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Chapter 5: Problem 5

A rocket propels itself rectilinearly by giving portions of its mass a constant fbackward) velocity \(U\) relative to its instantaneous rest frame. It continues to do so until it attains a velocity \(V\) relative to its initial rest frame. Prove that the ratio of the initial to the final rest mass of the rocket is given by $$ \frac{M_{\mathrm{i}}}{M_{\mathrm{f}}}=\left(\frac{c+V}{c-V}\right)^{\frac{c}{2 U}} $$

Short Answer

Expert verified
Answer: The ratio of the initial to the final rest mass of the rocket is given by $\left(\frac{c+V}{c-V}\right)^{\frac{c}{2U}}$.

Step by step solution

01

Understand the given data

We are given the backward velocity of the ejected mass \(U\) relative to the rocket's rest frame, the final velocity of the rocket \(V\), and the speed of light \(c\). Our job is to derive the ratio of the rocket's initial rest mass \(M_i\) to its final rest mass \(M_f\).
02

Define the velocity of ejected mass in the initial rest frame

To solve this problem, we first need to find the velocity of the ejected mass in the initial rest frame. According to the velocity addition formula in special relativity, the velocity of the ejected mass in the initial rest frame is given by: $$ w = \frac{U + V}{1 + \frac{UV}{c^2}} $$
03

Apply momentum conservation

For momentum conservation, the momentum of the ejected mass plus the final momentum of the rocket should be equal to the initial momentum of the rocket. Since the rocket is initially at rest, its momentum is 0. Thus, we can write the conservation of momentum as: $$ m_dw = (M_i - m_d) V $$ where \(m_d\) is the small mass ejected and \(w\) is the velocity of the ejected mass in the initial rest frame, as defined in Step 2.
04

Analyze the mass and energy simultaneous variation

As the rocket is giving away portions of its mass, the mass ejected \(m_d\) and the remaining mass can be related by the following differential equation: $$ \frac{dm_d}{M_i - m_d} = \frac{m_dw}{V (M_i - m_d)} $$
05

Integrate the differential equation

To find the mass ratio, we need to integrate the differential equation in Step 4. After integration, we obtain: $$ \ln \frac{M_i}{M_i - m_d} = -\frac{Vw}{c^2}\frac{M_i}{U} $$
06

Replace \(w\) with the given data

As we derived \(w\) in Step 2, we can replace the expression in the above equation and simplify it: $$ \ln \frac{M_i}{M_f} = \frac{(c+V)c}{2U} $$ where \(M_f = M_i - m_d\) is the final rest mass of the rocket.
07

Find the mass ratio

Now, we can find the mass ratio by exponentiating both sides of the equation in Step 6: $$ \frac{M_i}{M_f} = \left(\frac{c+V}{c-V}\right)^{\frac{c}{2U}} $$ So, the ratio of the initial to the final rest mass of the rocket is given by \(\left(\frac{c+V}{c-V}\right)^{\frac{c}{2U}}\).

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Most popular questions from this chapter

Consider a head-on elastic collision of a "bullet' of rest mass \(M\) with a stationary 'target' of rest mass \(m\). Prove that the post-collision \(\gamma\)-factor of the bullet cannot exceed \(\left(m^{2}+M^{2}\right) / 2 m M\). This means that for large bullet energies (with \(\gamma\)-factors much larger than this critical value), almost the entire energy of the bullet is transferred to the target. [Hint: if \(\mathbf{P}, \mathbf{P}^{\prime}\) are the pre-and post-collision four-momenta of the bullet, and \(\mathbf{Q}, \mathbf{Q}^{\prime}\) those of the target, show, by going to the \(\mathrm{CM}\) frame, that \(\left(\mathbf{P}^{\prime}-\mathbf{Q}\right)^{2} \geqslant 0\); in fact, in the CM frame \(\mathbf{P}^{\prime}-\mathbf{Q}\) has no spatial components.] The situation is radically different in Newtonian mechanics, where the pre- and post-collision velocities of the bullet are related by \(u / u^{\prime}=(M+m) /(M-m)\). Prove this.

Uniform parallel radiation is observed in two arbitrary inertial frames \(\mathbf{S}\) and \(\mathbf{S}^{\prime}\) in which it has frequencies \(v\) and \(v^{\prime}\) respectively. If \(p, g, \sigma\) denote, respectively, the radiation pressure, momentum density, and energy density of the radiation in \(S\), and primed symbols denote corresponding quantities in \(S^{\prime}\), prove \(p^{\prime} / p=g^{\prime} / g=\sigma^{\prime} / \sigma\) \(=v^{\prime 2} / v^{2} .[\) Hint: Exercise III \((17) .]\)

In an inertial frame \(\mathrm{S}\), two photons of frequencies \(v_{1}\) and \(v_{2}\) travel in the positive and negative \(x\)-directions respectively. Find the velocity of the CM frame of these photons. [Answer: \(v / c\) \(\left.=\left(v_{1}-v_{2}\right) /\left(v_{1}+v_{2}\right) .\right]\)

If a photon with four-momentum \(P\) is observed by two observers having four- velocities \(\mathrm{U}_{0}\) and \(\mathrm{U}_{1}\), prove that the observed frequencies are in the ratio \(\mathbf{U}_{0} \cdot \mathbf{P} / \mathbf{U}_{1} \cdot \mathbf{P}\). Hence rederive equation (17.3).

The position vector of the centre of mass of a system of particles in any inertial frame is defined by \(\mathbf{r}_{\mathrm{CM}}=\Sigma m r / \Sigma m\). If the particles suffer only collision forces, prove that \(\dot{\boldsymbol{r}}_{\mathrm{CM}}=u_{\mathrm{CM}}(\cdot \equiv \mathrm{d} / \mathrm{d} t)\); i.e. the centre of mass moves with the velocity of the CM frame. [Hint: \(\Sigma m\), \Sigmami are constant; \Sigmamir is zero between collisions, and at any collision we can factor out the r of the participating particles: \(r \Sigma \dot{m}=0 .]\)
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