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Universal gravitation
•  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 this week.
•  Old midterm exams are posted in D2L.
•  Files with clicker questions are posted in D2L.
1
Universal Gravitation: Potential energy
Gravity is a conservative force
and therefore has a potential
energy associated with it
Gm1m2
UG = −
r
Can use conservation of energy to
determine things like velocity at a
given height or the escape velocity.
Gm1m2
Total energy E = K +UG = m v + m v −
r
1
2
2
1 1
1
2
2
2 2
is conserved
Often one mass is much more massive and is basically
GMm
2
1
motionless and so the total energy is E = K +UG = 2 mv −
r
Negative total energy is a bound system (closed orbits or objects
eventually colliding). Total energy ≥ 0 is an unbound system.
2
Clicker question 1
Set frequency to BA
The gravitational potential energy of a
two mass system, where m << M is
plotted. Mass M is stationary at the
origin and mass m is at r=r0. Also
shown is the total energy Etot. Which
arrow represents the kinetic energy of
mass m?
A.  A
B.  B
C.  C
D.  None of these
UG = −
GMm
r
r0
B C
A
r
Etot
The total energy E is equal to the sum of the potential and
kinetic energies: E = K + U so K = E – U.
3
Escape velocity
We define escape velocity as the minimum speed needed to
escape a gravitational field (usually from the surface).
Escaping means being able to reach r = ∞. We found this is
possible if we have a total energy ≥ 0.
Total energy of 0 means UG + K = 0 , that is K = −UG = +
Gm1m2
r
For object m1 on the surface of a planet
1 m v 2 = Gm1M P
with mass MP and radius RP this translates to 2 1
RP
2GM P
Solving for the speed gives vescape =
RP
This equation actually works for any radius outside the planets
radius. Just replace RP with the distance from the planet’s center.
4
Clicker question 2
Set frequency to BA
Does escape velocity depend on launch angle?
That is, if a projectile is given an initial speed vo, is it
more or less likely to escape an airless, non-rotating
planet, if fired straight up than if fired at an angle?
A. Yes
B. No
We derived the escape velocity using conservation of
energy. All that is needed is for the projectile to have
enough kinetic energy such that the total energy is 0.
Kinetic energy only depends on the magnitude of the velocity
(squared), not the direction so the angle is irrelevant.
On rotating planets the escape velocity is the same
but the initial velocity is not zero so the angle matters.
On planets with air, take offs are more straight up
to reduce the energy loss due to air resistance.5
History of solar system understanding
1543: Nicholas Copernicus theorizes that the Earth orbits the sun.
1609 & 1619: Johannes Kepler uses 30 years of data collected
by Tycho Brahe to derive 3 laws for planetary motion
planet
1. Planets move in elliptical
orbits with sun at one focus
Sun
2. A line drawn between sun and planet sweeps
out equal areas during equal intervals of time
3. The square of a planet’s orbital period is
proportional to the cube of the semimajor-axis length
These laws can be derived from Newton’s 1687 gravity theory.
6
Circular orbits from a force perspective
All closed orbits are ellipses but we will analyze the
simpler case of circular orbits. Very good approximation
for planets and moons. Not so good for comets.
A small object m in orbit around a
large object M is called a satellite.
r
!
!
Newton’s 2nd law is Fnet = ma
M
!
v
!
FG
GMm
The only force is gravity: FG = r 2
2
v
The acceleration is only radial: a = r
GMm = m v2
Therefore: r 2
r
Solving for v gives vorbit =
Note that orbital velocity only depends on M/r.
GM
r
7
m
Orbital period
Orbital period T is the time it takes to complete one revolution.
Orbital speed can be determined by distance covered in one
revolution (circumference) divided by the period: vorbit
We already know that vorbit
GM
=
r
3/2
2π r
r
2π r
= 2π r
=
So T =
vorbit
GM
GM
2π r
=
T
!
v
r
M
!
FG
2
4
π
r3
Can also write as T =
GM
2
proving Kepler’s 3rd law (for circular orbits)
8
m
Orbits from an energy perspective
For a circular orbit we know vorbit =
can calculate the kinetic energy
1 2
1 GM GMm
K = mvorbit
= m
=
2
2
r
2r
GM
so we
r
GMm
Remember, potential energy is UG = −
r
!
v
r
M
!
FG
m
So kinetic energy is K = − 12 UG
Always before, kinetic and potential energy
were independent. For circular orbits, this is no longer true.
1
1
Furthermore, the total energy is E = K +UG = − 2 UG +UG = 2 UG
Since Ug < 0, total energy E < 0 which is the requirement
for a closed orbit (which of course a circular orbit is).
9
Gravitational potential energy
Graph of potential energy,
kinetic energy, and total
energy for circular orbits.
To get from a low orbit r1 to
a higher orbit r2 requires an
increase in energy.
While the kinetic energy
decreases, potential energy
increases (twice as much)
K
E
ΔE
UG
r1 r2
Work is needed to change the energy. Since work is force times
displacement, rockets are generally fired in (or against) the
direction of motion to change orbits rather than perpendicular.
10
Clicker question 3
Set frequency to BA
Two satellites are in stable circular orbits. Satellite 2 has a speed
of v2 and is twice as far from the center of the Earth as satellite 1,
which has a speed of v1. What is the ratio of their speeds?
A.  v1 / v2 = 1 / 2
B.  v1 / v2 = 1 / 2
Ecircular orbit = K orbit +UG = 12 UG
C.  v1 / v2 = 1
D.  v1 / v2 = 2
K orbit
E.  v1 / v2 = 2
v1 =
GM
GM
v2 =
r1 and
r2
GMm
=
2r
vorbit =
so
v1
GM
=
v2
r1
GMm
UG = −
r
GM
r
vescape =
2GM P
RP
2
4
π
T =
r3
GM
2
r2
r
= 2 = 2
GM
r1
11