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v=(gr2)^.5

i assume that g is directly related to r as g = Gm/r^2

in such I started to wonder when, if ever, will v = r as an INTEGER ONLY NOT DIMENSIONS?

v = ( 2 * g * r )^.5

v = ( 2 * G * ( m / r^2 ) * r )^.5

v = ( 2 * G * m / r )^.5

v * r^.5 = ( 2 * G * m )^.5

REMEMBER! v=r at some point NATURALLY

r^1.5 = ( 2 * G * m )^.5

(r^1.5)^(2/3) = ( 2 * G * m )^.5

r^1 = ( ( 2 * G * m )^.5 )^(2/3)

r = ( 2 * G * m )^(1/3)

then r =(G*m*2)^(1/3) ---- NOT IN DIMENSIONS ONLY TO FIND NUMERICAL VALUE!

now re-calc based on new r with previous calculations and for every planet in out solar system (including the sun) your v should be an equal integer as r. meaning with the above r = you then put that number in for r in the equation v=(2gr)^.5. by doing so you will result in the same numerical value for v as you have for r.

earth statistics:

m = 5.9742e24 mass of earth

r = ( 2 * G * m )^(1/3) NUMERICAL distance from center

G = 6.673e-11 gravitational constant

g = Gm/r^2 acceleration of gravity

v = (2gr)^.5 escape velocity

r = ( 2 * 6.673e-11 * 5.9742e24 )^(1/3) = 9.2728e4

v = ( 2 * (6.673e-11 * 5.9742e24 / 9.2728e4^2) * 9.2728e4 )^.5 = ?

whats your answer? I bet it's 9.2728e4

the premise is how can you relate this nifty radius-escape-velocity thing to something else celestially bound. first thing i noticed was the 11 km/s of escape velocity required for the surface of earth was about 10 times less than the radius-escape-velocity (when v=r) of earth (9.27278718604467680471147551902e4). actually 8.30. then i tested on the rest of our solar system and received a similar variance. about 10ish off from the Sun to Pluto and every planet in between.

(Radius-Escape-Velocity) / ~10 = (Surface-Escape-Velocity)

Sun's surface escape velocity times 10.4 = (Radius-Escape-Velocity)

Mercury's surface escape velocity times 8.31 = (Radius-Escape-Velocity)

Venus' surface escape velocity times 8.36 = (Radius-Escape-Velocity)

Earth's surface escape velocity times 8.29 = (Radius-Escape-Velocity)

our Moon's surface escape velocity times 9.01 = (Radius-Escape-Velocity)

Mars' surface escape velocity times 8.74 = (Radius-Escape-Velocity)

Jupiter's surface escape velocity times 10.6 = (Radius-Escape-Velocity)

Saturn's surface escape velocity times 11.8 = (Radius-Escape-Velocity)

Uranus' surface escape velocity times 10.6 = (Radius-Escape-Velocity)

Neptune's surface escape velocity times 10.2 = (Radius-Escape-Velocity)

and Pluto's surface escape velocity times 11.0 = (Radius-Escape-Velocity)

can others assist with any other patterns or relations? I'm just messing around and enjoy math and space. also can someone make sure i didn't fat finger it, please double check my work. I am wishing to define any parameters about this radius-escape-velocity. For instance if this is true then the radius-escape-velocity ALWAYS happens below the surface of a planet/star.

i am really hoping for an intelligent discussion as i am getting a lot of people on other sites that can't even begin to comprehend this. and please don't say... r doesn't equal (2gr)^.5. i know! but if you actually read what i posted then you'll find that is NOT what i am stating. sorry and thanks, i just like messing around with math and space so... is this a waste of time? probably, but its fun for me and I'm looking to see if someone else can have fun with me.

Thank you,

other men who seek one thing

often search to find it

but a man who seeks most anything

leaves other men behind him

-AutisticMoose

i assume that g is directly related to r as g = Gm/r^2

in such I started to wonder when, if ever, will v = r as an INTEGER ONLY NOT DIMENSIONS?

v = ( 2 * g * r )^.5

v = ( 2 * G * ( m / r^2 ) * r )^.5

v = ( 2 * G * m / r )^.5

v * r^.5 = ( 2 * G * m )^.5

REMEMBER! v=r at some point NATURALLY

r^1.5 = ( 2 * G * m )^.5

(r^1.5)^(2/3) = ( 2 * G * m )^.5

r^1 = ( ( 2 * G * m )^.5 )^(2/3)

r = ( 2 * G * m )^(1/3)

then r =(G*m*2)^(1/3) ---- NOT IN DIMENSIONS ONLY TO FIND NUMERICAL VALUE!

now re-calc based on new r with previous calculations and for every planet in out solar system (including the sun) your v should be an equal integer as r. meaning with the above r = you then put that number in for r in the equation v=(2gr)^.5. by doing so you will result in the same numerical value for v as you have for r.

earth statistics:

m = 5.9742e24 mass of earth

r = ( 2 * G * m )^(1/3) NUMERICAL distance from center

G = 6.673e-11 gravitational constant

g = Gm/r^2 acceleration of gravity

v = (2gr)^.5 escape velocity

r = ( 2 * 6.673e-11 * 5.9742e24 )^(1/3) = 9.2728e4

v = ( 2 * (6.673e-11 * 5.9742e24 / 9.2728e4^2) * 9.2728e4 )^.5 = ?

whats your answer? I bet it's 9.2728e4

the premise is how can you relate this nifty radius-escape-velocity thing to something else celestially bound. first thing i noticed was the 11 km/s of escape velocity required for the surface of earth was about 10 times less than the radius-escape-velocity (when v=r) of earth (9.27278718604467680471147551902e4). actually 8.30. then i tested on the rest of our solar system and received a similar variance. about 10ish off from the Sun to Pluto and every planet in between.

(Radius-Escape-Velocity) / ~10 = (Surface-Escape-Velocity)

Sun's surface escape velocity times 10.4 = (Radius-Escape-Velocity)

Mercury's surface escape velocity times 8.31 = (Radius-Escape-Velocity)

Venus' surface escape velocity times 8.36 = (Radius-Escape-Velocity)

Earth's surface escape velocity times 8.29 = (Radius-Escape-Velocity)

our Moon's surface escape velocity times 9.01 = (Radius-Escape-Velocity)

Mars' surface escape velocity times 8.74 = (Radius-Escape-Velocity)

Jupiter's surface escape velocity times 10.6 = (Radius-Escape-Velocity)

Saturn's surface escape velocity times 11.8 = (Radius-Escape-Velocity)

Uranus' surface escape velocity times 10.6 = (Radius-Escape-Velocity)

Neptune's surface escape velocity times 10.2 = (Radius-Escape-Velocity)

and Pluto's surface escape velocity times 11.0 = (Radius-Escape-Velocity)

can others assist with any other patterns or relations? I'm just messing around and enjoy math and space. also can someone make sure i didn't fat finger it, please double check my work. I am wishing to define any parameters about this radius-escape-velocity. For instance if this is true then the radius-escape-velocity ALWAYS happens below the surface of a planet/star.

i am really hoping for an intelligent discussion as i am getting a lot of people on other sites that can't even begin to comprehend this. and please don't say... r doesn't equal (2gr)^.5. i know! but if you actually read what i posted then you'll find that is NOT what i am stating. sorry and thanks, i just like messing around with math and space so... is this a waste of time? probably, but its fun for me and I'm looking to see if someone else can have fun with me.

Thank you,

other men who seek one thing

often search to find it

but a man who seeks most anything

leaves other men behind him

-AutisticMoose

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