Download Geosynchronous Orbit Laboratory

Title:
Geosynchronous Orbit Laboratory
Purpose:
To find the proportionality between the velocity and the radius of a mass in circular
motion under a constant centripetal force. To then extend this idea to satellites in geosynchronous
orbits to find the altitude necessary for a constant stable orbit.
(from
https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&cad=rja&uact=8&docid=cavx
YYPS7HbubM&tbnid=PlPJlEgZbo7QUM:&ved=0CAQQjB0&url=http%3A%2F%2Fwww.chegg.com%2Fhom
ework-help%2Fquestions-and-answers%2Fmass-m-5300-kg-suspended-string-length-l-1310-m-revolveshorizontal-circle-seefigure--tangq381492&ei=KlOrU76HDcbD8QHCp4D4Bg&bvm=bv.69620078,d.b2U&psig=AFQjCNHOszIlJsvLs0tboUj_n
guu1i7i6A&ust=1403823259783146)
https://www.google.com/url?sa=i&rct=j&q=&esrc=s&source=images&cd=&docid=frFqkEYVgSjSHM&tbn
id=w72Sa-tkBcueM:&ved=0CAQQjB0&url=http%3A%2F%2Fwww.docstoc.com%2Fdocs%2F143535891%2Fcircularmotion-monsterreview&ei=F1arU7qbOPDp8AHCl4DgBw&bvm=bv.69620078,d.b2U&psig=AFQjCNFHDonix6cjrZehVKzZxX
9MBnAjog&ust=1403823950962729
Procedure:
1) Divide your group so that each person has a job. Each group needs a A) Spinner B) Counter
C) Recorder D) Timer ; any extra people in the group can quietly help to be sure that the
counter is being accurate and that the spinner is keeping the angle of the string at 45 degrees.
2) The spinner ties the mass on to the string.
3) The spinner measures out one meter exactly of string and hangs the mass from the string.
4) The spinner holds the protractor vertically so the string is in the right spot to measure out an
angle when the mass is spun in a circle.
5) The counter, recorder, and timer check to see that the spinner is holding the protractor vertical
so that the angle can be accurately kept at 45 degrees while spinning.
6) The counter tells the spinner to spin until the string is at a 45 degree angle.
7) The counter counts out loud “One, Two, Three, Mark”.
8) The timer starts the stop watch when they hear the word “Mark “.
9) The counter counts out loud the number or revolutions.
10) The timer yells out the time when the counter counts the spinner going around ten revolutions.
11) The recorder writes down the time for 10 revolutions at the radius measured.
12) The spinner increases the radius by 0.05 meters and steps 5 through 12 in the procedure are
repeated until a radius of 2 meters is reached.
Materials:
1)
2)
3)
4)
5)
6)
3 meters of string
One 250 g mass
One giant protractor
Stop Watch
Calculator
A computer for writing up the Lab Report.
Source:
http://www.printactivities.com/ColoringPages/FigureSkating-ColoringPages/FigureSkating-Coloring-Pages.shtml
IF you find that you are getting a little too dizzy, then by all means your group
should rotate group job positions.
Source:
http://1.bp.blogspot.com/_BKY7hFNdCjw/TUVnCiSRAI/AAAAAAAAAHE/NAMcQ5EQ9-w/s1600/Dizzy+Girl.jpg
Data:
DATA TABLE
Time for 10 revolutions
(s)
Period (s)
String Length
(m)
Velocity (m/s)
Radius (m)
Circumference (m)
0
1
0
0.707106781
4.442882938
0
1.05
0
0.74246212
4.665027085
0
1.1
0
0.777817459
4.887171232
0
1.15
0
0.813172798
5.109315379
0
1.2
0
0.848528137
5.331459526
0
1.25
0
0.883883476
5.553603673
0
1.3
0
0.919238816
5.77574782
0
1.35
0
0.954594155
5.997891967
0
1.4
0
0.989949494
6.220036113
0
1.45
0
1.025304833
6.44218026
0
1.5
0
1.060660172
6.664324407
0
1.55
0
1.096015511
6.886468554
0
1.6
0
1.13137085
7.108612701
0
1.65
0
1.166726189
7.330756848
0
1.7
0
1.202081528
7.552900995
0
1.75
0
1.237436867
7.775045142
0
1.8
0
1.272792206
7.997189289
0
1.85
0
1.308147545
8.219333436
0
1.9
0
1.343502884
8.441477583
0
1.95
0
1.378858223
8.663621729
0
2
0
1.414213562
8.885765876
Graphs:
Velocity Vs Radius
10
9
8
Velocity (m/s)
7
6
5
4
3
2
1
0
0.50
0.70
0.90
1.10
1.30
1.50
1.70
1.90
Radius (m)
Questions:
1) What does your graph look like? Does it go in a straight line? Is it increasing or decreasing? Does
it curve up or down? Does it go through the origin? Does it have a y intercept? Does it have a x
intercept? Is the slope positive or negative (everywhere?)? Is the graph smooth or are the points
sporadic , erratic, and not seem to follow a rule?
2) If you had to guess, what type of function does your data look like? Does it look like any of these
functions? (from: http://ceadserv1.nku.edu/longa//classes/mat119/days/functionzoo.jpg)
Source: http://www.pinterest.com/swilson01/school-ideas/
My data looks like a _________________________________________function.
3) Draw a trend line through your data for the function that you think works best for your data. Do
this on your graph. Use the trend line in a spread sheet program.
4) Using your trend line, estimate what the velocity would be at a radius of 1.6 meters would be?
According to my trend line, at r = 1.6 meters, the velocity would be ______________________m/s
5) Using your trend line, estimate what the radius would be for a mass with a velocity of 0.5 m/s.
For a mass with a velocity of 0.5 m/s, the radius should be ____________________________meters
6) Write an equation for the Velocity (V) of the mass as a function of the radius (r).
V(r) = ________________________ (this should be just like f(x) is a function of x, except this is the
velocity as a function of r.)
7) The last page of this lab shows a some calculations taken from Wikipedia
http://en.wikipedia.org/wiki/Geostationary_orbit That show the relationship of the radius of
revolution (r), the planet mass (M), and the rate of rotation for the planet(ω). Use the equations
to find the calculated values for the altitude of a satellite (r), and the velocity (v) of a satellite in
geosynchronous orbit.
3
𝑟=√
𝐺𝑀 3 (6.67 ∗ 10−11 )(𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑎𝑟𝑡ℎ 𝑔𝑜𝑒𝑠 ℎ𝑒𝑟𝑒)
=√
= _______________________𝑚𝑒𝑡𝑒𝑟𝑠
𝜔2
(𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑔𝑜𝑒𝑠 ℎ𝑒𝑟𝑒)2
𝑣 = 𝜔𝑟 = (𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑔𝑜𝑒𝑠 ℎ𝑒𝑟𝑒) ∗ (𝑦𝑜𝑢𝑟 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑓𝑟𝑜𝑚 𝑎𝑏𝑜𝑣𝑒 𝑔𝑜𝑒𝑠 ℎ𝑒𝑟𝑒)
𝑚𝑒𝑡𝑒𝑟𝑠
= _______________
𝑠𝑒𝑐𝑜𝑛𝑑
M is the mass of Earth, M = 5.9736 × 1024 kg
𝜔 is the earth’s rotation rate, 𝜔 =
2𝜋 𝑅𝑎𝑑𝑖𝑎𝑛𝑠
𝜏 𝑆𝑒𝑐𝑜𝑛𝑑𝑠
2𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠
= 86,164 𝑠𝑒𝑐𝑜𝑛𝑑𝑠 = 7.29 ∗ 10−5 𝑟𝑎𝑑𝑠/𝑠𝑒𝑐
8) Do these values for V and r that you found above fit your equation that you found for the
Velocity (V) as a function of the radius (r)? Tell why your answer makes sense. What value do
you get when you plug the value of r above into the equation you derived from your graph for v.
These values of V and r DO or DON’T (circle one) fit my equation. This makes sense because
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
_____________________________________________________________________________________
What is the main difference between the two situations with the mass revolving in the laboratory and
the satellite revolving around the earth? Or asking the same question another way; what is different
about these two situations that causs such a difference in the proportionality of velocity V and radius r?
9) For your mass revolving at a 45 degree angle to the vertical, could you keep the same period of
rotation and increase the radius of rotation? Yes or no and explain why or why not.
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
10) Final concluding remarks: Tell what went right, what went wrong, if you were to do it again what
would you change, and what did you learn.
i.
What went right?
ii.
What went wrong?
iii.
What would you change?
iv.
What did you learn?
Derivation of geostationary altitude
In any circular orbit, the centripetal force required to maintain the orbit (Fc) is provided by the
gravitational force on the satellite (Fg). To calculate the geostationary orbit altitude, one begins
with this equivalence:
By Newton's second law of motion,[15] we can replace the forces F with the mass m of the object
multiplied by the acceleration felt by the object due to that force:
We note that the mass of the satellite m appears on both sides — geostationary orbit is
independent of the mass of the satellite.[c] So calculating the altitude simplifies into calculating
the point where the magnitudes of the centripetal acceleration required for orbital motion and the
gravitational acceleration provided by Earth's gravity are equal.
The centripetal acceleration's magnitude is:
where ω is the angular speed, and r is the orbital radius as measured from the Earth's center of
mass.
The magnitude of the gravitational acceleration is:
where M is the mass of Earth, 5.9736 × 1024 kg, and G is the gravitational constant, 6.67428 ±
0.00067 × 10−11 m3 kg−1 s−2.
Equating the two accelerations gives:
The product GM is known with much greater precision than either factor alone; it is known as
the geocentric gravitational constant μ = 398,600.4418 ± 0.0008 km3 s−2:
The angular speed ω is found by dividing the angle travelled in one revolution (360° = 2π rad) by
the orbital period (the time it takes to make one full revolution). In the case of a geostationary
orbit, the orbital period is one sidereal day, or 86,164.09054 seconds).[16] This gives:
The resulting orbital radius is 42,164 kilometres (26,199 mi). Subtracting the Earth's equatorial
radius, 6,378 kilometres (3,963 mi), gives the altitude of 35,786 kilometres (22,236 mi).
Orbital speed (how fast the satellite is moving through space) is calculated by multiplying the
angular speed by the orbital radius:
By the same formula we can find the geostationary-type orbit of an object in relation to Mars
(this type of orbit above is referred to as an areostationary orbit if it is above Mars). The
geocentric gravitational constant GM (which is μ) for Mars has the value of 42,828 km3s−2, and
the known rotational period (T) of Mars is 88,642.66 seconds. Since ω = 2π/T, using the formula
above, the value of ω is found to be approx 7.088218×10−5 s−1. Thus, r3 = 8.5243×1012 km3,
whose cube root is 20,427 km; subtracting the equatorial radius of Mars (3396.2 km) we have
17,031 km.
Derivation of the proportionality relationship between the velocity and the
altitude r of a satellite going around the earth.
∑ 𝐹⃗ = 𝑚𝑎⃗
Using Newton’s 2nd Law
In this case the only force on the body is the force of gravity and the
acceleration of the body is centripetal.
𝐹𝐺 = 𝐹𝐶
Where 𝐹𝐺 is the force of gravity on the satellite
and 𝐹𝐶 is the centripetal force.
Source:
𝐺
𝑚𝑀
𝑟2
http://www.grc.nasa.gov/WWW/k-12/rocket/TRCRocket/IMAGES/orbital.gif
=𝑚
𝑮𝑴
𝒓
𝒗= √
𝑣2
𝑟
= √𝑮𝑴 ∗
Where r is the distance from the center of earth. Simplifying and solving for v
𝟏
√𝒓
force is not a constant.
= (𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕) ∗
𝒗∝
𝟏
√𝒓
Therefore, v is proportional to
𝟏
;
√𝒓
here the centripetal
𝟏
√𝒓
For a geosynchronous orbit, the period is a constant and is one sidereal day. 𝜏 = 𝑝𝑒𝑟𝑖𝑜𝑑 𝑎𝑛𝑑 𝑣 =
2𝜋𝑟
.
𝜏
Substituting for the velocity.
2𝜋𝑟
𝜏
𝐺𝑀
𝑟
Squaring
𝐺𝑀
𝑟
Solving for r
=√
4𝜋2 𝑟 2
𝜏2
=
3
𝐺𝑀𝜏2
3
𝐺𝑀
𝑟 = √ 4𝜋2
𝑟 = √ 𝜔2
since 𝜔 =
2𝜋
𝜏
This is a constant value for a mass in geosynchronous orbit.
The resulting orbital radius is 42,164 kilometres (26,199 mi). Subtracting the Earth's equatorial radius,
6,378 kilometres (3,963 mi), gives the altitude of 35,786 kilometres (22,236 mi).
Derivation of the relationship between the velocity and the radius for a mass
on the end of a string rotating at a constant centripetal force such that the
angle with the vertical is 45 degrees.
Source: http://www.simplecapa.com/PDF/CAPA5_1110.pdf
Using Newton’s 2nd law in the horizontal direction
∑ 𝐹⃗ = 𝑚𝑎⃗
The only force on the body horizontally is the horizontal force of tension in the string and the
acceleration is centripetal and towards the center.
𝐹𝑇 𝑠𝑖𝑛𝜃 = 𝑚
𝑣2
𝑟
Using Newton’s 1st law in the vertical direction
∑ 𝐹⃗ = 0
The vertical component of the tension on the string is pulling up and gravity is pulling down.
𝐹𝑇 𝑐𝑜𝑠𝜃 − 𝐹𝐺 = 0
𝐹
𝐺
𝐹𝑇 = 𝑐𝑜𝑠𝜃
𝐹𝐺
𝑐𝑜𝑠𝜃
∗ 𝑠𝑖𝑛𝜃 = 𝑚
𝐹𝐺 = 𝑚
𝑣2
𝑟
This equation can be solved for 𝐹𝑇
Substitute this expression into the equation for the horizontal dimension
𝑣2
𝑟
At θ= 45 degrees, 𝑠𝑖𝑛𝜃 = 𝑐𝑜𝑠𝜃.
Which is the same derivation achieved for the satellite; 𝐹𝐺 = 𝐹𝐶 .
𝑚𝑔 = 𝑚
𝑣2
𝑟
Substitute for the weight at the earth’s surface, simplifying and solving for v.
𝒗 = √𝒈𝒓 = √𝒈 ∗ √𝒓 = (𝒂 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕) ∗ √𝒓
centripetal force is a constant.
𝒗 ∝ √𝒓
Therefore velocity is proportional to √𝒓; here the