SME
Classical Mechanics · Problem type
A Hohmann transfer between circular orbits
Use this for “A satellite in a circular orbit 7000 km from Earth’s center moves to one at 26,600 km on a Hohmann transfer. Find both burns, the total Δv and the coast time.”
Diagram
a
=
r₁ + r₂
2
=
6,700 + 42,164
2
a = 24,432 km
Δv₁
=
√(GM(
2
r₁
−
1
a
))
−
√(
GM
r₁
)
Δv₁ = 10.13 − 7.713 = 2.42 km/s
Δv₂
=
√(
GM
r₂
)
−
√(GM(
2
r₂
−
1
a
))
Δv₂ = 3.075 − 1.61 = 1.465 km/s
TOF
=
π√(
a
3
GM
)
=
5.279
h,
half
the
transfer
orbit
This is an example. Type your own values to start a new problem.
GM
Gravitational parameter
example · km³/s²
r₁
Inner orbit radius
example
r₂
Outer orbit radius
example
Δv₁
First burn
worked out (not typed)
Δv₂
Second burn
worked out (not typed)
Δv
Total Δv
worked out (not typed)
t
Transfer time
worked out (not typed)
Formulas
Δv₁
=
√(
GM
r₁
)
×
(√(2
×
r₂
r₁
+
r₂
)
−
1)
2.4195
=
√(
398,600
6,700
)
×
(√(2
×
42,164
6,700 + 42,164
)
−
1)
Δv₂
=
√(
GM
r₂
)
×
(1
−
√(2
×
r₁
r₁
+
r₂
))
1.4646
=
√(
398,600
42,164
)
×
(1
−
√(2
×
6,700
6,700 + 42,164
))
Δv = Δv₁ + Δv₂
3.8841 = 2.4195 + 1.4646
t
=
π
×
√(
((
r₁
+
r₂
) ÷ 2)³
GM
)
÷
3,600
5.2786
=
π
×
√(
((6,700 + 42,164) ÷ 2)³
398,600
)
÷
3,600
Assumptions
•
Both orbits are circles in one plane round a far heavier body (Earth: GM = 398,600 km³/s²); radii from its center.
•
Two short burns along the motion: one at r₁ onto an ellipse with rₚ = r₁ and rₐ = r₂, one at r₂ onto the outer circle.
•
Each Δv is the vis-viva speed on the ellipse less, or from, the circular speed √(GM ÷ r); the coast is half the ellipse’s period.
Step-by-step
WE KNOW
GM = 398,600 km³/s², r₁ = 6,700 km, r₂ = 42,164 km
FIND
first burn (Δv₁), second burn (Δv₂), total Δv (Δv), transfer time (t)
Step 1 · Find first burn (Δv₁)
Use Δv₁ = √(GM ÷ r₁) × (√(2 × r₂ ÷ (r₁ + r₂)) − 1)
The
ellipse’s
speed
at
perigee,
√(
GM
r₁
)
×
√(
2
r₂
÷
(
r₁
+
r₂
))
by
vis-viva,
less
the
inner
circle’s
speed.
Δv₁
=
√(
398,600
6,700
)
×
(√(2
×
42,164
6,700 + 42,164
)
−
1)
Δv₁
=
√(
398,600
6,700
)
×
(√(2
×
42,164
48,864
)
−
1)
Δv₁
=
√(
398,600
6,700
)
×
(√(
84,328
48,864
)
−
1)
Δv₁
=
√(
398,600
6,700
)
×
(
√
1.7258
−
1)
Δv₁
=
√
59.493
×
(1.3137
−
1)
Δv₁
=
√
59.493
×
0.3137
Δv₁ = 7.7132 × 0.3137
Δv₁ = 2.4195 km/s
Step 2 · Find second burn (Δv₂)
Use Δv₂ = √(GM ÷ r₂) × (1 − √(2 × r₁ ÷ (r₁ + r₂)))
The outer circle’s speed less the ellipse’s slower speed at apogee.
Δv₂
=
√(
398,600
42,164
)
×
(1
−
√(2
×
6,700
6,700 + 42,164
))
Δv₂
=
√(
398,600
42,164
)
×
(1
−
√(2
×
6,700
48,864
))
Δv₂
=
√(
398,600
42,164
)
×
(1
−
√(
13,400
48,864
))
Δv₂
=
√(
398,600
42,164
)
×
(1
−
√
0.27423
)
Δv₂
=
√
9.4536
×
(1
−
0.52367)
Δv₂
=
√
9.4536
×
0.47633
Δv₂ = 3.0747 × 0.47633
Δv₂ = 1.4646 km/s
Step 3 · Find total Δv (Δv)
Use Δv = Δv₁ + Δv₂
The two burns add up to the fuel budget.
Δv = 2.4195 + 1.4646
Δv = 3.8841 km/s
Step 4 · Find transfer time (t)
Use t = π × √(((r₁ + r₂) ÷ 2)³ ÷ GM) ÷ 3,600
Half
the
period
of
the
ellipse
with
a
=
r₁
+
r₂
2
(Kepler’s
third
law),
seconds
to
hours.
t
=
π
×
√(
((6,700 + 42,164) ÷ 2)³
398,600
)
÷
3,600
t
=
π
×
√(
(
48,864
2
)
3
÷
398,600)
÷
3,600
t
=
π
×
√
24,
432
3
÷ 398,600
3,600
t
=
π
×
√
14,584,013,549,568 ÷ 398,600
÷
3,600
t
=
π
×
√
36,590,000
3,600
t
=
π
×
6,049
3,600
t
=
19,000
3,600
t
=
5.2786
h
Check
Put the numbers back in. Both sides should match.
2.4195
=
√(
398,600
6,700
)
×
(√(2
×
42,164
6,700 + 42,164
)
−
1)
✓
1.4646
=
√(
398,600
42,164
)
×
(1
−
√(2
×
6,700
6,700 + 42,164
))
✓
3.8841 = 2.4195 + 1.4646 ✓
5.2786
=
π
×
√(
((6,700 + 42,164) ÷ 2)³
398,600
)
÷
3,600
✓
Topic check
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