Orbits and Applications
Finish Chapter 13 today,
Will cover part of Chapter 14 on Fluids on Friday
Review on Monday for Exam Next Thursday
Third Midterm will cover material from Chap. 10-13
Newton’s Law of Gravity
Newton’s Law of Gravitation can
be written as
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.
Key Fact: A uniform shell of matter exerts no net
gravitational force on a particle located inside it.
Johannes Kepler’s Planetary Laws
1. Planets move in elliptical
orbits with sun at one focus
planet
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
Kepler’s 3rd Law
Kepler's 3rd law: T2/R3=constant for all planets.
The closer a planet is to the Sun, the faster it will move.
Any spacecraft going around the Sun in an orbit smaller
than Earth's will also soon overtake and move away, and
will not keep a fixed station relative to Earth.
However, if a spacecraft is placed between Sun and
Earth, Earth's gravity pulls it in the opposite direction
and cancels some of the pull of the Sun. With a
weaker pull towards the Sun, the spacecraft then
needs less speed to maintain its orbit.
This is called a Lagrange Point!
There are 5 such points.
Lagrange Points
L1 used to monitor the sun.
L2 is planned location for
the James Webb telescope.
L3 could be used to monitor
sun spots and provide up 7
day advance notice.
L4 and L5 locations very
stable and possible
locations for future space
stations
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
, that is
For object m1 on the surface of a planet
with mass MP and radius RP this translates to
Solving for the speed gives
This equation works for any radius outside the planets radius.
Just replace RP with the distance from the planet’s center.
Clicker question 1
Set frequency to BA
The Moon does not fall to Earth because
A: It generates a gravitational force equal and opposite the earth's
pull.
B: The net force on it is zero.
C: It is beyond the main pull of Earth’s gravity.
D: It is being pulled by the Sun and planets as well as by Earth.
E: none of the above
Clicker question 1
Set frequency to BA
The Moon does not fall to Earth because
A: It generates a gravitational force equal and opposite the earth's
pull.
B: The net force on it is zero.
C: It is beyond the main pull of Earth’s gravity.
D: It is being pulled by the Sun and planets as well as by Earth.
E: none of the above
The moon falls to earth all the time! There is a large force, F =
GM(earth)M(moon)/R(earth to moon)^2 which pulls it towards us.
Acceleration towards the center, that's what you mean by "falling"!
Clicker question 2
Set frequency to BA
A planet in elliptical orbit around a star moves from the point
in its orbit furthest from the star (A) to the closest point (P).
The work done by the force of gravity during this movement
is A: 0
B: +
C: -
Clicker question 2
Set frequency to BA
A planet in elliptical orbit around a star moves from the point
in its orbit furthest from the star (A) to the closest point (P).
The work done by the force of gravity during this movement
is A: 0
B: +
C: -
The force of gravity from the Sun is centripetal (toward the
center). The force has a component
along the direction of
 
displacement.
W = F • Δr = F Δr cos θ > 0
€
Clicker question 3
Set frequency to BA
The planet executes one complete orbit starting from point A and
returning to A. The work done by the force of gravity during this orbit is:
A: 0
B: +
C: -
Clicker question 3
Set frequency to BA
The planet executes one complete orbit starting from point A and
returning to A. The work done by the force of gravity during this orbit is:
A: 0
B: +
C: -
Zero. By the work energy theorem, or just conservation of
energy.
Clicker question 4
Set frequency to BA
Kepler's 3rd law: T2/R3=constant for all planets.
The period T for the earth is 1 year (!) The distance of the earth
to the sun is called 1 A.U. Suppose an asteroid is in circular
orbit around the sun at a distance of 2 A.U. How long does it
take the asteroid to orbit the sun once?
A
2 years
B:
3 years
C:
2^(3/2) = 2.83 years
D:
2^(2/3) = 1.59 years
E:
Not enough information (An asteroid is not a planet
- can't figure it out from Kepler's third law.)
Clicker question 4
Set frequency to BA
Kepler's 3rd law: T2/R3=constant for all planets.
The period T for the earth is 1 year (!) The distance of the earth
to the sun is called 1 A.U. Suppose an asteroid is in circular
orbit around the sun at a distance of 2 A.U. How long does it
take the asteroid to orbit the sun once?
A
2 years
B:
3 years
C:
2^(3/2) = 2.83 years
D:
2^(2/3) = 1.59 years
E:
Not enough information (An asteroid is not a planet
- can't figure it out from Kepler's third law.)
T (period) goes like R^(3/2), so if you're 2 times as far, your year
is 2^3/2 times longer.
Orbits from an energy perspective
For a circular orbit we know
can calculate the kinetic energy
so we
m
Remember, potential energy is
M
So kinetic energy is
Always before, kinetic and potential energy
were independent. For circular orbits, this is no longer true.
Furthermore, the total energy is
Since Ug < 0, total energy E < 0 which is the requirement
for a closed orbit (which of course a circular orbit is).
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)
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.
Orbital modifications
Orbital mechanics is
a computationally
challenging field.
Shown is the Hohmann
transfer orbit, one of the
most fuel efficient ways
to change orbits.
Clicker question 5
Set frequency to BA
A rock, initially at rest with
respect to Earth and located far
away is released and accelerates
toward Earth. An observation
tower is built 3 Earth-radii
high to observe the rock as it plummets to Earth.
Neglecting friction, the rock’s speed when it hits the ground is
A:
twice
B:
three times C:
four times
D:
eight times E:
sixteen times
its speed at the top of the tower.
Clicker question 5
Set frequency to BA
A rock, initially at rest with
respect to Earth and located far
away is released and accelerates
toward Earth. An observation
tower is built 3 Earth-radii
high to observe the rock as it plummets to Earth.
Neglecting friction, the rock’s speed when it hits the ground is
A:
twice
B:
three times C:
four times
D:
eight times E:
sixteen times
its speed at the top of the tower.
Cons. of E with v=0, R=infinity, then 1/2 mv2 - G Mem/R = const. = 0
At “4R” we have 1/2 mv2 = GMm/(4R). At "R" it has1/2 mv2 = GMm/R,
i.e. 4x BIGGER KE. So 2x BIGGER v.
Clicker question 6
Set frequency to BA
An object moves along the x-axis . The potential energy U(x) vs. position
x is shown in the graph.
When the object is at x=A,
which of the statements must
be true?
When the object is at x=A, which of the statements must be true?
A: The velocity is negative, i.e. along the -x direction.
B: The acceleration is negative.
C: The total energy is negative.
D: The kinetic energy is positive.
E: None of these statements is always true.
Clicker question 6
Set frequency to BA
An object moves along the x-axis . The potential energy U(x) vs. position
x is shown in the graph.
When the object is at x=A,
which of the statements must
be true?
When the object is at x=A, which of the statements must be true?
A: The velocity is negative, i.e. along the -x direction.
B: The acceleration is negative.
C: The total energy is negative.
D: The kinetic energy is positive.
E: None of these statements is always true.
F = -dU(x)/dx, so the FORCE at point A is negative. (Therefore the
acceleration is negative). We know nothing about the velocity. The
total energy could be anything, really. K could be zero.
Clicker question 7
Set frequency to BA
A planet is in elliptical orbit around the Sun. The zero of
potential energy U is chosen at r = ∞, so
.
How does the magnitude of U, i.e. |U| compare to the KE?
A: |U| > KE B: |U| < KE
C: |U| =KE
D: depends on the
position in the orbit.
Clicker question 7
Set frequency to BA
A planet is in elliptical orbit around the Sun. The zero of
potential energy U is chosen at r = ∞, so
.
How does the magnitude of U, i.e. |U| compare to the KE?
A: |U| > KE B: |U| < KE
C: |U| =KE
D: depends on the
position in the orbit.
PE
The earth is bound, so K + U < 0.
So K < -U; -U= + G M m /r. Thus,
|U| > K. A diagram may help:
r
Etot
PE KE
Clicker question 8
Set frequency to BA
A satellite is in circular orbit around a planet that has a very tenuous
atmosphere extending up to the altitude of the satellite. Due to
atmospheric drag, the speed of the satellite…
A: increases
B: decreases
C: remains constant
Clicker question 8
Set frequency to BA
A satellite is in circular orbit around a planet that has a very tenuous
atmosphere extending up to the altitude of the satellite. Due to
atmospheric drag, the speed of the satellite…
A: increases
B: decreases
C: remains constant
Drag makes the satellite's radius get smaller; since v = Sqrt[ G M/r],
the speed will INCREASE. K+U = total energy, so K - GM m/R is the
total energy.