Re: Relativistic rocket formulae

On Nov 21, 6:31 pm, mluttg...@xxxxxxxxxx wrote:
On Nov 21, 5:04 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-





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On Nov 18, 2:08 pm, mluttg...@xxxxxxxxxx wrote:

On Nov 18, 7:39 pm, mluttg...@xxxxxxxxxx wrote:

On Nov 18, 7:06 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-

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[snip]

I am wondering if it can be derived from SR.

Can somebody help? Thanks!

Nobody has ever been able to help you in the past, so
no, probably not.

Dirk Vdm

Iow, you are unable to find such formula.
All you can do is parroting SR textbooks.
Hopefully, true SR experts will be eager to
meet this challenge!

The hallmark of Marcel Trollgens

Dirk Vdm

I am not interested in trolling, but only in the testing of
the limits of SR.
Anyhow, I think that the theoretical problem I posed
can be resolved by a mathematician well versed in SR, which is clearly
not your case.

Marcel Luttgens

Van de moortel, you proved your incompetence
by choosing alt.troll and alt.local.village.idiot,
the NGs where you belong.

I am not interested in trolling, but only in the testing of
the limits of SR.
Anyhow, I think that the theoretical problem I posed can be
resolved by a mathematician well versed in SR, which is clearly
not your case.

Marcel Luttgens- Hide quoted text -

- Show quoted text -

For actual accelerationIf a body has constant acceleration in its own
rest frame, the

velocity at time T, with respect to the initial rest frame, will be

V= aT'
_____________
SQRT( 1 + ((aT'/c)^2))

the body's time, T' = c/a ln (aT/c + SQRT(1 +aaTT/cc).

An equivalent formula, easier to work with on the calculator on your
computer, is
T= sinh aT' , X = cosh aT'

where a is the acceleration in light years/year,
T is earth time, and X-1 is distance traveled from earth's
persepctive.

sinh is the hyperbolic sine, cosh is the hyperbolic cosine.

0.97g is about 1 light year/year, so letting a rocket accelerate at 1
light year/year for 1 "rocket year,
T= sinh 1 = 1.175 earth years.
The rocket will have traveled cosh(1) -1 = 1.54-1 =0.54 light years.

And its velocity will be 0.707 light speed.
A. McIntire- Hide quoted text -

- Show quoted text -

If a body has constant acceleration in its own rest frame, the

velocity at time T, with respect to the initial rest frame, will be

V= aT
_____________
SQRT( 1 + ((aT/c)^2))

let v be the velocity in frame S, let v' be the velocity in frame S',
which is moving at velocity V with respect to frame S.

using the addtion of velocities rule, v= (v' + V)/(1+(v'V/c*c))
t= (t'
+(Vx'/c*c))/SQRT(1-(vv/cc)

then dv= dv'(1-(VV/cc))/((1+(v'V/cc))^2)
dt=dt'(1+(Vv'/cc))/(SQRT(1-((V^2)/(c^2))

dv/dt= (dv'/dt')((1- (VV/cc))^3)/((1+(Vv'/cc)) where dv'/dt' is
the acceleration in S'.
Let S' be the instantaneous co-moving system of point P, then v'=0,
V=v.
If dv'/dt'=a, then dv/dt=a((1-(VV/cc)^3).

let v=0 when t=0.
then (dv/((1-(V^2/c^2))^3))=adt
integrate both sides and get
v/SQRT(1-(V^2/c^2)) =at

which can be rearranged to give v=dx/dt=at/SQRT(1 +(aatt/cc)), Q.E.D

Sorry, but I do not find the formula allowing
to solve the following problem:

A rocket has an initial velocity v = 0.95056 c.
What would be its velocity after a trip of
1 ly, during which it is subject to a constant
deceleration of 9.81 m/s^2 ?
Note that in this problem, the acceleration
is negative.

Marcel

No SR expert can solve this elementary problem?

Here is a clue: the residual valocity = 0.8894 c

"Sorry, but I do not find the formula allowing
to solve the following problem"

I told you that you were trolling.
It turns that you weren't even aware of it.
You were even lying.
If you wouldn't be so stupid, it could make you think.
A stupid liar... an essentially harmless combination.

Dirk Vdm

What is your solution, wise guy?

Marcel Luttgens

.

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