Centre of mass and rockets

Every calculation so far has treated cars, blocks and planets as points, which is obviously false, and this lesson supplies the theorem that makes it legitimate anyway.

Carried in: the second law in the form $\sum \vec{F} = d \vec{p} / d t$ , the third law, and the fact that the total momentum of a system is unchanged by internal forces.

A system of particles

Take $N$ particles with masses $m_{i}$ at positions ${\vec{r}}_{i}$ , total mass $M = \sum m_{i}$ . Define the centre of mass as the mass weighted average position:

\vec{R} = \frac{1}{M} \sum_{i} m_{i} {\vec{r}}_{i}

In components, $X = \sum m_{i} x_{i} / M$ and likewise for $Y$ . Nothing physical has been claimed yet; this is a definition, and its justification is what follows from it.

Differentiate twice with respect to time. Since the masses are constant,

M \vec{R} = \sum_{i} m_{i} {\vec{r}}_{i} \Rightarrow M \dot{\vec{R}} = \sum_{i} m_{i} {\vec{v}}_{i} = \vec{P} \Rightarrow M \ddot{\vec{R}} = \sum_{i} m_{i} {\vec{a}}_{i} = \sum_{i} {\vec{F}}_{i}

The last sum is over every force on every particle, and it splits into internal forces, which particles exert on each other, and external ones. By the third law the internal forces occur in equal and opposite pairs, so they cancel exactly, leaving

\sum {\vec{F}}_{ext} = M {\vec{A}}_{cm}

This is the result the course has been assuming since its third lesson. The centre of mass of any system moves exactly as a single particle of the total mass would, driven by the external forces alone. The system may be a spinning wrench, an exploding shell, a crowd of people or a galaxy; the internal complexity is irrelevant to that one point.

It also says $\vec{P} = M {\vec{V}}_{cm}$ , so the momentum of a whole system is the total mass times the velocity of its centre of mass, which is why momentum conservation is the statement that the centre of mass keeps moving uniformly when no external force acts.

Finding it

For a few discrete masses the definition is arithmetic. For a continuous body the sum becomes an integral,

X = \frac{1}{M} \int x d m

and the work is in expressing $d m$ in terms of $d x$ . For a uniform rod of length $L$ and mass $M$ , $d m = (M / L) d x$ and the integral gives $X = L / 2$ , as symmetry already said. For a rod whose density grows linearly along its length, $d m = c x d x$ , and

X = \frac{\int_{0}^{L} c x \cdot x d x}{\int_{0}^{L} c x d x} = \frac{L^{3} / 3}{L^{2} / 2} = \frac{2 L}{3}

Symmetry is the shortcut worth taking whenever it is available: the centre of mass of a uniform body lies on every plane of symmetry it has, which locates it instantly for a sphere, a cube or a cylinder. It need not lie inside the body at all. The centre of mass of a ring is at its centre, where there is no material, and the centre of mass of a high jumper arched over the bar in a Fosbury flop passes underneath the bar while every part of the jumper passes over it, which is the entire reason the technique won the 1968 Olympics.

Example. Three particles lie in a plane: 2.0 kg at the origin, 3.0 kg at (4.0, 0) and 5.0 kg at (6.0, 3.0), in metres. Where is the centre of mass?

The total mass is 10.0 kg. Then $X = [2 (0) + 3 (4) + 5 (6)] / 10 = 42 / 10 = 4.20$ m and $Y = [2 (0) + 3 (0) + 5 (3)] / 10 = 15 / 10 = 1.50$ m. The centre of mass is at (4.20, 1.50) m, which is not at any particle and is pulled towards the 5.0 kg mass, as a weighted average should be.

Now you. Three particles: 1.0 kg at the origin, 2.0 kg at (3.0, 4.0) and 4.0 kg at (5.0, 0), in metres. Find the centre of mass.

Answer

$M = 7.0$ kg, $X = [0 + 6 + 20] / 7 = 3.71$ m and $Y = [0 + 8 + 0] / 7 = 1.14$ m.

What the theorem buys

Three familiar facts become one line each.

A hammer thrown spinning across a room follows a complicated path in every part except one: its centre of mass traces a clean parabola, because gravity is the only external force. Photographs of this are the standard demonstration, and they show a point moving smoothly through a body that is tumbling wildly around it.

A shell fired on a parabola and exploding in flight has fragments flying in all directions, but the explosion is internal, so the centre of mass of the fragments carries on along the original parabola until the first piece lands.

And the Earth does not orbit the Sun. Both orbit their common centre of mass, and the same is true of the Earth and the Moon: with $m_{M} / (m_{E} + m_{M}) = 7.342 \times 10^{22} / 6.045 \times 10^{24}$ and a separation of $3.844 \times 10^{8}$ m, the barycentre is $4.67 \times 10^{6}$ m from the Earth's centre, which is 1700 km beneath the surface. The Earth wobbles about that point once a month, and the same wobble applied to stars is one of the ways planets around other stars are detected.

Example. A shell is fired from level ground at 80 m s⁻¹ at 50° and explodes into two equal fragments at the top of its flight. One fragment drops vertically from rest and lands directly below the burst. Where does the other land?

The undisturbed range would be $R = v_{0}^{2} \sin 2 θ / g = 6400 \sin 100^{\circ} / 9.81 = 642.5$ m, and the burst is at the halfway point horizontally, 321.2 m from the launch. The explosion is internal, so the centre of mass continues on the original parabola and reaches 642.5 m at the moment the pieces land, both of which land together since they fall from the same height. With equal masses, the centre of mass is midway between them, so $(321.2 + x) / 2 = 642.5$ , giving $x = 963.7$ m.

Now you. The same experiment with a shell fired at 60 m s⁻¹ at 45°. Where does the second fragment land?

Answer

$R = 3600 \sin 90^{\circ} / 9.81 = 367.0$ m, the burst is above 183.5 m, and $(183.5 + x) / 2 = 367.0$ gives $x = 550.5$ m.

The zero momentum frame

Since $\vec{P} = M {\vec{V}}_{cm}$ , a frame moving with the centre of mass is one in which the total momentum is exactly zero. It is an inertial frame whenever no external force acts, so all the mechanics of this course works in it, and it makes the energetics of a collision transparent.

Split the kinetic energy of a system into two parts: the energy of the whole moving together, $\frac{1}{2} M V_{cm}^{2}$ , and the energy of the internal motion relative to the centre of mass. Momentum conservation fixes ${\vec{V}}_{cm}$ once and for all, so the first part is untouchable, and a collision can only spend the second. That is the reason a perfectly inelastic collision is the maximum loss case: in the zero momentum frame both bodies end at rest, and every joule that could be lost has been.

The car collision of the previous lesson makes the point numerically. A 1200 kg car at 20 m s⁻¹ striking a stationary 900 kg car gives $V_{cm} = 11.43$ m s⁻¹, so the untouchable energy is $\frac{1}{2} (2100) ({11.43}^{2}) = 137$ kJ of the original 240 kJ. The internal energy, and therefore the absolute maximum that could go into crushing metal, is $240 - 137 = 103$ kJ, which is precisely the loss that was calculated there. The shortcut is $\frac{1}{2} μ v_{rel}^{2}$ with the reduced mass $μ = m_{1} m_{2} / (m_{1} + m_{2}) = 514$ kg, giving $\frac{1}{2} (514) (400) = 103$ kJ directly.

Systems that shed mass

Now the case the second law handles badly. A rocket accelerates by throwing mass backwards, so its mass falls as it goes, and writing $F = m a$ with a changing $m$ is a well known route to a wrong answer. The fix is to apply momentum conservation to a fixed collection of matter: the rocket plus the fuel it is about to eject, considered together over a short interval.

At time $t$ the rocket has mass $m$ and velocity $v$ , so the momentum of the whole is $m v$ . In the interval $d t$ it ejects a mass $| d m |$ backwards at a speed $v_{e x}$ relative to itself, and the rocket's mass becomes $m + d m$ , with $d m$ negative. Afterwards the momentum is $(m + d m) (v + d v)$ for the rocket plus $(- d m) (v - v_{e x})$ for the exhaust. Setting the two totals equal in the absence of external forces and cancelling $m v$ :

m d v + v d m + d m d v = - d m v + d m v_{e x} + v d m

Discarding the second order term $d m d v$ and simplifying leaves

m d v = - v_{e x} d m

which separates at once:

\int_{v_{0}}^{v} d v = - v_{e x} \int_{m_{0}}^{m} \frac{d m}{m} \Rightarrow Δ v = v_{e x} \ln \frac{m_{0}}{m}

This is the rocket equation, derived by Konstantin Tsiolkovsky in 1897 and, unknown to him, by William Moore in 1813. Its form is the whole difficulty of spaceflight. The velocity gained depends on the logarithm of the mass ratio, so every extra increment of speed costs exponentially more propellant, and the exhaust speed multiplies everything.

Real rockets, and why they are built in stages

The quantity engineers quote is the specific impulse $I_{s p} = v_{e x} / g_{0}$ , in seconds. The Saturn V's first stage F-1 engines burned kerosene and oxygen at $I_{s p} = 263$ s at sea level, an exhaust speed of 2579 m s⁻¹; its hydrogen burning second stage reached $I_{s p} = 421$ s in vacuum, or 4129 m s⁻¹, which is why hydrogen was worth the trouble of keeping it at 20 K.

Example. A rocket with an exhaust speed of 3000 m s⁻¹ burns until its mass has fallen to one fifth of its starting value. What speed does it gain in free space?

$Δ v = 3000 \ln 5 = 3000 \times 1.609 = 4828$ m s⁻¹. Notice how little the fifth of the mass that remains has bought: reaching low Earth orbit needs about 9400 m s⁻¹ once gravity and drag losses are included, and with the same engine that would require a mass ratio of $e^{9400 / 3000} = 23.0$ , meaning the empty vehicle, engines, tanks and payload together would have to be 4.4 per cent of the launch mass. No material is good enough, which is why rockets are staged: the tanks that have been emptied are thrown away so that they need not be accelerated further.

Now you. A stage with an exhaust speed of 2500 m s⁻¹ has a mass ratio of 8. What is its ideal velocity change?

Answer

$Δ v = 2500 \ln 8 = 2500 \times 2.079 = 5199$ m s⁻¹.

The derivation assumed no external forces, which is why the result is called the ideal velocity change. A launch from the ground also fights gravity throughout the burn, losing roughly $g$ times the burn duration, and fights atmospheric drag for the first minute. Those two together account for the gap between the 7800 m s⁻¹ needed for orbital speed and the 9400 m s⁻¹ actually budgeted.

Everything that is left

The centre of mass theorem is a licence and a limitation in the same sentence. It says that one point of any body moves in a way that eight lessons of this course can already predict. It says nothing whatever about the motion of the body around that point.

That is not a small omission. A wrench thrown across a room tumbles; a wheel rolls; a diver somersaults; a planet spins. In each case the centre of mass does something simple and the interesting behaviour is elsewhere. Worse, the missing motion has its own conserved quantity, which is how a skater speeds up by pulling their arms in without anything pushing them.

Describing that motion needs new variables, because position and velocity do not capture an orientation, and it needs a new measure of inertia, because a body's resistance to being spun depends on where its mass sits and not merely how much of it there is. That is the next lesson.