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Conservation of momentum and angular momentum for systems, the center of mass, and the rocket equation $\Delta v = v_{\text{ex}}\ln(m_0/m_f)$ with thrust and gravity losses.
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By the end of this lesson you will be able to apply momentum conservation to systems, locate and follow a center of mass, derive and use the rocket equation, and find a rocket's thrust and acceleration.
From Physics C you know impulse and momentum, the center of mass of a set of particles, collisions, and torque and angular momentum for rotation about a fixed axis. The first lesson of this course used the third law to show that internal forces cancel. This lesson turns that into the conservation laws of a system, and applies them to the case every other method struggles with: a body that throws its own mass away.
| Term | What it means |
|---|---|
| Total momentum | $\vec{P} = \sum m_\alpha\dot{\vec{r}}_\alpha$, changed only by external forces. |
| Center of mass | $\vec{R} = \sum m_\alpha\vec{r}_\alpha/M$, which moves as a particle of mass $M$ under the external force. |
| Exhaust speed | $v_{\text{ex}}$, the speed of the ejected gas relative to the rocket. |
| Thrust | $-\dot{m}v_{\text{ex}}$, the forward force from ejecting mass. |
| Mass ratio | $m_0/m_f$, launch mass over burnout mass. |
| Angular momentum | $\vec{\ell} = \vec{r} \times \vec{p}$ for a particle; $\vec{L}$ is the sum over a system. |
| Torque | $\vec{\Gamma} = \vec{r} \times \vec{F}$, the rate of change of angular momentum. |
Sum Newton's second law over the particles of a system. The internal forces cancel in pairs by the third law, leaving
$$\dot{\vec{P}} = \vec{F}^{\text{ext}}, \qquad M\ddot{\vec{R}} = \vec{F}^{\text{ext}}.$$
The total momentum changes only by external force, and the center of mass $\vec{R} = \sum m_\alpha\vec{r}_\alpha/M$ moves exactly as a single particle of mass $M$ would. With no external force, momentum is conserved.
A rocket applies this to a system that includes its own exhaust. In a time $dt$ it ejects mass $-dm$ backward at speed $v_{\text{ex}}$ relative to itself. Momentum before and after is the same, which gives $m\,dv = -v_{\text{ex}}\,dm$. The rocket gains speed as it loses mass. Integrating,
$$v - v_0 = v_{\text{ex}}\ln\frac{m_0}{m},$$
the rocket equation. Dividing $m\,dv = -v_{\text{ex}}\,dm$ by $dt$ gives $m\dot{v} = -\dot{m}v_{\text{ex}}$: the thrust is the mass ejected per second times the exhaust speed.
Another way: picture
Picture an astronaut on frictionless ice with a pile of bricks, throwing them backward one at a time. Each throw pushes her forward a little, and each push matters more because she has fewer bricks left to carry. A rocket does the same with gas molecules, trillions per second. The last bits of fuel are the most effective, because by then the rocket is light.
Another way: steps
At time $t$ the rocket has mass $m$ and speed $v$, so momentum $mv$. At $t + dt$ it has ejected mass $-dm$ (a positive amount, since $dm < 0$) at speed $v - v_{\text{ex}}$ in the ground frame, and itself has mass $m + dm$ and speed $v + dv$. The total momentum is
$$(m + dm)(v + dv) + (-dm)(v - v_{\text{ex}}) = mv + m\,dv + dm\,v_{\text{ex}},$$
dropping the product of two small quantities. Setting this equal to $mv$ with no external force gives $m\,dv = -v_{\text{ex}}\,dm$. Separating, $dv = -v_{\text{ex}}\,dm/m$, and integrating from $m_0$ to $m$ gives the rocket equation. It was first written by Konstantin Tsiolkovsky in 1903 and, independently, by Robert Goddard in Massachusetts, whose 1926 flight at Auburn was the first liquid-fueled rocket.
The rocket equation says the speed gain grows only as the logarithm of the mass ratio. To gain one exhaust speed the rocket must be $63$ percent fuel; to gain two, $86$ percent; to gain three, $95$ percent. Low Earth orbit needs about $9.4$ km/s once gravity and drag losses are included, and chemical engines have exhaust speeds of $2.5$ to $4.5$ km/s, so a single-stage rocket would have to be more than $90$ percent fuel, leaving almost nothing for tanks, engines and payload.
Staging escapes the tyranny. Dropping empty tanks and engines partway means the later burns do not accelerate dead weight, and each stage's gains add. That is why every orbital launcher has at least two stages, and why reusing the first stage, as SpaceX does by landing Falcon 9 boosters, is where the savings are.
Near the ground, gravity acts during the burn: $m\dot{v} = -\dot{m}v_{\text{ex}} - mg$. The rocket lifts off only if its thrust exceeds its weight, and its initial acceleration is $(\text{thrust} - \text{weight})/m$, often only a fraction of $g$. As fuel burns, the mass falls and the acceleration grows, so a launch starts slowly and finishes hard: the Space Shuttle throttled its engines to keep the crew below $3g$.
Gravity also costs speed. Integrating with gravity included gives $v = v_{\text{ex}}\ln(m_0/m) - gt$: every second spent climbing loses $g\,t$ of speed, the gravity loss. A fast burn loses less, which is the one place burn rate matters. In free space the rate has no effect at all, as the equation without $g$ shows.
Because $M\ddot{\vec{R}} = \vec{F}^{\text{ext}}$, the center of mass follows a simple path whatever the system does inside. A diver tumbling off a board in Indianapolis has her center of mass on a parabola; a firework shell that bursts in the air has the center of mass of its fragments continue on the shell's parabola until the first fragment lands.
For a continuous body the sum becomes an integral, $\vec{R} = \frac{1}{M}\int\vec{r}\,dm$. The Earth–Moon system's center of mass lies $4700$ km from Earth's center, inside the Earth, and both bodies orbit it each month; the Earth's wobble around that point is how astronomers detect planets around other stars, by the tiny back-and-forth motion of the star.
The same argument works for angular momentum. For a particle, $\vec{\ell} = \vec{r} \times \vec{p}$ and $\dot{\vec{\ell}} = \vec{r} \times \vec{F}$, the torque, since $\dot{\vec{r}} \times \vec{p} = 0$. For a system, the internal torques cancel in pairs provided the internal forces act along the lines joining the particles, the strong form of the third law, leaving $\dot{\vec{L}} = \vec{\Gamma}^{\text{ext}}$.
A useful split: the total angular momentum is the angular momentum of the center of mass moving as a particle, plus the angular momentum of the motion about the center of mass. The Earth has both, orbital and spin, and each is nearly conserved separately. A cat dropped upside down, with zero angular momentum, still lands on its feet by twisting parts of its body in opposite directions, never violating conservation.
Checking an answer. A center of mass must lie within the span of the masses. A rocket's gain must be zero when no fuel is burned, since $\ln 1 = 0$, and must equal $v_{\text{ex}}\ln 2 \approx 0.69\,v_{\text{ex}}$ when half the mass is burned. Thrust in newtons is kg/s times m/s. And an answer in which the speed gain depends on the burn time, in free space, has gone wrong somewhere.
Rocket engineers quote an engine's performance as its specific impulse, $I_{sp} = v_{\text{ex}}/g_0$ with $g_0 = 9.81$ m/s², measured in seconds. It is the number of seconds one unit of propellant weight can hold up one unit of thrust, and it is simply the exhaust speed in disguise: a kerosene and liquid oxygen engine near $I_{sp} = 310$ s has $v_{\text{ex}} \approx 3.0$ km/s, and a hydrogen and oxygen engine near $450$ s has $4.4$ km/s.
The exhaust speed is set by chemistry and nozzle design. Burning hydrogen releases more energy per kilogram of exhaust, and its light water molecules come out faster, which is why the upper stages of the Saturn V and the Space Launch System burn hydrogen. But liquid hydrogen is bulky and must be kept near $20$ K, so first stages, which need high thrust more than high exhaust speed, often burn denser kerosene or methane. Choosing a propellant is a trade between the logarithm, which rewards exhaust speed, and the tanks, which reward density.
A launch from Florida must reach about $7.8$ km/s horizontally for low Earth orbit, and loses roughly another $1.5$ km/s to gravity and drag on the way up. Launching east gains about $0.4$ km/s from the Earth's rotation at that latitude, which is why American launch sites are on the east coast and launches head out over the Atlantic.
A two-stage kerosene rocket has exhaust speeds near $3$ km/s at sea level and $3.4$ km/s in vacuum. The first stage, with a mass ratio near $4$, gives about $3$ km/s once gravity losses are paid; the upper stage, with a mass ratio near $5$, gives about $5.5$ km/s. Together they reach orbit with a payload that is only about four percent of the launch mass. Every kilogram of payload needs about twenty-five kilograms of rocket, almost all of it fuel, which is what the logarithm in the rocket equation costs.
An ion engine accelerates xenon ions through an electric field to exhaust speeds of $20$ to $50$ km/s, ten times a chemical rocket's. Its thrust is tiny, a fraction of a newton, about the weight of a sheet of paper, because it ejects only milligrams per second. But in free space the rocket equation does not care: NASA's Dawn spacecraft, launched from Florida in 2007, gained over $11$ km/s from its ion engines while burning only about $400$ kg of xenon.
With $v_{\text{ex}} = 30$ km/s, a mass ratio of $1.4$ gives $30\ln 1.4 = 10$ km/s. A chemical rocket would need a mass ratio of $e^{10/3.4} = 19$ for the same gain. The price is time: Dawn's engines ran for years in total, which is fine for a probe cruising to the asteroid belt, and useless for escaping the Earth's gravity, where the thrust must exceed the weight.
It seems that a more powerful engine should give a faster rocket. In free space it does not: the rocket equation contains the exhaust speed and the masses, and no time. Burning twice as fast doubles the thrust and halves the burn time, and the final speed is the same. What does make a difference is the exhaust speed, which is why engineers prize engines with high specific impulse, and the mass ratio.
A second error is to think the rocket pushes against the air or the ground. It pushes against its own exhaust, which is why rockets work best in the vacuum of space, where there is no air to get in the way.
Earth has $5.97 \times 10^{24}$ kg and the Moon $7.35 \times 10^{22}$ kg, $3.84 \times 10^8$ m apart. Put the origin at Earth's center.
$x_E = 0, \quad x_M = 3.84 \times 10^8\ \text{m}$
Measuring from one body makes its term vanish.
Write the center of mass.
$X = \dfrac{m_Ex_E + m_Mx_M}{m_E + m_M}$
The mass-weighted average.
Substitute the masses and positions.
$X = \dfrac{7.35 \times 10^{22} \times 3.84 \times 10^8}{6.04 \times 10^{24}}$
Only the Moon's term survives in the numerator.
Evaluate the position.
$X = 4.67 \times 10^6\ \text{m}$
About $4700$ km from Earth's center.
Compare with Earth's radius.
$4.67 \times 10^6 < 6.37 \times 10^6\ \text{m}$
The balance point is inside the Earth, about three-quarters of the way out.
A probe with $v_{\text{ex}} = 3.0$ km/s burns until its mass is a quarter of its launch mass. Write momentum conservation for an instant.
$m\,dv = -v_{\text{ex}}\,dm$
From balancing momentum before and after ejecting $-dm$.
Separate the variables.
$dv = -v_{\text{ex}}\dfrac{dm}{m}$
Divide both sides by $m$.
Integrate from launch to burnout.
$\Delta v = -v_{\text{ex}}\ln\dfrac{m_f}{m_0} = v_{\text{ex}}\ln\dfrac{m_0}{m_f}$
The antiderivative of $1/m$ is $\ln m$.
Substitute the mass ratio.
$\Delta v = 3.0 \times \ln 4$
$m_0/m_f = 4$.
Evaluate the speed gain.
$\Delta v = 3.0 \times 1.386 = 4.16\ \text{km/s}$
More than the exhaust speed, because the ratio exceeds $e$.
A first stage burns $13{,}000$ kg/s with $v_{\text{ex}} = 2.6$ km/s. Find the thrust.
$F = |\dot{m}|v_{\text{ex}} = 13{,}000 \times 2600 = 3.38 \times 10^7\ \text{N}$
Mass per second times exhaust speed.
The launch mass is $2.90 \times 10^6$ kg. Find the weight.
$W = mg = 2.90 \times 10^6 \times 9.8 = 2.84 \times 10^7\ \text{N}$
Near the Earth's surface.
Compare thrust and weight.
$\dfrac{F}{W} = \dfrac{3.38}{2.84} = 1.19$
Greater than one, so it lifts off.
Find the initial acceleration.
$a = \dfrac{F - W}{m} = \dfrac{3.38 \times 10^7 - 2.84 \times 10^7}{2.90 \times 10^6} = 1.86\ \text{m/s}^2$
Newton's second law with gravity as an external force.
Find the acceleration after half the mass has burned.
$a = \dfrac{3.38 \times 10^7 - 1.42 \times 10^7}{1.45 \times 10^6} = 13.5\ \text{m/s}^2$
The same thrust on half the mass: the launch speeds up as it goes.
Find the speed lost to gravity in a $150$ s burn.
$\Delta v_{\text{grav}} = gt = 9.8 \times 150 = 1470\ \text{m/s}$
Every second spent climbing against gravity costs $g$ of speed, which a faster burn would reduce.
Write the rocket equation.
$\Delta v = v_{\text{ex}}\ln\dfrac{m_0}{m_f}$
Free space.
Substitute the mass ratio.
$\Delta v = 4.0 \times \ln 2$
Half the mass burned means $m_0/m_f = 2$.
Evaluate the speed gain.
Two identical probes in deep space carry the same fuel and the same engines' exhaust speed. One burns all its fuel in $17$ s, the other in $170$ s. How do their final speed gains compare?
Complete the worked solution: at liftoff a rocket of mass $20$ tonnes burns $400$ kg of propellant per second with exhaust speed $3$ km/s. With $g = 10$ m/s², find the thrust and the weight in kN, and the initial acceleration in m/s².
Multiply the mass flow rate by the exhaust speed.
$F = |\dot{m}|v_{\text{ex}} =$ f
The momentum carried off per second is the force on the rocket.
Multiply the mass by the gravitational field.
$W = Mg =$ w
Tonnes times meters per second squared gives kilonewtons.
Divide the net force by the mass.
$a = \dfrac{F - W}{M} =$ a
Newton's second law at the instant of liftoff.
Check that the rocket actually rises.
$\text{thrust greater than weight} \Rightarrow a > 0$
Otherwise it would sit on the pad burning fuel.
Match each quantity to its expression.
| $-\dot{m}v_{\text{ex}}$ | $v_{\text{ex}}\ln(m_0/m)$ | $\sum m_\alpha\vec{r}_\alpha/M$ | $\vec{\Gamma}^{\text{ext}}$ | |
|---|---|---|---|---|
| thrust | ||||
| speed gained | ||||
| center of mass | ||||
| rate of change of angular momentum |
Three carts on a straight track have masses $1$ kg, $2$ kg and $7$ kg, at $x = 2$ m, $7$ m and $13$ m, moving at $4$ m/s, $-4$ m/s and $2$ m/s. Fill in the total momentum in kg m/s, the velocity of the center of mass in m/s, and its position in m.
| value | |
|---|---|
| total momentum | |
| center-of-mass velocity | |
| center-of-mass position |
A rocket in free space starts at rest with mass $50$ tonnes, and its exhaust speed is $4$ km/s. Write its remaining mass $m$, in tonnes, as a formula in the speed $v$ it has gained, in km/s.
Answer:
A single-stage rocket with exhaust speed $2.5$ km/s must gain $5$ km/s in free space. What percentage of its launch mass must be fuel?
Answer: percent fuel
After separation over the Atlantic, the upper stage of a rocket launched from Cape Canaveral ignites with a total mass of $110$ tonnes and burns until it is $25$ tonnes. Its engine's exhaust speed is $3.4$ km/s. Ignoring gravity during the short burn, what speed does the stage gain, in km/s?
Answer: km/s gained by the stage
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A rocket in free space starts at rest with mass $87$ tonnes, and its exhaust speed is $2$ km/s. Write its remaining mass $m$, in tonnes, as a formula in the speed $v$ it has gained, in km/s.
Answer:
You can use momentum conservation for systems and rockets. Explain to someone why rockets are built in stages.
18. Your turn: a rocket in free space with $v_{\text{ex}} = 4.0$ km/s burns half its mass. Find its speed gain., step 3
$\Delta v = 4.0 \times 0.693 = 2.77\ \text{km/s}$
About $0.69$ of the exhaust speed.