SME
Orbital Mechanics
Orbital elements
Use this for “A transfer orbit has perigee radius 6778 km and apogee radius 42,164 km. Find a, e, T, h, vₚ and vₐ.”
Diagram
a
=
rₚ + rₐ
2
=
6,778 + 42,164
2
a = 24,471 km
v
=
√(μ(
2
r
−
1
a
))
v
=
√(398,600
×
(
2
6,778
−
1
24,471
))
v = 10.07 km/s
That is the vis-viva equation, energy per kilogram kept along the orbit.
Fastest at perigee, slowest at apogee: energy is shared between speed and height.
This is an example. Type your own values to start a new problem.
rₚ
Perigee radius
example · km
rₐ
Apogee radius
example · km
a
Semi-major axis
calculated · km
e
Eccentricity
calculated
T
Period
calculated
h
Specific angular momentum
calculated · km²/s
vₚ
Speed at perigee
calculated · km/s
vₐ
Speed at apogee
calculated · km/s
Formulas
a
=
rₚ + rₐ
2
24,471
=
6,778 + 42,164
2
e
=
rₐ − rₚ
rₐ + rₚ
0.72302
=
42,164 − 6,778
42,164 + 6,778
T
=
2π
×
√(
a
3
398,600
)
38,100
=
2π
×
√(
24,
471
3
398,600
)
h
=
√(398,600
×
a
×
(1
−
e
2
))
68,230
=
√(398,600
×
24,471
×
(1
−
0.72302
2
))
vₚ
=
h
rₚ
10.07
=
68,230
6,778
vₐ
=
h
rₐ
1.618
=
68,230
42,164
Formulas are worked in s. The step-by-step shows the conversions.
Assumptions
•
Two-body motion: Earth is a point mass with μ = 398,600 km³/s², and nothing else pulls.
•
Radii are from Earth’s centre, not altitudes: add 6378 km to an altitude.
•
h is the angular momentum per kilogram, r × v; it is the same all round the orbit.
Step-by-step
WE KNOW
rₚ = 6,778 km, rₐ = 42,164 km
FIND
semi-major axis (a), eccentricity (e), period (T), specific angular momentum (h), speed at perigee (vₚ), speed at apogee (vₐ)
Step 1 · Find semi-major axis (a)
Use a = (rₚ + rₐ) ÷ 2
The
long
axis
runs
from
perigee
to
apogee,
so
it
is
rₚ
+
rₐ;
a
is
half
of
it.
a
=
6,778 + 42,164
2
a
=
48,942
2
a
=
24,471
km
Step 2 · Find eccentricity (e)
Use e = (rₐ − rₚ) ÷ (rₐ + rₚ)
The
focus
sits
ae
from
the
centre,
and
rₐ
−
rₚ
=
2ae
while
rₐ
+
rₚ
=
2
a
.
e
=
42,164 − 6,778
42,164 + 6,778
e
=
35,386
48,942
e
=
0.72302
Step 3 · Find period (T)
Use T = 2π × √(a³ ÷ 398,600)
Kepler’s
third
law:
the
period
depends
on
a
alone,
not
on
e
.
T
=
2π
×
√(
24,
471
3
398,600
)
T
=
38,096.8262
s
Step 4 · Find specific angular momentum (h)
Use h = √(398,600 × a × (1 − e²))
The
semi-latus
rectum
p
=
a
(1
−
e
2
)
is
h
2
μ
,
so
h
=
√
μp
.
h
=
√(398,600
×
24,471
×
(1
−
0.72302
2
))
h
=
√(398,600
×
24,471
×
(1
−
0.5228))
h
=
√
398,600 × 24,471 × 0.4772
h
=
√((9.754
×
10
9
)
×
0.4772)
h
=
√
4.655 ×
10
9
h
=
68,230
km²/s
Step 5 · Find speed at perigee (vₚ)
Use vₚ = h ÷ rₚ
At
perigee
the
velocity
is
square
to
the
radius,
so
h
=
rₚ
×
vₚ.
vₚ
=
68,230
6,778
vₚ = 10.07 km/s
Step 6 · Find speed at apogee (vₐ)
Use vₐ = h ÷ rₐ
At
apogee
the
velocity
is
also
square
to
the
radius,
so
h
=
rₐ
×
vₐ.
vₐ
=
68,230
42,164
vₐ = 1.618 km/s
Step 7 · Convert the answers
Change each answer back to the unit you chose.
T = 38,100 s = 10.58 h (1 h = 3600 s)
Check
Put the numbers back in. Both sides should match.
24,471
=
6,778 + 42,164
2
✓
0.72302
=
42,164 − 6,778
42,164 + 6,778
✓
38,100
=
2π
×
√(
24,
471
3
398,600
)
✓
68,230
=
√(398,600
×
24,471
×
(1
−
0.72302
2
))
✓
10.07
=
68,230
6,778
✓
1.618
=
68,230
42,164
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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