Subfield: Mechanics

4. The fundamental rocket equation

In this problem, we will investigate the acceleration of rockets.

Rocket repulsion

In empty space, an accelerating rocket must "throw" something backward to gain speed from repulsion. Assume there is zero gravity.

The rocket ejects fuel from its tail to propel itself forward. From the rocket's frame of reference, the fuel is ejected at a constant (relative) velocity $\mathbf{u}$. The rate of fuel ejection is $\mu=d m / d t<0$, and this rate is constant until the fuel runs out.
<img_4415>

Figure 1: Rocket (and fuel inside) with mass $m$, ejecting fuel at rate $\mu=d m / d t$ with relative velocity $\mathbf{u}$.

(Note: the $m$ in $\mu=d m / d t$ denotes the mass of rocket plus fuel, not the mass of an empty rocket.)
Context question:
(a) Consider a small time interval $\Delta t$ in the rocket's frame. $\Delta t$ is small enough that the rocket's frame can be considered an inertial frame (i.e., the frame has no acceleration). The amount of fuel ejected in this time is $\Delta m=|\mu| \Delta t$. (See Figure 2.)
<img_4337>

Figure 2: Rocket gaining speed $\Delta \mathbf{v}$ during interval $\Delta t$ by ejecting mass $\Delta m$ with relative velocity u.

Write down the momentum-conservation relation between $\mathbf{u}, \Delta \mathbf{v}, \Delta m$ and $m$ (the total mass of the rocket and the fuel inside at the moment). What is the acceleration $\mathbf{a}=d \mathbf{v} / d t$ (independent of reference frame) in terms of $\mathbf{u}, \mu$ and $m$ ?
Context answer:
\boxed{$\mathbf{u} \Delta m+m \Delta \mathbf{v}=0$ , $\mathbf{a}=-\frac{\mu}{m} \mathbf{u}$}


Context question:
(b) Suppose that the rocket is stationary at time $t=0$. The empty rocket has mass $m_{0}$ and the rocket full of fuel has mass $9 m_{0}$. The engine is turned on at time $t=0$, and fuel is ejected at the rate $\mu$ and relative velocity $\mathbf{u}$ described previously. What will be the speed of the rocket when it runs out of fuel, in terms of $u=|\mathbf{u}|$ ?



(Hint: useful integral formula: $\int_{a}^{b} \frac{1}{x} d x=\ln \left(\frac{b}{a}\right)$, and $\ln (10) \approx 2.302, \ln (9) \approx 2.197$.)
Context answer:
$u \ln 10$


Context question:
(c) Discuss the factors that limit the final speed of the rocket.
Context answer:
\boxed{证明题}


Extra Supplementary Reading Materials:

Chemical rockets 

Nearly all rockets get energy from the chemical reaction of the fuel (burning with the oxidant) they carry. This is called a combustion reaction. For example, the combustion reaction for hydrogen is

$$
2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}
\tag{8}
$$
Context question:
(a) Suppose that the product of combustion is just one species of molecule $\left(\mathrm{H}_{2} \mathrm{O}\right.$, for example) with mass $m_{p}$ and average kinetic energy $E$. What is the upper limit of exhaust speed $u$ when these molecules are ejected from the rocket?
Context answer:
\boxed{$\sqrt{\frac{2 E}{m_{p}}}$}


Context question:
(b) To manufacture fast rockets, would you recommend using hydrogen as the fuel? Assume that we only use oxygen as oxidant. If so, please give your reasoning. If not, what fuel would you suggest and why? (There are no numerical values here, so please give the best reasoning based on your own knowledge. Any reasonable answer is acceptable.)
Context answer:
开放性回答


Question: (c) The enthalpy of combustion of the reaction $2 \mathrm{H}_{2}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}$ is approximately $15.76 \times 10^{6} \mathrm{~J} / \mathrm{kg}$. That is, for every kilogram of water $\left(\mathrm{H}_{2} \mathrm{O}\right)$ produced, the energy released by the reaction is $15.76 \times 10^{6} \mathrm{~J}$. Assuming that all the energy released can be converted to the kinetic energy of the water molecules, what is their exhaust speed $u$ ? (Hint: $\sqrt{15.76 \times 10^{6}} \approx 3970$ )

Solution:
Plug in the numbers:

$$
u=\sqrt{\frac{2 E}{m_{p}}} \approx \sqrt{2} \times \sqrt{15.76 \times 10^{6} \mathrm{~J} / \mathrm{kg}}=\sqrt{2} \times 3970 \mathrm{~m} / \mathrm{s} \approx 5614 \mathrm{~m} / \mathrm{s}
\tag{18}
$$

Final answer: $\sqrt{2} \times 3970$ m/s