The relativistic photon rocket equation

Chris wrote:

Hello,

I got a bit of a problem, my navigation computer has a power supply problem
and I need to do some manual calculations, perhaps could you help.

The force on a rocket is the mass change per second times the velocity of
the exhaust, cdm/dt

The force is balanced by the reaction due to acceleration m(dv/dt)

so cdm/dt=m(dv/dt) that is the ordinary rocket equation.

But the momentum according to the special theory of relativity is:

m(v/(1-(v/c)^2)^(1/2)) so the rate of change of momentum is:

m(d/dt(v/(1-(v/c)^2)^(1/2))
m(duv)=m(vdu-udv) u=v v=(1-(v/c)^2)^(-1/2)

rate of change of momentum
is:=(dv/dt)(1-(v/c)^2)^(-1/2)-v(-(1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2)dv/dt

so making the rocket equation
cdm=((1-(v/c)^2)^(-1/2)-v((1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2))dv Integrating
this gives the mass ratio for a particular velocity in 3-space that you can
measure.

To find the 4-velocity you just use V=v/(1-(v/c)^2)(1/2)

The journey time is D/V where D is the ordinary distance measured at rest.

Oh by the way we all live in hyperspace, as we go faster things get mixed
up. The faster we go the quicker we get there!

Well go on then work it out and tell me how I get home. How much hydrogen do
I have to collect from Jupiter and manufacture into mercury to give enough
mass ratio, assuming my payload is 20,000 Tonnes. It is 1000 light years
and I've got ten days.

Thanks, Chris.

You should submit this question to the saucerheads, they know
everything.

--
Official Associate AFA-B Vote Rustler
Official Overseer of Kooks and Saucerheads in alt.astronomy
Co-Winner, alt.(f)lame Worst Flame War, December 2005

"I am a sean being from another planet."
-- Darla aka Dr. Why aka Dr. Yubiwan aka Silouen aka ...

Art Deco wrote:
Chris wrote:

Hello,

I got a bit of a problem, my navigation computer has a power supply problem
and I need to do some manual calculations, perhaps could you help.

The force on a rocket is the mass change per second times the velocity of
the exhaust, cdm/dt

The force is balanced by the reaction due to acceleration m(dv/dt)

so cdm/dt=m(dv/dt) that is the ordinary rocket equation.

But the momentum according to the special theory of relativity is:

m(v/(1-(v/c)^2)^(1/2)) so the rate of change of momentum is:

m(d/dt(v/(1-(v/c)^2)^(1/2))
m(duv)=m(vdu-udv) u=v v=(1-(v/c)^2)^(-1/2)

rate of change of momentum
is:=(dv/dt)(1-(v/c)^2)^(-1/2)-v(-(1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2)dv/dt

so making the rocket equation
cdm=((1-(v/c)^2)^(-1/2)-v((1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2))dv Integrating
this gives the mass ratio for a particular velocity in 3-space that you can
measure.

To find the 4-velocity you just use V=v/(1-(v/c)^2)(1/2)

The journey time is D/V where D is the ordinary distance measured at rest.

Oh by the way we all live in hyperspace, as we go faster things get mixed
up. The faster we go the quicker we get there!

Well go on then work it out and tell me how I get home. How much hydrogen do
I have to collect from Jupiter and manufacture into mercury to give enough
mass ratio, assuming my payload is 20,000 Tonnes. It is 1000 light years
and I've got ten days.

Thanks, Chris.

You should submit this question to the saucerheads, they know
everything.


He'll never make it!

Double-A

"Double-A" wrote in message
oups.com...

Art Deco wrote:
Chris wrote:

Hello,

I got a bit of a problem, my navigation computer has a power supply
problem
and I need to do some manual calculations, perhaps could you help.

The force on a rocket is the mass change per second times the velocity
of
the exhaust, cdm/dt

The force is balanced by the reaction due to acceleration m(dv/dt)

so cdm/dt=m(dv/dt) that is the ordinary rocket equation.

But the momentum according to the special theory of relativity is:

m(v/(1-(v/c)^2)^(1/2)) so the rate of change of momentum is:

m(d/dt(v/(1-(v/c)^2)^(1/2))
m(duv)=m(vdu-udv) u=v v=(1-(v/c)^2)^(-1/2)

rate of change of momentum
is:=(dv/dt)(1-(v/c)^2)^(-1/2)-v(-(1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2)dv/dt

so making the rocket equation
cdm=((1-(v/c)^2)^(-1/2)-v((1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2))dv
Integrating
this gives the mass ratio for a particular velocity in 3-space that you
can
measure.

To find the 4-velocity you just use V=v/(1-(v/c)^2)(1/2)

The journey time is D/V where D is the ordinary distance measured at
rest.

Oh by the way we all live in hyperspace, as we go faster things get
mixed
up. The faster we go the quicker we get there!

Well go on then work it out and tell me how I get home. How much
hydrogen do
I have to collect from Jupiter and manufacture into mercury to give
enough
mass ratio, assuming my payload is 20,000 Tonnes. It is 1000 light
years
and I've got ten days.

Thanks, Chris.

You should submit this question to the saucerheads, they know
everything.


He'll never make it!

Double-A

Can he make it back to the hospital in which he lives, at least?

"Ugly Bob" wrote in message
...

"Double-A" wrote in message
oups.com...

Art Deco wrote:
Chris wrote:

Hello,

I got a bit of a problem, my navigation computer has a power supply
problem
and I need to do some manual calculations, perhaps could you help.

The force on a rocket is the mass change per second times the velocity
of
the exhaust, cdm/dt

The force is balanced by the reaction due to acceleration m(dv/dt)

so cdm/dt=m(dv/dt) that is the ordinary rocket equation.

But the momentum according to the special theory of relativity is:

m(v/(1-(v/c)^2)^(1/2)) so the rate of change of momentum is:

m(d/dt(v/(1-(v/c)^2)^(1/2))
m(duv)=m(vdu-udv) u=v v=(1-(v/c)^2)^(-1/2)

rate of change of momentum
is:=(dv/dt)(1-(v/c)^2)^(-1/2)-v(-(1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2)dv/dt

so making the rocket equation
cdm=((1-(v/c)^2)^(-1/2)-v((1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2))dv
Integrating
this gives the mass ratio for a particular velocity in 3-space that you
can
measure.

To find the 4-velocity you just use V=v/(1-(v/c)^2)(1/2)

The journey time is D/V where D is the ordinary distance measured at
rest.

Oh by the way we all live in hyperspace, as we go faster things get
mixed
up. The faster we go the quicker we get there!

Well go on then work it out and tell me how I get home. How much
hydrogen do
I have to collect from Jupiter and manufacture into mercury to give
enough
mass ratio, assuming my payload is 20,000 Tonnes. It is 1000 light
years
and I've got ten days.

Thanks, Chris.

You should submit this question to the saucerheads, they know
everything.


He'll never make it!

Double-A

Can he make it back to the hospital in which he lives, at least?

Are you SharonB or what?

"Real Friendly Neighborhood Vote Ranger" wrote in
message
...

"Ugly Bob" wrote in message
...

"Double-A" wrote in message
oups.com...

Art Deco wrote:
Chris wrote:

Hello,

I got a bit of a problem, my navigation computer has a power supply
problem
and I need to do some manual calculations, perhaps could you help.

The force on a rocket is the mass change per second times the velocity
of
the exhaust, cdm/dt

The force is balanced by the reaction due to acceleration m(dv/dt)

so cdm/dt=m(dv/dt) that is the ordinary rocket equation.

But the momentum according to the special theory of relativity is:

m(v/(1-(v/c)^2)^(1/2)) so the rate of change of momentum is:

m(d/dt(v/(1-(v/c)^2)^(1/2))
m(duv)=m(vdu-udv) u=v v=(1-(v/c)^2)^(-1/2)

rate of change of momentum
is:=(dv/dt)(1-(v/c)^2)^(-1/2)-v(-(1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2)dv/dt

so making the rocket equation
cdm=((1-(v/c)^2)^(-1/2)-v((1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2))dv
Integrating
this gives the mass ratio for a particular velocity in 3-space that
you can
measure.

To find the 4-velocity you just use V=v/(1-(v/c)^2)(1/2)

The journey time is D/V where D is the ordinary distance measured at
rest.

Oh by the way we all live in hyperspace, as we go faster things get
mixed
up. The faster we go the quicker we get there!

Well go on then work it out and tell me how I get home. How much
hydrogen do
I have to collect from Jupiter and manufacture into mercury to give
enough
mass ratio, assuming my payload is 20,000 Tonnes. It is 1000 light
years
and I've got ten days.

Thanks, Chris.

You should submit this question to the saucerheads, they know
everything.


He'll never make it!

Double-A

Can he make it back to the hospital in which he lives, at least?

Are you SharonB or what?

I'm what.

"Ugly Bob" wrote in message
. ..

"Real Friendly Neighborhood Vote Ranger" wrote in
message
...

"Ugly Bob" wrote in message
...

"Double-A" wrote in message
oups.com...

Art Deco wrote:
Chris wrote:

Hello,

I got a bit of a problem, my navigation computer has a power supply
problem
and I need to do some manual calculations, perhaps could you help.

The force on a rocket is the mass change per second times the
velocity of
the exhaust, cdm/dt

The force is balanced by the reaction due to acceleration m(dv/dt)

so cdm/dt=m(dv/dt) that is the ordinary rocket equation.

But the momentum according to the special theory of relativity is:

m(v/(1-(v/c)^2)^(1/2)) so the rate of change of momentum is:

m(d/dt(v/(1-(v/c)^2)^(1/2))
m(duv)=m(vdu-udv) u=v v=(1-(v/c)^2)^(-1/2)

rate of change of momentum
is:=(dv/dt)(1-(v/c)^2)^(-1/2)-v(-(1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2)dv/dt

so making the rocket equation
cdm=((1-(v/c)^2)^(-1/2)-v((1/2)(1-(v/c)^2)^(-3/2)(-2v/c^2))dv
Integrating
this gives the mass ratio for a particular velocity in 3-space that
you can
measure.

To find the 4-velocity you just use V=v/(1-(v/c)^2)(1/2)

The journey time is D/V where D is the ordinary distance measured at
rest.

Oh by the way we all live in hyperspace, as we go faster things get
mixed
up. The faster we go the quicker we get there!

Well go on then work it out and tell me how I get home. How much
hydrogen do
I have to collect from Jupiter and manufacture into mercury to give
enough
mass ratio, assuming my payload is 20,000 Tonnes. It is 1000 light
years
and I've got ten days.

Thanks, Chris.

You should submit this question to the saucerheads, they know
everything.


He'll never make it!

Double-A

Can he make it back to the hospital in which he lives, at least?

Are you SharonB or what?

I'm what.

Are you twhat?