- #1

shinobi20

- 267

- 19

- Homework Statement
- Show the relativistic rocket equation, in contrast to the Newtonian case.

- Relevant Equations
- ##E' = \frac{E}{\sqrt{1- v^2 / c^2}}##

##p' = \frac{p}{\sqrt{1- v^2 / c^2}}##

Show that, according to relativistic physics, the final velocity ##v## of a rocket accelerated by its rocket motor in empty space is given by

##\frac{M_i}{M} = \Big ( \frac{c+v}{c-v} \Big) ^ \frac{c}{2 v_{ex}}##

where ##M_i## is the initial mass of the rocket at launch (including the fuel), ##M## is the final mass at burnout, and ##v_{ex}## is the velocity, relative to the rocket, with which the exhaust gases emerge from the nozzle of the rocket motor. (Hint: If the rocket ejects a small mass ##dm## of exhaust, conservation of energy and momentum in the instantaneous rest frame ##x'y'z't'## of the rocket imply

##c^2 dM = \frac{-c^2 dm}{\sqrt{1- v_{ex}^2 / c^2}}, \quad \quad M dv' = \frac{v_{ex} dm}{\sqrt{1- v_{ex}^2 / c^2}}##.

Furthermore, the velocity addition formula implies

##v + dv = \frac{v + dv'}{1 + v dv' / c^2}\Bigg)##

Solution.

I am not sure why the problem went to the instantaneous rest frame of the rocket and from there applied the conservation of energy and momentum, but given that, I will deduce the energy and momentum relations given above.

At some instantaneous time ##t##, the total mass is ##M+dm## and the velocity is zero. After the rocket spews out exhaust with mass ##dm## and velocity ##v_{ex}##, the rocket with mass ##M## moves to the right with velocity ##dv'##, all these with respect to the original instantaneous rest frame of the rocket.

##(M+dm)0 = M dv' - v_{ex} dm \quad \rightarrow \quad M dv' = v_{ex} dm##

for the relativistic case,

##M dv' = \frac{v_{ex} dm}{\sqrt{1- v_{ex}^2 / c^2}}##

The energy of the rocket initially is ##(M+dm) c^2##, and so when the exhaust left the rocket the total energy of the system (rocket plus exhaust) must be the same, but their changes are related by,

##c^2 dM = - c^2 dm##

for the relativistic case,

##c^2 dM = \frac{- c^2 dm}{\sqrt{1- v_{ex}^2 / c^2}}##.

What I am confused about is why the problem stated the velocity addition in that way. I believe ##v## is the velocity of the rocket (the original instantaneous rest frame of the rocket) with respect to the laboratory frame so that when the rocket spewed out the exhaust, it gained a velocity ##dv'## with respect to the instantaneous rest frame of the rocket and so with respect to the lab frame, the velocity of the rocket after exhaust is ##v+dv## in which ##dv## is how the lab frame viewed ##dv'##. I kind of understand and kind of not understanding this. Also, can anyone give me a hint on how to start deriving the original question of showing the mass ratio?

***I know there are other ways of deriving this relation, however, I want to stick with the given problem and probably gain the intuition it wants me to gain from how the problem was stated.

##\frac{M_i}{M} = \Big ( \frac{c+v}{c-v} \Big) ^ \frac{c}{2 v_{ex}}##

where ##M_i## is the initial mass of the rocket at launch (including the fuel), ##M## is the final mass at burnout, and ##v_{ex}## is the velocity, relative to the rocket, with which the exhaust gases emerge from the nozzle of the rocket motor. (Hint: If the rocket ejects a small mass ##dm## of exhaust, conservation of energy and momentum in the instantaneous rest frame ##x'y'z't'## of the rocket imply

##c^2 dM = \frac{-c^2 dm}{\sqrt{1- v_{ex}^2 / c^2}}, \quad \quad M dv' = \frac{v_{ex} dm}{\sqrt{1- v_{ex}^2 / c^2}}##.

Furthermore, the velocity addition formula implies

##v + dv = \frac{v + dv'}{1 + v dv' / c^2}\Bigg)##

Solution.

I am not sure why the problem went to the instantaneous rest frame of the rocket and from there applied the conservation of energy and momentum, but given that, I will deduce the energy and momentum relations given above.

At some instantaneous time ##t##, the total mass is ##M+dm## and the velocity is zero. After the rocket spews out exhaust with mass ##dm## and velocity ##v_{ex}##, the rocket with mass ##M## moves to the right with velocity ##dv'##, all these with respect to the original instantaneous rest frame of the rocket.

##(M+dm)0 = M dv' - v_{ex} dm \quad \rightarrow \quad M dv' = v_{ex} dm##

for the relativistic case,

##M dv' = \frac{v_{ex} dm}{\sqrt{1- v_{ex}^2 / c^2}}##

The energy of the rocket initially is ##(M+dm) c^2##, and so when the exhaust left the rocket the total energy of the system (rocket plus exhaust) must be the same, but their changes are related by,

##c^2 dM = - c^2 dm##

for the relativistic case,

##c^2 dM = \frac{- c^2 dm}{\sqrt{1- v_{ex}^2 / c^2}}##.

What I am confused about is why the problem stated the velocity addition in that way. I believe ##v## is the velocity of the rocket (the original instantaneous rest frame of the rocket) with respect to the laboratory frame so that when the rocket spewed out the exhaust, it gained a velocity ##dv'## with respect to the instantaneous rest frame of the rocket and so with respect to the lab frame, the velocity of the rocket after exhaust is ##v+dv## in which ##dv## is how the lab frame viewed ##dv'##. I kind of understand and kind of not understanding this. Also, can anyone give me a hint on how to start deriving the original question of showing the mass ratio?

***I know there are other ways of deriving this relation, however, I want to stick with the given problem and probably gain the intuition it wants me to gain from how the problem was stated.