Watch a launch from Sriharikota on television. For the first few seconds the rocket seems to hang above the pad, climbing slowly on a column of fire. Then it rises faster and faster, and within two minutes it is a bright dot. The engines burn at a steady rate all the while. Yet the rocket's speed does not grow steadily.
Systems that lose or gain mass
A rocket is a body whose mass keeps falling. Every second it throws out tonnes of hot gas. The rocket alone is an open system: mass leaves it, and that mass carries momentum away. The law \(\vec F = m\vec a\), with one fixed \(m\), cannot be used for it. Instead, follow a closed system, the same matter throughout: the rocket together with the gas it throws out in a short time.
Far from any planet, no external force acts on this closed system. So its momentum at \(t + \Delta t\) equals its momentum at \(t\). The derivation below writes both and compares them. The exhaust speed \(u\) is the speed of the gas relative to the rocket, which the engine fixes, whatever the rocket's own speed.
\[ P(t) = mv \]
Why. at time \(t\), the rocket and the fuel still inside it have mass \(m\) and velocity \(v\).
\[ P(t + \Delta t) = (m - \Delta m)(v + \Delta v) + \Delta m\,(v - u) \]
Why. the rocket is lighter by \(\Delta m\) and faster by \(\Delta v\); the gas it threw out moves at \(u\) backwards relative to the rocket.
\[ m\,\Delta v - \Delta m\,\Delta v - u\,\Delta m = 0 \]
Why. set \(P(t + \Delta t) = P(t)\) and expand. The terms \(mv\), \(-\Delta m\,v\) and \(\Delta m\,v\) cancel.
\[ m\,\Delta v = u\,\Delta m \]
Why. \(\Delta m\,\Delta v\) is a product of two small quantities and vanishes in the limit.
\[ m\,dv = -u\,dm \]
Read it. the rocket's own mass changes by \(dm = -\Delta m\), a negative amount, as it burns.
Why can \(F = ma\) not be applied, with a fixed \(m\), to a rocket alone?
Thrust
Divide \(m\,dv = -u\,dm\) by \(dt\). The left side becomes \(m\,dv/dt\), the rocket's mass times its acceleration. The right side becomes \(u(-dm/dt)\). Here \(-dm/dt\) is the rate at which the rocket burns fuel, a positive number in kg/s. So the rocket accelerates as if pushed by a force of size \(u(-dm/dt)\). This force is the thrust of the escaping gas.
The same bookkeeping works for a body that gains mass, like a cart catching sand. Let \(\vec v_{\text{rel}}\) be the velocity of the incoming or outgoing mass relative to the body. Then the thrust is \(\vec v_{\text{rel}}\,dm/dt\), with \(dm/dt\) positive when the body gains mass. Add any real external force, such as gravity, and the result is one equation for every open system.
A \(2000\,\text{kg}\) rocket stands on its pad. Its engine throws out gas at \(40\,\text{kg/s}\), at \(1500\,\text{m/s}\) relative to the rocket. Take \(g = 10\,\text{m/s}^2\). Find the thrust and the initial acceleration. Find also the least burn rate at which it could leave the pad.
Thrust: \(F = u(-dm/dt) = 1500 \times 40 = 60\,000\,\text{N}\), upward, opposite to the gas.
Weight: \(2000 \times 10 = 20\,000\,\text{N}\). So \(2000\,a = 60\,000 - 20\,000\), and \(a = 20\,\text{m/s}^2\).
To leave the pad the thrust must exceed the weight: \(1500 \times (-dm/dt) > 20\,000\), so the burn rate must exceed \(13.3\,\text{kg/s}\).
When half the mass is gone, the same thrust acts on \(1000\,\text{kg}\) against a weight of \(10\,000\,\text{N}\), and \(a = 50\,\text{m/s}^2\). The acceleration grows as the rocket lightens.
A \(500\,\text{kg}\) rocket hovers at rest just above its pad. Its exhaust speed is \(2500\,\text{m/s}\) relative to the rocket. Take \(g = 10\,\text{m/s}^2\). At what rate must it burn fuel, in kg/s?
A \(3000\,\text{kg}\) rocket on its pad throws out gas at \(50\,\text{kg/s}\), at \(1200\,\text{m/s}\) relative to the rocket. Take \(g = 10\,\text{m/s}^2\). Find its upward acceleration at lift-off, in m/s².
A rocket needs no air to push against. The rocket pushes the gas backwards, and the gas pushes the rocket forwards; both bodies are part of the rocket system. The chapter on Forces and Newton's Laws told how Robert Goddard was mocked for this in 1920. A rocket works best in empty space, where no air drags on it.
The rocket equation
The account \(m\,dv = -u\,dm\) holds at every instant of the burn. Each kilogram burnt adds a little speed, and it adds more when the rocket is already light. Adding all the small gains is an integral. The derivation below does it for a rocket in free space that starts from rest with mass \(m_0\). One step is left for you to choose.
\[ dv = -u\,\dfrac{dm}{m} \]
Why. divide the account \(m\,dv = -u\,dm\) by the mass \(m\).
\[ \int_0^v dv = -u\int_{m_0}^{m}\dfrac{dm}{m} \]
Why. integrate. The speed runs from \(0\) to \(v\) while the mass runs from \(m_0\) down to \(m\).
?
Your turn. the integral of \(1/m\) is \(\ln m\). Which line is it?
\[ v = u\ln\dfrac{m_0}{m} \]
Read it. \(\ln m_0 - \ln m = \ln(m_0/m)\), which is positive since \(m < m_0\).
Read the curve in Fig 5.2 with the equation. The mass falls by the same \(15\,\text{kg}\) every second. But the last \(15\,\text{kg}\) is a bigger share of a light rocket than the first \(15\,\text{kg}\) is of a heavy one. The same thrust acts on less and less mass, so the acceleration keeps growing. Each halving of the mass adds the same speed, \(u\ln 2 \approx 0.693\,u\). The mass ratio matters only through its logarithm.
A rocket in free space starts from rest with an exhaust speed of \(2.5\,\text{km/s}\). It burns until its mass is a quarter of the starting mass. Use \(\ln 2 = 0.693\). Find its final speed. What mass ratio would make it move as fast as its own exhaust?
\(v = u\ln(m_0/m) = 2500 \times \ln 4 = 2500 \times 2 \times 0.693 \approx 3465\,\text{m/s}\), about \(3.5\,\text{km/s}\).
To reach \(v = u\), the equation needs \(\ln(m_0/m) = 1\), so \(m_0/m = e \approx 2.72\). Nothing special happens then: the gas still leaves at \(u\) relative to the rocket.
The answer does not depend on the burn rate. A slow burn and a quick burn in free space give the same final speed.
A rocket starts from rest in free space with an exhaust speed of \(3000\,\text{m/s}\). What is its speed when it has burnt half its starting mass, in m/s? Use \(\ln 2 = 0.693\).
Climbing against gravity
A rocket rising from the Earth feels its weight as a real external force. The equation of motion becomes \(m\,dv/dt = u(-dm/dt) - mg\). Divide by \(m\) and integrate over a steady burn that lasts a time \(t\), with \(g\) taken as constant. The gravity term adds \(-gt\) to the result.
A \(1200\,\text{kg}\) rocket rises vertically, burning \(15\,\text{kg/s}\) for \(60\,\text{s}\) with an exhaust speed of \(2000\,\text{m/s}\). Take \(g = 10\,\text{m/s}^2\) and ignore drag. Find its speed at burnout and its acceleration at lift-off and at burnout.
The fuel burnt is \(15 \times 60 = 900\,\text{kg}\), so the mass at burnout is \(300\,\text{kg}\) and \(m_0/m = 4\). The thrust is \(15 \times 2000 = 30\,000\,\text{N}\).
\(v = u\ln(m_0/m) - gt\), and \(ma = F_{\text{thrust}} - mg\) at each instant.
\(v = 2000\ln 4 - 10 \times 60 = 2773 - 600 = 2173\,\text{m/s}\). At lift-off \(a = (30\,000 - 12\,000)/1200 = 15\,\text{m/s}^2\). At burnout \(a = (30\,000 - 3000)/300 = 90\,\text{m/s}^2\).
The rocket reaches about \(2170\,\text{m/s}\), \(600\,\text{m/s}\) less than in free space. Its acceleration grows sixfold during the burn.
The loss \(gt\) grows with the time of the burn. A rocket that burns the same fuel more slowly spends longer holding itself up against gravity, and it ends slower. So launchers burn their fuel quickly. They also throw away empty tanks and engines as they go. The next stage then starts from a much smaller \(m_0\), and its mass ratio does real work.
In free space the final speed of a rocket does not depend on how quickly it burns its fuel. Explain why the burn rate does matter for a rocket launched from the ground.
In free space \(v = u\ln(m_0/m)\), which contains no time, so a slow burn and a quick burn end at the same speed. Near the Earth gravity acts during the whole burn and removes \(gt\) from the speed. A slow burn lasts longer, so gravity takes more. Burning quickly reduces the time and so the loss.
A rocket must reach about \(9\,\text{km/s}\) to go into orbit, with drag and gravity losses included. Its engines have an exhaust speed of about \(3\,\text{km/s}\). Estimate what percentage of its lift-off mass a single-stage rocket would have to burn.
Konstantin Tsiolkovsky was a largely deaf schoolteacher in the small Russian town of Kaluga. In 1903 he published a paper on the exploration of space by reaction devices. In it he derived the equation of this lesson, \(v = u\ln(m_0/m)\). He saw that orbit needs a mass ratio too large for one rocket to reach. In 1929 he proposed trains of rockets, each dropped when its fuel was spent: the stages of every modern launcher.
ISRO's Polar Satellite Launch Vehicle has four stages, stacked one above the other. The first stage lifts the whole vehicle off the pad and is dropped when its fuel is spent. Each later stage starts light, with only the stages above it to carry. So each one gains a good mass ratio, and their speeds add. This is the rocket equation turned into engineering.
Now return to the launch at Sriharikota. The engines burn at a steady rate, so the thrust stays nearly constant. But the rocket's mass falls quickly, and the same thrust acts on less and less mass. The acceleration therefore grows through the burn, as the example found. The early seconds are slow also because gravity takes \(g\) from every second of the climb.
- Applying \(F = ma\) with a fixed mass to a rocket.Use \(M\,dv/dt = F_{\text{ext}} + v_{\text{rel}}\,dm/dt\), with the mass at that instant.
- Taking the exhaust speed as a speed over the ground.\(u\) is relative to the rocket. The engine fixes it, whatever the rocket's speed.
- Writing \(\ln(m/m_0)\) and getting a negative speed.The mass ratio is \(m_0/m\), the starting mass over the present mass, which is greater than one.
- Forgetting the \(-gt\) for a vertical launch.Gravity acts during the whole burn. In free space only is \(v = u\ln(m_0/m)\).