# Learning Objectives

### Learning Objectives

By the end of this section, you will be able to do the following:

• State Newtons third law of motion
• Explain the principle involved in propulsion of rockets and jet engines
• Derive an expression for the acceleration of the rocket
• Discuss the factors that affect the rockets acceleration
• Describe the function of a space shuttle

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 8.16 shows a rocket accelerating straight up. In part (a), the rocket has a mass $mm size 12{m} {}$ and a velocity $vv size 12{v} {}$ relative to Earth, and hence a momentum $mvmv size 12{ ital "mv"} {}$. In part (b), a time $ΔtΔt size 12{Δt} {}$ has elapsed in which the rocket has ejected a mass $ΔmΔm size 12{} {}$ of hot gas at a velocity $veve size 12{v rSub { size 8{e} } } {}$ relative to the rocket. The remainder of the mass $m−Δmm−Δm size 12{ left (m - right )} {}$ now has a greater velocity $v+Δvv+Δv size 12{ left (v+Δv right )} {}$. The momentum of the entire system (rocket plus expelled gas) has actually decreased because the force of gravity has acted for a time $ΔtΔt size 12{Δt} {}$, producing a negative impulse $Δp=−mgΔtΔp=−mgΔt size 12{Δp= - ital "mg"Δt} {}$. 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 in a vacuum.

By calculating the change in momentum for the entire system over $ΔtΔt size 12{Δt} {}$, and equating this change to the impulse, the following expression can be shown to be a good approximation for the acceleration of the rocket.

8.99 $a=vemΔmΔt−g.a=vemΔmΔt−g. size 12{a= { {v" lSub { size 8{e} } } over {m} } { {Δm} over {Δt} } - g} {}$

The rocket is that part of the system remaining after the gas is ejected, and $gg size 12{g} {}$ is the acceleration due to gravity.

### Acceleration of a Rocket

Acceleration of a rocket is

8.100 $a=vemΔmΔt−g,a=vemΔmΔt−g, size 12{a= { {v" lSub { size 8{e} } } over {m} } { {Δm} over {Δt} } - g,} {}$

where $aa size 12{a} {}$ is the acceleration of the rocket, $veve size 12{v rSub { size 8{e} } } {}$ is the exhaust velocity, $mm size 12{m} {}$ is the mass of the rocket, $ΔmΔm size 12{Δm} {}$ is the mass of the ejected gas, and $ΔtΔt size 12{Δt} {}$ is the time in which the gas is ejected.

Figure 8.16 (a) This rocket has a mass $mm size 12{m} {}$ and an upward velocity $vv size 12{v} {}$. The net external force on the system is $−mg−mg size 12{ size 11{ - ital "mg"}} {}$, if air resistance is neglected. (b) A time $ΔtΔt size 12{Δt} {}$ 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, $veve size 12{v rSub { size 8{e} } } {}$, the greater the acceleration is. The practical limit for $veve size 12{v rSub { size 8{e} } } {}$ is about 2.5 × 103 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 $Δm/ΔtΔm/Δt size 12{Δm/Δt} {}$ in the equation. The quantity $(Δm/Δt)ve(Δm/Δt)ve size 12{ $$Δm/Δt$$ v rSub { size 8{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, $m,m, size 12{m} {}$ of the rocket. The smaller the mass is (all other factors being the same), the greater the acceleration. The rocket mass $mm size 12{m} {}$ decreases dramatically during flight because most of the rocket is fuel to begin with, so acceleration increases continuously, reaching a maximum just before the fuel is exhausted.

### Factors Affecting a Rocket’s Acceleration

• The greater the exhaust velocity, $ve,ve, size 12{v rSub { size 8{e} } } {}$ 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 8.8Calculating Acceleration: Initial Acceleration of a Moon Launch

A Saturn V’s mass at liftoff was 2.80 × 106 kg, its fuel-burn rate was 1.40 × 104 kg/s, and the exhaust velocity was 2.40 × 103 m/s. Calculate its initial acceleration.

Strategy

This problem is a straightforward application of the expression for acceleration because $aa size 12{a} {}$ 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

8.101 $a=vemΔmΔt−g=2.40×103m/s2.80×106kg1.40×104kg/s−9.80m/s2= 2.20m/s2.a=vemΔmΔt−g=2.40×103m/s2.80×106kg1.40×104kg/s−9.80m/s2= 2.20m/s2.$

Discussion

This value is fairly small, even for an initial acceleration. The acceleration does increase steadily as the rocket burns fuel, because $mm size 12{m} {}$ decreases while $veve size 12{v rSub { size 8{e} } } {}$ and $ΔmΔtΔmΔt size 12{ { {Δm} over {Δt} } } {}$ remain constant. Knowing this acceleration and the mass of the rocket, you can show that the thrust of the engines was $3.36×107N3.36×107N size 12{3 "." "36" times "10" rSup { size 8{7} } N} {}$.

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

8.102 $v=velnm0mr,v=velnm0mr, size 12{v=v rSub { size 8{e} } "ln" { {m rSub { size 8{0} } } over {m rSub { size 8{r} } } } ,} {}$

where $lnm0/mrlnm0/mr size 12{"ln" left (m rSub { size 8{0} } /m rSub { size 8{r} } right )} {}$ is the natural logarithm of the ratio of the initial mass of the rocket $m0m0 size 12{ left (m rSub { size 8{0} } right )} {}$ to what is left $mrmr size 12{ left (m rSub { size 8{r} } right )} {}$ after all the fuel is exhausted. (Note that $vv size 12{v} {}$ 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 $11.2×103m/s11.2×103m/s size 12{"11" "." 2 times "10" rSup { size 8{3} } "m/s"} {}$, and assuming an exhaust velocity $ve=2.5×103m/sve=2.5×103m/s size 12{v rSub { size 8{e} } =2 "." 5 times "10" rSup { size 8{3} } "m/s"} {}$.

8.103 $lnm0mr=vve=11.2×103m/s2.5×103m/s=4.48.lnm0mr=vve=11.2×103m/s2.5×103m/s=4.48. size 12{"ln" { {m rSub { size 8{0} } } over {m rSub { size 8{r} } } } = { {v} over {v rSub { size 8{e} } } } = { {"11" "." 2 times "10" rSup { size 8{3} } "m/s"} over {2 "." 5 times "10" rSup { size 8{3} } "m/s"} } =4 "." "48"} {}$

Solving for $m0/mrm0/mr size 12{m rSub { size 8{0} } /m rSub { size 8{r} } } {}$ gives

8.104 $m0mr=e4.48=88.m0mr=e4.48=88. size 12{ { {m rSub { size 8{0} } } over {m rSub { size 8{r} } } } =e rSup { size 8{4 "." "48"} } ="88" "." } {}$

Thus, the mass of the rocket is

8.105 $mr=m088.mr=m088. size 12{m rSub { size 8{r} } = { {m rSub { size 8{0} } } over {"88"} } "." } {}$

This result means that only 1/88 of the mass is left when the fuel is burned, and 87/88 of the initial mass was fuel. Expressed as percentages, 98.9 percent of the rocket is fuel, while payload, engines, fuel tanks, and other components make up only 1.10 percent. Taking air resistance and gravitational force into account, the mass $mrmr size 12{m rSub { size 8{r} } } {}$ remaining can only be about $m0/180m0/180 size 12{ size 11{m rSub { size 8{0} } /"180"}} {}$. It is difficult to build a rocket in which the fuel has a mass 180 times everything else. The solution is to use 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 8.17). 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 a double advantage: the initial velocity is significantly above zero and the rocket can avoid most of the atmosphere’s resistance.

Figure 8.17 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.

Figure 8.18