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


Context 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: $\sqrt{15.76 \times 10^{6}} \approx 3970$ )
Context answer:
\boxed{$\sqrt{2} \times 3970$}

Question: (d) Now, consider the effects of gravity. Given a rocket of mass $m_{0}$ and mass ejection rate $\mu$, what ejection speed $u$ would be required to launch satellites in the Earth's gravitational field, of strength $g$ ? (An estimate, ignoring numerical factors, is acceptable.) What about launching space probes to other planets or out of the solar system? Can you guess why most rockets are multistage?

Solution:
There are multiple ways of thinking about this problem.

- (Easiest possible answer) To launch the rocket from earth we need at least the force from repulsion to be stronger than the gravity. That is $a=\mu u / m_{0}>g$. (We get rid of numerical factors on the total mass.)



- A nicer perspective: The rocket needs to get to gain enough mechanical energy (kinetic plus potential energy) from the repulsion process. It is known (from multiple ways) that the energy of an object with mass $m_{0}$ orbiting a planet in a orbit whose length of the semimajor axis is $a$ is $-G M m_{0} / 2 a=-g R^{2} m_{0} / 2 a$, where $R$ is the radius of the planet ( $G M$ the standard gravitational parameter). Therefore, the minimum energy is at $a=R$ ( $a$ cannot be lower), which is $-g R m_{0} / 2$. Besides that, we know that before launching, when the rocket is on the ground, its potential energy (kinetic energy being zero) is $-G M m_{0} / R=-g R m_{0}$.

Therefore, the minimum mechanical energy needed is $g R m_{0} / 2$. This situation corresponds to "launching" the rocket horizontally. The rocket does not go high into the sky but flies right above the ground. As a result, as a not so bad approximation, the launching process is not affected much by the gravity. The kinetic energy of the rocket after its fuel used up, $m_{0} v_{f}^{2} / 2$ should be greater than $g R m_{0} / 2$. Therefore, we get a result:

$$
v_{f}^{2}>g R
\tag{19}
$$

Any answer that gets rid of numerical factors like $u^{2} \geq g R$ is acceptable. If a student remember the value of the first cosmic speed of the earth: $v_{o}=\sqrt{g R}=7.9 \mathrm{~km} / \mathrm{s}$, and states that $v_{f}$, or a multiple of $u$ (like $2.302 u$ we got above), should be greater than $v_{o}$ is also great.

- For launching our rocket to other planets, we need to escape from the gravitational field of the earth. That is, the final mechanical energy of the rocket should be greater than zero:

$$
v_{f}^{2}>2 g R
\tag{20}
$$

It is also great if a student remember that the escape speed of the each is $v_{e}=\sqrt{2 g R}=11.2$ and states that a multiple of $u$ should be larger than $v_{e}$.

- The reason why rockets are multistage is that We see from previous parts that the ejection speed of the lightest fuel with $100 \%$ energy conversion rate is only $u=5614 \mathrm{~m} / \mathrm{s}$, and the final speed $v_{f}$ of a very light rocket (mass ratio between fuel and rocket itself being 9 ) is just $\approx 2.3 u=12.9 \mathrm{~km} / \mathrm{s}$, which is only at the same magnitude of the first cosmic speed and the escape speed.

In real world, rockets cannot be so ideal. Therefore, a single stage rocket can hardly be launched to orbit around the earth, not to mention escaping. Multistage rockets drop part of its mass to further increase the mass ratio between the remaining fuel and itself in order to make itself to the sky.



Final answer: $\sqrt{g R}$ , $\sqrt{2 g R}$