| t Time | y exact | y Euler DELt=1 | y Euler DELt=0.01 |
|---|---|---|---|
| 0 | 0.00 | 0.00 | 0.00 |
| 1 | 1.96 | 2.00 | 1.96 |
| 2 | 3.84 | 3.92 | 3.84 |
| 3 | ***** | ***** | ***** |
| 4 | ***** | ***** | ***** |
| 5 | ***** | ***** | ***** |
| 6 | ***** | ***** | ***** |
| 7 | ***** | ***** | ***** |
| 8 | ***** | ***** | ***** |
| 9 | ***** | ***** | ***** |
| 10 | ***** | ***** | ***** |
| 20 | ***** | ***** | ***** |
| 30 | ***** | ***** | ***** |
| 40 | ***** | ***** | ***** |
| 50 | ***** | ***** | ***** |
| 60 | ***** | ***** | ***** |
| 70 | ***** | ***** | ***** |
| 80 | ***** | ***** | ***** |
| 90 | ***** | ***** | ***** |
| 100 | ***** | ***** | ***** |
| 200 | 49.98 | 49.99 | 49.98 |
| t Time | x Amt in T1 | y Amt in T2 |
|---|---|---|
| 0 | 0.00 | 50.00 |
| 1 | ****** | ****** |
| 2 | ****** | ****** |
| 3 | ****** | ****** |
| 4 | ****** | ****** |
| 5 | ****** | ****** |
| 6 | ****** | ****** |
| 7 | ****** | ****** |
| 8 | ****** | ****** |
| 9 | ****** | ****** |
| 10 | ****** | ****** |
| 20 | ****** | ****** |
| 30 | ****** | ****** |
| 40 | ****** | ****** |
| 50 | ****** | ****** |
| 60 | ****** | ****** |
| 70 | ****** | ****** |
| 80 | ****** | ****** |
| 90 | ****** | ****** |
| 100 | ****** | ****** |
| 200 | ****** | ****** |
| 300 | ****** | ****** |
| K Speed |
a Semimajor Axis |
e Eccentricity |
P Period |
a3 / P2 Constant |
|---|---|---|---|---|
| 28000 | 6600 | 0.00 | . | . |
| 36000 | . | . | . | . |
| 38000 | . | . | . | . |
| 39000 | . | . | . | . |
Kepler's Laws
Extension of problem 3. Add to your program in problem 3 the
gravitational acceleration of the moon. For the moon
It desires to go into circular synchronous orbit (magenta) 42160 km above
the
center of the earth. To do this, it must increase its velocity (bottom red
point) to go into
the elliptical transfer orbit (blue) and then increase its velocity again
(top red point) at the
circular synchronous orbit to go into that orbit. Determine by simulation
(this means try various velocities until you find the correct one) needed
to enter the transfer orbit and then the velocity needed to enter the
circular synchronous orbit. Your work in project 3 should be helpful to
this problem. The black half of the transfer orbit is not
traveled. A synchronous orbit is
23 hours 56 minutes measured in
mean solar time (24 h sidereal
time = 23 h 56 m MST).This site has a good but short biography of Issac Newton plus links to the giants
upon whose shoulders he stood: Descartes, Galileo, Copernicus, Kepler, Brahe, Halley, Euclid, Archimedes
"
If I have been able to see further, it was only because I stood on the shoulders of
giants."
Letter to Robert Hooke
|
|
|
| Year | World
Population Millions |
|---|---|
| 1650 | 545 |
| 1750 | 728 |
| 1800 | 906 |
| 1850 | 1171 |
| 1900 | 1608 |
| 1950 | 2509 |
| 1960 | 3010 |
| 1970 | 3611 |
| 2000 | 6000 |
|
|
|
|
| Food M | Food N | Requirements | |
|---|---|---|---|
| Calcium | 30 units/oz | 10 units/oz | 360 units |
| Iron | 10 units/oz | 10 units/oz | 160 units |
| Vitamin A | 10 units/oz | 30 units/oz | 240 units |
| Cholesterol | 8 units/oz | 4 units/oz | Minimize |
| Shipping Costs | . | Store 1 | Store 2 |
|---|---|---|
| Warehouse A | 5 $/ton | 8 $/ton |
| Warehouse B | 6 $/ton | 10 $/ton |
| Shipping Costs | . | Store 1 | Store 2 |
|---|---|---|
| Warehouse A | 8 $/ton | 5 $/ton |
| Warehouse B | 6 $/ton | 10 $/ton |
|
****** |
|
***** |
|
||||||||||||||||||||||||||||||
| Transition | Probability |
|---|---|
| G0 to G0 | .85 |
| G0 to G1 | .10 |
| G0 to G2 | .05 |
| G1 to G0 | .15 |
| G1 to G1 | .75 |
| G1 to G2 | .10 |
| G2 to G0 | .10 |
| G2 to G1 | .30 |
| G2 to G2 | .60 |