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Universal gravitation
•  CAPA due today.
•  Today will finish up with
the hinge problem I
started on Wednesday.
•  Will start on Gravity.
Hinge Problem from Wednesday
Hinge Problem cont.
∑F
∑F
x
= 0 = FNx − T cosθ
y
= 0 = T sin θ − W1 − W 2 − FNy = 0
Guessed
wrong on
direction
L
∑ τ = 0 = T sinθ • L − W 2 • L − W1 • 2
W1
W1
T = (W 2 + ) /sin θ FNx = (W 2 + )cot θ FNy = −W1 /2
2
2
Things to keep in mind
Force of gravity acts at the center of mass
Tension force must be in the same direction as the rope
There are often multiple choices for a reasonable axis
about which to calculate torques. All of them are OK
and you should still be able to solve the problem. A
good choice just makes the problem easier.
Clicker question 1
Set frequency to BA
A mass M is placed on a very light board supported at the ends, as
shown. The free-body diagram shows directions of the forces, but
not their correct relative sizes.
FL
M
(2/3)L
FR
Mg
L/3
What is the ratio FR/FL?
A: 2/3
B: 1/3
E: some other answer
C: 1/2
D: 2
Clicker question 1
Set frequency to BA
A mass M is placed on a very light board supported at the ends, as
shown. The free-body diagram shows directions of the forces, but
not their correct relative sizes.
FL
M
(2/3)L
FR
Mg
L/3
What is the ratio FR/FL?
A: 2/3
B: 1/3
C: 1/2
D: 2
E: some other answer
Sum of the Torques = 0, or
- FL*(2/3L ) + FR*(L/3) = 0, so FR/FL = 2/1.
Newton’s Law of Gravity
Newton and Einstein are generally thought
to be the two greatest physicists ever.
Not only did Newton come up with the three laws
of motion and invent calculus, he was the first to
realize that the force associated with things falling
was also responsible for astronomical phenomena.
Newton’s Law of Gravitation can be written as
Between any two masses (here m1 & m2) there is an attractive
force proportional to the product of the masses and inversely
proportional to the square of the distance between them.
Gravitational Force
is the force of gravity which is felt by each mass
and directed towards the other mass.
Newton figured out the 1/r2 dependence assuming that the
celestial objects and the Earth were point particles.
By inventing integral calculus he could prove that for a mass
m2, outside a spherical mass m1, the force of gravity was as
if all of the mass m1 was in the center of the sphere.
Therefore for any two spherically symmetric objects, the
distance r that enters into the force of gravity is the distance
between the centers of the spheres.
Newton’s Shell Theorem
A uniform spherical shell of matter attracts a particle
that is outside the shell as if all of the shell’s mass
were concentrated at its center.
M
Let ρ =
2
4 πR
dA = (2πRsin θ )Rdθ
dM = ρdA =
€
1
M sin θ dθ
2
1 GmM
€dF = GmdM
cos α sin θdθ
€2 cos α =
2
s
2
s
Find F = GMm /r
€
2
Force rules
is the force of gravity with
Newton’s 2nd law still works. The net force on an object
determines the object’s acceleration:
Remarkably, the mass in Newton’s 2nd law (called the inertial
mass) is the same as the mass in the law of gravitation (called
the gravitational mass). Einstein figured out (230 years later)
that this “coincidence” could be explained by assuming space
and time were curved (in the theory of general relativity).
Remember, force is still a vector and the law of superposition
still works. To find the net gravitational force on an object,
determine the magnitude and direction of the force from all
other masses and then add these forces together.





F1,net = F12 + F13 + F14 + ......+ F1n
Clicker question 2
Set frequency to BA
Two asteroids in inter-galactic space are a distance r = 20 km apart.
Asteroid 2 has 10 times the mass of asteroid 1. The magnitudes of
the accelerations of asteroids 1 and 2 are a1 and a2, respectively.
What is the ratio a1/a2?
r = 20 km
A. 1/100
m1
m2
B. 1/10
C. 1
D. 10
E. 100
Clicker question 2
Set frequency to BA
Two asteroids in inter-galactic space are a distance r = 20 km apart.
Asteroid 2 has 10 times the mass of asteroid 1. The magnitudes of
the accelerations of asteroids 1 and 2 are a1 and a2, respectively.
What is the ratio a1/a2?
r = 20 km
A. 1/100
m1
m2
B. 1/10
C. 1
The force on m1 is the
D. 10
same as the force on m2:
E. 100
Acceleration is force divided by mass
and
which gives us
so
Comments about Earth
The density of the Earth is higher than most of the
other planets in our solar system. Sources vary when
it comes to the density of the Earth. ~5.5 g/cm3
Inner Core: solid, ~13
g/cm3 mainly Fe + Ni
Outer Core: Thought to
be mainly responsible
for earth’s magnetic
field.~11g/cm3
Lower Mantle: Silicon,
magnesium, Oxygen
~3.5g/cm3
Earth is not a sphere + rotating!
The equator radius is larger than the polar radius by
21 km! Means gravitational acceleration is larger at
6
poles than at equator!
R = 6.37 ×10 m
∑F =
2
FN − mag = m(−v /R)
2
FN = ma
−
m
ω
R
€ g
g = ag − ω R
€
€
2
g = 9.8m /s − 0.034m /s
€
€
dθ 2π radians
ω=
=
dt
24 hr
2
2
Force of gravity on Earth
How does
correspond to our new force
?
If we consider mass 2 to be the Earth (ME) and r to be the radius
of the Earth (RE) then we can write
Using known values we can find that
So, on the surface of the Earth, the force of
gravity between the Earth and an object m1 is
We can only use
if the distance above the
surface is very small compared to the radius.
Clicker question 3
Used
to find
Set frequency to BA
near the Earth’s surface
Planet X has the same mass as the Earth, but ½ the radius due to
its higher density. What is the acceleration of gravity on Planet X?
A. ¼ g
B. ½ g For Earth
C. g
D. 2 g
E. 4 g
Clicker question 3
Used
to find
Set frequency to BA
near the Earth’s surface
Planet X has the same mass as the Earth, but ½ the radius due to
its higher density. What is the acceleration of gravity on Planet X?
A. ¼ g
B. ½ g For Earth
C. g
D. 2 g
For Planet X
E. 4 g
The higher density of Krypton (being made
of Kryptonite) makes the force of gravity at
the surface stronger, meaning Superman
must be stronger to do any old normal thing.
Clicker question 4
Set frequency to BA
A rock is released from rest in space beyond
the orbit of the Moon. The rock falls toward
the Earth and crosses the orbit of the Moon.
At this point, the acceleration of the rock is…
A. greater
B. smaller
C. the same as
the acceleration of the Moon.
Moon
Earth
rock
Clicker question 4
Set frequency to BA
A rock is released from rest in space beyond
the orbit of the Moon. The rock falls toward
the Earth and crosses the orbit of the Moon.
At this point, the acceleration of the rock is…
A. greater
B. smaller
Moon
Earth
rock
C. the same as
the acceleration of the Moon.
If the Moon and the rock are a distance r from the center of the
Earth then the acceleration of either mass can be determined by
independent of whether it is the
Moon or a rock
Note, the speeds are probably not the same but the accelerations are!
Gravitational potential energy
When we used
we found a potential energy of
What is the potential energy associated with the force
.
?
To make sense, potential energy should increase as the distance
increases and be smallest when the objects are closest together.
A while ago we learned force is the derivative of potential energy.
The potential energy
gives the force
when you take the derivative with respect to r.
Gravitational potential energy
Potential energy increases
(less negative) as the
separation increases.
This is what we wanted.
Maximum potential energy is 0 when r approaches infinity.
Since two objects cannot share the same space, r > 0. The
minimum potential energy is when the objects are touching.
Earth’s gravitational potential energy
Potential energy due
to Earth’s gravity is
where r is the distance
from the center and h
is the height above the
surface of the Earth.
RE = radius of Earth = 6380 km
Distance from center of the Earth (km)
Suppose a rock is released from rest at r = 30000 km. Initially it
only has potential energy. It will start falling, converting potential
energy to kinetic energy. The total energy remains the same.
The rock cannot go past r = 30000 km because it would have
negative kinetic energy at that point (which is impossible).
Effect of total energy on trajectory
If the total energy were 0
then it is possible for the
object to make it to r = ∞.
We can identify three
basic scenarios for a total
energy
which is
positive, negative, or 0.
RE = radius of Earth = 6380 km
Distance from center of the Earth (km)
For total energy < 0 the object is bound by the gravitational field
(and orbits are ellipses). Examples are planets around the sun.
For total energy of 0 the object is barely unbound (parabolic orbit).
For total energy > 0, object is unbound with a hyperbolic
orbit.