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Class Notes Physics I, 10/30/98
Energy and Orbits
We note that John Glenn at the age of 77 began his second trip into space yesterday and
has now been in orbit for the better part of a full day.
The quiz problem: Assuming that Senator and astronaut Glenn has a total body mass of 70
kg, that his orbit is at an altitude of 300 kilometers above the surface of the earth and
that the speed required for an orbit at this altitude is 7500 m/s, how much total energy
did his mass gain between blastoff and the achievement of orbit? (Note: the initial speed
of the rocket actually gives in an initial kinetic energy > 0; however for the purposes
of this problem this kinetic energy can be regarded as negligible.)
It should be clear that any mass placed in orbit under these conditions will experience
an increase in both potential and kinetic energy. The kinetic energy of a 70 kg mass at a
speed of 7500 meters/second is easily found to be approximately 1.97 * 10^9 Joules = 1.97
G J (gigaJoules) as shown below. To raise the 70 kg object in altitude of 300 kilometers,
or 300,000 meters, requires a potential energy increase of 210 * 10^6 Joules = 210 M J
(megaJoules), as shown below.
http://youtu.be/mmntDMEERJI
In order to achieve orbit, an object requires a certain minimal velocity. Suppose for
example that we build a tower high enough that an object shot from the tower will
experience no significant air friction at that altitude. We might load a gun or a cannon,
in a direction parallel to surface of the earth at the base of the tower, in an attempt to
achieve orbit. If we do not farther projectile that a very great velocity, it will act
pretty much like a projectile rolling off the edge of the table in the horizontal
direction and follow a parabolic path to the ground. This is shown by the shortest dotted
red line in the figure below. If we then fire the projectile somewhat faster, it might do
significantly further, like the second red dotted line in the figure below. Note that this
projectile traverses a noticable part of the curvature of the earth. If we fire a
projectile very very fast, it will take off like the red streak in a nearly horizontal
direction, its path curving only slightly toward the center of the earth.
In order to achieve a circular orbit, we have to achieve just the right velocity. The
blue dotted line depicts the path of an object fire data velocity greater than that of
either of the red dotted lines, but less than that of the red streak. This particle tends
to follow a parabolic path, but as it falls the Earth curves away from it at the same rate
that it falls and it maintains the same distance from the Earth. This object will then
travel in a circular path all the way around the Earth (assuming that the Earth is
perfectly spherical, which is almost but not exactly so).
These paths are shown in the figure below. The paths starting with the red streak
actually becomes an orbit, though not a circular one. As the object travels away from the
Earth, it is pulled back toward the center of the Earth, though with a decreasing force.
An object fired at a great enough velocity could in fact complete the escape the pull of
the Earth and travel away forever. Suppose that the present object is not fired at such a
great velocity. Then as it travels away from the Earth, the pull of the Earth is a force
with two components, one parallel to the direction of motion and the other perpendicular
to that direction. The component parallel to motion tends to slow the object down,
decreasing its kinetic energy as the potential energy of the object increases. The
component perpendicular to motion tends to bend the path toward the Earth. As a result the
object follows a path somewhat like the one shown, slowing down as it moves further and
further from the Earth. At some point it stops moving away from the Earth and starts
falling back, increasing its speed as its potential energy changes back to kinetic. The
path well in fact be an ellipse with one focus at the center of the Earth.
http://youtu.be/FxqfLg3SrQA
Suppose now that a 70 kg individual desires to climb a hypothetical tower from the
surface of the Earth, which of course lies at one Earth radius, denoted rE, from the
center of the Earth, to a point twice this distance from the center of the Earth. We posed
the following problem: if the individual is able to perform 1 kilowattHour of work per
day, how long will it take to climb the tower?
We first note that a kilowattHour is 1000 watts * 1 hour = 1000 Joules / sec * 3600 sec
= 3.6 mJ. We then proceed to estimate the work required to climb the tower.
As the 70 kg individual climbs the tower, the work done through any change `dr in
altitude will be equal to the product of the weight of the person and `dr. The force
exerted by gravity decreases with distance from the center of the Earth. The total work
will be the product of the average force exerted against gravity multiplied by the total
vertical distance.
Our first estimate will be based on the forces at the surface of the Earth and at the
highest point of the climb, at double the Earth radius. At the surface of the Earth the
'vertical' force required to climb at a constant speed is equal to the weight of the
person, which is mg = 70 kg (9.8 m/s^2) = 680 N (approx.).
We can think of the gravitational influence of the Earth at a point a distance r >
rE as being spread out over the surface of a sphere of radius r, much as the light from a
point source is spread over larger and larger spheres as it moves out from the source. In
the case of light, the greater the area over the light is spread, the less light falls per
unit area, and the less illumination objects experience. In the case of gravity, when the
influence of gravity is spread out over greater and greater areas, the less acceleration
results.
Since a doubling of the radius of a sphere, whose area is 4 `pi r^2, results in an area
increase by factor 2^2 = 4, the acceleration of gravity will therefore decreased by factor
1 / 2^2 = 1/4 to 2.5 m/s^2 (approximately), as indicated below.
Having found the acceleration gravity at the beginning and end of the climb, we can
find the weight of the 70 kg individual at both of these positions. At the beginning of
the climb, the acceleration of gravity is 9.8 m/second/second, so the weight is 70 kg *
9.8 m/s/s = 680 Newtons (approximately). At the end of the climb, where the acceleration
is approximately 2.5 m/second/second, a similar calculation shows is that the weight is
170 Newtons.
The figure below shows a graph of force F vs. distance r from the center of the Earth.
We see that the 680 Newton force is associated with distance rE, corresponding to position
at the surface of the Earth, and that the 170 Newton force is associated with the distance
2rE, a distance rE above the service of the Earth.
If the force falls off linearly, then the total work downward correspond to the area
under the trapezoid formed by the dotted red line. However, the force does not fall off
linearly, being proportional to 1 / r^2. We can see that the graph of such a function must
be decreasing at a decreasing rate for positive values of r, and so must follow a curve
closer to that of the green dotted line.
We first calculate the trapezoidal approximation, corresponding to the red dotted line.
Noting, as indicated in the figure above, that the radius of the Earth is rE = 6.4 * 10^6
meters, we calculate `dW = Fave `dr. Taking the average force as the average of the
initial and final forces, as we do in a trapezoidal approximation, we find that the
average forces 425 Newtons; multiplying this by `dr = rE, we obtain `dW = 2.7 * 10^9
Joules = 2.7 G J.
If we use proportionality to find
the force function F(r), we could then use calculus to find the exact value of the area
under the curve and hence of the work `dW. Lacking calculus skills, we can still make a
reasonable estimate based on the graph. If the area under the trapezoidal approximation is
2.7 G J, then a reasonable estimate of the area under the curve might be 2 G J.
http://youtu.be/CvqjtZG5A-k
We could easily extend our
trapezoidal approximations to be greater and greater distances. For example, between 2 rE
and 3 rE we see that g falls from 1/2^2 * 9.8 m/s^2 = 2.5 m/s^2 t= 1 / 3^2 * 9.8 m/s^2 =
1.1 m/s^2, resulting an approximate force of 76 N. This is indicated in the figure below.
A trapezoidal approximation of the area under this new part of the force vs. distance
curve results in an approximate `dW of .8 G J, as indicated below.
We could easily extend our trapezoidal approximations to greater and greater distances.
For example, between 2 rE and 3 rE we see that g falls from 1/2^2 * 9.8 m/s^2 = 2.5 m/s^2
t= 1 / 3^2 * 9.8 m/s^2 = 1.1 m/s^2, resulting an approximate force of 76 N. This is
indicated in the figure below. A trapezoidal approximation of the area under this new part
of the force vs. distance curve results in an approximate `dW of .8 G J, as indicated
below.
The figure below shows roughly estimated approximate areas under the actual curve,
based on trapezoidal approximations. The last three areas, corresponding to .3 G J, .2 G J
and .1 G J have not been validated; you should calculate these two see if the estimates
are reasonably close.
It turns out that if this figure is extended as far as we wish, there is an upper bound
on the total area, so the total area under the curve turns out to be finite, even if the
curve extends to infinite distance. This means that if we have enough energy, we can at
least theoretically climb as far as we wish (provided the tower is 'high' enough and
provided we live long enough, both of which are pretty questionable assumptions).
Your instructor estimated that the total area under the curve would probably be less
than double the area under the first segment of the curve. That is, the total energy
required to reach to infinite distance is less than double the energy required to climb
from the surface of the Earth to distance 2 rE from the center of the Earth. Your
instructor also had the wisdom to put a question mark next to this estimate, indicated by
A < 4 G J (A stands for total area; 4 GJ is double the 2 GJ estimate for the first
interval), and commented that this estimate might not be slow but that the total area
would almost certainly be < 5 GJ.
http://youtu.be/eBv23_TYRGA
(Note: A subsequent
calculation using calculus results indicated that the total area would be about 4.3 G J.)
We can obtain a formula for the
gravitational field at distance r from the center of Earth. Assuming that the
gravitational influence is spread over the surfaces of concentric spheres, we see that at
radius r the influence is spread over a larger sphere than that at the surface of the
planet.
Since the areas of the spheres
are proportional to the squares of their radii, so that the area ration (A / Ae) is the
square of the ratio (r / rE) of the radii, we can therefore say that the gravitational
field strength is inversely proportional to the square of the ratio of the radii, so that
g(r) / 9/8 m/s^2 = 1 / (r / rE)^2, as indicated below.
Solving this proportionality for
g(r), we obtain g(r) = 9.8 m/s^2 * (rE / r) ^ 2. We note that whenever r is greater
than rE, this gives us a gravitational acceleration that is less than that at the surface
of the Earth, as we would expect.
Thus on a mass of 70 kg, we have
the force function F(r) = 70 kg * (9.8 m/s^2) * (rE / r) ^ 2 = 680 N * (rE / r) ^ 2.
Students who know calculus will see that the total work done to 'climb' from
distance rE (at Earth's surface) to distance r is easily found by integrating this force
function, as indicated in the picture preceding the last. Performing the integration
we find that the total work is 680 N * rE^2 (1 / rE - 1 / r).
As r becomes sufficiently large, 1 / r becomes insignificant, and we see that
work 680 N * rE^2 (1 / rE) = 680 N * rE = 4 * 10^9 Joules is an upper limit on the work
required to move away from Earth, no matter how far we go. This means that with a KE
a bit in excess of 4 G J at the surface of the Earth a 70 kg object could, ignoring air
resstance, completely escape the gravitational influence of Earth. We could set .5
mv^s equal to this energy and determine the velocity required. This velocity is
called escape velocity.
http://youtu.be/6y2fzDj87YQ
http://youtu.be/wUgaKT3Nhqc
http://youtu.be/DyapAzB5o-A
http://youtu.be/AI_6AApCmTs
http://youtu.be/ESsoJDneGas
http://youtu.be/M8YFU1E9m74
http://youtu.be/gKJCv-a4Xlk
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