Jump to content

stargazine_ep34_banner.thumb.jpg.28dd32d9305c7de9b6591e6bf6600b27.jpg

Recommended Posts

I have discovered a fundamentally new type of relationship of planetary orbits, which can be expressed by the following formula:

Ratio = SQRT(((R1+R2)/(2*R1))^3)

Substituting in this formula values of orbital radius of the Earth R1=1a.u. (by definition) and R2=0,723 – orbital radius of the Venus, we get 4 / 5 - relations tiny integers. If we substitute values R1=5,203 and R2=0,723 (the Jupiter and the Venus), we get 3 / 7. Substituting values R1=9,539 (the Saturn) and R2=30,06 (the Neptune), we obtain 2.99 , which close to integer - 3. This golden formula works not only for the planets, but also for satellites of giant planets.

Link to post
Share on other sites

At first sight it seems, that the formula is not meaningful. However it at all so. The formula has a precise physical sense. That it to comprehend, we shall consider in the beginning widely widespread phenomenon - orbital resonance http://en.wikipedia.org/wiki/Orbital_resonance.

According to Kepler's third law of the attitude of the periods of planets it is possible to find under the following formula:

T2/T1 = SQRT ((R2/R1) ^3)

For example, having substituted in formula R1=5,203 (The Jupiter) and R2=9,539 (The Saturn), we shall receive value 2,48 extremely close to 2,5=5/2.

My formula turns to the formula of orbital resonances if we replace R2 on (R1+R2)/2.

Expression (R1+R2)/2 is a semi-major axis of the elliptic orbit of an intermediate particle. We shall designate it R2. Then my formula will coincide with the formula for usual orbital resonances.

Link to post
Share on other sites

didnt Kepler have some idea of plantetary orbital ratios that he tried to demonstrate in some sort of elaborate silver drinking vessel that he made for the king of somewhere only he didn't have a budget for silver so he made it out of paper. Or something? :o is this like that?

Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
  • Recently Browsing   0 members

    No registered users viewing this page.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue. By using this site, you agree to our Terms of Use.