Investigation of time relativity with respect to Tsiolkovski formula

Tsiolkovski Formula

m1 v 2 − v1 = v p .Log e m2

∆v = v p .LN . mm12

∆v

m1 v =e p > m2

m p = m1 − m2

>

m p = m1 − m1 .e

∆v −   vp m p = m1 .1 − e   

∆v

m1 v =e p m2

Why is this rate of change of mass ratio so important What is the corelation with

∆v r vp

Using an external reference point of an observer on the surface As the Intertial time reference. ∆v v r 2 − v r1 = vp vp

 L T Local =   L T  Earth

  =   

 L   TEarth   TEarth     =  TLocal   L   TLocal 

−

∆v vp

 L T =  Local  L T  Earth

   L  T T  ∆v v r 2 − v r1 Earth  = =  Earth  =   vp vp   TLocal   L   TLocal    Then what if at a certain point the time locally at the rocket [TLocal ] slowed relative to the reference time on Earth [TEarth ].

At the Tsiolkovski Point where a specific mass flowrate and rocket velocity creates an energy condition where

[TLocal ] = [TEarth ] and

therefore the value of e

After this point the value of e ∆v r increases. vp

TEarth TLocal

TEarth TLocal

= 2.7182 .

will increase as [TLocal ] slows and the relative value of

That is a condition exists where the value of

∆v r  TEarth  =  ≥1 vp T Local  

If a similar dimensional analysis of specific impulse is undertaken then an expression for

Impulse Specific Impulse Isp in seconds

I sp

Fth = m& p .g

secs

 M L  1 .T 2 Local Becomes   M . L 2 T  Earth TEarth

     

2 2  M   L  T   T  T  T  =    2   E   E  =  E2   E   1   TL   M   L   T L   1 

Then a reduction in [TLocal ] would also alter Isp

For this condition to exist the rate of change of mass of the projectile has to create its own space time horizon. John Gwynn

The Water Rocket Explorer