Classical-Mechanics · Unit 13 · Video 8 · Interactive Practice

The Tyranny of the Logarithm: Why Rockets Stage and Burn Fast

IKey Formulas

Formula	Name	What you need
$v_{r,x,f} = v_{r,x,i} + u \ln R, \qquad R = \dfrac{m_{r,i}}{m_{r,d}}$	Rocket equation in free space	Exhaust speed $u$ and the mass ratio $R$
$m_r(t) = m_{r,i} - b\,t, \qquad v_{r,x}(t) = v_{r,x,i} + u \ln\dfrac{m_{r,i}}{m_{r,i} - b\,t}$	Constant burn rate	The ejection rate $b = -\,dm_r/dt$
$v_{r,x,f} = u \ln R_1 + u \ln R_2 = u \ln (R_1 R_2)$	Two-stage burn	Each stage ratio, measured after its own jettison
$v_{r,x}(t_f) = u \ln R - g\,t_f$	Vertical launch, constant $g$	Burn time $t_f$

Key Insight: Speed adds, but mass ratio multiplies — the logarithm turns a sum of stage gains into a product of stage ratios, and $t_f$ appears nowhere in the answer until gravity starts charging by the second.

IIThe Logarithmic Price of Speed

Each extra $u$ of speed costs another whole factor of $e$ in the mass ratio.

Propellant on the pad (×10⁷ kg) 2.46

💡 Structure sets the ceiling: tanks and engines can never weigh nothing, so a single stage cannot climb this curve indefinitely — the only way further right is to stop carrying part of $m_{r,d}$ .

IIIStaging — Two Ratios Multiply

Same fuel, same $u$ : what does cutting the empty first-stage tank loose at $150\ \text{s}$ buy?

Time since ignition (s) 0

IVThe Gravity Tax

Gravity charges by the second: $u \ln R$ is fixed, while $g\,t_f$ grows with every second of burn.

Burn time (s) 510

💡 Only the component of $\vec{g}$ along the velocity is taxed, so a real launch pitches over as early as it can — the full $-g\,t_f$ here is the worst case of a purely vertical climb.

VQuiz Questions

Problem 1 · Mass Ratio to Final Speed

Given: a probe at rest in deep space with total mass $m_{r,i} = 6.0\times10^{5}\ \text{kg}$ , of which $4.8\times10^{5}\ \text{kg}$ is propellant, ejected at $u = 2500\ \text{m/s}$ — find the final speed once the tanks are dry.

✅ Correct! The dry mass is

1.2\times10^{5}\ \text{kg}

, so

R = 5

and

v_{r,x,f} = 2500\ln 5 = 2500(1.6094) \approx 4020\ \text{m/s}

— a rocket that is

80\%

propellant buys only

1.6\,u

❌ That is $uR$ , not $u \ln R$ . Integrating

dv_{r,x} = -u\,dm_r/m_r

produces a logarithm, so speed grows with

\ln R

: five times the mass ratio is worth

1.609\,u

, not

5u

❌ Wrong logarithm.

\int dm_r/m_r

gives the natural log:

\ln 5 = 1.609

, while

\log_{10} 5 = 0.699

❌ The ratio is upside down. You evaluated

u\ln(m_{r,d}/m_{r,i})

; flipping the ratio inside a log flips the sign, and the rocket must speed up, so the larger mass goes on top.

Show solution

Step 1 — dry mass. Everything except propellant:

m_{r,d} = 6.0\times10^{5} - 4.8\times10^{5} = 1.2\times10^{5}\ \text{kg}

Step 2 — mass ratio.

R = \frac{m_{r,i}}{m_{r,d}} = \frac{6.0\times10^{5}}{1.2\times10^{5}} = 5.0

Step 3 — rocket equation (starting from rest, $v_{r,x,i} = 0$ ):

v_{r,x,f} = u \ln R = (2500)(\ln 5) = (2500)(1.6094) = 4023\ \text{m/s}

Sanity check: $m_{r,i} > m_{r,d}$ , so $R > 1$ , so $\ln R > 0$ and the rocket speeds up.

Problem 2 · Does the Burn Time Matter?

Given: two identical probes in free space, each starting from rest with identical propellant and identical $u$ . Probe A empties its tanks at a constant rate in $60\ \text{s}$ ; probe B does so in $600\ \text{s}$ — compare their final speeds.

✅ Correct! The integration ran over mass, not time:

\Delta v = u\ln(m_{r,i}/m_{r,d})

contains no

t

at all. In free space, burn time is invisible in the answer — it only reappears through

-g\,t_f

once there is a gravitational field.

❌ Nothing in $u \ln R$ scales with $b$ . Ten times the burn rate gives ten times the thrust for one tenth the time — the accumulated

\int u\,dm_r/m_r

is identical.

❌ Larger thrust, but for proportionally less time. A's acceleration is ten times bigger and lasts one tenth as long; both probes pass through exactly the same sequence of masses.

❌ Longer, but ten times weaker. B's thrust

ub

is one tenth of A's, so the extra

540\ \text{s}

buys nothing: the same propellant leaves at the same

u

Show solution

Separating variables gives one variable per side:

dv_{r,x} = -u\,\frac{dm_r}{m_r}

Integrating from ignition to burnout runs over mass, from $m_{r,i}$ to $m_{r,d}$ :

v_{r,x,f} - v_{r,x,i} = u\ln\frac{m_{r,i}}{m_{r,d}} = u\ln R

No time limit ever entered. With a constant burn rate $b$ the intermediate history does depend on $b$ ,

v_{r,x}(t) = u\ln\frac{m_{r,i}}{m_{r,i} - b\,t}

but at burnout $b\,t_f = m_{r,i} - m_{r,d}$ for both probes, so both arrive at the same $u \ln R$ . This is exactly why the $510\ \text{s}$ given in the video's single-stage example never appeared in the arithmetic.

Problem 3 · Two Stages, One Product

Given: a rocket at rest in free space with total mass $1.00\times10^{5}\ \text{kg}$ . Stage one burns $6.0\times10^{4}\ \text{kg}$ of propellant, after which the empty stage-one structure ( $1.0\times10^{4}\ \text{kg}$ ) is jettisoned; stage two then burns the remaining $2.0\times10^{4}\ \text{kg}$ of propellant. Both stages exhaust at $u = 3000\ \text{m/s}$ .

What is the product $R_1 R_2$ ?

What is the final speed?

✅ Correct!

R_1 = 2.5

and

R_2 = 3.0

, so

v_{r,x,f} = 3000\ln 7.5 = 6045\ \text{m/s}

— versus

3000\ln 5 = 4828\ \text{m/s}

if the same propellant were burned in one stage, a gain of about

1215\ \text{m/s}

from dropping

1.0\times10^{4}\ \text{kg}

at the right moment.

❌ That is the single-stage ratio.

5.0 = 1.00\times10^{5} / 2.0\times10^{4}

carries the empty stage-one structure all the way to burnout; jettisoning it removes that dead mass from stage two's denominator.

❌ Ratios multiply, they do not add. Speeds add —

u\ln R_1 + u\ln R_2

— and the logarithm converts that sum into

u\ln(R_1R_2)

, a product.

❌ Check the two ratios separately.

R_1

uses the mass before and after the first burn;

R_2

starts from the mass remaining after the jettison.

❌ That is the no-jettison result.

3000\ln 5 = 4828\ \text{m/s}

is what this rocket gets if the empty stage-one structure rides along to the end.

❌ You took $u\ln(R_1 + R_2)$ . The sum of the two speed gains is

u\ln R_1 + u\ln R_2 = u\ln(R_1R_2)

, so the ratios multiply inside the logarithm.

❌ Not quite. Once you have the product

R_1R_2

, one logarithm finishes the job:

v_{r,x,f} = u\ln(R_1R_2)

Show solution

Stage one. Burning $6.0\times10^{4}\ \text{kg}$ leaves

m_{r,1,d} = 1.00\times10^{5} - 6.0\times10^{4} = 4.0\times10^{4}\ \text{kg}, \qquad R_1 = \frac{1.00\times10^{5}}{4.0\times10^{4}} = 2.5

Jettison. Dropping $1.0\times10^{4}\ \text{kg}$ of empty structure leaves $m_{r,2,i} = 3.0\times10^{4}\ \text{kg}$ .

Stage two. Burning the last $2.0\times10^{4}\ \text{kg}$ leaves

m_{r,2,d} = 3.0\times10^{4} - 2.0\times10^{4} = 1.0\times10^{4}\ \text{kg}, \qquad R_2 = \frac{3.0\times10^{4}}{1.0\times10^{4}} = 3.0

Add the speeds, multiply the ratios.

v_{r,x,f} = u\ln R_1 + u\ln R_2 = u\ln(R_1R_2) = 3000\ln 7.5 = (3000)(2.0149) = 6045\ \text{m/s}

Compare. One stage with the same $8.0\times10^{4}\ \text{kg}$ of propellant ends at $m_{r,d} = 2.0\times10^{4}\ \text{kg}$ , so $R = 5$ and $v_{r,x,f} = 3000\ln 5 = 4828\ \text{m/s}$ . The $1215\ \text{m/s}$ difference is the price of accelerating $1.0\times10^{4}\ \text{kg}$ of empty tank through the whole second burn.

Problem 4 · Burnout Speed Against Gravity

Given: a rocket rising vertically from rest through a constant field $g = 9.8\ \text{m/s}^2$ , with $u = 2600\ \text{m/s}$ , mass ratio $R = 6.0$ , and a burn lasting $t_f = 120\ \text{s}$ — find the speed at burnout.

✅ Correct!

u\ln R = 2600(1.7918) = 4659\ \text{m/s}

and

g\,t_f = (9.8)(120) = 1176\ \text{m/s}

, leaving

3483\ \text{m/s}

— gravity took a quarter of the prize in only two minutes.

❌ That is the free-space answer.

u\ln R = 4659\ \text{m/s}

is only the first term; standing in a field costs an extra

-g\,t_f

❌ Sign error. With

+x

upward the weight is

-m_r(t)g

, so the gravity term subtracts:

u\ln R - g\,t_f

❌ The tax is not halved.

\int_0^{t_f}(-g)\,dt' = -g\,t_f

exactly;

\tfrac{1}{2}g\,t_f^2

is a displacement, not a speed.

❌ Not quite. Evaluate the two terms separately:

u\ln R

first, then subtract

g\,t_f

Show solution

With $+x$ upward, $F_{ext,x} = -m_r(t)g$ , and separating variables gives two independent pieces:

dv_{r,x} = -g\,dt - u\,\frac{dm_r}{m_r(t)}

Integrating from launch to burnout:

v_{r,x}(t_f) = u \ln R - g\,t_f

Term 1 (the prize):

u\ln R = (2600)(\ln 6.0) = (2600)(1.7918) = 4659\ \text{m/s}

Term 2 (the tax):

g\,t_f = (9.8)(120) = 1176\ \text{m/s}

v_{r,x}(t_f) = 4659 - 1176 = 3483\ \text{m/s}

Only the second term knows about time. Burning the same propellant in $60\ \text{s}$ instead would leave $4659 - 588 = 4071\ \text{m/s}$ — the same mass ratio, $588\ \text{m/s}$ cheaper. That is the whole argument for a short, violent first stage.

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