Universal gravitation
•  LA applications due Monday: lacentral.colorado.edu
•  Midterm #2 on Thursday at 7:30pm in same rooms
•  Midterm covers Chapters 5-8 (up through Monday’s
lecture). Includes CAPA due Tuesday and the tutorial
homework due next week.
•  Old midterm exams are posted in D2L
–  Fall 2014 midterm #2 correlates well with the material we
have covered.
–  Fall 2008 material was a different order so consider problems
1,2,3,5,6,8,10,11,12,13,14 and long problems 1, 2 from exam
2 and 7,9,10,13,14 and a-d of long problem 1 from exam 3.
–  There will be more universal gravitation than on old exams.
•  A file with all clicker questions will be in D2L by tonight.
1
Force of gravity on Earth
Gm1m2
F
=
mg
How does g
correspond to our new force FG =
?
2
r
If we consider mass 2 to be the Earth (ME) and r to be the radius
of the Earth (RE) then we can write F = m GM E
G
1
RE2
Using known values we can find that
−11
2
2
24
6.67
×10
N
⋅
m
/
kg
5.97
×10
kg
GM E
2
=
=
9.8
m/s
2
6
RE2
6.38 ×10 m
(
)(
(
)
)
So, on the surface of the Earth, the force of
GM E
F
=
m
= m1g
1
gravity between the Earth and an object m1 is G
2
R
E
We can only use Fg = mg if the distance above the
2
surface is very small compared to the radius.
Clicker question 1
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 forces on asteroids 1 and 2 are F1 and F2, respectively. What is
the ratio F1/F2?
r = 20 km
Gm1m2
FG =
A. 1/100
m1
m2
r2
B. 1/10
m2 = 10m1
C. 1
D. 10
The force on m1 is the
Gm1m2
same as the force on m2: FG = r 2
E. 100
This is also a consequence of Newton’s 3rd law.
3
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?
A. 1/100
B. 1/10
C. 1
D. 10
E. 100
FG =
Gm1m2
r2
r = 20 km
m1
m2
m2 = 10m1
The force on m1 is the
Gm1m2
same as the force on m2: FG = r 2
F
Acceleration is force divided by mass a = m
so
FG
FG
a1 FG m2 m2
and a2 =
a1 =
=
⋅
=
= 10
which gives us
m2
m1
a2 m1 FG m1
4
Clicker question 3
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.
You can ignore the force between Moon and rock.
Moon
Earth
rock
Gm1m2
FG =
r2
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
F 1 GM E m GM E independent of whether it is the
a= =
= 2
2
Moon or a rock
m m r
r
Note, the velocity directions and magnitudes may not be the
same but the accelerations are the same!
5
Gravitational potential energy
When we used Fg = mg we found a potential energy of U g = mgy .
Gm1m2
What is the potential energy associated with the force FG =
?
2
r
To make sense, potential energy should increase as the distance
increases and be smallest when the objects are closest together.
We recently learned that force is the derivative of potential energy.
Gm1m2
Gm1m2
gives the force FG =
r2
r
when you take the derivative with respect to r.
The potential energy UG = −
6
Gravitational potential energy
Potential energy increases
(less negative) as the
separation increases.
This is what we wanted.
UG = −
Gm1m2
→ 0 as r → ∞
r
Gm1m2
UG = −
→ − ∞ as r → 0
r
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.
7
Earth’s gravitational potential energy
Potential energy due
to Earth’s gravity is
Gm1mE
Gm1mE
UG = −
=−
r
RE + h
where r is the distance
from the center of the
Earth and h is the
height above the
surface of the Earth.
K
UG + K
UG
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 (UG+K) stays the same.
The rock cannot go past r = 30000 km because it would have
negative kinetic energy at that point (which is impossible). 8
Effect of total energy on trajectory
If the total energy is ≥ 0
then it is possible for the
object to make it to r = ∞.
We can identify three
basic scenarios for a total
energy (UG + K ) which is
positive, negative, or 0.
UG + K
K
UG
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.
9