# 8.7: Introduction to Rocket Propulsion

- Page ID
- 1543

Rockets range in size from fireworks so small that ordinary people use them to immense Saturn Vs that once propelled massive payloads toward the Moon. The propulsion of all rockets, jet engines, deflating balloons, and even squids and octopuses is explained by the same physical principle—Newton’s third law of motion. Matter is forcefully ejected from a system, producing an equal and opposite reaction on what remains. Another common example is the recoil of a gun. The gun exerts a force on a bullet to accelerate it and consequently experiences an equal and opposite force, causing the gun’s recoil or kick.

Making Connections: Take-Home Experiment —Propulsion of a Balloon

- Hold a balloon and fill it with air. Then, let the balloon go. In which direction does the air come out of the balloon and in which direction does the balloon get propelled? If you fill the balloon with water and then let the balloon go, does the balloon’s direction change? Explain your answer.

Figure shows a rocket accelerating straight up. In part (a), the rocket has a mass \(m\) and a velocity \(v\) relative to Earth, and hence a momentum \(mv\) In part (b), a time \(\Delta t\) has elapsed in which the rocket has ejected a mass \(\Delta m\) of hot gas at a velocity \(v_e\) relative to the rocket. The remainder of the mass \((m - \Delta m)\) now has a greater velocity \((v + \Delta v)\). The momentum of the entire system (rocket plus expelled gas) has actually decreased because the force of gravity has acted for a time \(\Delta t\), producing a negative impulse \(\Delta p = -mg\Delta t\). (Remember that impulse is the net external force on a system multiplied by the time it acts, and it equals the change in momentum of the system.) So, the center of mass of the system is in free fall but, by rapidly expelling mass, part of the system can accelerate upward. It is a commonly held misconception that the rocket exhaust pushes on the ground. If we consider thrust; that is, the force exerted on the rocket by the exhaust gases, then a rocket’s thrust is greater in outer space than in the atmosphere or on the launch pad. In fact, gases are easier to expel into a vacuum.By calculating the change in momentum for the entire system over \(\Delta t\), and equating this change to the impulse, the following expression can be shown to be a good approximation for the acceleration of the rocket.

\[a = \dfrac{v_e}{m} \dfrac{\Delta m}{\Delta t} - g,\]

where \(a\) is the acceleration of the rocket, \(v_e\) is the escape velocity, \(m\) is the mass of the rocket, \(\ Delta m\) is the mass of the ejected gas, and \(\Delta t\) is the time in which the gas is ejected.

** Figure \(\PageIndex{1}\):** *(a) This rocket has a mass \(m\) and an upward velocity \(v\). The net external force on the system is \(-mg\), if air resistance is neglected. (b) A time \(\Delta t\) later the system has two main parts, the ejected gas and the remainder of the rocket. The reaction force on the rocket is what overcomes the gravitational force and accelerates it upward.*

A rocket’s acceleration depends on three major factors, consistent with the equation for acceleration of a rocket **. **First, the greater the exhaust velocity of the gases relative to the rocket, \(v_e\), the greater the acceleration is. The practical limit for \(v_e\) is about \(2.5 \times 10^3 \space m/s\) for conventional (non-nuclear) hot-gas propulsion systems. The second factor is the rate at which mass is ejected from the rocket. This is the factor \((\Delta m/\Delta t)v_e\), with units of newtons, is called "thrust.” The faster the rocket burns its fuel, the greater its thrust, and the greater its acceleration. The third factor is the mass \(m\) of the rocket. The smaller the mass is (all other factors being the same), the greater the acceleration. The rocket mass \(m\) decreases dramatically during flight because most of the rocket is fuel to begin with, so that acceleration increases continuously, reaching a maximum just before the fuel is exhausted.

Factors Affecting a Rocket’s Acceleration

- The greater the exhaust velocity \(v_e\) of the gases relative to the rocket, the greater the acceleration.
- The faster the rocket burns its fuel, the greater its acceleration.
- The smaller the rocket’s mass (all other factors being the same), the greater the acceleration.

Example \(\PageIndex{1}\): Calculating Acceleration: Initial Acceleration of a Moon Launch

A Saturn V’s mass at liftoff was \(2.80 \times 10^6 \space kg\), its fuel-burn rate was \(1.40 \times 10^4 \times kg/s\), and the exhaust velocity was \(2.40 \times 10^3 m/s\). Calculate its initial acceleration.

**Strategy**

This problem is a straightforward application of the expression for acceleration because * a* is the unknown and all of the terms on the right side of the equation are given.

**Solution**

Substituting the given values into the equation for acceleration yields

\[a = \dfrac{v_e}{m} \dfrac{\Delta m}{\delta t} - g\]

\[= \dfrac{2.40 \times 10^3 \space m/s}{2.80 \times 10^6 \space kg}(1.40 \times 10^4 \space kg/s) - 9.8 \space m/s^2\]

\[= 2.20 \space m/s^2.\]

**Discussion**

This value is fairly small, even for an initial acceleration. The acceleration does increase steadily as the rocket burns fuel, because \(m\) decreases while \(v_e\) and \(\frac{\Delta m}{\Delta t} \) remain constant. Knowing this acceleration and the mass of the rocket, you can show that the thrust of the engines was \(3.36 \times 10^7 \space N.\)

To achieve the high speeds needed to hop continents, obtain orbit, or escape Earth’s gravity altogether, the mass of the rocket other than fuel must be as small as possible. It can be shown that, in the absence of air resistance and neglecting gravity, the final velocity of a one-stage rocket initially at rest is

\[v = v_e \space ln \dfrac{m_0}{m_r},\]

where \(ln (m_0/m_r) \)

**Figure \(\PageIndex{2}\):***The space shuttle had a number of reusable parts. Solid fuel boosters on either side were recovered and refueled after each flight, and the entire orbiter returned to Earth for use in subsequent flights. The large liquid fuel tank was expended. The space shuttle was a complex assemblage of technologies, employing both solid and liquid fuel and pioneering ceramic tiles as reentry heat shields. As a result, it permitted multiple launches as opposed to single-use rockets. (credit: NASA)*

Phet Explorations: Lunar Lander

Can you avoid the boulder field and land safely, just before your fuel runs out, as Neil Armstrong did in 1969? Our version of this classic video game accurately simulates the real motion of the lunar lander with the correct mass, thrust, fuel consumption rate, and lunar gravity. The real lunar lander is very hard to control.

Phet Explorations: Lunar Lander

Can you avoid the boulder field and land safely, just before your fuel runs out, as Neil Armstrong did in 1969? Our version of this classic video game accurately simulates the real motion of the lunar lander with the correct mass, thrust, fuel consumption rate, and lunar gravity. The real lunar lander is very hard to control.

* Figure \(\PageIndex{3}\): *Lunar Lander

# Summary

- Newton’s third law of motion states that to every action, there is an equal and opposite reaction.
- Acceleration of a rocket is \(a = \frac{v_e}{m} \frac{\Delta m}{\Delta t} - g.\)
- A rocket’s acceleration depends on three main factors. They are
- The greater the exhaust velocity of the gases, the greater the acceleration.
- The faster the rocket burns its fuel, the greater its acceleration.
- The smaller the rocket's mass, the greater the acceleration.

## Contributors

Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).