Download Energy, Angular momentum and orbits

Last lecture Dr. Fallscheer defined circular motion, force and energy and took you through applications of
escape velocity.
Given the following graphs for the Force acting on a tungsten slug assuming it starts at rest:
Describe what is happening to the object during each portion of the graph. Make graphs of acceleration,
velocity and position. Describe a physical scenario that matches these graphs.
Linear Momentum:
* The linear momentum of an object is a measure of the object's "umph"
* Momentum is a vector quantity. This means that the momentums of two objects have to be added
vectorially in order to find the total momentum of a system consisting of the two objects.
As the force and acceleration graphs are related, so are the velocity and momentum graphs.
Impulse:
* The force-time "crack" applied to an object due to the action of a force for a duration of time.
Impulse is the area under the force vs time graph.
Impulse-Momentum Theorem:
* Any impulse acting on a system changes the system's momentum.
* For any individual body we can also write the impulse-momentum theorem as,
where Fave is the average net force acting on the body. More generally,
* Stated in words. The final momentum of any object (or collection of objects) must equal to its
initial momentum plus any impulse imparted to the object (or collection of objects).
Connection to Newton's Second Law:
Rewriting kinematics in terms of momentum.
Naively F = ma. However this is a simplification. Since F = dp /dt and p = mv. Write a
complete form of Newton’s 2nd law.
F = mdv /dt + v dm/dt. We recognize that the first term is our familiar F = ma. The second
term only exists if the mass is a function of time.
Since F = dp /dt or for small time intervals, we would write F = (pf – pi) / t
We could then write an expression for momentum as:
pf = pi + Favt
We could rewrite our kinematic equation as :
xf = xi + pav / m *dt
Center of Mass
Last time we defined the center of mass.
We can find the center of mass of a body as:
Center of Mass Velocity:
* The center of mass velocity of a system of particles is the average velocity of all the particles
weighted relative to their mass.
or
vcm 
1
M
N
mv
i 1
i i
Total Momentum and Center of Mass Velocity:
* The total momentum of all the particles in a system is equal to the momentum of a single particle
with a mass m = m1 + m2 + ...mN moving at the velocity of the center of mass vcm.
* Since momentum conserved, the velocity of the center of mass will always be constant.
Work and Energy
Definition of Work in Physics (Thanks to Michael Vawter)

Work is the measure of a quantity that is capable of accomplishing Macroscopic Motion of a
System due to the action of a Force over a Distance.

When a force acts to move an object, we say that Work was done on the object by the
Force.
Work done by a Constant Force:
Work Done By a Variable Force
If the force is not constant along the path of the object we need to calculate the product of the force and
displacement over very tiny intervals and then add them up. This is exactly what the integration over
differential small intervals of length can accomplish.
Power:
* The rate at which work is being done on/by a system.
* The rate of energy transfer into or out of a system.
Definition:
Average Power
Instantaneous Power
Applications to orbital dynamics:
Acceleration of Gravity
At any distance r from a large circular body of mass M a smaller body of mass m will experience
a gravitational force.
Acceleration of Gravity Near the Surface of a Large Body
Assume that a large body of mass M and radius R has a radically uniform mass distribution. The
gravitational force on a small mass m at height h above the surface of the large body can be
expressed as,
If h << R, then
If we neglect air resistance, then this is the net force acting on the mass m. Applying Newton's
Second Law, Fnet = ma,
Derivation of Gravitational Potential:
The gravitational potential energy is defined as the negative of the work it takes to move an object
from some reference point to a height h above the surface of the Earth. Near the surface of the earth
the force of gravity is nearly constant.
Given the expression for general gravitation as
and the definition of work as
Write an expression for the potential energy associated with moving from the surface of the earth
to some small height h above the surface. Note, we will assume that h is small relative to the
radius of the earth.
If we do not neglect the force of gravity's dependence upon distance r, we get the following integral
to evaluate,
If we let r = RE + y then dr = dy evaluate this definite integral.
Thus we can see that the change in potential energy is the difference between two negative numbers
which turns out to be a positive number.
Let us start with the definition of gravitational potential energy and show that it gives the same
equation as above. Starting with
the change in potential is
Reference Level
Had we defined the base level for the potential energy to be the surface of the body, the potential
energy could be defined as
Observe that the change in potential energy PE would still be the same, except our definition of
potential energy would also depend upon the size of the body as well as its mass. Moreover it will
not be zero at the base level - the surface of the body.
Gravitational Mechanical Energy of an Orbiting Body
The objective here is to show that the total mechanical energy of an orbiting body can be expressed
as a function of the body’s orbital parameters.
Start with the properties of ellipses:
Given the following definition:
The eccentricity of an ellipse, usually denoted by ε or e, is the ratio of the distance between the
two foci, to the length of the major axis or e = c/a
If we assume that a star is located at the left focal point, redraw the diagram above with the
perihelion and aphelion distances labeled as rp and ra. Write rp and ra as a function of a and e.
rp = a – c
ra = a + c
e = c /a => c = e * a
so
rp = a (1 – e)
ra = a (1 + e)
The diagram below gives some properties of ellipses.
r = Distance of the planet from the Star (SI: m)
 = Angular location of the Planet from measured from Perihelion (SI: rad)
a = Semimajor Axis "Mean Size of the Orbit" (SI: m)
c = Distance of the Star/Focus from Center (SI: m)
e = Eccentricity "Measures the out-of-roundness of the Orbit" (dimensionaless ratio)
rp = Perihelion distance "Closest Approach to Star" (SI: m)
ra = Apohelion distance "Maximum Separation from Star" (SI: m)

The origin is located at the center of the primary body - the Sun - for the above equations.
Angular Momentum:
The angular "persistence or umph" of a rotating system
Definition:
For a group of interacting particles or a rigid body.
This is the vector cross product of each particle's radius vector from the point of rotation crossed with
the momentum of the particle. For a freely rotating group particles the point of rotation would be the
center of mass of the group.
For a single particle the magnitude of a particle's angular momentum can be expressed as:
Derivation of the Angular Momentum of Orbiting Body.
From the fundamental definition of angular momentum applied
to a point mass of the orbiting body.
L = m r vT
L = angular momentum
m = mass of the orbiting body
r = radial distance of the orbiting body from the primary body.
= the velocity of the orbiting body perpendicular to the radial
vT
direction
There are two points where we know that the velocity is purely tangential (ie v lines up with vT in
the diagram above). Based on the diagram above, what are those two points called? (assuming the
sun is our central body) Write an equation for each of these points as function of a and e.
perihelion and aphelion.
Next we apply the conservation of angular momentum at these turning points in the body’s orbit.
We pick these two points since the body’s velocity will be perpendicular to the radius vector so that
the body’s tangential velocity and total spacial velocity will be one and the same. The body must
have the same angular momentum at these two locations if angular momentum is conserved.
Given L = m r vT Write an expression for the conservation of angular momentum between
perihelion and aphelion:
If an object is in orbit, what direction does for the force on that object due to the central body
point? How is this direction related to the displacement of the object? How much work does this
force do on the body?
Since the force is perpendicular to the displacement, the work is zero so Energy is conserved.
If we now apply the conservation of energy we can find an explicit expression for the perihelion
velocity. If gravitational mechanical energy is conserved then the mechanical energy at perihelion
must also equal the mechanical energy at aphelion. At ra and the rp, the object will have both
kinetic and potential energy. Recalling the definition of KE as ½ mv2 and
write
an expression that relates the energy at apogee to the energy at perigee using only vp, va, G, M, a
and e.
Keep in mind that angular momentum is also conserved for the orbiting object for the same reasons
that energy is conserved. Using what we derived earlier:
Write an expression for va as a function of vp, substitute and then simplify to write an expression
for vp as a function of G, M, a and e.
This shows that the velocity at closest approach is a function of the parameters a and e as well as
the mass of the primary body M. We could have solved for the velocity at apogee and we would
have found a similar expression.
With either of these points we can now find the mechanical energy which will have to also be the
mechanical energy at any other point in the body’s orbit. Recall that the sum of the kinetic and
potential energy at perihelion can be written as:
Substitute the expression that we derived for vp into the expression above and solve for the total
mechanical energy as a function of m, G, M and a.
a)

This result shows that it is the mechanical energy which determines the size – semimajor
axis of the orbit.
It is possible for the mechanical energy to be positive. In these case the body will no longer
be in a bound orbit. If ME = 0 the body will be in a parabolic escape orbit. If ME > 0 the
body will be in a hyperbolic orbit. Dr Fallsheer will talk more about these cases this
afternoon.
Now we can determine the velocity at any point along the orbit as a function of r. Using the result
for the total ME of a bound orbit
Equate these two relationships and solve for the velocity.


This is the general relationship for the velocity of any orbiting body as a function of its
distance from the central body.
It is the total mechanical energy which determine the shape a bodies orbit - in particular the
orbit's semimajor axis a.
Unbound Hyperbolic Orbits:
If the velocity of an object is increased enough so that the total ME becomes positive, then the
object will move along a hyperbolic orbit. Such orbits are unbound in that the orbiting body will
never return, and will have an eccentricity greater than one. The equations describing hyperbolic
orbits are basically the same except for a few signs associated with a.
Parabolic Escape Orbits:
If the total ME energy of a body is zero, then the object will be on an escape trajectory. Here the
eccentricity of the orbit becomes one, and the semimajor axis goes to infinity. An elliptical orbit
with an infinite semimajor axis of infinity is a parabola. Dr Fallsheer talked about escape velocity
yesterday.
Derivation of the Angular Momentum of an Orbiting Body.
There are two points where we know that the velocity is purely
tangential and those are at perihelion and aphelion.
We also know the velocity at these points what we just derived
from conservation of energy and angular momentum.
Using the definition of angular momentum L = m r vT, and the velocity we derived above,
write an expression for the angular momentum L at the aphelion as a function of a, G, M
and e.
Since the angular momentum is the same at any point in a body's orbit we have found the angular
momentum in terms of the orbital parameters.
Geostationary Satellite Problem
A 2100 kg satellite is to be placed into a
circular, geostationary orbit(r = 4.2242e7 m
from the center of the earth) above the
equator. It will be launched on an elliptical
section as shown:
a)
Find the velocity for a circular
orbit when r = rc = a. (Hint think
about the force required to keep
the satellite in orbit.
The circular satellite velocity is given by
b) What is the Energy required to just put the satellite into orbit? First calculate the initial potential
energy at the surface of the earth, then the final mechanical energy. Use these to find the energy
required to orbit.
From the conservation of energy, the energy required is equal to the difference between the final mechanical
energy of the satellite when it is in orbit less the satellite's initial mechanical energy before it is launched.
The initial mechanical energy of the satellite is just its gravitational potential at the Earth's surface, if we neglect
the Earth's rotational motion.
The satellite's final mechanical energy is given by
Thus
This means that it takes at a very minimum 57.7 MJ per kg of pay load. This does not include any drag losses or
the energy required by the booster rockets carrying the satellite.
If we could perfectly convert energy to boost, it costs about $0.15 per kilowatt hour (or 3.6 MJ)
# Year Specific Size
Strength
(MYuri)
1 2000 27.7
110 um x
1336 nm
2 2001 72.7
1 um x
12 nm
3 2002 44.1
∞
4 2003 46
∞
5 2005 3.74
-
6 2005 3.6
-
7 2005 2.8
2 gram,
2m
2 gram,
2m
8 2006 3
Description
"Strength and Breaking Mechanism of Multi-walled Carbon Nanotubes Under Tensile
Load",
Min-Feng Yu et al, Science Magazine [2000] Measurement of individual Multi-walled Nanotubes.
"Direct mechanical measurement of the tensile strength and elastic modulus of multiwalled carbon nanotubes",
Demczyk et al. Materials Science and Engineering A [2002] Another Measurement of individual Multi-walled Nanotubes
"Atomistic Simulations of Nanotube Fracture",
T. Belytschko et al, PHYSICAL REVIEW B [2002] Computer Modeling of Single-Walled Nanotubes, showing failure mechanisms.
"Mechanical properties of carbon nanotubes: theoretical predictions and experimental
measurements"
Ruoff et al., C. R. Physique 4 [2003] Summary of experimental results and modeled prediction
Specified strength for Zylon fiber, by Toyobo Inc.
Strength: 5.8 GPa, Density: 1.55 g/cc.
Specified strength for Spectra 2000 fiber, by Honeywell Inc.
Strength: 3.5 GPa, Density: 0.97 g/cc.
Centaurus entry, best entry in 2005 tether competition, using spectra 2000 fiber.
1300 lbs, 1.9 grams
AstroAranea entry, best entry in 2006 tether competition, using Zylon fiber [Est.].
1350 lbs, 2 grams
9 2007 2.9
10 2007 9
11 2008 40
14
30
2 gram,
2m
um x mm
House Tether, using Zylon fiber and M77 adhesive.
1770 lbs, 2.7 grams
"High-Performance Carbon Nanotube Fiber", Krzysztof Koziolet al, Science Magazine,
2007 Measurement of the strongest of a sample of mm-long pure aggregated carbon nanotube
fibers.
Sparse
"The extraordinary reinforcing efficiency of single-walled carbon nanotubes in oriented
CNT
poly(vinyl alcohol) tapes",
Composite Wang et al. IOP Nanotechnology vol. 18 –
inferred strength of SWNTs from a 1% CNT reinforced plastic tape.
1E8 m
Nominal value for Space Elevator ribbon, Taper Ratio=5.0
x mm
Tether Mass Ratio = 144
Power with giant LASERS!
On a different note:
https://www.youtube.com/watch?v=pD_yQZ4iNjY
c)
Suppose the satellite was launched vertically with
an initial speed that would just allow it to reach a
maximum altitude equal to that of the geostationary
satellite. Determine the initial launch speed, assuming it
was launched vertically from the ground at the equator.
HINT: There are two ways we could solve this problem:
One way is to observe that once the satellite is
launched, it will be in orbit about the center of
mass of the Earth. So if we could find the orbital
parameters al and el, we could use these to find the
velocity of the rocket when its distance is exactly equal to the radius of the Earth.
Find vo so that the aphelion radius of the satellite's orbit when it launched is equal to the semimajor axis
when it is in circular orbit, ra = rc.
Orbital Motion Solution:
This problem is equivalent to that of launching an ICBM missile with a given initial speed v at an initial angle
. Here the initial angle  is nearly vertical.
Looked at as an orbital problem, we know the
maximum distance of the satellite from the
center of the Earth to be 4.2242x107 m. This is
the apogee distance ra of the satellite's orbit
after it is launched. We also know the apogee
velocity from applying the conservation of
angular momentum to be va = vT,f = 70.030
m/s. For an elliptical orbit we know that
(note my typo, should be ra!)
This gives us two equations with two unknowns al and el. Solve for a1 and e1 given that ra = 4.2242e7, G =
6.67e-11 and M = 6e24kg.
Solving for al in the first equation and placing it into the second we get
Solving for el
Since rp = a and vc2= GM/a, this expression is also equivalent to
We could probably have predicted that if the satellite was launched straight up, the orbit was going to be nearly
a straight line with an eccentricity of approximately one.
Next we find al
Using the orbital velocity equation we can calculate the satellite's speed at the surface of the earth when it was
launched and r = RE.
velocity.
Using the values we just determined and find the required launch
Note, this is only slightly smaller than the escape velocity from the earth of 11.2 km/s!
To find the angle of launch as observed from a frame in space moving with the Earth but not rotating with it, we
have to find how fast the earth is rotating at the equator. Calculate this based on the radius of the earth and the
length of a day.
Now calculate the launch angle from trigonometry.
VPython simulation of Apollo Flight
(General Gravitation and Keplers Laws)
1 Newton's Law of Gravitation
Before we start the simulation we need to understand the concept of gravitation force given by Sir Isaac Newton
in 1665.
Newton was the first to show that the force that holds the Moon in its orbit is the same force that makes an
apple fall. Newton concluded that not only does the Earth attract an apple and the Moon but every body in
universe attracts every other body, this phenomena of bodies to move towards each other is called gravitation.
Quantitatively, Newton proposed a law, which is famously called as Newton's Law of Gravitation. This law
states that every particle attracts any other particle with a gravitational force whose magnitude is given by
F
Gm1m2
rˆ
2
r
here m1 and m2 are the mass of the particles, r = r2r1 is the distance between them, r12 is a vector of unit length
pointing from r1 to r2 and G is the gravitational constant, whose value is known to be
G = 6.67 ×1011 N m2/kg2
The force F acting on the two particles is opposite in direction and equal in magnitude.
2 Simulating Flight Path of Apollo 13
To help us visualize and make this tutorial more interesting we will simulate the flight path of Apollo 13 when
subjected to the Earth's gravitational force. We will need to calculate the parking orbit (orbit of the space craft
to revolve around the Earth) of Apollo 13.
For successfully implementing the simulation we will need to know the acceleration, position and velocity of
Apollo 13, which can be respectively found by equation (1.1) and the equations of motion learnt in the
projectile tutorial.
2.1 Initial attributes and display window
In every computer we need to specify some basic variables without which a computer program fails to run. In
this and the next section we will fill in these attributes,
from visual import *
G=6.67e-11
dt=10
Here G is the gravitational constant and dt is the time interval.
2.2 Attributes of Apollo 13 and Earth
Now we specify the attributes concerning the earth and apollo 13 needed for the simulation. In order to
simplify the problem we will assume that the Apollo spacecraft starts in a circular orbit at a height of 173 km
above the earth. Setup the initial position of the Apollo so it is located 173 km above the earth in the x
direction. From newton’s laws derive the required velocity for the spacecraft so it will remain in a stable
circular orbit.
This is done as:
###################################################
#
Declaration of Variables
#
###################################################
#Declaring variables for earth
earth= sphere(pos=(0,0,0), color=color.green)
earth.mass=5.97e24 #mass of the earth
earth.p = vector(0,0,0)*earth.mass
earth.radius = 6378.5e3 #radius of the earth
#Declaring variables for apollo
apollo=sphere(pos=(??????,0,0),radius = 100e3,color=color.red)
apollo.mass = 31.3e3 #mass of apollocraft
apollo.p=vector(0,???????,0)*apollo.mass
apollo.radius= 100e3
#Creating the trails for graphing the orbital path of Apollo
apollo.orbit=curve(pos=[apollo.pos],color=apollo.color,radius = 1e4)
2.4 Acceleration of Apollo 13
To calculate the position and velocity of Apollo 13 as it moves around the earth we will use a momentum
formulation to determine how the motion of the spacecraft evolves over time.
The momentum of an object is determined by the forces on it, which in turn depends on its position compared to
other objects. For example the force of gravity between two objects in space is given by equation (1). The
change in momentum of the body can be calculated by
F
Gmearth mapollo
r
2
p  Fdt 
rˆ
Gmearth mapollo
r
2
* dtrˆ 
Gmearth mapollo
r
3
* dtr
In Vpython we can calculate the acceleration of the two planets can be calculated as:
####################################################
#
Loop for moving the Apollo 13 around earth
#
####################################################
while (1==1):
rate(1000)
# Calculating the change in p of apollo
dist = apollo.pos - earth.pos
force = 6.67e-11 * apollo.mass * earth.mass * dist / mag(dist)**3
apollo.p = apollo.p - force*dt
Here mag gives the magnitude of the distance between the two objects.
In our program we will need to insert this in the while loop part of the program because the acceleration of
Apollo 13 will constantly change with respect to time. The condition while (1==1): specifies the simulation to
continue forever, as the condition 1==1 can never become false. We can exit the program by just closing down
the display window.
2.5 Position calculations
We will determine the new position of Apollo 13 at some later point in time. This is done in the form of
#Calculating the updated position of the Apollo
apollo.pos = apollo.pos + apollo.p/apollo.mass * dt
We also need to update the orbit curve and the label:
apollo.orbit.append(pos=apollo.pos)
apollo.name.pos = apollo.pos
Now we have managed to complete our simulation of Apollo 13, around the earth, due to earth's gravitational
force.
NOTE: We have ignored the force exerted by Apollo 13 on the earth because of its small magnitude. This is
fine for a spacecraft, but in general we cannot ignore the forces moons exert on planets and vice versa.
If you have time, add labels to the earth and Apollo and experiment with some different orbits.