Can anyone help mess around with escape velocities?

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The discussion revolves around the relationship between escape velocity and radius, specifically when both can be equal integers. The user presents calculations using gravitational constants and mass to derive a formula for escape velocity based on radius. They observe a consistent pattern across various celestial bodies, noting that the escape velocity is approximately ten times less than the derived radius-escape-velocity. A debate ensues regarding the validity of equating these two quantities, emphasizing the importance of dimensional analysis in physics. The user seeks further insights and verification of their calculations while expressing a passion for exploring mathematical concepts related to space.
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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
 
Last edited:
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autisticmoose said:
in such my question is when, if ever, will v = r as an INTEGER ONLY NOT DIMENSIONS?

Whenever you want. Since the dimensions don't match, by suitable choice of units you can make this happen whenever you like. Simple example - if v is in units of escape velocities, and r is in units of radii, it happens automatically.
 
really? really? did you even read my post?

look we have 1.1e4 km/s of escape velocity on Earth at surface. the surface is at a radius of 6.378e6. one is smaller than the other! well... the closer to the center of the planet with less radius then the more escape velocity you get. THERE IS A POINT NATURALLY THAT THEY WILL BE EQUAL INTEGERS! it is not all the time only at a certain radius from the center! people! come on! the math is up there, didn't you do the Earth example? what was your answer?!
 
Yes, I read it. It's not physics, since in physics you cannot equate two quantities with different dimensions. I was trying to explain why.
 
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