#$&*
course PHY 241
Nov 21, 2010. 18:00
Note that the data program is in a continual state of revision and should be downloaded with every lab.This experiment uses the program located at
http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/grav_field_simulation_v1b_no_filing.exe
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Experiment 26. From a simulation we find that orbital velocity is inversely proportional to the square root of orbital radius; potential energy increase from one orbit to another is double the kinetic energy decrease so we have to speed up to slow down.
Note: The editing of these instructions into their new format began at just about the time one of the late Shuttle flights landed on July 31, 2009. At this time there is active discussion about the future of the space program, including a debate on whether we should return to the Moon, aim for Mars (or its moons), focus on deeper space and Lagrange points, or some combination of the above.
Click here for instructions using the former DOS version of the program, which is located under Simulations on the 'real' Physics I homepage. This program should only be used if the program downloaded from the Sup Study ... site does not work.
Instructions for program grav_field_simulation.exe for Experiment 26:
The program will be found at the Sup Study ... site under Course Documents > Downloads > Physics I. Download and/or run it in order to see the buttons and boxes described here:
The array of boxes and buttons at the right side of the screen contains information about the planet, the satellite and the time scale of the simulation.
Planet mass is the mass of the planet in multiples of the mass of the Earth. The default values assume that the planet is Earth, so the default planet mass is 1. You can enter any planet mass you wish. For example the Moon has a mass about 0.0123 times that of the Earth, the Sun has a mass which is about 340,000 times that of the Earth. If you wanted to simulate and orbit around the Sun or the Moon you would enter 0.0123 or 340,000 in this box.
Planet radius is given in multiples of the radius of the Earth. Since the default planet is Earth the planet radius has default value 1. If you wanted to simulate the Moon you might enter 0.26, which represents the fact that the Moon has a radius about 0.26 times that of the Earth. If you what and to simulate the Sun you might enter 1100, since the Sun has a radius about 1100 times that of the Earth.
Time factor is the factor but which the simulation is speed up. The default value of the time factor is 1,000, which means that everything runs about 1000 times faster than actual. This means, for example, that a low-Earth orbit will take place in about six seconds rather than the actual approximate time of 6,000 seconds.
Screen scale is the distance from the center of the picture to the edges, in Earth radii. The default value is 3, which works well for low and moderate Earth orbits. However if you are trying to investigate orbits which move further than 3 Earth radii from the center of the planet you need to adjust the screen scale accordingly or the satellite or projectile might not show up on the screen.
Initial distance is the distance of your satellite or projectile from the center of the Earth. This distance is set to 1.02, which is around the minimum distance at which it is possible to orbit at least a few times without encountering significant atmosphere. You can set it for any distance you wish. [ Note that this simulation ignores atmospheric drag and will work just fine for orbits inside the atmosphere. In fact it ignores any sort of interference at all so orbits low enough to encounter mountains will work just find here. Not only that, but this program implicitly assumes that all the mass of the planet is concentrated at its center and even allows orbits inside the surface of the planet. The only problem arises if you get very very close to the center of the planet, in which case the simulation breaks down and spits the satellite or projectile out at very high velocity in a straight line (which is just an anomaly of the simulation and would not really happen in any circumstance). ]
Initial angular position is the angle in radians made with the positive x axis (which is directed toward the right, as is standard for many applications) by a line segment from the center of the planet to the initial position of your satellite or projectile. Note that there are approximately six radians (actually 2 pi, closer to 6.28 radians) around a circle.
The impulse of the 'burn' is actually impulse per kg. Recall that the impulse of a force acting on an object, which is the product F `dt of the average force and time interval during which it acts, gives the change in the momentum of the object. It follows that the impulse kg is in fact the change in the velocity of the object. Note that we are here assuming that the 'burn' does not significantly change the mass of the object; this is not always the case with actual satellites and certainly is not the case with a rocket boosting a satellite into orbit. The default impulse is 8000, which will give the satellite or projectile a velocity of 8000 m/s, a bit in excess of the velocity required to achieve a circular low-Earth orbit.
To deliver an impulse you first choose the magnitude of the impulse, then click on the Forward, Backward, To Right or To Left button.
The direction of the initial impulse depends on the goal of the simulation. If we wish to achieve a circular orbit then because of the geometry of a circle (at every point the circle is perpendicular to the radial line from the center to that point) the impulse must be at a right angle to the initial angular position; otherwise circularity is in the first instant violated. Since a right angle is 1/4 of the angle around a circle, the right angle is 2 pi / 4 radians = pi / 2 radians, or approximately 1.57 radians. On the other hand if we wish to shoot a projectile 'straight up' from the surface of the Earth we must 'fire' it in the direction directly away from the center of the planet, which means that we must 'fire' along the radial line from the center to our starting point. This means that the initial direction must be the same as the initial angular position.
Clock time is displayed as the simulation runs. Clock time is the actual simulation time since the 'run' started.
Circle radius is the radius of a circular orbit you might be trying to achieve, in Earth radii. If the number in this box is not zero then when you click Run Simulation the program will place a red circle of this radius, centered at the center of the planet, on the screen.
Realtime interval is the 'real world' time in minutes since the simulation began.
Speed is the speed of the satellite or projectile in meters/second.
The Run Simulation button is used to begin the simulation. When the simulation is begun the planet will show in blue the center of the screen and the satellite will show in white.
The first eight buttons in the rightmost column are used to deliver an impulse to the satellite or projectile.
The top four buttons deliver the impulse forward, i.e., in the direction of velocity of the object, or backward in the direction directly opposite that of the object's velocity, or to the right (defined to be at a right angle to the right as perceived by an individual facing the direction of motion) or to the left.
The default impulse is 0. The magnitude of the impulse is chosen by clicking one of the next four buttons. Once clicked this impulse is 'set' until another impulse button is clicked, so that it is possible with successive clicks to deliver any reasonable chosen impulse.
The Pause Simulation button, as you might expect, allows you to pause the simulation. There are two reasons you might want to do this. One is to simply have a look at the numbers in the boxes, another might be to change the numbers and restart the simulation without erasing the existing screen.
The Continue button will continue the program after a pause; if you haven't changed anything in the boxes the program simply picks up where it left off.
The Run (don't clear) button restarts the simulation after a pause, without erasing the existing screen.
The 2d planet button will create a second planet (e.g., the Moon), but first you have to go down to text boxes at the bottom of this column (just above the Apply button) and enter the necessary information. The default message in each box will tell you what you need to know, but those messages are repeated here. Note that only the first word or two of each message actually shows in the box.
The first of the two text boxes contains the message 'second planet mass as multiple of Earth mass (Moon = .0123)', telling you to enter the mass of the second planet, for example .0123 if you mean the Moon. Enter just the number with no punctuation (except a decimal place if required) and no letters or words.
The second box contains the message 'second planet dist as multiple of Earth radius (Moon = 60.2)', meaning that you should enter the number 60.2 if you want to simulate the Moon, or any appropriate number if you want to set up some other situation.
Be sure you have the correct numerical information in these boxes. If you don't the program is likely to crash.
After entering this information you can click on the 2d planet button, which will give you a message telling you that the 2d planet has been created. If the simulation is already running the second planet should appear, provided the screen scale can accommodate it. If not, or if you wish to make other changes, you may change the screen scale and any other information you wish then click on Run Simulation.
First investigations:
Without changing anything click on Run Simulation. You will see a blue circle representing the Earth and a moving white dot representing the successive positions of a satellite. The satellite completes its orbit 1000 times as fast as an actual satellite, due to the default time factor.
While the simulation is running click on 'Impulse 100' then on Forward and see what happens to the orbit.
Click on Forward a few more times and see what happens to the orbit.
Click on 'Impulse 1000' then on 'forward' and see what happens. Then click on 'right' and on 'left' and see what happens.
below describe how these buttons affect the behavior of the satellite, and how they contrast with one another: ----->>>>>>>>
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Your answer (start in the next line):
Impulse 100 + forward results in a larger elliptical circle than the original orbit. As you continue to click forward the orbit continues to increase at the same rate as before. Each time you add the impulse its effects occur immediately. The same holds true for Impulse 1000 + forward except about 10x the effect of Impulse 100.
The total effect of adding any impluse forward is a slower orbit per unit of time.
The effect of backward is a tighter orbit closer to the earth. Of course it is produced in the same way as forward.
effect of impulse button forward, right, left:
Forward is explained above.
Right vs. Left Impulse of any scale pushes out ir to the right and in toward the earth if to the left.
your brief discussion/explanation:
When adding any impulse forward it only effects the tangential direction of velocity by accelerating the satellite. When the impulse is backward it is apposing the motion of the tangential velocity.
When adding impulse to the right it pushes the satellite out away from the earth in the normal direction. It does not increase the radius of the orbit though it simply shifts it. Of course to the left moves the orbit to the left it shifts the orbit to the left.
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Any time the simulation gets out of control you can start over by clicking on Run Simulation. See if you can figure out the most efficient way to move from the default orbit into a 'higher' circular orbit.
Do this:
See how efficiently (i.e., in how few clicks) you can get the satellite to the edge of the screen using 100 (kg m/s) / kg impulses, starting from the default orbit.
below describe what you think is the most efficient way to get off the screen using (100 kg m/s) / kg impulses. Be specific in your description of how you time your clicks, and how many clicks it actually takes. Remember you are trying to minimize the number of clicks.
----->>>>>>>>18 clicks
Your answer (start in the next line):
18 clicks
most efficient way to get off the screen (minimize number of clicks): to continue adding impulses at the same point in orbit ST its orbit expanded the its eliptisity of the orbit.
your brief discussion/explanation:
Hypothesis:
By adding consistent impulses in one direction at the same time it effects only one part [the other end of the orbit. This seems to me to be the most efficient way. [No test were taken to prove otherwise].
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Start the simulation over by clicking on Run Simulation:
The program shows you the speed of the satellite. The speed is in the neighborhood of 8000, give or take a bit, and is in meters / second. The speed changes slightly during the orbit, and the average speed changes slightly from orbit to orbit.
What is the range of speeds for the default orbit? Does the orbit appear perfectly circular? If you can discern a deviation from a circular orbit, is the speed greater or less when the satellite is furthest from the center of the Earth?
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Your answer (start in the next line):
From 7.324 km/s to 7.999 km/s it is slightly elliptical. The speed is slower at the greatest distance from the radius out from the center of the earth.
Note** I cannot till a difference in the change in average speed in the different orbits. It seems that the min and max speeds are relatively consistent => the ave speed should be constant.
range of speeds, deviation from circularity of default orbit: explained above
your brief discussion/explanation: Hope it makes sense; it does to me.
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An orbit can be circular or elliptical. The three figures below depict ellipses. Assume in each figure that the ellipse is centered at the origin of an x-y coordinate system, with the x axis horizontal and the y axis vertical on the page.
Clearly the shapes of these figures differ with the third being circular or nearly circular and the first being furthest from circular. The further an ellipse is from a circle the more eccentric we say it is. The first ellipse is the most eccentric, being about twice as long in the x direction as in the y direction. The third ellipse is the least eccentric, with its length along the x axis being about the same as its length along the y axis. If the lengths along the two axes are equal then the ellipse is also said to be a circular.
Starting from the default orbit, if you give the satellite a 1000 (kg m/s) / kg impulse in the forward direction, what it its its velocity immediately before and immediately after after the change?
What is the KE of a 1 kg mass at the speed you observed immediately before, and immediately after the change? Your results give you the KE per kilogram of mass.
Report velocity immediately before, and velocity immediately after the impulse, in first line of the box below. In the second line give KE per kg of mass immediately before and immediately after the impulse. In the third line describe the shape of the resulting orbit, and also specify whether the center of the Earth in the simulation appears to coincide with the center of the orbit or to be closer to one side of the orbit than to the other.
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Your answer (start in the next line):
velocity before impulse and after 1000 (kg m/s) / kg impulse: before = about 8000 m/s
after = about 9000m/s
KE/kg immediately before and after impulse: before = 32 kg * Mm^2 / s^2
After = about 40.5 kg * Mm^2 / s^2
shape of orbit, center of earth vs. center of orbit:
before near circular shifted ;slightly elliptical.
After near circular but expanded [slightly more elliptical] and shifted.
your brief discussion/explanation:
Due to the greater amount of KE it has more energy to overcome a greater distance away from the earth.
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Click Run Simulation again and repeat, this time using a backward impulse of 1000 kg m/s. Note the shape of the new orbit and the maximum and minimum velocities.
Give the max and min velocities in the first line, max and min KE per kg in the second, and in the third line describe the shape of the resulting orbit in terms similar to those used previously.
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Your answer (start in the next line):
(backward impulse) max and min orbital velocities: 7.3 Mm / s, 8.0 Mm/s
max and min KE/kg: 26.6 kg*Mm^2/s^2, 32 kg *Mm^2/s^2
shape of orbit: more-so circular but shifted to the right w/ respect to the earths position.
your brief discussion/explanation: It may be that as we slow it down the energy is lessened => the radius of the orbital is smaller. Due to this the effects of gravity are increased => It would hit the earth, but according to the model it does not slow; this may be du the attraction of gravity.
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Click Run Simulation again and repeat, this time using an impulse of 1000 kg m/s to the left. Note the shape of the new orbit and the maximum and minimum velocities.
Give the max and min velocities in the first line, max and min KE per kg in the second, and in the third line describe the shape of the resulting orbit.
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Your answer (start in the next line):
(left impulse) max and min orbital velocities: 7.3 km / s, 8.0 km/s
max and min KE/kg: 26.6 kg*Mm^2/s^2, 32 kg *Mm^2/s^2
shape of orbit: about circular based on the orbital around the earth.
your brief discussion/explanation: I calculated the KE /kg my multiplying our unit mass of 1 kg * v^2/2.
The reasoning behind the consistency is that the only effect was in the normal direction which would be like shifting a position while maintaining a consistent ave speed with the test above the initial run.
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Which direction resulted in the greatest change in KE per kg from immediately before to immediately after impulse? Which resulted in the greatest difference between the maximum and the minimum KE of the resulting orbit? What was this difference in KE per kg, and how did you obtain your result?
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Your answer (start in the next line):
direction resulting in greatest change in KE/kg: a elliptical more-so shape where the min speed was on the outermost portion of the circuit. And the greatest speed was on the inside.
direction resulting in greatest difference between max and min KE/kg of orbit: impulse forward =an increase in vmax + a decrease in v-min => greatest change.
difference between max and min KE/kg: 40.5 Mm^2/s^2 10.58 Mm/s^2 => about 30 Mm^2/s^2 or the change in speed appears to be a power function.
your brief discussion/explanation:
the four to one relationship is quite feasible for we do have a power function. BC kg*v^2/2
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Starting again from the default orbit, click twice in rapid succession to provide two 1000 (kg m/s)/kg impulses and observe the velocity immediately afterwards. Use the direction that maximizes KE increase. Compare the KE increase that resulted from the first 1000 (kg m/s) / kg impulse (which you reported above) with the KE increase due to the second. Explain in detail how you calculated the KE increase due to the second click.
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Your answer (start in the next line):
two quick impulses, compare KE increase with that of single click:
two clicks yields a gross increase in KE of (9^2 - 7^2)/2 or 16x relation. This should be true BC the first relation was four to one so.. the relation for an increase of two units will be the same relation- four to one=].
of course this is true due to the power relation
your brief discussion/explanation:
I do not know what else there is to explain. [besides what I know not]
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If after applying the first 1000 (m/s)/kg impulse you apply a second identical impulse, at what point in the orbit should it be applied in order to obtain the orbit with the greatest possible maximum KE within the orbit?
Investigate this question and answer below, describing what you did and how it led you to your answer:
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Your answer (start in the next line):
at what point in orbit will second impulse result in greatest orbital KE: If each impulse is added at the exact same point the KE will be the greatest. If this gives the greatest KE it will also give the greatest `ds and the most elliptical shape.
your brief discussion/explanation:**note to test my hypothisis I added the impulse every 3.14 rad. This gave the most circular model and made the speed vary the least. Then I tried every 1.57 [adding a impulse of 1000]. This enhanced the `dKE but I supposed for the bust rewsults why not try at about the same point in the orgit. Of course giving ea additional impulse where the initial impulse was given maximized or optimized the `dv and inherently the `dKE.
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Change the initial impulse:
Note that the default impulse of 8000, which gives the satellite an initial velocity of 8000 m/s, gives an orbit which is not quite circular. Change the 8000 to 9000 and click the Run Simulation button to see how the orbit changes as a result of the greater initial velocity. Then change the impulse to 7000 and click the Run Simulation button to see what happens. Then see if you can find the initial velocity that gives you a good circular orbit. (A 'good' circular orbit would be one in which the velocity doesn't change by more that about 100 m/s from one point of the orbit to another. If you can do better than that with a few minutes' extra effort, fine; if you can revise a good strategy for efficiently refining your attempts you can get very good precision).
Report in the first line the velocity that gives you a good circular orbit. In the second line report the shapes of the orbits for impulses of 9000 (kg m/s) / kg and for 7000 (kg m/s) / kg. Starting in the third line describe your strategy for obtaining a good circular orbit:
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Your answer (start in the next line):
initial velocity for good circular orbit at default position: Initial impulse = 7825 N/(kg*`dt)
for 9000 elliptical /w respect to the horizontal. For 7000: Elliptical /w respect to vertical.
your brief discussion/explanation: the faster you go from the initial burn impulse the more it elongates for an horizontal stretched ellipsoid [to the left]. The slower it goes below the constant position required for an initial speed the more it is compressed on the horizontal which creates a smaller ellipsoid in the opposite direction [to the right].The second staement is not true I trake it back it seems perfectly circular on the interior of the planet.
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Achieve circular orbits at 1.5 and 2 Earth radii:
Change number in the init box to 1.5 and click on Run Simulation. What happens to the shape of the orbit?
Change the initial impulse by a reasonable amount, predict what will happen to the orbit, and click again on Run Simulation.
Use trial and error to find the initial impulse required at 1.5 Earth radii to achieve a circular orbit, as nearly as you can reasonably achieve it. A reasonable result would be one in which the velocity doesn't change by more than 1% during an orbit.
Report below the initial impulse required to achieve a circular orbit at 1.5 Earth radii. In the second line give the maximum and minimum velocities observed in that orbit and the difference in these velocities as a percent of the average of the two velocities. If your strategy for obtaining good results has improved you may report this starting in the next line.
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Your answer (start in the next line):
impulse for circular orbit at 1.5 Earth radii:
same as the above speed: about 7828.
max and min velocities, percent difference: max=7835, min=7824
the change in velocity is 11m/s, the percent change with respect to the initial speed is 11/7828
= 0.1 percent
your brief discussion/explanation:
Shape Change /w diff radius = I can till no diff in the effects with the same initial impulses.
I used the guess method for the speed [also slowing the time ratio down to help.
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Repeat for init distance equal to 2 Earth radii.
Report below the initial impulse required to achieve a circular orbit at 2.0 Earth radii. In the second line give the maximum and minimum velocities observed in that orbit and the difference in these velocities as a percent of the average of the two velocities.
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Your answer (start in the next line):
impulse for circular orbit at 2.0 Earth radii: about 7825
max and min velocities, percent difference:all exactly the same as before. => 0.1 percent error
your brief discussion/explanation:
I simply checked the speed without adjusting the prior tracks and they went exactly the same with the same change of speed.
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'Shoot' a projectile 'straight up' from the surface: To shoot 'straight up' you shoot straight out along a radial line (see direction of the initial impulse in the description of the program above).
First set the number in the Circle Radius box to 2.
To position the projectile on the surface set Initial Distance at 1, which places the projectile at 1 Earth radius from the center.
Set Initial Angular Position to 1, which will place the projectile at the 1-radian position.
Set Direction of Initial Impulse also to 1, which will 'shoot' the projectile in the 1 radian direction. This will 'shoot' the projectile straight out from the Earth, which from the perspective of an observer at that position will appear to be 'straight up'.
Set the Initial Impulse to 6000.
Click on Run Simulation. See how far out from the planet the projectile goes before 'falling back'.
See also how long the projectile 'rises' before beginning to fall back.
Repeat for Initial Angular Positions of 1 2, 3, 4, 5 and 6 radians, with respective initial impulses of 3000, 5000, 7000, 8000 and 9000. Be sure in every case to set the Direction of Initial Impulse so that the projectile 'shoots' straight away from the Earth.
Report below in the first line the maximum distance reached at with initial impulse 4000 (kg m/s)/kg, and the time required before the projectile began to fall back. In subsequent lines report the same for 5000, 6000, 7000, 8000 and 9000 (kg m/s)/kg impulses. In the first line below your data, give a brief explanation of the meaning of the quantities you have reported.
STUDENT QUESTION:
The question from the above box is from the lab Motion in the Gravitational field of the Earth from assignment 29. My question is how to determine the distance traveled? Is it an actual number, such as 7000 meters and if so, does the program show this value, or do we need to calculate it from the velocity and time?
INSTRUCTOR RESPONSE
Good question. There are ways to do this with the program showing the information it does. However for the present it is sufficient to give a good estimate of the distance.
Presumably you have set the 'circle radius' box to 2, as instructed, so that the red circle represents two Earth radii. The greatest distance from the center, for at least the first few impulses, will be between 1 and 2 Earth radii (the starting position is 1 Earth radius). You could estimate how much of the distance between the surface and the radius-2 circle was covered. For example the 4000 kg m/s impulse will take the projectile which is pretty obviously more that 10%, but less than 25%, of the distance. I'll leave it to you to be more specific, but it's clear that the maximum distance is somewhere between 1.10 and 1.25 Earth radii (and I believe it's much closer to one of those numbers than the other). Similar estimates should work for different impulses.
It isn't hard to improve on these estimates, and it doesn't take much time, so if your estimates don't give you satisfying results, you can do the following. Once an impulse and the angles are set, it takes only a couple of seconds to resent the 'circle radius', and only a few trials to nail it down nicely. I just tried this to test it out. An impulse of 4000 didn't get very close to the radius 2 circle; I changed the circle radius to 1.1 and found that the projectile overshot the new circle by a little bit; then tried 1.15 and appear to have gotten lucky. This took less than a minute. So in 5-10 minutes you could get pretty good information, accurate to a least +- .01 Earth radius..
There's no need at this point to figure out how many meters or how many kilometers the projectile moved from the surface. You can simply report results in Earth radii. If you later need to know how many meters or kilometers are involved, just use the fact that Earth radius is about 6.38 * 10^6 meters or 6.38 * 10^3 kilometers.
----->>>>>>>>1 2, 3, 4, 5 and 6 radians
Your answer (start in the next line):
max distance with 4000 (kg m/s) / kg impulse, time to max distance:
infinite =}. Ok so when I did it for the 5 rad it was not happy first it went the greatest distance then the ellipse became smaller then after a few revolutions it shot off never to return. It was 5 rad that went 1.14 rad from the earth or traveled .14 rad. The real time that elapsed was 28 -29 s
max distance with 5000 (kg m/s) / kg impulse, time to max distance:
The distance max is constant at, the time I believe is constant as well. This is true for each portion of the trial. 12.5 = about the average real time min. the max height is
1.26 radii
1.26 1.41 1.66 2.15 2.85
max distance with 6000 (kg m/s) / kg impulse, time to max distance:
16.7 min = average time
Max distance = 1.41 earth radii
max distance with 7000 (kg m/s) / kg impulse, time to max distance:
`dx = 1.66, `dt = 27 min
max distance with 8000 (kg m/s) / kg impulse, time to max distance:
`dx = 2.15, `dt = 37 min
max distance with 9000 (kg m/s) / kg impulse, time to max distance:
`dx = 2.85, `dt = 60.5
interpret meanings: The greater the initial impulse the farther it travels away from the earth.
Also it takes longer to go a greater distance.
your brief discussion/explanation: The greater the initial impulse the higher as object goes. As the object travels farther and farther from the center of the earth the less attraction there is => a longer suspension time in space / the atmosphere and a greater time id the same diminutions.
#$&*
Which initial impulse got the projectile closest to the radius-2 red circle. Estimate the impulse you would need to exactly reach that circle without 'overshooting' it and test your estimate.
Give your results in narrative format below:
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Your answer (start in the next line):
About an intial impulse of 7900 kg * m/s /kg
initial impulse closest to radius-2 circle: 7890
estimate optimal impulse to radius-2 circle:
is this not the same 7890? do not really understand the question here.
your brief discussion/explanation:
Because an initial impulse of 8000 results in a 2.15 radii a little less=> about an impulse of 7900 kg * m/s /kg. The optimum angle to shoot is the same as that of the initial impulse given it shoots it straight up..
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The Experiment:
To start with you will determine the velocity required for a circular orbit at a distance of 1.2 Earth radii.
Set Initial Distance at 1.2, set the initial angular position to 0 and the direction of the initial impulse to 1.5708. Leave the remaining settings as they are and click Run Simulation.
Determine from the shape of the resulting orbit whether the initial velocity is too high or too low and change the number in the Initial Velocity box to a value you believe will bring the orbit closer to a circular shape. The click Run Simulation.
Repeat until you have achieved a good circular orbit (e.g., velocity not varying by more than 1%) and record the velocity.
What is the velocity of the circular orbit at distance 1.2? Report in the first line. How accurate do you think your velocity is--within +- 1 m/s, within +- 10 m/s, within +-100 m/s, etc.? Report in the second line, and in the third line indicate the basis for the accuracy you have claimed:
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Your answer (start in the next line):
velocity of circular orbit at 1.2 Earth radii: Initail Impulse = 7200
how accurate: +- 36 m/s
basis for accuracy estimate: Letting the simulation continue to run allows for a good estimation of the changing velocity. The min velocity = 7200 and the max = 7272 m/s.
I found the diff in the two and divided by two this yeilds the `dv ave. We can then rewrite the velocity as +- 36 m/s
your brief discussion/explanation:
explained above
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Now find the velocity with which a projectile would have to be 'shot' from the surface of the Earth, ignoring air resistance, to get 'up' to the altitude of the orbit you have just created.
Pick a reasonable initial velocity.
Position the initial angular position at 0 radians and make an appropriate selection for the direction of the initial velocity (see the initial introduction to the program above).
Set the Time Factor to about 200 in order to give yourself time to judge when the object has reached is maximum altitude.
Pause the simulation before the object has much time to fall back to Earth. If the projectile falls back to Earth then through to the center you might find the result interesting, but though harmless to the simulation this would not be useful for the experiment.
Repeat but change your initial angular position to 1 radian (and of course change the initial direction of the 'shot' accordingly), and adjust your velocity to get closer to your goal. Adjust the time of the simulation to allow the projectile to stop moving outward and begin falling back. Continue changing the initial angular position and your angular velocity until you manage to just reach the circular orbit you created in the first part of the experiment.
Give your result in the first line below. In the second line estimate the accuracy of your result, and the basis for your estimate. In the first subsequent line indicate the meanings of the quantities you have reported.
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Your answer (start in the next line):
initial velocity to reach your original circular orbit: 8500m/s
accuracy estimate and basis:10 percent
interpretation: the closer to earth the greater the initial impulse must be. Also the best way to get a circular orbit is to give an object an impulse once out in space ST it goes in a circular orbit.
your brief discussion/explanation: I could not get a perfect circular orbit wo adding an impulse after already in orbit. Otherwise whenever launching something I need to plan on it gening slightly eliptical.
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Repeat this procedure for the orbital radius you have been specifically assigned.
If you have not been assigned an orbital radius use 1.6 + .1 * (total number of letters in you first name) + .01 * (number of letters in your middle name) + .001 * (number of letters in your last name).
Determine the velocity necessary to maintain that orbit and the velocity required to achieve the altitude of that orbit.
Report the orbital radius, the velocity necessary to maintain a circular orbit of that radius, and the velocity required to achieve that altitude in the first line below. In the second line estimate the accuracy of your result, and the basis for your estimate:
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Your answer (start in the next line):
your specific orbital radius:2.565
The speed initially is about 9500 m/s. This causes the orbit to extend all the way out to the radius. If I launched it from the initial position equal to the radius then I could get it to be circular, but I am not sure how to get a circular orbit solely from the launch at earth.
accuracy and basis: This is not accurate for it is not what is wanted.
your brief discussion/explanation:
I am not clear on the initial firing position. If it is from the earth I am near convinced it cannot be launched into a perfectly circular orbit.
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Now achieve an elliptical orbit that just skims the surface of the Earth.
Starting from a position at 2 Earth radii from the center, adjust your initial velocity until you have an orbit that at its 'lowest' point just touches the Earth's surface. Observe everything you can about the motion of the satellite in this orbit, including the maximum and minimum velocity.
Report your observations below:
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Your answer (start in the next line):
observations for elliptical orbit from 2 Earth radii to skimming surface:
Initial impulse = about 4575 kg*m/s/kg, v min = 4.5 km/s, v max=9 km/s, real time for orbit = 156.6 min.
your brief discussion/explanation: As it comes [the object] approaches the earth it speed up and as it leaves it slows. The earth has a great effect on its KE. In fact the v min is exactly half of the v max.
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Looks great. We should have fun discussing this tomorrow.
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