The purpose of this program is to simulate the movement of three objects m0, m1 and m2 by using Newton' Law under 13 different conditions. The simulations are subdivided in 7 groups.
test group m0 m1 m2 ellipse oblateness 1 1 100000 1000 1 2 2 10000 1000 10 3 2 10000 1000 10 m1 4 3 1000 1000 10 5 3 1000 1000 10 m2 6 3 1000 1000 10 m1 7 4 1000 1000 10 8 4 1000 1000 10 m2 9 5 1000 .1 20 10 5 1000 .1 20 m1 11 6 100 .01 0 m1 0 12 6 100 .01 0 m1 1 13 7 100 100 1
The movement of the objects of the tests 1,2,4,7,9 and 13 involve straight lines and circles. They are part of CHAPTER 2. The movement of the tests 3,5,6,8,10,11 and 12 involve ellipses and are more complex. They are part of CHAPTER 5
Test 1 consists of:
One star m0. Mass of m0 = 100000 One planet m1. Mass of m1 = 1000 One moon m2. Mass of m2 = 1 Distance between m0 and m1 is large: 5000 Distance between m1 and m2 is small: 10
Moon m2 revolves around the planet m1. Planet m1 has no speed in the y direction i.e. planet m1 will move in a straight line towards the star m0.
From the Test Selection Display
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Test 2 consists of:
One big star m0. Mass of m0 = 10000 One smaller star m1. Mass of m1 = 1000 A planet m2 revolving around the star m1. Mass of m2 = 1 Distance between m0 and m1 is large: 500. Distance between m1 and planet is small: 10.
| | .............................. | | m0 | ............ | | m1 m2
This simulation is similar as m0 representing our sun, m1 the earth and m2 our moon.
What the simulation shows is that the revolution time of the moon is constant and approximate 6.28 seconds.
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Test 3 is identical as test 2.
This simulation is similar as :
What the simulation shows is that the revolution time of the moon is not
constant:
The revolution time of the moon m2 increases (i.e. the moon moves slower)
when the speed of the planet m1 increases.
The revolution time of the moon m2 decreases (i.e. the moon moves faster)
when the speed of the planet m1 decreases.
After one full revolution of Mercury the revolution time of the moon is the
same.
The results are made visible in figure 3 for delta time of 0.00002
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Test 4 consists of:
One double star m0 and m1. m0 = 1000, m1 = 1000 A planet m2 revolving around the star m1. m2 = 10 Distance between m0 and m1 is large i.e. 500. Distance between m2 and double star is small i.e. 50.
| | .............................. | | m0 | ............ | | m1 m2
This simulation shows the behaviour of our sun through our galaxy. m0 represent our galaxy, m1 is the sun and m2 is the earth.
What the simulation shows is that the revolution time of the earth is constant and approximate 24.82 seconds.
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Test 5 is identical as Test 4.
This simulation is almost identical as test 3. The major difference it that there are 2 planets similar as Mercury. One of the planets has a moon which services as a clock.
The results are made visible in figure 5.
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Test 6 is identical as test 4.
This simulation shows the behaviour of our sun through our galaxy. m0 represent our galaxy, m1 is the sun and m2 is the planet Mercury.
The simulation shows that the major axis of the trajectory of the planet m2 moves forward.
The following forward angles are observed after one, two and three revolutions of m0 and m1:
The number in brackets indicates the number of revolutions of m2 (Mercury)
This means after approximate 120 revolutions of m0 and m1 that the major axis of the planet m2 has also made one complete revolution.
The results are made visible in figure 6.
Test 7 consists of
One double star m0 and m1. m0 = 1000, m1 = 1000 A planet m2 revolving around the double star. m2 = 10 Distance between m0 and m1 is small i.e. 160. Distance between m2 and double star is large i.e. 500.
| | .............................. | | ............ m2 | | m0 m1
What the simulation shows is that the revolution time of the planet m2 is constant and approximate 1547 seconds.
Test 8 is identical as test 7.
This simulation shows that the major axis of the trajectory of the planet m2 moves forward.
Observed values are : 6.2, 13.1, 19.9, 26.7, 33.0, 38.7 degrees.
The results are made visible in figure 8
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Test 9 consists of:
one star m0 m0 = 1000 one planet m1 m1 = .1 distance m0 and m1 = 110 one planet m2 m2 = 10 distance m0 and m2 = 200
| | .............................. | | ................... m2 | | m0 m1
The purpose of this test is to show how the revolution time of one planet m1 is influenced by an outer planet m2. The trajectory of both planets is a circle.
The tests consists of two parts.
This simulation is similar as m0 being our Sun, m1 the Earth and m2 one of the outer planets.
What the simulation shows is that the revolution time of the Earth is constant. approximate 226 seconds.
What the simulation shows is that the revolution time of the Earth is constant. approximate 229 seconds.
Comparing part 1 with part 2 shows that when a planet has a close companion, the revolution time becomes shorter i.e. its speeds increases.
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Test 10 is identical as test 9. m0 is our Sun
The simulation shows that the major axis of the trajectory of the planet m1 moves forward.
The angle is irregular. After each revolution those values are:
The results are made visible in figure 10
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The purpose of this demonstration is to test if the shape of m0 has
any influence on the trajectory of m1.
In this demonstration the oblateness of m0 = 0 i.e. round
Test 11 consists of:
one star m0 m0 = 100 one planet m1 m1 = .01 distance m0 and m1 = 60
| | .............................. | | m0 m1This simulation is similar as m0 being our Sun and m1 the planet Mercury.
The simulation shows no forward movement of the major axis of the planet when oblateness = 0
The purpose of this demonstration is to test if the shape of m0 has
any influence on the trajectory of m1.
In this demonstration the oblateness of m0 = 1 i.e. not round
This simulation shows that the major axis of the trajectory of the planet m1 moves forward.
For oblateness of 1 the angles are respectively : 67.2, 134.4 etc.
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Test 13 consists of
The purpose of this test is to see how one planet behaves when it is attracted towards a binary star system.
The result of the demonstration show that the planet is finally ejected with high speed.
In order to see all the combinations in a range over 180 degrees perform the following test:
What this demonstrations "proves" that (only?) from a binary system high objects can be ejected at high speeds. This could be two stars from which a third star is ejected or two galaxies from which a third galaxy is ejected. The speed of this third galaxy is not in agreement with Hubble's Law i.e. that the speed has a linear relation with its distance.
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The program uses two standard features:
From the Parameter Selection Display the following parameters can be changed:
0 = Select test display1 = Set standard parameters.
2 = Screen mode. Valid values are 7,8,9 and 12. Standard value = 9 3 = Directory name. Standard name is C:\NOW\FIG
4 = Wait time in second. Physical wait time between each simulation cycle. Standard value = 0 5 = Speed of light. Standard value is 10
6 = Delta time in seconds between each calculation cycle. Standard value is .02 7 = Origin Body. Standard value = 0
8 = Eccentricity. Standard value = 0.3
9 = Distance between m0 and m1. Standard value = 1000
10 = Display condition. -1 means once each revolution of Mercury x means after each x calculation cycles
11 = Save condition 0 means no file save 1 means file save of results
12 = End Condition -1 no end x means after x revolutions of Mercury
13 = Sub Test. Sub test are used to select a specific command file 0 = no sub test 1 = Test with special condition 2 4 = Test with Mercury and Venus 5 = Test with Mercury, Venus and Earth. 6 = Test with Mercury and all the planets except Pluto.
14 = Oblatness of Sun. Standard value = 1
15 = Angle alpha for test 13. Standard value = 0
16 = # of calculation cycles saved. Standard value = 0 0 means no calculation value saved.
m0-------->a0 a1<-------m1 <--------------------r-------------------> <----------r0------->X<--------r1-------->
The force between the two masses m0 and m1 is described by Newton's Law (G=1):
m0 * m1 F = ------- = m0 * a0 = m1 * a1 r²m1 m0 a0 = -- a1 = -- r² r²
r0 is distance m0 to center of mass X r1 is distance m1 to center of mass X
r0 + r1 = r r0 * m0 = r1 * m1 r0 * m0 = (r - r0) * m1 r0 * (m0 + m1) = r * m1 r * m1 r * m0 r0 = ------- r1 = ------- m0 + m1 m0 + m1For a circle with radius r0
v0² v0² * (m0 + m1) m1 a0 = --- = --------------- = -- r0 r * m1 r²
m1² v0² = ------------- r * (m0 + m1)
1 v0 = m1 * sqrt { ------------- } r * (m0 + m1)
1 v1 = m0 * sqrt { ------------- } r * (m0 + m1)
For example : m0 = 1000 m1 = 0 distance = 10
1 1000 v0 = 0 v1 = 1000 * sqrt { --------- } = ---- = 10 10 * 1000 100
2 * pi * r 2 * pi * 10 T0 = 0 T1 = ---------- = ----------- = 2 * pi = 6.28 v1 10
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