Jet Propulsion and the Tsiolkovsky Rocket Equation

A rocket speeds up by throwing out hot gas: it pushes off neither the air nor the ground but its own exhaust. Launch it in open space or from the pad, watch the fuel run down and the speed grow — and catch the moment the rocket outruns its own gas. In space the momentum of the rocket and the momentum of the ejected gas always add up to zero.

→p₁ — momentum of cart 1 →p₂ — momentum of cart 2 →Σp — total momentum ◉centre of mass of the system

Presets

m₁ mass of cart 1

v₁ velocity of cart 1

m₂ mass of cart 2

v₂ velocity of cart 2

“+” — to the right, “−” — to the left. Cart 1 is on the left, cart 2 on the right

Collision type

Slow motion

Presets

M cannon mass

m shell mass

v shell speed

Relative to the ground, right after leaving the barrel

α barrel angle °

μ friction coefficient (wheels braked)

Slow-motion shot

Momentum and velocity arrows

Centre of mass

Test yourself

The calculation is exact: the motion is given by formulas of time, and the moments of impact, shot, lift-off and burnout are found from equations, so the numbers match the formulas and do not depend on the playback speed. The impact and the shot are instantaneous, there is no air drag, g = 9.8 m/s².

Before the impact

p₁ = m₁v₁

p₂ = m₂v₂

Σp = p₁ + p₂

v_c = Σp / (m₁ + m₂)

velocity of the centre of mass — the same before and after the impact

After the impact

v = (m₁v₁ + m₂v₂) / (m₁ + m₂)

E_k before

E_k′ after

Q = m₁m₂(v₁ − v₂)² / 2(m₁ + m₂)

turned into heat

Q / E_k

V = m v cosα / M

recoil speed right after the shot

m v cosα

horizontal momentum of the shell

−M V

momentum of the cannon

m v sinα

vertical momentum — given by the support

E_sh / E_c = M / (m cos²α)

E_sh = mv²/2

E_c = MV²/2

s = V² / (2μg)

recoil distance

t = V / (μg)

recoil time

μ = (m₀ − m_f) / t_burn

fuel flow rate

F = μu

thrust

Now

m = m₀ − μt

v − u

speed of the just-ejected gas relative to the launch point

The rocket launches from the Earth: with u = 3 km/s and m₀/m = 5 it reaches 4.04 km/s at burnout (80 s) — 0.78 km/s less than u·ln 5 ≈ 4.83 km/s. The difference is the gravity loss gt = 9.8·80 = 784 m/s. Make the burn long (300 s): the thrust becomes less than the weight, and the rocket stands on the pad until it has burnt part of its fuel. Then switch to space: there the final speed is exactly u·ln(m₀/m) and does not depend on the burn time. And try to bring a single stage to the first cosmic velocity of 7.9 km/s.

KEY FORMULAS

Conservation of momentum

$$\vec p = m\vec v, \qquad m_1\vec v_1 + m_2\vec v_2 = m_1\vec v_1^{\,\prime} + m_2\vec v_2^{\,\prime}$$

In a closed system the total momentum before an interaction equals the total after it

▶ Detailed Explanation

Momentum is mass times velocity, a vector pointing along the velocity. When two bodies interact, the forces they exert on each other are equal and opposite (Newton's third law) and act for the same time. So one body gets the momentum FΔt, the other −FΔt, and the sum does not change.

Signs matter: for motion along one line choose a positive direction and give velocities pointing the other way a minus sign. In the simulator the arrows p₁, p₂ are the momenta of the bodies and the golden arrow Σp is their sum: during the impact, the shot and the rocket burn in space it changes neither its length nor its direction.

Thrust

$$F = \mu u, \qquad \mu = \frac{\Delta m}{\Delta t}$$

μ is the mass flow rate (gas mass per second), u the gas speed relative to the rocket

▶ Detailed Explanation

During Δt the rocket throws out gas of mass Δm = μΔt and gives it the momentum Δm·u backwards (in the rocket's frame). By Newton's third law the gas pushes the rocket forward with the force F = Δm·u/Δt = μu. No support is needed: the rocket pushes off its own gas.

In the simulator m₀ = 100 t and the flow rate is μ = (m₀ − m_f)/t_burn, where m_f is the mass of the rocket without fuel. On a launch from the Earth the rocket lifts off only when F > mg; if the thrust is less than the weight, it waits on the pad until its mass drops to F/g. The acceleration grows as the fuel burns: a = F/m − g.

Tsiolkovsky rocket equation

$$v = u\,\ln\frac{m_0}{m}, \qquad v = u\,\ln\frac{m_0}{m} - g\,t$$

m₀ is the initial mass, m the current mass; the second form is a vertical launch with constant g, lifting off at t = 0

▶ Detailed Explanation

Let a rocket of mass m throw out gas of mass dm at a speed u relative to itself. The momentum of the system is conserved: m·dv = u·dm, so dv = u·dm/m. Adding up such small steps while the mass falls from m₀ to m gives v = u·ln(m₀/m). The speed does not depend on how long the engine works — only on u and the mass ratio.

At m₀/m = e ≈ 2.72 the rocket's speed equals u; beyond that the rocket outruns its exhaust, and the gas behind it (speed v − u) already moves forward. To reach 7.9 km/s with u = 3 km/s you need m₀/m = e^2.63 ≈ 14: 93 % of the initial mass is fuel. That is why rockets have several stages. On a launch from the Earth gravity takes away gt during the burn — the longer the burn, the larger the loss (if the rocket does not lift off at once, t is counted from lift-off).

Velocity of the centre of mass

$$\vec v_c = \frac{m_1\vec v_1 + m_2\vec v_2 + \dots}{m_1 + m_2 + \dots} = \frac{\sum \vec p}{\sum m}$$

With no external forces the centre of mass of a system moves uniformly in a straight line

▶ Detailed Explanation

The numerator is the total momentum of the system, the denominator its mass. Since Σp does not change, neither does the velocity of the centre of mass. Internal forces cannot move the centre of mass: when a person walks to the stern of a boat, the boat moves forward; when a cannon fires, the centre of mass of 'cannon + shell' stays where it was.

In the simulator the centre of mass is the golden marker. In a collision it glides at the same speed before and after the impact, and the stuck carts move along with it. For a rocket in space the centre of mass of 'rocket + exhaust gas' stays at the launch point, even though the rocket itself flies hundreds of kilometres away.

How a rocket moves. A rocket throws gas backwards at a speed u relative to itself. By conservation of momentum the gas carries momentum backwards, and the rocket gets the same momentum forwards: every second the engine ejects a mass μ and creates a thrust F = μu. No support is needed, so a rocket accelerates in empty space as well — where a propeller or a wheel has nothing to push against.

The Tsiolkovsky equation. Let a rocket of mass m throw out a portion of gas dm. The momentum of the system is conserved: m·dv = u·dm, so dv = u·dm/m. Adding up such steps while the mass falls from m₀ to m gives v = u·ln(m₀/m). The speed depends only on the exhaust speed and the mass ratio, not on how fast the fuel burns: a long burn gives the same speed, just with a smaller acceleration.

The rocket outruns its gas. At m₀/m = e ≈ 2.72 the rocket's speed equals u. Beyond that the rocket flies faster than its own exhaust, and the gas just thrown out (its speed relative to the launch point is v − u) already moves forward, following the rocket. To reach 7.9 km/s with u = 3 km/s you need m₀/m ≈ 14: 93 % of the initial mass is fuel. That is why rockets have stages and drop their empty tanks.

Launch from the Earth. On the Earth gravity pulls the rocket down: v = u·ln(m₀/m) − gt, and the longer the engine burns, the larger the “gravity losses” gt. If the thrust is less than the weight (μu < m₀g), the rocket stands on the pad until its mass drops to F/g — lift-off happens at t_lo = (m₀ − F/g)/μ. After burnout the rocket coasts upwards and slows down to the height H = h_f + v_f²/(2g).

The same law works in cannon recoil — a single “shot” instead of a continuous jet — and in an inelastic collision. The model is ideal: m₀ = 100 t, constant g = 9.8 m/s², no air drag, vertical flight.

Jet Propulsion and the Tsiolkovsky Rocket Equation

KEY FORMULAS

Conservation of momentum

Thrust

Tsiolkovsky rocket equation

Velocity of the centre of mass

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