C. Energy and Momentum · Momentum and CollisionsRockets and Propulsion: Momentum in Action
C1.1 C1.2 C2.5About 65 minutes
Learning goals- I can explain rocket propulsion using conservation of momentum and Newton's third law.
- I can calculate thrust from the rate of mass ejection and the exhaust speed.
- I can find a rocket's acceleration at lift-off and use the rocket equation to explain staging.
- I can propose improvements to propulsion technology and assess its social and environmental impact.
Words to know
thrust
The forward force on a rocket or jet from its exhaust: F = (Δm ÷ Δt) × v_ex.
exhaust speed (v_ex)
The speed of the exhaust gas relative to the rocket.
mass ratio
The starting mass of a rocket (with fuel) divided by its final (empty) mass.
rocket equation
Δv = v_ex ln(m₀ ÷ m_f): the change in velocity a rocket can gain by burning all its fuel.
staging
Building a rocket in sections that are dropped when their fuel is used up.
Pushing on nothing
In 1920, a newspaper editorial famously mocked the rocket pioneer Robert Goddard, claiming that a rocket could not work in the vacuum of space because there would be nothing for it to push against. The editors were wrong, and conservation of momentum shows why. A rocket does not push on the air or the ground. It pushes on its own exhaust.
This lesson applies momentum to one of the technologies your curriculum names: rocket propulsion. You will calculate thrust, see why rockets are mostly fuel and why they are built in stages, and consider ways to improve rockets and the effects of launches on society and the environment.
How a rocket works
A rocket engine burns fuel and blasts the hot exhaust gas out of its nozzle at very high speed. Think of the rocket and its fuel as one system, initially at rest in space:
- The total momentum starts at zero.
- The engine pushes each bit of gas backward. By Newton's third law, the gas pushes the rocket forward.
- These are internal forces, so the total momentum stays zero: the backward momentum of the exhaust is balanced by the forward momentum of the rocket.
A rocket is really a continuous explosion: like the carts pushed apart by a spring in the lesson on explosions and recoil, but happening thousands of times a second.
A rocket in space. In each short time interval, a small mass of gas Δm is pushed backward at the exhaust speed. The rocket gains an equal amount of forward momentum, so the total momentum of the system does not change.Thrust
Apply the impulse–momentum theorem to the exhaust. In a time Δt, the engine gives a mass Δm of gas a backward velocity v_ex (the exhaust speed, relative to the rocket). The force needed is F = Δp ÷ Δt = (Δm · v_ex) ÷ Δt. By the third law, the gas pushes the rocket forward with the same force, called the thrust:
F_thrust = (Δm ÷ Δt) × v_ex
So thrust depends on how much mass is thrown out each second and how fast it is thrown. This is Newton's second law in its momentum form, F = Δp ÷ Δt, the form that works when mass changes.
At launch, the rocket rises only if the thrust is greater than its weight: F_net = F_thrust − mg.
Worked exampleExample 1: Thrust of an engine
A rocket engine burns fuel and ejects exhaust at 1300 kg/s with an exhaust speed of 2500 m/s. Find the thrust.
- F = (Δm ÷ Δt) × v_ex = (1300 kg/s)(2500 m/s) = 3.25 × 10⁶ N.
F ≈ 3.3 × 10⁶ N, roughly the weight of 330 tonnes.
Worked exampleExample 2: Lift-off
A 2.0 × 10⁵ kg rocket has the engine from Example 1. Find its acceleration at lift-off.
- Weight: mg = (2.0 × 10⁵)(9.8) = 1.96 × 10⁶ N.
- F_net = 3.25 × 10⁶ − 1.96 × 10⁶ = 1.29 × 10⁶ N [up].
- a = F_net ÷ m = 1.29 × 10⁶ ÷ 2.0 × 10⁵ ≈ 6.5 m/s² [up].
a ≈ 6.5 m/s² [up]. As fuel burns, m decreases while the thrust stays about the same, so the acceleration increases during the flight.
Worked exampleExample 3: One puff at a time
A 1000 kg space probe (including fuel) is at rest. Its thruster fires a 10 kg burst of gas backward at 2000 m/s. Find the probe's velocity afterward.
- Momentum before = 0.
- After: (10)(−2000) + (990)v = 0, so v = 20 000 ÷ 990 ≈ 20.2 m/s.
About 20 m/s forward. Each burst adds a little speed, and the probe gets lighter each time, so later bursts add slightly more.
Why rockets are mostly fuel
As the rocket burns fuel, it gets lighter, so each kilogram of exhaust changes its velocity a bit more. Adding up all these small changes needs calculus (MCV4U and first-year university). The result is the rocket equation, published by the Russian scientist Konstantin Tsiolkovsky in 1903:
Δv = v_ex ln(m₀ ÷ m_f)
where m₀ is the starting mass (with fuel), m_f is the final mass (empty), and ln is the natural logarithm. You do not need to derive it, but notice what it says:
- Δv grows only with the logarithm of the mass ratio. Doubling the fuel does not double the speed.
- To reach low Earth orbit (about 7.8 km/s, plus extra to overcome air resistance and gravity on the way up) with v_ex ≈ 3 km/s, the mass ratio must be well over 10: a rocket that is more than 90% fuel.
- This is why rockets use stages: when a stage's tanks are empty, the whole stage is dropped, so the remaining fuel does not have to keep accelerating dead weight.
The rocket equation as a graph. The speed gained (as a multiple of the exhaust speed) grows only slowly as the mass ratio increases. Reaching orbital speed with an exhaust speed of 3.0 km/s needs a mass ratio of about 13.5, ignoring air resistance and gravity losses.Worked exampleExample 4: Using the rocket equation
A rocket stage has an exhaust speed of 3000 m/s and a mass ratio m₀ ÷ m_f = 5.0. Find the change in velocity it can provide in space. What mass ratio would 7.8 km/s need with one stage?
- Δv = v_ex ln(m₀ ÷ m_f) = 3000 ln 5.0 ≈ 3000 × 1.609 ≈ 4830 m/s.
- For 7800 m/s: ln(m₀ ÷ m_f) = 7800 ÷ 3000 = 2.6, so m₀ ÷ m_f = e^2.6 ≈ 13.5.
About 4.8 km/s. A single stage would need a mass ratio of about 13.5 (over 92% fuel), which is very hard to build. Staging solves this.
Other momentum engines
- Jet engines take in air, compress it, burn fuel in it, and blast it out faster than it came in. Thrust ≈ (mass of air per second) × (exhaust speed − incoming speed).
- Propellers and boat paddles push air or water backward; the air or water pushes the plane or canoe forward.
- Squid and octopuses squirt water out through a funnel to dart away.
- Ion thrusters on some space probes accelerate charged atoms to very high exhaust speeds (tens of kilometres per second). Their thrust is tiny, but they use very little propellant and can run for months.
Improving rocket technology (C1.1)
Engineers improve rockets by raising v_ex (better fuels, such as liquid hydrogen with liquid oxygen, and better nozzles), reducing the empty mass (lighter tanks and composite materials), using staging, and reusing boosters: some companies now land first stages and fly them again, which cuts the cost and the waste of each launch.
Impacts on society and the environment (C1.2)
Rockets put satellites in orbit that people rely on every day: weather forecasting, communications, navigation (GPS) and Earth observation. Canada's RADARSAT Constellation Mission, for example, uses radar satellites to monitor sea ice, ships and floods. But launches also have costs: rocket exhaust releases carbon dioxide, water vapour and, for some fuels, soot directly into the upper atmosphere; spent stages and debris are left in orbit or fall into the ocean; and launch sites affect local wildlife and communities. Evaluating a launch program means weighing these benefits and costs.
Just for funSome squid are rocket scientists. They squirt water out of a funnel to shoot backward, and a few kinds have been seen launching themselves out of the sea and gliding through the air for several metres, fins spread like wings.
Hands-on activityBalloon rocket on a string
You need: a long balloon (round ones work too), a drinking straw, about 5 m of smooth string, tape, a tape measure, a phone stopwatch (or slow-motion video), and a kitchen scale. Grown-up note: balloon pieces are a choking hazard for young children, and some people are allergic to latex.
- Thread the string through the straw. Tie the string tightly between two chair backs about 4 m apart, level with the floor.
- Blow up the balloon, pinch the neck closed (do not tie it), and tape it under the straw with the neck pointing toward one end.
- Release the neck. Time how long the balloon takes to travel a measured distance, and how far it goes.
- Repeat with the balloon blown up less and more. Then try adding a paper clip of mass to the straw.
- Explain the motion using momentum: what is being thrown backward, and what gets the forward momentum?
- Improve it (C1.1): change one thing at a time (a narrower neck using a short straw taped in it, a smoother string, a lighter straw) to get the balloon to go farther. Record what worked and explain why using thrust = (Δm ÷ Δt) × v_ex and friction.
What to look for: the balloon shoots away from the escaping air. More air means it accelerates for longer. A narrower nozzle often makes the air leave faster (bigger v_ex) but for a longer time; friction between the straw and the string is usually what limits the distance.
Your turn
1An engine ejects 50 kg of exhaust per second at 3000 m/s. Find the thrust.
F = (50)(3000) = 1.5 × 10⁵ N.
2A toy water rocket pushes out 0.20 kg of water per second at 15 m/s. Find its thrust.
3A 5.0 × 10⁵ kg rocket has a thrust of 7.6 × 10⁶ N at lift-off. Find its acceleration.
Weight = (5.0 × 10⁵)(9.8) = 4.9 × 10⁶ N. F_net = 2.7 × 10⁶ N. a = 2.7 × 10⁶ ÷ 5.0 × 10⁵ = 5.4 m/s² [up].
4A 100 kg astronaut (with jet pack) at rest fires 0.50 kg of gas backward at 50 m/s. Find the astronaut's speed.
(0.50)(50) = (99.5)v, so v ≈ 0.25 m/s.
5What rate of mass ejection gives a thrust of 1.0 × 10⁶ N with an exhaust speed of 2500 m/s?
Δm ÷ Δt = F ÷ v_ex = 1.0 × 10⁶ ÷ 2500 = 400 kg/s.
6Explain why a rocket can accelerate in the vacuum of space.
The rocket pushes its own exhaust backward, and the exhaust pushes the rocket forward (third law). Total momentum is conserved: the exhaust's backward momentum equals the rocket's forward momentum. No air is needed.
7Use the rocket equation to find Δv for a stage with v_ex = 2500 m/s and a mass ratio of 3.0.
Δv = 2500 ln 3.0 ≈ 2500 × 1.099 ≈ 2.7 × 10³ m/s.
More practice: warm-up
1What does a rocket push against?
2Write the formula for thrust.
F = (Δm ÷ Δt) × v_ex: mass ejected per second times exhaust speed.
3What must be true for a rocket to lift off the ground?
Its thrust must be greater than its weight.
4Why do rockets use stages?
Empty stages are dropped, so the remaining fuel does not have to accelerate useless mass.
5Name one benefit of satellites launched by rockets.
Any one: weather forecasting, communications, navigation (GPS), Earth observation such as monitoring sea ice.
More practice: on your own
1A 1500 kg lander hovers just above the Moon's surface, where g = 1.62 N/kg. Its exhaust speed is 3000 m/s. Find the thrust needed and the rate at which it burns fuel.
Hovering: thrust = weight = (1500)(1.62) = 2430 N (2.4 × 10³ N). Δm ÷ Δt = 2430 ÷ 3000 ≈ 0.81 kg/s.
2A 500 kg probe at rest ejects 5.0 kg of gas backward at 1200 m/s. Find its velocity afterward.
(5.0)(1200) = (495)v, so v ≈ 12 m/s forward.
3A jet engine on a plane flying at 250 m/s takes in 80 kg of air per second and expels it at 600 m/s relative to the plane. Ignoring the mass of fuel, find the thrust.
Each second, the air's speed relative to the plane increases from 250 m/s to 600 m/s. F = (80)(600 − 250) = 2.8 × 10⁴ N.
4Explain why doubling the amount of fuel in a rocket does not double the change in velocity it can produce.
Extra fuel must itself be accelerated by the earlier burning. In the rocket equation, Δv depends on ln(m₀ ÷ m_f), which grows more slowly than the mass ratio, so doubling the fuel adds much less than double the speed.
5What mass ratio is needed for Δv = 9.0 km/s with v_ex = 4.5 km/s?
ln(m₀ ÷ m_f) = 9.0 ÷ 4.5 = 2.0, so m₀ ÷ m_f = e² ≈ 7.4.
Challenge
1Use the impulse–momentum theorem to derive F_thrust = (Δm ÷ Δt)v_ex.
In time Δt, a mass Δm of exhaust goes from moving with the rocket to moving backward at v_ex relative to it, so its momentum changes by Δm·v_ex (backward). The engine must exert F = Δp ÷ Δt = (Δm ÷ Δt)v_ex on the gas. By the third law, the gas exerts the same size of force forward on the rocket.
2A 100 kg rocket (80 kg empty plus 20 kg of fuel) is at rest in space. Exhaust leaves at 1000 m/s relative to the rocket. Compare its final speed if (a) all 20 kg is ejected in one burst, (b) two 10 kg bursts are used, and (c) the rocket equation (continuous burning).
(a) After the burst: 0 = 80V + 20(V − 1000), so V = 200 m/s. (b) First: 0 = 90V₁ + 10(V₁ − 1000), V₁ = 100 m/s. Second: 90(100) = 80V₂ + 10(V₂ − 1000), so 9000 + 10 000 = 90V₂ and V₂ ≈ 211 m/s. (c) Δv = 1000 ln(100 ÷ 80) ≈ 223 m/s. Splitting the burn into smaller pieces does better, because later pieces push a lighter rocket; continuous burning is the limit.
3An ion thruster ejects xenon at 30 km/s with a thrust of 0.10 N. Find the mass of xenon used per second, and how long it takes to change a 1000 kg probe's velocity by 1.0 km/s (ignore the small mass change).
Δm ÷ Δt = 0.10 ÷ 30 000 ≈ 3.3 × 10⁻⁶ kg/s (about 3 mg per second). Impulse needed = (1000)(1000) = 1.0 × 10⁶ N·s, so Δt = 1.0 × 10⁶ ÷ 0.10 = 1.0 × 10⁷ s ≈ 116 days. It uses only about 33 kg of xenon.
Check your understandingQuiz
1. A rocket can accelerate in empty space because:
2. An engine ejects 200 kg of exhaust each second at 3000 m/s. Its thrust is:
3. A rocket can lift off the launch pad only when:
4. Why are large rockets built in stages?
5. According to the rocket equation, the change in velocity a rocket can achieve depends on:
6. Which is an environmental cost of rocket launches?
For parents and teachersBig idea (C1.1, C1.2, C2.5): a rocket is a continuous explosion. It throws exhaust backward and gains equal forward momentum; thrust = (mass ejected per second) × (exhaust speed). The rocket equation shows why rockets are mostly fuel and use stages. The lesson also asks your teen to propose improvements and weigh the benefits and costs of launches.
How to help: Build the balloon rocket together (watch younger children around balloons). Challenge your teen to improve the design and to explain each improvement with momentum or friction.
Common difficulties:
- Thinking rockets push on the air. They push on their own exhaust and work best in a vacuum.
- Forgetting the weight at lift-off. Net force = thrust − weight.
- The rocket equation uses the natural logarithm (ln), not log base 10. It is introduced as a preview; its derivation needs calculus.
- Relative velocities in Challenge 2: the exhaust speed is relative to the rocket, so the ground-frame exhaust velocity is V − 1000.