SME
Orbital Mechanics
Orbital maneuvers
Use this for “A satellite in a circular orbit of radius 6778 km moves to geostationary radius, 42,164 km, by a Hohmann transfer. Find both burns and the time of flight.”
Diagram
a
=
r₁ + r₂
2
=
6,778 + 42,164
2
a = 24,470 km
Δv₁
=
√(μ(
2
r₁
−
1
a
))
−
√(
μ
r₁
)
Δv₁ = 10.07 − 7.669 = 2.398 km/s
Δv₂
=
√(
μ
r₂
)
−
√(μ(
2
r₂
−
1
a
))
Δv₂ = 3.075 − 1.618 = 1.457 km/s
TOF
=
π√(
a
3
μ
)
=
5.291
h,
half
the
transfer
orbit
This is an example. Type your own values to start a new problem.
r₁
Inner orbit radius
example · km
r₂
Outer orbit radius
example · km
a
Transfer semi-major axis
calculated · km
v₁
Inner circular speed
calculated · km/s
vₚ
Transfer speed at perigee
calculated · km/s
vₐ
Transfer speed at apogee
calculated · km/s
v₂
Outer circular speed
calculated · km/s
Δv₁
First burn
calculated · km/s
Δv₂
Second burn
calculated · km/s
TOF
Time of flight
calculated
Δv
Total Δv
worked out (not typed)
Formulas
a
=
r₁
+
r₂
2
24,471
=
6,778 + 42,164
2
v₁
=
√(
398,600
r₁
)
7.6686
=
√(
398,600
6,778
)
vₚ
=
√(398,600
×
(
2
r₁
−
1
a
))
10.0661
=
√(398,600
×
(
2
6,778
−
1
24,471
))
vₐ
=
√(398,600
×
(
2
r₂
−
1
a
))
1.6182
=
√(398,600
×
(
2
42,164
−
1
24,471
))
v₂
=
√(
398,600
r₂
)
3.0747
=
√(
398,600
42,164
)
Δv₁
=
vₚ
−
v₁
2.3975 = 10.0661 − 7.6686
Δv₂
=
v₂
−
vₐ
1.4565 = 3.0747 − 1.6182
Δv = Δv₁ + Δv₂
3.854 = 2.3975 + 1.4565
TOF
=
π
×
√(
a
3
398,600
)
19,048.4131
=
π
×
√(
24,
471
3
398,600
)
Formulas are worked in s. The step-by-step shows the conversions.
Assumptions
•
Coplanar circular orbits round Earth, μ = 398,600 km³/s², well inside its sphere of influence (about 925,000 km).
•
Impulsive burns along the motion; the cheapest two-burn transfer while r₂ ÷ r₁ < 11.94.
•
Vis-viva gives the ellipse’s speeds: v² = μ(2 ÷ r − 1 ÷ a).
•
The time of flight is half the transfer ellipse’s period; radii are from Earth’s center.
Step-by-step
WE KNOW
r₁ = 6,778 km, r₂ = 42,164 km
FIND
transfer semi-major axis (a), inner circular speed (v₁), transfer speed at perigee (vₚ), transfer speed at apogee (vₐ), outer circular speed (v₂), first burn (Δv₁), second burn (Δv₂), total Δv (Δv), time of flight (TOF)
Step 1 · Find transfer semi-major axis (a)
Use a = (r₁ + r₂) ÷ 2
The
ellipse
touches
both
circles,
so
its
major
axis
is
r₁
+
r₂
.
a
=
6,778 + 42,164
2
a
=
48,942
2
a
=
24,471
km
Step 2 · Find inner circular speed (v₁)
Use v₁ = √(398,600 ÷ r₁)
Circular speed: gravity supplies exactly the pull the circle needs.
v₁
=
√(
398,600
6,778
)
v₁
=
√
58.81
v₁
=
7.6686
km/s
Step 3 · Find transfer speed at perigee (vₚ)
Use vₚ = √(398,600 × (2 ÷ r₁ − 1 ÷ a))
Vis-viva
on
the
transfer
ellipse
at
perigee,
r
=
r₁
.
vₚ
=
√(398,600
×
(
2
6,778
−
1
24,471
))
vₚ
=
√(398,600
×
(0.00029507
−
(4.0865
×
10
-5
)))
vₚ
=
√
398,600 × 0.0002542
vₚ
=
√
101.32
vₚ = 10.0661 km/s
Step 4 · Find transfer speed at apogee (vₐ)
Use vₐ = √(398,600 × (2 ÷ r₂ − 1 ÷ a))
Vis-viva
on
the
transfer
ellipse
at
apogee,
r
=
r₂
.
vₐ
=
√(398,600
×
(
2
42,164
−
1
24,471
))
vₐ
=
√(398,600
×
((4.743
×
10
-5
)
−
(4.086
×
10
-5
)))
vₐ
=
√(398,600
×
(6.570
×
10
-6
))
vₐ
=
√
2.619
vₐ = 1.6182 km/s
Step 5 · Find outer circular speed (v₂)
Use v₂ = √(398,600 ÷ r₂)
Circular speed: gravity supplies exactly the pull the circle needs.
v₂
=
√(
398,600
42,164
)
v₂
=
√
9.454
v₂
=
3.0747
km/s
Step 6 · Find first burn (Δv₁)
Use Δv₁ = vₚ − v₁
The first burn speeds the craft up from the circle onto the ellipse.
Δv₁ = 10.0661 − 7.6686
Δv₁ = 2.3975 km/s
Step 7 · Find second burn (Δv₂)
Use Δv₂ = v₂ − vₐ
At apogee the craft is too slow for the outer circle; the second burn makes up the gap.
Δv₂ = 3.0747 − 1.6182
Δv₂ = 1.4565 km/s
Step 8 · Find total Δv (Δv)
Use Δv = Δv₁ + Δv₂
The two burns add up to the transfer’s fuel budget.
Δv = 2.3975 + 1.4565
Δv = 3.854 km/s
Step 9 · Find time of flight (TOF)
Use TOF = π × √(a³ ÷ 398,600)
Half the ellipse’s period by Kepler’s third law.
TOF
=
π
×
√(
24,
471
3
398,600
)
TOF
=
√(
14,653,965,039,111
398,600
)
×
π
TOF
=
√
36
,760,000
×
π
TOF = 6,063 × π
TOF = 19,048.4131 s
Step 10 · Convert the answers
Change each answer back to the unit you chose.
TOF = 19,048.4131 s = 5.2912 h (1 h = 3600 s)
Check
Put the numbers back in. Both sides should match.
24,471
=
6,778 + 42,164
2
✓
7.6686
=
√(
398,600
6,778
)
✓
10.0661
=
√(398,600
×
(
2
6,778
−
1
24,471
))
✓
1.6182
=
√(398,600
×
(
2
42,164
−
1
24,471
))
✓
3.0747
=
√(
398,600
42,164
)
✓
2.3975 = 10.0661 − 7.6686 ✓
1.4565 = 3.0747 − 1.6182 ✓
3.854 = 2.3975 + 1.4565 ✓
19,048.41
=
π
×
√(
24,
471
3
398,600
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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Combined circularize-and-turn burn
Use this for “At apogee of a transfer orbit a craft moves at 1.618 km/s. One burn reaches a 3.075 km/s circle and turns the plane 28.5°. Find Δv.”