University Physics I: Mechanics · Problem type

The rocket equation

Use this for “A 12,000 kg rocket burns fuel until it weighs 4,000 kg, its exhaust leaving at 3,000 m/s. How much speed does it gain?”

Chart

Δv(m) = 2,500 ln(5,000/m)01,0002,0003,0004,0005,00005001,0001,5002,0002,5003,0003,5004,0004,5005,000Mass left m (kg)Change in speed Δv (m/s)(2,000, 2,291)
Δv(2,000) ≈ 2,291
This is an example. Type your own values to start a new problem.
Δv (m/s)
=
u (m/s)
×
ln
m₀ (kg)
mf (kg)
R
Mass ratio m₀ ÷ mf
worked out (not typed)

Formulas

R
=
m₀
mf
2.5
=
5,000
2,000
Δv
=
u
×
ln(
R
)
2,290.7268 = 2,500 × ln(2.5)

Assumptions

•
No gravity or air drag acts during the burn (a rocket in deep space), so the momentum of the rocket and its exhaust together is kept.
•
The exhaust leaves at a steady speed u measured from the rocket, straight backward.
•
Throwing back a small mass dm gives m dv = u dm; adding these up as the mass falls from m₀ to mf gives Δv = u ln(m₀ ÷ mf).

Step-by-step

WE KNOW
u = 2,500 m/s, m₀ = 5,000 kg, mf = 2,000 kg
FIND
mass ratio m₀ ÷ mf (R), change in speed (Δv)
Step 1 · Find mass ratio m₀ ÷ mf (R)
Use R = m₀ ÷ mf
The mass ratio is the start mass over the end mass: how many times lighter the rocket gets.
R
=
5,000
2,000
R
=
2.5
Step 2 · Find change in speed (Δv)
Use Δv = u × ln(R)
The speed gained is the exhaust speed times the natural log of the mass ratio.
Δv = 2,500 × ln(2.5)
Δv = 2,500 × 0.9163
Δv = 2,290.727 m/s
Check
Put the numbers back in. Both sides should match.
2.5
=
5,000
2,000
✓
2,290.727 = 2,500 × ln(2.5) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

Related lessons

© 2026 SME. All rights reserved.