View Single Post
  #10  
Old January 19th 18, 03:57 AM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default Riedt’s constant Y = 5022635

On Wednesday, January 17, 2018 at 3:39:24 PM UTC+8, Martin Brown wrote:
On 16/01/2018 23:55, Peter Riedt wrote:
On Wednesday, January 17, 2018 at 5:11:46 AM UTC+8, Paul B. Andersen wrote:
Den 15.01.2018 23.23, skrev Peter Riedt:
Riedt’s constant Y = 5022635

The value of Y connects the nine major planets by their semi major axis
divided by their orbital velocity and divided by their orbital period in earth years.

rm vm opy Y=rm/vm/opy
MER 57,909,231,029 47872.33119 0.2408416 5022635
VEN 108,209,525,401 35020.93000 0.6151856 5022635
EAR 149,598,319,494 29784.86500 0.9999988 5022635
MAR 227,943,771,564 24130.71224 1.8807278 5022635
JUP 778,342,761,465 13056.07169 11.8693454 5022635
SAT 1,426,714,892,866 9643.82000 29.4548244 5022635
URA 2,870,633,540,862 6798.95700 84.0627938 5022635
NEP 4,498,393,012,162 5432.26100 164.8713207 5022635
PLU 5,906,438,090,764 4737.55900 248.2214844 5022635



For any object moving in a circle, we have:

v = 2π⋅r/Ts
whe
v is the orbital speed in metres per second
r is the radius of the circle in metres
Ts is the period in seconds

Riedt’s constant Y = r/v/Ty
where Ty is the period in year

Y = r/(2π⋅r/Ts)/Ty = Ty/(2π⋅Ts)
Ty/Ts = 31556952
Y = 5022445

What's the point with your constant?


--
Paul

https://paulba.no/


What's the point with restating my formula?


He is pointing out that it is a boring numerical identity with a value
that depends only on the measurement units used and nothing else.

--
Regards,
Martin Brown


What else could there be to make it more exciting?