SME
Orbital Mechanics
Interplanetary transfers
Use this for “A probe leaves a 300 km parking orbit (rₚ = 6678 km) on a Hohmann transfer from Earth’s orbit, 1.496 × 10⁸ km, to Mars’s, 2.279 × 10⁸ km. Find v∞, the burn and the flight time.”
Diagram
a
=
r₁ + r₂
2
a
=
1.496 ×
10
8
+ 2.279 ×
10
8
2
a
=
1.888 ×
10
8
km
v∞
=
√(μ(
2
r₁
−
1
a
))
−
√(
μ
r₁
)
v∞ = 32.73 − 29.78 = 2.943 km/s
TOF
=
π√(
a
3
μ
)
=
258.8
days,
half
the
transfer
orbit
Departure:
Δv
=
√(v∞²
+
2μ
rₚ
)
−
√(
μ
rₚ
)
=
11.32
−
7.726
=
3.59
km/s
This is an example. Type your own values to start a new problem.
r₁
Earth’s orbit radius
example · km
r₂
Target planet’s orbit radius
example · km
v∞
Hyperbolic excess speed
calculated · km/s
rₚ
Parking orbit radius
example · km
vc
Parking orbit speed
calculated · km/s
Δv
Departure burn
calculated · km/s
TOF
Time of flight
calculated
Formulas
v∞
=
√(1.327
×
10
11
×
(
2
r₁
−
2
r₁
+
r₂
))
−
√(
1.327 ×
10
11
r₁
)
2.9433
=
√(1.327
×
10
11
×
(
2
1.496 ×
10
8
−
2
1.496 ×
10
8
+ (2.279 ×
10
8
)
))
−
√(
1.327 ×
10
11
1.496 ×
10
8
)
vc
=
√(
398,600
rₚ
)
7.7258
=
√(
398,600
6,678
)
Δv
=
√(v∞²
+
2
×
398,600
rₚ
)
−
vc
3.5897
=
√(
2.9433
2
+
2
×
398,600
6,678
)
−
7.7258
TOF
=
π
×
√(
((
r₁
+
r₂
) ÷ 2)³
1.327 ×
10
11
)
2.2364 ×
10
7
=
π
×
√(((1.496
×
10
8
+
(2.279
×
10
8
))
÷
2)³
÷
(1.327
×
10
11
))
Formulas are worked in s. The step-by-step shows the conversions.
Assumptions
•
Circular, coplanar planet orbits; patched conics: only the Sun pulls between spheres of influence.
•
μSun = 1.327 × 10¹¹ km³/s² and μEarth = 398,600 km³/s²; radii from each body’s center.
•
One impulsive burn in the parking orbit sends the craft out on a hyperbola leaving at v∞.
•
For an inner planet r₂ < r₁ and v∞ is negative: the craft leaves against Earth’s motion.
Step-by-step
WE KNOW
r₁ = 1.496 × 10⁸ km, r₂ = 2.279 × 10⁸ km, rₚ = 6,678 km
FIND
hyperbolic excess speed (v∞), parking orbit speed (vc), departure burn (Δv), time of flight (TOF)
Step 1 · Find hyperbolic excess speed (v∞)
Use v∞ = √(1.327 × 10¹¹ × (2 ÷ r₁ − 2 ÷ (r₁ + r₂))) − √(1.327 × 10¹¹ ÷ r₁)
Vis-viva
gives
the
transfer’s
speed
at
r₁
;
take
away
the
planet’s
own
circular
speed
round
the
Sun.
v∞
=
√(1.327
×
10
11
×
(
2
1.496 ×
10
8
−
2
1.496 ×
10
8
+ (2.279 ×
10
8
)
))
−
√(
1.327 ×
10
11
1.496 ×
10
8
)
vₜ
=
√(1.327
×
10
11
×
(
2
1.496 ×
10
8
−
2
1.496 ×
10
8
+ (2.279 ×
10
8
)
))
=
32.73
v₁
=
√(
1.327 ×
10
11
1.496 ×
10
8
)
=
29.78
v∞ = 32.73 − 29.78
v∞ = 2.9433 km/s
Step 2 · Find parking orbit speed (vc)
Use vc = √(398,600 ÷ rₚ)
The parking orbit’s circular speed round Earth.
vc
=
√(
398,600
6,678
)
vc
=
√
59.69
vc = 7.7258 km/s
Step 3 · Find departure burn (Δv)
Use Δv = √(v∞² + 2 × 398,600 ÷ rₚ) − vc
Energy on the hyperbola gives its speed at rₚ; take away the circular speed already there.
Δv
=
√(
2.9433
2
+
2
×
398,600
6,678
)
−
7.7258
Δv
=
√(8.663
+
2
×
398,600
6,678
)
−
7.7258
Δv
=
√(8.663
+
797,200
6,678
)
−
7.7258
Δv
=
√
8.663 + 119.38
−
7.7258
Δv
=
√
128.043
−
7.7258
Δv = 11.316 − 7.7258
Δv = 3.5897 km/s
Step 4 · Find time of flight (TOF)
Use TOF = π × √(((r₁ + r₂) ÷ 2)³ ÷ (1.327 × 10¹¹))
Half
the
transfer
orbit’s
period,
whose
a
=
r₁
+
r₂
2
.
TOF
=
π
×
√(((1.496
×
10
8
+
(2.279
×
10
8
))
÷
2)³
÷
(1.327
×
10
11
))
TOF
=
√(
((3.775 ×
10
8
) ÷ 2)³
1.327 ×
10
11
)
×
π
TOF
=
√(
(1.8875 ×
10
8
)
3
÷
(1.327
×
10
11
))
×
π
TOF
=
√(
6.725 ×
10
24
1.327 ×
10
11
)
×
π
TOF
=
√
5.068 ×
10
13
×
π
TOF = 7,119,000 × π
TOF
=
2.2364 ×
10
7
s
Step 5 · Convert the answers
Change each answer back to the unit you chose.
TOF = 2.2364 × 10⁷ s = 258.8398 days (1 day = 86400 s)
Check
Put the numbers back in. Both sides should match.
2.9433
=
√(1.327
×
10
11
×
(
2
1.496 ×
10
8
−
2
1.496 ×
10
8
+ (2.279 ×
10
8
)
))
−
√(
1.327 ×
10
11
1.496 ×
10
8
)
✓
7.7258
=
√(
398,600
6,678
)
✓
3.5897
=
√(
2.9433
2
+
2
×
398,600
6,678
)
−
7.7258
✓
2.2364 ×
10
7
=
π
×
√(((1.496
×
10
8
+
(2.279
×
10
8
))
÷
2)³
÷
(1.327
×
10
11
))
✓
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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Sphere of influence
Use this for “Earth orbits 1.496 × 10⁸ km from the Sun and has 3.003 × 10⁻⁶ of its mass. How far out does Earth’s pull take over from the Sun’s?”