Science > Physics > Hey Potter--What about my question relating to tompotter.us/units.html
| Topic: |
Science > Physics |
| User: |
"Sam Wormley" |
| Date: |
08 Aug 2003 01:17:41 AM |
| Object: |
Hey Potter--What about my question relating to tompotter.us/units.html |
Hey Potter--What about my question relating to http://tompotter.us/units.html
Looking at
http://tompotter.us/units.html
"It should be noted that the prime function of the constant "C" is to
differentate between time periods and time intervals, and the prime
function of the constant "G" is to differentate between objects
perceived to be fixed in media (Time and space) and objects perceived
to be varying in media. For example, in the Sun/Earth system, the
Earth is perceived to be varying in media, and it gets the 365.25
days and 93,000,000 miles, while the Sun is perceived to be fixed in
media, and it gets the "G" to balance the equation.
mass(Sun) * G = space(Earth)^3 / time period(common)^2
Expressing space as a time interval we have:
mass(Sun) * G / C^3 = time interval(Earth)^3 / time period(common)^2
Note that G/C^3 is a universal time per mass constant and that mass
can be expressed more fundamentally as a time.
The holding of one object as fixed in media comes about due to the
limitations of two body math".
Kepler's Third Law
http://scienceworld.wolfram.com/physics/KeplersThirdLaw.html
Kepler's Third Law can be in the form of n^2 a^3 = G M, where where n
is the mean motion of an orbiting body, a its semimajor axis G the
gravitational constant, and M the mass. If a is measured is
astronomical units, T is measured in years (so that n = 2 pi/T is
measured in inverse years, and and the central mass is the sun's
mass.
The more general for (independent of units) is the square of the
orbital period, T^2 = (2 pi)^2 a^3 / G M
Tom--I think what you are "trying" to say is totally wrong and that you
do not understand Kepler's Third Law. Can you clarify your equations
for me?
.
|
|
|
Related Articles |
|
|
