SME
Propulsion
Rocket propulsion
Use this for “An engine with Iₛₚ = 300 s burns a 50 t rocket down to 15 t. What is its Δv? What propellant fraction gives 3000 m/s?”
Diagram
Δv
=
300
×
9.81
×
ln(
50
15
)
Δv = 3,543 m/s
Propellant
fraction:
1
−
15
50
=
0.7
This is an example. Type your own values to start a new problem.
Δv (m/s)
=
Iₛₚ (s)
×
g
×
ln(
m₀ (t)
÷
mf (t)
)
R
Mass ratio
calculated
ζ
Propellant fraction
calculated
Formulas
R
=
m₀
mf
3.3333
=
50
15
Δv
=
Iₛₚ
×
9.81
×
ln(
R
)
3,543.292 = 300 × 9.81 × ln(3.3333)
ζ
=
1
−
1
R
0.7
=
1
−
1
3.3333
Assumptions
•
Ideal Δv: no gravity or drag losses on the way.
•
The exhaust speed vₑ is constant through the burn.
•
g = 9.81 m/s² defines specific impulse, so vₑ = Iₛₚ × g.
Step-by-step
WE KNOW
Iₛₚ = 300 s, m₀ = 50 t, mf = 15 t
FIND
mass ratio (R), ideal velocity change (Δv), propellant fraction (ζ)
Step 1 · Find mass ratio (R)
Use R = m₀ ÷ mf
Divide the starting mass by the burnout mass.
R
=
50
15
R
=
3.333333
Step 2 · Find ideal velocity change (Δv)
Use Δv = Iₛₚ × 9.81 × ln(R)
Integrating
m
dv
=
−vₑ
dm
from
m₀
down
to
mf
gives
vₑ
ln(
m₀
mf
),
with
vₑ
=
Iₛₚ
×
g.
Δv = 300 × 9.81 × ln(3.333333)
Δv = 300 × 9.81 × 1.204
Δv = 2,943 × 1.204
Δv = 3,543.292 m/s
Step 3 · Find propellant fraction (ζ)
Use ζ = 1 − 1 ÷ R
The
share
of
the
starting
mass
that
is
propellant;
mf
m₀
is
1
over
the
mass
ratio.
ζ
=
1
−
1
3.333333
ζ
=
1
−
0.3
ζ
=
0.7
Check
Put the numbers back in. Both sides should match.
3.333333
=
50
15
✓
3,543.292 = 300 × 9.81 × ln(3.333333) ✓
0.7
=
1
−
1
3.333333
✓
Topic check
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