- #1

Arman777

Gold Member

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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.

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

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.