# Finding the velocity of a relativistic rocket

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Homework Statement:
I am trying to solve a relativistic rocket equation. The rocket dumbes fuel ##N## times, in the ##-x## direction, with a mass of ##\varepsilon m_0##, where ##m_0## is the rest mass of the rocket. The rocket only moves in the ##+x## direction and initally its at rest. The question asks the final speed of the rocket.
Relevant Equations:
Momentum conservation
Since initally the rocket at rest I wrote

$$\vec{p}^R_i = (m_0, 0,0,0)$$

and at final situation

$$\vec{p}^R_f = [m_0-\varepsilon m_0 N](\gamma, v\gamma, 0 ,0)$$
$$\vec{p}^F_f = [\varepsilon m_0 N](\gamma', -u\gamma', 0 ,0)$$

After equating them I get

$$1 = \gamma - \varepsilon N\gamma + \varepsilon N\gamma'$$
$$0 = v\gamma - \varepsilon N\gamma v - \varepsilon N\gamma'u$$

Where ##v## is the speed of the rocket, ##u## is the speed of the fuel, ##\gamma = \gamma(v)##, ##\gamma' = \gamma(u)##.

Now the question has three parts

##(a)ε = 1/2, N = 1##
##(b) ε = 1/4, N = 2 ##
##(c) ε \ll 1, N \gg 1, εN = 1/2##

For me it seems all of them are equal. But I cannot be sure. Either they are indeed equal since in all cases ##\varepsilon N = 1/2##, or I made a mistake in the calculations and they cannot be equal.

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If I fire my rocket in a short pulse, what is the velocity of my exhaust in my original rest frame? If I fire another short pulse, what is the velocity of my second exhaust pulse in my original rest frame?

Arman777
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If I fire my rocket in a short pulse, what is the velocity of my exhaust in my original rest frame? If I fire another short pulse, what is the velocity of my second exhaust pulse in my original rest frame?
It will be different I guess. Let me call the velocity of the exhaust pulse, ##u_1## and the velocity of the rocket ##v_1##. I can write another momentum conservation equation.

$$\vec{p}^R_i = [m_0-\varepsilon m_0](\gamma, v\gamma, 0 ,0)$$
$$\vec{p}^F_i = [\varepsilon m_0](\gamma', -u\gamma', 0 ,0)$$

and

$$\vec{p}^R_f = [m_0-2\varepsilon m_0 ](\gamma_1, v_1\gamma_1, 0 ,0)$$
$$\vec{p}^F_f = [2\varepsilon m_0 ](\gamma' + \gamma'_1, -u\gamma' -u_1\gamma'_1, 0 ,0)$$

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What are each of those gammas in terms of the ##u##s and ##v##s?

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What are each of those gammas in terms of the ##u##s and ##v##s?
##\gamma_\alpha = \frac{1}{\sqrt{1-v_\alpha^2}}## and ##\gamma'_\alpha = \frac{1}{\sqrt{1-u_\alpha^2}}##

such that

##\gamma = \frac{1}{\sqrt{1-v^2}}##, ##\gamma_1 = \frac{1}{\sqrt{1-v_1^2}}##
##\gamma' = \frac{1}{\sqrt{1-u^2}}##, ##\gamma'_1 = \frac{1}{\sqrt{1-u_1^2}}##

mitochan
In your equation mass is conserved. Not mass but energy as a component of 4-momentum vector is conserved. So let mass of ejected fuel be ##\epsilon_f\ m##.

Initial 4-momentum vector is
$$m(1,0,0,0)$$
where m has dimension of energy or mc^2.
For first ejection of fuel with relative velocity of given v to the rocket, the rocket gets speed of u
Rocket 4-momentum:
$$m(1-\epsilon)\gamma_u(1,u,0,0),u>0$$
Fuel 4-momentum:
$$m\epsilon_f \gamma_v (1,v,0,0),v<0$$
where u and v are dimensionless velocity divided by c.

Conservation of 4-vector
$$(1-\epsilon)\gamma_u+\epsilon_f \gamma_v=1$$
$$(1-\epsilon)\gamma_u u+ \epsilon_f \gamma_v v=0$$

By deleting ##\epsilon_f## we get the relation
$$(1-\epsilon)(v-u)\gamma_u=v$$
which gives u(##\epsilon##,v) implicitly.

For the second ejection similarly

$$(1-2\epsilon)(v-u_2)\gamma_{u_2}=v(1-\epsilon)$$

The speed of rocket in original IFR where the rocket was at rest after the second fuel ejection is
$$U_2=\frac{U_1+u_2}{1+U_1 u_2}$$ where
$$U_1 = u$$ which of cource is the speed of rocket in original IFR where the rocket was at rest after the first fuel ejection

For the third one

$$(1-3\epsilon)(v-u_3)\gamma_{u_3}=v(1-2\epsilon)$$
The speed of rocket in original IFR where the rocket was at rest after the second fuel ejection is
$$U_3=\frac{U_2+u_3}{1+U_2 u_3}$$

So in general

$$(1-n\epsilon)(v-u_n)\gamma_{u_n}=v[1-(n-1)\epsilon]$$ or
$$(1-\frac{u_n}{v})\gamma_{u_n}=1+\frac{1}{1-n\epsilon}$$
The speed of rocket in original IFR where the rocket was at rest after the n th fuel ejection is
$$U_n=\frac{U_{n-1}+u_n}{1+U_{n-1} u_n}$$

After all the ejection

$$(1-N\epsilon)(v-u_N)\gamma_{u_N}=v[1-(N-1)\epsilon]$$
The speed of rocket in original IFR where the rocket was at rest after the last N th fuel ejection is
$$U_N=\frac{U_{N-1}+u_N}{1+U_{N-1} u_N}$$

I am not confident on the above calculation so would appreciate corrections.

Last edited:
Arman777
Gold Member
In your equation mass is conserved. Not mass but energy as a component of 4-momentum vector is conserved. So let mass of ejected fuel be ##\epsilon_f\ m##.

Initial 4-momentum vector is
$$m(1,0,0,0)$$
where m has dimension of energy or mc^2.
For first ejection of fuel with relative velocity of given v to the rocket, the rocket gets speed of u
Rocket 4-momentum:
$$m(1-\epsilon)\gamma_u(1,u,0,0)$$
Fuel 4-momentum:
$$m\epsilon_f \gamma_v (1,v,0,0)$$
where u and v are dimensionless velocity divided by c.

Conservation of 4-vector
$$(1-\epsilon)\gamma_u+\epsilon_f \gamma_v=1$$
$$(1-\epsilon)\gamma_u u+ \epsilon_f \gamma_v v=0$$

By deleting ##\epsilon_f## we get the relation
$$(1-\epsilon)(v-u)\gamma_u=v$$
which gives u(##\epsilon##,v) implicitly.

For the second ejection similarly

$$(1-2\epsilon)(v-u_2)\gamma_{u_2}=v(1-\epsilon)$$

The speed of rocket in original IFR where the rocket was at rest after the second fuel ejection is
$$U_2=\frac{U_1+u_2}{1+U_1 u_2}$$ where
$$U_1 = u$$ which of cource is the speed of rocket in original IFR where the rocket was at rest after the first fuel ejection

For the third one

$$(1-3\epsilon)(v-u_3)\gamma_{u_3}=v(1-2\epsilon)$$
The speed of rocket in original IFR where the rocket was at rest after the second fuel ejection is
$$U_3=\frac{U_2+u_3}{1+U_2 u_3}$$

So in general

$$(1-n\epsilon)(v-u_n)\gamma_{u_n}=v[1-(n-1)\epsilon]$$
The speed of rocket in original IFR where the rocket was at rest after the n th fuel ejection is
$$U_n=\frac{U_{n-1}+u_n}{1+U_{n-1} u_n}$$

After all the ejection

$$(1-N\epsilon)(v-u_N)\gamma_{u_N}=v[1-(N-1)\epsilon]$$
The speed of rocket in original IFR where the rocket was at rest after the last N th fuel ejection is
$$U_N=\frac{U_{N-1}+u_N}{1+U_{N-1} u_N}$$

I am not confident on the above calculation so would appreciate corrections.
I am not sure that is the correct answer. The question has three parts for ##\varepsilonN = 1/2##. But I guess similar ideas can be applied.

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Hey all, I realized my mistake and here is the final equations that I find. So initally I have a rest rocket and its easy to write is momentum.

$$\vec{p}_i = (m_0, 0, 0 , 0)$$.

Later on I thought that each of the fuel - particles has a mass $\varepsilon m_0$ mass and the rocket has a final velocity $v$. However each of these fuel particles will not have same velocity. So I think I can write something llke this for final momentum

$$\vec{p}_f^R = (m_0 - N\varepsilon m_0)(\gamma, v\gamma, 0,0)$$
$$\vec{p}_f^F = (\varepsilon m_0)(\sum_{i=1}^N \gamma'_i, -\sum_{i=1}^N u_i\gamma'_i, 0,0)$$ where ##\gamma'_i = (1-u_i^2)^{-1/2}## and ##\gamma = (1-v^2)^{-1/2}##

So If I set $$K = \sum_{i=1}^N \gamma'_i$$ and $$L=\sum_{i=1}^N u_i\gamma'_i$$ I can write,

$$1 = (1-N\varepsilon)\gamma + \varepsilon K$$
$$0 = (1-N\varepsilon)v\gamma - \varepsilon L$$

From here I can write ##v## as

$$v = \frac{\varepsilon L}{(1-N\varepsilon)\gamma} = \frac{\varepsilon L}{(1-K\varepsilon)}$$

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@Arman777 : Just to make sure everyone is interpreting the problem in the same way, would you mind typing the problem statement exactly as given (word for word)?

Is this a problem from a textbook?

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A rocket of rest mass ##m_0## is at rest in an inertial frame O. It ejects a ﬁxed fraction of it's initial mass ##εm_0##, N times in −x direction with relative speed u. What will be its ﬁnal speed if
(a)..
(b)..
(c)..
Is this a problem from a textbook?
Nope its not

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For me, there is some ambiguity in the phrase "with relative speed u". Does the ejected mass have a speed of u relative to the frame of the rocket before the mass is ejected or does the ejected mass have a speed of u relative to the frame of the rocket after the mass is ejected?

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For me, there is some ambiguity in the phrase "with relative speed u". Does the ejected mass have a speed of u relative to the frame of the rocket before the mass is ejected or does the ejected mass have a speed of u relative to the frame of the rocket after the mass is ejected?
How can it have a relative speed before its ejected ? When its ejected it will have a relative velocity $u$

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How can it have a relative speed before its ejected ? When its ejected it will have a relative velocity $u$
In what reference frame is ##u## measured? Is it the "initial" rest frame of the rocket (just before ejection) or the "final" rest frame of the rocket (just after ejection)? Either interpretation would be ok with me, but it wasn't clear (to me), from the wording of the problem, which interpretation is intended.

It appears to me that you are setting up your equations under the assumption that ##u## is the speed of the ejected material in the initial rest frame of the rocket. So, u is not the speed of the ejected material relative to the rocket after the material is ejected.

It is also important to note that if ##m_0## is the initial rest mass of the rocket and if mass ##\varepsilon m_0## is the rest mass of the material ejected, then the rest mass of the rocket after ejection is not ##m_0 - \varepsilon m_0##. The rest mass of the rocket after ejection is an unknown that can be determined from the conservation equations.

Arman777
Gold Member
It appears to me that you are setting up your equations under the assumption that uuu is the speed of the ejected material in the initial rest frame of the rocket.
Well yes
It is also important to note that if m0m0m_0 is the initial rest mass of the rocket and if mass εm0εm0\varepsilon m_0 is the rest mass of the material ejected, then the rest mass of the rocket after ejection is not m0−εm0m0−εm0m_0 - \varepsilon m_0.
How so ?

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Suppose you have two blocks each of rest mass ##m_0## traveling at equal speeds ##v## toward each other so that they collide head-on, stick together, and come to rest. The rest mass of the final object is not ##2 m_0##. The initial kinetic energy of the two blocks contributes to the final rest mass. Rest mass is not conserved (in general) in an inelastic collision.

The ejection of mass from the rocket is like an inelastic collision in reverse.

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Suppose you have two blocks each of rest mass ##m_0## traveling at equal speeds ##v## toward each other so that they collide head-on, stick together, and come to rest. The rest mass of the final object is not ##2 m_0##. The initial kinetic energy of the two blocks contributes to the final rest mass. Rest mass is not conserved (in general) in an inelastic collision.

The ejection of mass from the rocket is like an inelastic collision in reverse.
I see. So what kind of another equation I could use .. ?

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I see. So what kind of another equation I could use .. ?
I think @mitochan has the basic equations that you need in post #6. Note, he has interpreted the problem such that ##\varepsilon M_0## is the amount by which the rest mass of the rocket is decreased when a mass ejection takes place. So, after the first ejection, the rest mass of the rocket will be ##M_0 - \varepsilon M_0##. The rest mass of the ejected material will not be ##\varepsilon M_0##, but it may be written as ##\varepsilon_f M_0## where ##\varepsilon_f## is some unknown number which @mitochan eliminates during the algebraic manipulations.

I have not checked the details of @mitochan's work. When it comes to part (c), I believe you should get the standard result for the relativistic rocket where the ejected mass can be considered as being ejected continuously.

When I first read the problem statement in post #1, I interpreted ##\varepsilon M_0## as the rest mass of the ejected mass. In that case, the mass of the rocket after the first ejection would not be ##M_0 - \varepsilon M_0##. But after reading your re-wording of the problem statement in post #10, I favor @mitochan's interpretation.

Gold Member
In your equation mass is conserved. Not mass but energy as a component of 4-momentum vector is conserved. So let mass of ejected fuel be ##\epsilon_f\ m##.

Initial 4-momentum vector is
$$m(1,0,0,0)$$
where m has dimension of energy or mc^2.
For first ejection of fuel with relative velocity of given v to the rocket, the rocket gets speed of u
Rocket 4-momentum:
$$m(1-\epsilon)\gamma_u(1,u,0,0),u>0$$
Fuel 4-momentum:
$$m\epsilon_f \gamma_v (1,v,0,0),v<0$$
where u and v are dimensionless velocity divided by c.

Conservation of 4-vector
$$(1-\epsilon)\gamma_u+\epsilon_f \gamma_v=1$$
$$(1-\epsilon)\gamma_u u+ \epsilon_f \gamma_v v=0$$

By deleting ##\epsilon_f## we get the relation
$$(1-\epsilon)(v-u)\gamma_u=v$$
which gives u(##\epsilon##,v) implicitly.

For the second ejection similarly

$$(1-2\epsilon)(v-u_2)\gamma_{u_2}=v(1-\epsilon)$$

The speed of rocket in original IFR where the rocket was at rest after the second fuel ejection is
$$U_2=\frac{U_1+u_2}{1+U_1 u_2}$$ where
$$U_1 = u$$ which of cource is the speed of rocket in original IFR where the rocket was at rest after the first fuel ejection

For the third one

$$(1-3\epsilon)(v-u_3)\gamma_{u_3}=v(1-2\epsilon)$$
The speed of rocket in original IFR where the rocket was at rest after the second fuel ejection is
$$U_3=\frac{U_2+u_3}{1+U_2 u_3}$$

So in general

$$(1-n\epsilon)(v-u_n)\gamma_{u_n}=v[1-(n-1)\epsilon]$$ or
$$(1-\frac{u_n}{v})\gamma_{u_n}=1+\frac{1}{1-n\epsilon}$$
The speed of rocket in original IFR where the rocket was at rest after the n th fuel ejection is
$$U_n=\frac{U_{n-1}+u_n}{1+U_{n-1} u_n}$$

After all the ejection

$$(1-N\epsilon)(v-u_N)\gamma_{u_N}=v[1-(N-1)\epsilon]$$
The speed of rocket in original IFR where the rocket was at rest after the last N th fuel ejection is
$$U_N=\frac{U_{N-1}+u_N}{1+U_{N-1} u_N}$$

I am not confident on the above calculation so would appreciate corrections.
I talked to my teacher and he said that the velocity of the each fuel particles will have a velocity of ##u## after its ejected. He also said that we need to use energy conservation beacuse the rest mass of the rocket will change. Is your answer encapsulates that. If not then please do not share the whole solution I prefer tips...