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Sunday, May 22, 2011

Ejs Open Source Newton's Mountain Projectile Orbits Model java applet « on: May 20, 2011, 11:44:32 pm »

Ejs Open Source Newton's Mountain Projectile Orbits Model java applet
update: 14 june 2913
http://weelookang.blogspot.sg/2011/05/ejs-open-source-newtons-mountain.html
with firing and retarding thrusters to get in circular orbit or otherwise

https://dl.dropboxusercontent.com/u/44365627/lookangEJSworkspace/export/ejs_NewtonsMountainwee06.jar
author: timberlake and lookang




http://weelookang.blogspot.sg/2011/05/ejs-open-source-newtons-mountain.html
white out background for inserting  pictures into tutorials with y = 4.23E07 m and vi = 3070.98 m/s
https://dl.dropbox.com/u/44365627/lookangEJSworkspace/export/ejs_NewtonsMountainwee.jar

author: timberlake and lookang

 white out background for inserting  pictures into tutorials with y = 4.23E07 m and vi = 3070.98 m/s


new interface with controls all at the bottom


 27 May 2011 version with better Newton's mountain picture thanks to leongster, x control selectable

23 May 2011 version with multiple representation of versus time and radius from center of Earth
« on: May 20, 2011, 11:44:32 pm »


Ejs Open Source Newton Mountain Model java applet by Todd Timberlake remixed by lookang
Ejs Open Source Newton's Mountain Projectile Orbits Model java applet
reference:
http://www.compadre.org/osp/items/detail.cfm?ID=9391
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=398.msg7167#msg7167


Simulation above is kindly hosted by NTNUJAVA Virtual Physics Laboratory by Professor Fu-Kwun Hwang
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=2204.0
alternatively, go direct to http://www.phy.ntnu.edu.tw/ntnujava/index.php?board=28.0
Collaborative Community of EJS (Moderator: lookang) and register , login and download all of them for free :) This work is licensed under a Creative Commons Attribution 3.0 Singapore License
Author: Todd K. Timberlake, this remixed version is by lookang


General Description by Todd K. Timberlake (ttimberlake@berry.edu)
This simulation illustrates the motion of a projectile launched from the top of a VERY tall mountain on Earth. The diagram shown in the simulation is taken from Newton's A Treatise on the System of the World, which he wrote after the Principia, but the basic idea is found in the Principia itself. Newton concluded that a projectile launched horizontally with sufficient speed would orbit Earth rather than crashing to Earth's surface. Thus the motion of a projectile fired on Earth was not qualitatively different from that of the moon orbiting Earth.
The simulation allows the user to adjust the initial speed and launch angle of the projectile using sliders. The projectile is launched from the top of the mountain (note that the mountain shown in the diagram is ridiculously tall - such a mountain would stick out above Earth's entire atmosphere). The motion of the projectile is calculated using Newton's Second Law of motion and Newton's Universal Law of Gravitation.
The user has the option to allow the projectile to pass through Earth. The Earth is treated either as a homogeneous sphere (the default) or as a point mass located at its center (if selected in the Model Options menu). The homogeneous sphere is more realistic, but using the point mass can help illustrate the "true" shape of the projectile's path before it hits Earth.

Newton's Mountain
Display Options Menu
Show Center of Earth: show a red dot at Earth's center.
Show Velocity Vector: show a blue arrow representing the projectile's velocity.
Show Force Vector: show a red arrow representing the force on the projectile (or the projectile's acceleration, since its mass is an arbitrary constant).
Trace Projectile's Path: trace the path followed by the projectile.
Show Newton's Diagram: show the mountain diagram in the background.
Model Options Menu
Let Projectile Pass Through Earth: allows the projectile to pass into and through Earth, treating Earth as either a homogeneous sphere or a point mass.
Treat Earth As Point Mass: model Earth as a point mass. If not selected, Earth is modeled as a homogeneous sphere.
Visual Elements
The background is a diagram from Newton's A Treatise on the System of the World showing Earth with a VERY tall mountain at the top. The lines on the diagram show the paths of projectiles launched from the mountain top, or satellites (moons) orbiting Earth.
Green disk: the projectile.
Red spot: the center of Earth, labeled C in the diagram.
Blue arrow: represents the projectile's current velocity.
Red arrow: represents the force on the projectile, (or the projectile's current acceleration).
Magenta trace: trace of the object's path.
Controls
Play/Pause: start and stop the simulation.
Step: advance the simulation by one time step.
Reset: reset the simulation to its initial state.
Clear Traces: clears all traces of the object's path.
Initialize: read in values and place the projectile back on the mountain top. This will NOT clear any existing traces, but it will start a new trace unconnected to the old traces. Note that any changes to the parameter values will not take effect until the Initialize button is clicked.
Initial Speed: sets the initial speed of the object (in km/s).
Launch Angle: sets the launch angle for the projectile (in degrees). The angle is measured counterclockwise with zero degrees corresponding to a horizontal launch to the right.


changes by lookang:
made sliders and fields conform to my design preference
redo for SI units for closer to sg syllabus instead of arbitrary units originally made
22May2011
add earth picture
add panel of x,y,vx,vy,ax,ay for calculating values to coincident with Earth data
add trace instead of the older trail that can display older data
rotate newton's diagram by 5 degree to coincident with x=0, instead of the older 0.1*R
add time
change to acceleration instead of force



question.
what is the velocity of the projectile for circular motion to occur if it is launch horizontally at the surface of the Earth?

hint: use circular motion equation to find the v!

enjoy!

answer:
7920.05 m/s

try it with the this customized with Earth data simulation :)

A newton's mountain with editing done by leongster, thanks bro!



added more content:


Orbits round the Earth
An object projected horizontally near the Earth’s surface follows a parabolic trajectory. As the velocity of projection increases, there will be an occasion in which the trajectory follows the curvature of the Earth’s surface. If air resistance is negligible, the object will orbit round the Earth and will never meet the Earth’s surface.
speed of cannonball at 0 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory that hits Earth surface.

speed of cannonball at 1 000 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory that hits Earth surface.

speed of cannonball at 2 000 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory that hits Earth surface.

speed of cannonball at 3 000 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory that hits Earth surface.

speed of cannonball at 4 000 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory that hits Earth surface.

speed of cannonball at 5 000 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory that hits Earth surface.

speed of cannonball at 6 000 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory that hits Earth surface.

speed of cannonball at 7 000 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory that does NOT hit the Earth surface.

speed of cannonball at 7 300 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory orbit that does NOT hit the Earth surface that is a approximately a circular orbit.

speed of cannonball at 8 000 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory orbit that does NOT hit the Earth surface.

speed of cannonball at 9 000 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory orbit that does NOT hit the Earth surface.

speed of cannonball at 11 200 km/hr launched horizontally from newton's very tall mountain, resulting in trajectory that does NOT hit the Earth surface and escape from the Earth's gravitational pull


plot of kinetic energy  E_{K}, gravitational potential energy,  E_{P} and total Energy  Eversus distance away from centre of earth, r at R= Re

plot of kinetic energy  E_{K}, gravitational potential energy,  E_{P} and total Energy  Eversus distance away from centre of earth, r at R= Re, R= 2*Re

plot of kinetic energy  E_{K}, gravitational potential energy,  E_{P} and total Energy  Eversus distance away from centre of earth, r at R= Re, R= 2*Re, R=3*Re

plot of kinetic energy  E_{K}, gravitational potential energy,  E_{P} and total Energy  Eversus distance away from centre of earth, r at R= Re, R= 2*Re, R=3*Re and lastly R = geostationary radius


Question01

An earth satellite of mass 200 kg lost energy slowly through atmospheric resistance and fell from an orbit of radius 8.0 x 106 m to 7.8 x 106 m. Calculate the changes in the potential, kinetic and total energies of the satellite during this transition period.
Answers: ( -2.57E8 J, 1.28E8 J, -1.28E8 J ) where E denotes x10

Solution:
a)
\Delta PE\ = - G \frac{M m}{r_2}\ - ( - G \frac{M m}{r_1}\ ) \Delta PE\ = - 6.67x10^{-11} \frac{(6.0x10^{24}) (200)}{7.8x10^6}\ - ( - 6.67x10^{-11} \frac{(6.0x10^{24}) (200)}{8.0x10^6}\ ) \Delta PE\ = -2.57x10^8 J
b)
Assuming the orbits are in circular motion as it decays from r_1 = 8.0 x 10^6 to r_2 = 7.8 x 10^6
equation  G \frac{M m}{r^2}\ =  \frac{m v^2}{r}\ can be assumed to be valid to approximate this motion
Thus, \Delta KE\ =\frac{1}{2}\ m v_2^2 - \frac{1}{2}\ m v_1^2
\Delta KE\ =\frac{1}{2}\ mG \frac{M}{r_2}\ - \frac{1}{2}\ mG \frac{M}{r_1}\

\Delta KE\ =\frac{1}{2}\ (200)(6.67x10^{-11}) \frac{6.0x10^{24}}{7.8x10^6}\ - \frac{1}{2}\ (200)(6.67x10^{-11}) \frac{6.0x10^{24}}{8.0x10^6}\
\Delta KE\ = 1.28x10^8 J
c)
since  TE = PE + KE
 TE = -2.57x10^8 + 1.28x10^8
 TE = - 1.28x10^8 J

check:
This answer  TE = - 1.28x10^8 J can also be verified from the formula relationship on page Newton's Mountain Projectile Orbits which is not intended to be memorized.

Using the simulation:


using R = 8.0X10^6 , v = 7072.8 m/s for circular motion orbit, you can get for satellite m = 1kg, KE = 2.501X10^7 J, PE = -5.002X10^7 J and TE = -2.501X10^7 J


using R = 7.8X10^6 , v = 7162.9 m/s for circular motion orbit, you can get for satellite m = 1kg, KE = 2.565X10^7 J, PE = -5.131X10^7 J and TE = -2.565X10^7 J
 Thus,  
ΔKE = 200(2.565X10^7 - 2.501X10^7) = 1.28X10^8 J

ΔPE = 200(-5.131X10^7 - (-5.002X10^7)) = -2.58X10^8 J

ΔTE = 1.28X10^8 + (-2.58X10^8) = 1.30X10^8 J

where the slight difference is due to carry over error from the significant figures of the computer model. 





reference:
http://commons.wikimedia.org/wiki/File:R_%3D_geo_Re_2012-10-08_1809.png#.7B.7Bint:filedesc.7D.7D








also contributed to wikimedia media commons
http://en.wikipedia.org/wiki/Newton%27s_cannonball#The_experiment


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10:15, 8 October 2012R=Re 2012-10-08 1808.png (file)151 KBLookang(User created page with UploadWizard)
10:15, 8 October 2012R = geo Re 2012-10-08 1809.png (file)47 KBLookang(User created page with UploadWizard)
10:15, 8 October 2012R = 3Re 2012-10-08 1809.png (file)79 KBLookang(User created page with UploadWizard)
10:15, 8 October 2012R = 2Re 2012-10-08 1809.png (file)127 KBLookang(User created page with UploadWizard)
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