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.

Figure 1 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 $\mathrm{\Delta }t$ has elapsed in which the rocket has ejected a mass $\mathrm{\Delta }m$ of hot gas at a velocity $v_e$ relative to the rocket. The remainder of the mass $(m-\mathrm{\Delta }m)$ now has a greater velocity $(v+\mathrm{\Delta }v).$ The momentum of the entire system (rocket plus expelled gas) has actually decreased because the force of gravity has acted for a time $\mathrm{\Delta }t,$ producing a negative impulse $\mathrm{\Delta }p=-mg\mathrm{\Delta }t.$ (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 $\mathrm{\Delta }t,$ and equating this change to the impulse, the following expression can be shown to be a good approximation for the acceleration of the rocket.

“The rocket” is that part of the system remaining after the gas is ejected, and $g$ is the acceleration due to gravity.

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 $2.5\times {10}^3\text{ m/s}$ 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 $\mathrm{\Delta }m/\mathrm{\Delta }t$ in the equation. The quantity $(\mathrm{\Delta }m/\mathrm{\Delta }t)v_e,$ 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.

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

where $\text{ln}(m_0/m_r)$ is the natural logarithm of the ratio of the initial mass of the rocket $(m_0)$ to what is left $(m_r)$ after all of the fuel is exhausted. (Note that $v$ is actually the change in velocity, so the equation can be used for any segment of the flight. If we start from rest, the change in velocity equals the final velocity.) For example, let us calculate the mass ratio needed to escape Earth’s gravity starting from rest, given that the escape velocity from Earth is about $11.2\times {10}^3\text{ m/s},$ and assuming an exhaust velocity $v_e=2.5\times {10}^3\text{ m/s}.$

Solving for $m_0/m_r$ gives

Thus, the mass of the rocket is

This result means that only $1/88$ of the mass is left when the fuel is burnt, and $87/88$ of the initial mass was fuel. Expressed as percentages, 98.9% of the rocket is fuel, while payload, engines, fuel tanks, and other components make up only 1.10%. Taking air resistance and gravitational force into account, the mass $m_r$ remaining can only be about $m_0/180.$ It is difficult to build a rocket in which the fuel has a mass 180 times everything else. The solution is multistage rockets. Each stage only needs to achieve part of the final velocity and is discarded after it burns its fuel. The result is that each successive stage can have smaller engines and more payload relative to its fuel. Once out of the atmosphere, the ratio of payload to fuel becomes more favorable, too.

The space shuttle was an attempt at an economical vehicle with some reusable parts, such as the solid fuel boosters and the craft itself. (See Figure 2) The shuttle’s need to be operated by humans, however, made it at least as costly for launching satellites as expendable, unmanned rockets. Ideally, the shuttle would only have been used when human activities were required for the success of a mission, such as the repair of the Hubble space telescope. Rockets with satellites can also be launched from airplanes. Using airplanes has the double advantage that the initial velocity is significantly above zero and a rocket can avoid most of the atmosphere’s resistance.

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.