PhET simulation if you want to see that!
If the Earth and the Sun are farther apart, they have more
potential energy, because they are attracted to each other.
If they are closer together, they have less PE. To make up
for losing that PE as Earth approaches the sun, it speeds
up, so the total energy will remain the same.
Kepler’s Second Law
•
That’s probably very confusing,
so let’s just try a numerical
example, comparing the PE at
100 units of distance vs. 10
units of distance. Assume that
G, ME, m are held constant in
each case.
• 𝑃𝐸 =
𝐺𝑀𝑒 𝑚
−
100
• 𝑃𝐸 = −
𝐺𝑀𝑒 𝑚
10
= − 0.01 G𝑀𝐸 𝑚
= − 0.1 G𝑀𝐸 𝑚
Do you see how PE
decreases (gets more
negative) when you go
from r= 100 to r=10?
Kepler’s Third Law
The square of the orbital period, T of any planet
is proportional to cube of the average distance
from the Sun to the planet, r.
 𝑻𝟐 = 𝑲𝒓𝟑 << write these eq. on your card


Where 𝑲 =
𝟒𝝅𝟐
𝑮𝑴𝒔
For orbit around the Sun specifically,
 KS = 2.97x10-19 s2/m3
 K is independent of the mass of the planet doing
the orbiting (it’s a function of the sun only)

Kepler’s Third Law:
(TA / TB
2
)
(T=period)
= (rA/rB
3
)
Example of using Kepler’s Third





If the average distance from the sun for Venus is
1.082E8 m and for Saturn is 1.434E9 m, and the
period of revolution for Saturn is 29.42 Earth
years, what is the period for Venus?
let A=Venus, B=Saturn
(TA/TB)2=(rA/rB)3
TA = ((rA/rB)3 x TB2)1/2
TVenus = ((1.082E8/1.434E9)3 x 29.422)1/2
TVenus = 0.61 Earth years
Kepler’s Third Law application
Ms = Mass such as the
Sun or other celestial
body that has
something orbiting it.
 Mp = Mass of planet
 Assuming a circular
orbit is a good
approximation
 Eccentricity is low for
many planets.

Eccentricity, e
Mercury
Venus
.206
.0068
Earth
.0167
Mars
.0934
Jupiter
.0485
Saturn
.0556
Uranus
.0472
Neptune
.0086
Kepler’s Third Law Derivation
𝑴 𝒔 𝑴𝒑
𝑴𝒑 𝒗
𝑮
=
𝟐
𝒓
𝒓
𝟐𝝅𝒓 𝟐
(
)
𝑴𝒔
𝑻
𝑮 𝟐 =
𝒓
𝒓
𝟐
𝟒𝝅
𝑻𝟐 =
𝒓𝟑
𝑮𝑴𝒔
𝑻𝟐 = 𝑲𝒓𝟑
𝟐
< Set force due to gravity equal
to centripetal force
< if orbit is assumed to be
circular, then v = distance/time
= (2πr/T) where T= period and
v = 2πr/T.
< collect r’s all on right, T’s on
left, items in parenthesis are
constants which are grouped to
be called “K”
Example problem (like HW)

What will be the period of a satellite that is orbiting
between the orbits of Earth and Mars which has a
distance of 1.98 x 1011 m from the sun? Express your
answer in years.
 𝑇2

=
𝑇=
4𝜋2
𝐺𝑀𝑠
𝑟3
4𝜋2 (1.98 𝑥 1011 )3
(6.67 𝑥 10−11 )(1.9889 𝑥1030 )
 4.8𝑥107
𝑠𝑒𝑐 𝑥
1 ℎ𝑜𝑢𝑟
3600 𝑠𝑒𝑐
𝑥
= 4.8 x 107 sec
1 𝑑𝑎𝑦
24 ℎ𝑜𝑢𝑟
𝑥
1 𝑦𝑒𝑎𝑟
365 𝑑𝑎𝑦𝑠
= 1.52 𝑦𝑒𝑎𝑟𝑠
Example problem:
Geosynchronous Orbit


Satellite dishes do not have to change direction in order to
stay focused on a signal from a satellite. This means that
the satellite always has to be found at the same location
with respect to the Earth’s surface. For this to occur, the
satellite must be at a height such that its revolution period is
the same as that of Earth, 24 hours. At what height must a
satellite be to achieve this?
The force that produces the centripetal acceleration of the
satellite is the gravitational force, so
𝑮𝑴𝑬 𝒎
𝒎𝒗𝟐
=
𝟐
𝒓
𝒓
Ex: Geosynchronous Orbit
𝐺𝑀𝐸 𝑚

𝑟2
=
𝑚𝑣 2
𝑟
Where ME is mass of the Earth, m
is mass of the satellite, r is orbital radius from center
of Earth.


2𝜋𝑟
The velocity of the satellite is 𝑣 =
. Substituting
𝑇
this in for v in the first equation and solving for r,
𝒓=
𝑻𝟐 𝑮𝑴𝑬 𝟏/𝟑
( 𝟐 )
this is the radius that will cause a
𝟒𝝅
geosynchronous orbit
Professor Lewin – MIT


http://www.youtube.com/watch?v=MJYQGPl3MNI
4’ video on why gravitational force falls away as
1/R^2
Gravity Visualized

https://www.youtube.com/watch?v=MTY1Kje0yLg
9 minute video on visualizing gravity

Ties to PhET simulation “My Solar System”

Space video on what happens when you try to wring
water out of a wet wash cloth (great video even if it’s
not exactly on the topic of gravity, rather lack of
gravity)
http://apod.nasa.gov/apod/ap130424.html

