#$&*
course Phy 231
4/22/11 about 3:30pm
simulation of motion in the gravitational field of the earth#$&*
Phy 231
Your 'simulation of motion in the gravitational field of the earth' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
#$&* Your intial comment: **
#$&* Effect of Impulse 100, impulse 1000, right, left, forward, backward. **
5 hours
#$&* The most efficient way to get off the screen using (100 kg m/s) / kg impulses. **
This experiment uses the program located at
http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/grav_field_simulation_v1b_no_filing.exe
Copy this document into a word processor or text editor.
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Copy the document into a text editor (e.g., Notepad; but NOT into a word processor or html editor, e.g., NOT into Word or FrontPage).
Highlight the contents of the text editor, and copy and paste those contents into the indicated box at the end of this form.
Click the Submit button and save your form confirmation.
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):
Upon hitting “impulse 100” then forward the orbit of the object goes further away from the earth for about half of the revolution. When you hit forward repeatedly the orbit of the object goes further and further out from the earth. “Impulse 1000” makes the orbit of the object increase way out from the earth for more than half of the revolution.
effect of impulse button forward, right, left:
your brief discussion/explanation
#$&*
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.
----->>>>>>>>
Your answer (start in the next line):
31 clicks, I timed my clicks at to be at ½ a rotation starting from the point where the orbit of the object was furthest out from the earth
most efficient way to get off the screen (minimize number of clicks):
#$&*
Start the simulation over by clicking on Run Simulation:
The program shows you the speed of the satellite. The speed is in the neighborhood of 8000, give or take a bit, and is in meters / second. The speed changes slightly during the orbit, and the average speed changes slightly from orbit to orbit.
What is the range of speeds for the default orbit? Does the orbit appear perfectly circular? If you can discern a deviation from a circular orbit, is the speed greater or less when the satellite is furthest from the center of the Earth?
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Your answer (start in the next line):
about 7320m/s to about 8000m/s. The orbit doesn’t appear to be perfectly circular. The speed is greater when the object is orbiting nearer to the earth’s center, less when furthest from the center of the earth.
range of speeds, deviation from circularity of default orbit:
your brief discussion/explanation:
#$&*
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):
about 7320m/s to about 8000m/s before giving it a 1000 impulse, and about 4620m/s to about 9000m/s after.
3660 to 4000 before, 2310 to 4500 after; these are the KE values per kg of mass
The resulting orbit still looks circular, the center of the earth in the simulation appears to be closer to one side of the orbit than the other.
velocity immediately before and after 1000 (kg m/s) / kg impulse:
KE/kg immediately before and after impulse:
shape of orbit, center of earth vs center of orbit:
your brief discussion/explanation:
#$&*
Click Run Simulation again and repeat, this time using a backward impulse of 1000 kg m/s. Note the shape of the new orbit and the maximum and minimum velocities.
Give the max and min velocities in the first line, max and min KE per kg in the second, and in the third line describe the shape of the resulting orbit in terms similar to those used previously.
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Your answer (start in the next line):
About 6900m/s to about 10500m/s
3450 to 5250 KE
The resulting orbit still looks circular, just inside the earth on all but a small part of the rotation.
(backward impulse) max and min orbital velocities:
max and min KE/kg:
shape of orbit:
your brief discussion/explanation:
#$&*
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):
About 6400m/s to 8900m/s
3200 to 4450 KE
The resulting orbit is still circular but shifted to the left of the original orbit
(left impulse) max and min orbital velocities:
max and min KE/kg:
shape of orbit:
your brief discussion/explanation:
#$&*
@& These last few orbits might well appear circular; however the are actually elliptical. The reason the center of the Earth looks closer to one side than the other is that the center of the Earth is at one focus ot the ellipse.
These ellipses are still close enough to circular that your eye will accept them as circular; this is especially the case because your computer screen doesn't necessarily depict circles perfectly and your eye is used to accepting slightly elliptical shapes as circular.*@
Which direction resulted in the greatest change in KE per kg from immediately before to immediately after impulse? Which resulted in the greatest difference between the maximum and the minimum KE of the resulting orbit? What was this difference in KE per kg, and how did you obtain your result?
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Your answer (start in the next line):
The forward direction gives the greatest change and the greatest difference in min and max KE. 2190KE/kg obtained by taking the max value of KE and subtracting the min value of KE from it.
direction resulting in greatest change in KE/kg:
direction resulting in greatest difference between max and min KE/kg of orbit:
difference between max and min KE/kg:
your brief discussion/explanation:
#$&*
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):
KE increase from the 1st was 2190KE/kg, the 2nd was 1585KE/kg. Obtained by KE=1/2mv which I calculated for the min and max values for each, then I subtracted the min from the max and get the amount of increase for each.
two quick impulses, compare KE increase with that of single click:
your brief discussion/explanation:
#$&*
If after applying the first 1000 (m/s)/kg impulse you apply a second identical impulse, at what point in the orbit should it be applied in order to obtain the orbit with the greatest possible maximum KE within the orbit?
Investigate this question and answer below, describing what you did and how it led you to your answer:
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Your answer (start in the next line):
By clicking the impulse 1000 button when the orbit is nearest the earth then waiting for a complete rotation from this point and clicking it again.
at what point in orbit will second impulse result in greatest orbital KE:
your brief discussion/explanation:
#$&*
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):
7830m/s
Shape of 9000kg m/s is circular but goes further out from the earth on the left side, 7000kg m/s is circular but cuts into the left side of the earth
My strategy for finding a good circular orbit was starting at 8000kg m/s then +-250kg m/s, then starting from 7750kg m/s going +10kg m/s until I went past the point of the best orbit. I then backed up 5kg m/s and went at 5kg m/s increments until obtaining the best circular orbit.
initial velocity for good circular orbit at default position:
your brief discussion/explanation:
#$&*
Achieve circular orbits at 1.5 and 2 Earth radii:
• Change number in the init box to 1.5 and click on Run Simulation. What happens to the shape of the orbit?
Change the initial impulse by a reasonable amount, predict what will happen to the orbit, and click again on Run Simulation.
Use trial and error to find the initial impulse required at 1.5 Earth radii to achieve a circular orbit, as nearly as you can reasonably achieve it. A reasonable result would be one in which the velocity doesn't change by more than 1% during an orbit.
Report below the initial impulse required to achieve a circular orbit at 1.5 Earth radii. In the second line give the maximum and minimum velocities observed in that orbit and the difference in these velocities as a percent of the average of the two velocities. If your strategy for obtaining good results has improved you may report this starting in the next line.
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Your answer (start in the next line):
6460m/s
6443m/s to 6461m/s , .28%
impulse for circular orbit at 1.5 Earth radii:
max and min velocities, percent difference:
your brief discussion/explanation:
#$&*
• 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.
----->>>>>>>>
Your answer (start in the next line):
5591m/s
5587m/s to 5595m/s, .14%
impulse for circular orbit at 2.0 Earth radii:
max and min velocities, percent difference:
your brief discussion/explanation:
#$&*
'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 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.
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Your answer (start in the next line):
1.3 earth radii, 1.02s
1.4 earth radii, 1.07s
1.6 earth radii, 1.1s
1.8 earth radii, 1.4s
2.025 earth radii, 2.2s
2.3 earth radii, 3.6s
These are the distances the object went out from the earth with each initial impulse listed with their times to reach this max distance.
max distance with 4000 (kg m/s) / kg impulse, time to max distance:
max distance with 5000 (kg m/s) / kg impulse, time to max distance:
max distance with 6000 (kg m/s) / kg impulse, time to max distance:
max distance with 7000 (kg m/s) / kg impulse, time to max distance:
max distance with 8000 (kg m/s) / kg impulse, time to max distance:
max distance with 9000 (kg m/s) / kg impulse, time to max distance:
interpret meanings:
your brief discussion/explanation:
#$&*
• 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):
8000kg m/s impulse got the closest, 7900kg m/s hits the radius-2 red circle.
initial impulse closest to radius-2 circle:
estimate optimal impulse to radius-2 circle:
your brief discussion/explanation:
#$&*
The Experiment:
To start with you will determine the velocity required for a circular orbit at a distance of 1.2 Earth radii.
Set Initial Distance at 1.2, set the initial angular position to 0 and the direction of the initial impulse to 1.5708. Leave the remaining settings as they are and click Run Simulation.
Determine from the shape of the resulting orbit whether the initial velocity is too high or too low and change the number in the Initial Velocity box to a value you believe will bring the orbit closer to a circular shape. The click Run Simulation.
Repeat until you have achieved a good circular orbit (e.g., velocity not varying by more than 1%) and record the velocity.
What is the velocity of the circular orbit at distance 1.2? Report in the first line. How accurate do you think your velocity is--within +- 1 m/s, within +- 10 m/s, within +-100 m/s, etc.? Report in the second line, and in the third line indicate the basis for the accuracy you have claimed:
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Your answer (start in the next line):
7218m/s
Within +-1m/s
By adding and subtracting 1m/s from 7218m/s and running the simulation over, this shown me that the best circular orbit is 7218m/s
velocity of circular orbit at 1.2 Earth radii:
how accurate:
basis for accuracy estimate:
your brief discussion/explanation:
#$&*
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):
1725m/s
Within +-100m/s because it took a lot more adjustment in velocity to see a change in the distance that the projectile went from the earth. I couldn’t see much change unless I adjusted it by 100m/s or so.
The 1st quantity is the velocity required at a 45 degree angle from the earth to launch a projectile out to the 1.2earth radii point.
initial velocity to reach your original circular orbit:
accuracy estimate and basis:
interpretation:
your brief discussion/explanation:
#$&*
Repeat this procedure for the orbital radius you have been specifically assigned.
If you have not been assigned an orbital radius use 1.6 + .1 * (total number of letters in you first name) + .01 * (number of letters in your middle name) + .001 * (number of letters in your last name).
Determine the velocity necessary to maintain that orbit and the velocity required to achieve the altitude of that orbit.
Report the orbital radius, the velocity necessary to maintain a circular orbit of that radius, and the velocity required to achieve that altitude in the first line below. In the second line estimate the accuracy of your result, and the basis for your estimate:
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Your answer (start in the next line):
2.355, 5153m/s, 350m/s
Within +-1m/s for the maintaining of the orbital radius, and about +-100m/s for the altitude.
your specific orbital radius:
accuracy and basis:
your brief discussion/explanation:
#$&*
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):
The initial velocity is 4575m/s, the velocity is from 4575m/s to about 9080m/s with the greatest velocity being at the point where the object skims the surface of the earth and the smallest velocity being when the object is furthest from the earth
observations for elliptical orbit from 2 Earth radii to skimming surface:
your brief discussion/explanation:
#$&*
Analysis
If you wish to submit your data before conducting your analysis, you may do so, and submit the analysis later. The analysis requires good understanding of the Introductory Problem Sets and other assigned materials.
For each radius you investigated determine, by setting centripetal acceleration equal to gravitational acceleration, the 'actual' velocity for that circular orbit. Refer to the Introductory Problem Sets, your text and other assigned materials. The radii include 1.2 Earth radii, 1.5 Earth radii, 2 Earth radii, and the number you were assigned (or is determined by the number of letters in your name).
• In the first line give the orbital velocity you calculate for 1.2 Earth radii;
• in the second give the same for 1.5 Earth radii;
• in the third give the same for 2.0 Earth radii;
• in the fourth give your assigned orbital radius and the orbital velocity you calculate for that radius.
• Starting in the fifth line explain how you obtained your results
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Your answer (start in the next line):
7230m/s
6466m/s
5600m/s
2.355, 5160m/s
Obtained by using the equation v= sqrt((9.8m/s^2 * rE^2)/r), rE= to 6.4 * 10^6m
Orbital velocity for 1.2 Earth radii:
same for 1.5 Earth radii:
same for 2.0 Earth radii:
assigned orbital radius and calculated orbital velocity:
how did you obtain these results:
your brief discussion/explanation:
#$&*
Compare with the velocities you observed using the simulation. For each orbital radius, give your calculated velocity and your observed velocity, the difference between these quantities and the percent difference. Report in narrative format below, being sure to explain the quantities you report:
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Your answer (start in the next line):
7230m/s, 7218m/s, 12m/s, .17%
6466m/s, 6460m/s, 6m/s, .09%
5600m/s, 5591m/s, 9m/s, .16%
5160m/s, 5153m/s, 7m/s, .14%
1st column #s are the calculated velocities, 2nd column are the observed velocities, 3rd column are the differences between the 2, and the 4th column is the percent difference.
comparison with values obtained in simulation:
your brief discussion/explanation:
#$&*
According to your observations:
• How much KE per kg is required in your initial 'shot' to get to the 'altitude' of each orbit? How much does the PE per kg change as it 'climbs' to each orbit (i.e., what is the difference between the PE of a 1 kg mass at the surface and at the position of the orbit)? Note that PE calculations are discussed in the Introductory Problem Sets.
• What is the KE per kg in each orbit?
• By how much does the PE therefore change between your 'assigned' circular orbit and a circular orbit at 2.0 Earth radii?
• By how much does the KE change between your 'assigned' circular orbit and a circular orbit at 2.0 Earth radii?
• How are the PE and KE changes related?
Give the PE change and the KE change in the first line below, in comma delimited format. Use + for a change in which the quantity at 2.0 Earth radii is greater that at your 'assigned' radius, - for a change in which the quantity at 2.0 Earth radii is less. Starting in the second line give a complete summary of your results and how you determined them.
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Your answer (start in the next line):
125440000=deltaPE, 26049762= deltaKE ??? I’m not sure how to calculate this. . . is using KE=1/2mv^2 and mg(delta h)= PE right?
@& m g h applies to objects in a uniform gravitational field.
The gravitational field of the Earth is not uniform over significant changes in r.
The change in r is significant here. For eacmple a 10% change in r corresponds to about a 20% change in the gravitational field.
You have to use the correct expression for gravitational PE. Relative to inifinity, the gravitational PE at distance R is equal and opposite to the integral from infinity to R of the gravitational force G M m / r^2. That's a straghtforward integral, yielding the result that PE = - G M m / r.*@
@& You will need to recaculate the PE change.*@
&&&&PE= (-6.67*10^-11Nm^2/kg^2)(6*10^24kg)(1kg/(2(6.4*10^6m))=
-3.13*10^13 Joules&&&&
@& 10^-11 * 10^24 / 10^6 is 10^7, nowhere near 10^13.
In other words, right calculation for PE but watch your exponents. I suspect you just accidentally left that 10^6 in the denominator out of your actual calculation*@
@& You have calculated the PE at the 'higher' point. You haven't yet calculated the PE at 1.0 Earth radii, so you don't yet have the change in PE.*@
PE and KE change between assigned orbit and 2.0 Earth radii:
complete summary of calculated results:
your brief discussion/explanation:
#$&*
According to theory:
• How much KE per kg is required in your initial 'shot' to get to the 'altitude' of each orbit? How much does the PE of the projectile change as it 'climbs' to each orbit?
• What is the KE per kg in each orbit?
• By how much does the PE therefore change between the two circular orbits you investigated?
• By how much does the KE change between the two circular orbits you investigated?
• How are the PE and KE changes related?
Give the PE change and the KE change in the first line below, in comma delimited format. Use + for a change in which the quantity at 2.0 Earth radii is greater that at your 'assigned' radius, - for a change in which the quantity at 2.0 Earth radii is less. Starting in the second line give a complete summary of your results and how you determined them.
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Your answer (start in the next line):
125440000=deltaPE, 26136450= deltaKE ??? I’m not sure how to calculate this. . . is using KE=1/2mv^2 and mg(delta h)= PE right?
theoretical PE and KE changes assigned orbit to orbit at 2.0 Earth radii:
summary of results and methods:
your brief discussion/explanation:
#$&*
@& You should refigure your `dPE in light of the preceding.*@
For the elliptical orbit between 2.0 Earth radii and 1.0 Earth radii, based on your observations:
• How does the observed velocity of the satellite at the 'initial' position 2 Earth radii from the center compare to the velocity as it 'skims' the surface?
• What happens to the KE between the extreme furthest distance from center and the closest approach to center?
Report in narrative format below, being sure to explain everything:
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Your answer (start in the next line):
Skim velocity= 4575m/s, initial 2earth radii velocity= 5591m/s, difference of 1016m/s
The KE at the extreme furthest distance will be smallest, and it will increase as it gets closer to the center.
comparison of observed velocity at 2.0 Earth radii and 1.0 Earth radii, within elliptical orbit:
how does KE change:
@& You can quantify the change in KE (i.e., calculate 1/2 m v^2 for each and find the change).*@
&&&&KE= .5(1kg)(4575m/s)^2= 10465312.5Joules (for skim vel.)
KE= .5(1kg)(5591m/s)^2= 15629640.5Joules (for v0)
dKE= 5164328&&&&
your brief discussion/explanation:
#$&*
What happens to the KE between the extreme furthest distance from center and the closest approach to center?
Based on the applicable formulas for PE, by how much the PE change between a position 2.0 Earth radii from the center and 1.0 Earth radii from the center?
• Are your observations for KE and your calculation for PE consistent with the conservation of energy?
Report in narrative format below, again being sure to include explanation.
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Your answer (start in the next line):
The KE will decrease as the object gets further away from the earth, PE will increase as the object gets further from the earth.
(elliptical orbit) theoretical PE change compared to KE change, is conservation of energy illustrated and how:
@& Again you need to caclculate specific results.*@
&&&&-3.13*10^13 Joules- (-6.3*10^13Joules)= 3.17*10^13Joules is dPE
My observations are not consistent with the conservation of energy because the dKE + dPE + dW does not equal 0.&&&&
your brief discussion/explanation:
#$&*
For the elliptical orbit, based on theory:
• By how much should the KE of the satellite change from 'initial' position 2 Earth radii from the center and the point where it 'skims' the surface?
• What therefore should be the velocity of the object as it 'skims' the surface, assuming that the initial velocity you gave it was accurate? How does this velocity compare to the velocity you observed?
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Your answer (start in the next line):
30757877.5Joules= delta KE
11600m/s? difference of 2520m/s?
(elliptical orbit) how much KE change in theory:
theoretical velocity skimming surface, based on observed initial velocity:
@& I could be wrong but those don't appear to be consistent with the correct calculations.
A change of 2500 m/s from 11 600 m/s would imply a starting orbital velocity of about 9 000 m/ s, much higher than the orbital velocity required at that distance.*@
comparison with observations:
your brief discussion/explanation:
#$&*
Speculate on what sorts of strategies are required to get a Space Shuttle into a circular orbit at an 'altitude' of 400 km.
Rocket fuel doesn't carry enough energy to even get its own mass into orbit. How do we get a Shuttle into orbit?
Give your speculations below:
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Your answer (start in the next line):
The shuttle would have to be shot at a great enough velocity from the earth, so that it travels in a hor direction from the surface of the earth and eventually the earths pull will be overcome enough for the object to be in orbit.
strategies to get shuttle into circular orbit:
your brief discussion/explanation:
@& It can't be 'shot' at that speed. All its energy comes from its fuel. It requires multiple stages to get to orbit.*@
#$&*
Speculate on what strategies are required to get a spacecraft to the Moon, which is about 60 Earth radii away, and back. Note that you can if you wish experiment with this situation by following the instructions for 2d Planet; just be careful to set your original parameters so that the Moon is visible.
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Your answer (start in the next line):
Having a velocity great enough to get into orbit, then as the shuttle gets further from the earth the KE of the shuttle decrease and the velocity decreases which means that less energy is required once it gets into orbit for it to reach the moon.
strategy to get to Moon:
your brief discussion/explanation:
#$&*
*#&!
@& Everything else looks good, but your energy calculations in the last part of the experiement are not all correct. You are using the wrong expression for gravitational PE.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you.
&#
*@"
Self-critique (if necessary):
------------------------------------------------
Self-critique rating:
@& This might not all post, but here's a brief summary of my comments:
Your expression for calculating PE was right, but 10^-11 * 10^24 / 10^6 = 10^7, nowhere near 10^13. I suspect you accidentally left the 10^6 out of your calculation.
To get the change in PE you have to calculate PE at the two distances and subtract.
Your KE calculations appear to be correct.
Recalculate your PE change for the elliptical orbit. It should be negative, and the KE change should be equal and opposite the PE change.
Between two circular orbits, the KE change is half the magnitude of the PE change, so you can't get from one orbit to the other without either expending or dissipating a lot of energy.
*@