|
|

Planets and Satellite Orbits: Conservation of Momentum
27 July 2006
|
A couple of interesting facts:
The effective gravity at the equator is lower than the effective gravity at earth's poles
The distance between planets and their Suns are expanding.
The orbit time around a sun is expanding.
Effective Gravity
Effective Gravity at the equator is higher due to the moment arm of the surface relative to the axis of rotation. The gravity of a planet can be estimated as
gravity=gravitational_Constant*planet_massKG/(planet_radiusMeters^2)
However, at the equator rotational momentum reduces the effective gravity on an object. For this reason, space launches require less energy if launched at the equator as opposed to other latitudes.
Expanding Solar System
Planet to Sun Distances Increase Over Time because the sun radiaton of energy results in a slight loss of mass [e=mc^2]. Conservation of energy implies that as the sun emits a particle of mass, the momentum of the particle relative to the sun's rotational axis must increase. So in order to conserve angular momentum, the sun's rotation must slow slightly. [Seemingly similar to the electron quantum states]
In reality, the sun is not a closed system. As a sun loses mass, conservation of momentum can also be accomplished by planets gaining mass and planets increasing orbit distances or axis momentum. For this reason it seems logical that over time the universe will expand to a point where all matter is distributed to the corners of the universe and motion is non-existent. However, the laws observed today cannot be extrapulated to such simple conclusions and a 4 dimensional closed system may not be reality.
Use the following calculator to examine conservation of momentum as well as a number of other planet orbiting characteristics. If you have time, see if you can calculate the effective gravity (at various) latitudes and the earth's time to orbit millions of years ago when the earth and moon may have been joined. (Could planet's have energy states similar to electrons?)
|
|
Earth Moon System: Conservation of Momentum
A couple more interesting facts:
Rocks found on the moon are composed of the same elements as earth's crust.
Natural Prehistoric Nuclear Reactors Existed
If the earth and moon where joined billions of years ago, conservation of momentum would imply a rotational rate of 25 seconds per revolution. (Not possible)
Conservation of Momentum Calculation
Angular momentum is calculated by multiply mass by velocity by distance.
| |
Mass (kg) |
Days To Revolve (earth center) |
Radius (km) (earth center) |
Circumference (km) |
Angular Velocity (km/hr) |
Angular Momentum (kgm^2/s) |
| Moon |
7.32E22 |
27.322 |
384000 |
2412743 |
3637.11 |
2.83869E34 |
| Earth |
5.97E24 |
1 |
5061* |
31799 |
1324.97 |
1.11277E34 |
| Earth/Moon |
6.05E24 |
0.000289*** |
5095** |
32010 |
4617479 |
3.95146E34 |
Notes: Effective radius is the half volume radius. [4/3 PI R^3]
*5061 is 50% of 6376
**5095 is 50% for 6376+1737 [earth+moon]
***0.000289 days/revolution equals 25 seconds per revolution.
If earth and moon were joined billions of years ago, it is unlikely that the earth could spin at a rotational rate of once every 25 seconds. This rate exceeds the escape velocity of the planet. The Kinetic energy could have been increased over time once via natural nuclear reactions in the earth (volcanoes would vent gas). Over time, the planet's surface might reach escape velocity near the equator. When this event occurred, the moon might break off from the earth. As the moon moved farther and farther away from earth, the earth days would lengthen.
Gravity: Physical Size Is Everything
The equation for gravity is:
Gravity = 6.67E-14 * mass / (radius^2)
where units are grams and meters.
Gravity is therefore driven by mass density. A mass with
an infinitely small radius may create a nearly infinite gravitational force at it's surface.
This type of characteristic may be present in black holes. However, the tendency for
oblateness competes with gravity and may decrease densities over time.
Mass and Radius Define Orbit Characteristics
Kepler's third law is:
T^2 = (4 * PI^2) * (R^3) / (gravitational_Constant * M)
which reduces to:
T = 769337.588 * SQRT(R^3 / M) where units are seconds, meters, and kgs.
In additon to defining the orbit period, Kepler's third law indirectly defines orbital velocity:
Orbital_Velocity = (2* PI * R) / T = 8.16701E-06 / SQRT(R / M)
For our solar system, M is the planet's mass plus the sun's mass in kg, and R is the radius of the planet from the sun in meters. Since the mass of the Sun overshadows the mass of any planet, for the sun's solar system the equations can be simplified to:
Days_To_Orbit_Sun = 6.31214E-15 * (Distance_To_Sun_In_Meters)^1.5
Orbital_Velocity_Meters_Per_Second = 1.15E+10 / SQRT(Distance_To_Sun_In_Meters)
| |
Distance From Sun (meters) |
Orbit Period Around Sun (days) |
Orbit Period Around Sun (years) |
Orbital Velocity (meters per second) |
| Mercury |
5.85E10 |
89 |
0.2 |
47547 |
| Venus |
1.08E11 |
224 |
0.6 |
34993 |
| Earth |
1.5E11 |
367 |
1 |
29693 |
| Mars |
2.28E11 |
687 |
1.9 |
24084 |
| Jupiter |
7.80E11 |
4348 |
11.9 |
13021 |
| Saturn |
1.43E12 |
10805 |
29.5 |
9613 |
| Uranus |
2.88E12 |
30899 |
84.3 |
6773 |
| Neptune |
4.51E12 |
60436 |
164.8 |
5416 |
| Pluto |
5.92E12 |
90966 |
248.1 |
4726 |
|
|
|