SME
Orbital Mechanics · Problem type
Geostationary orbit
Use this for “What orbit radius and altitude give a period of one sidereal day, 86,164 s, and how fast does that satellite move?”
Diagram
Orbital
speed:
v
=
√(
μ
r
)
=
√(
398,600
42,160
)
=
3.075
km/s
Period:
T
=
2π√(
r
3
μ
)
=
2π
×
√(
42,
160
3
398,600
)
=
86,200
s
(23.9
h)
Gravity
is
the
centripetal
force:
μ
r
2
=
v
2
r
,
so
the
satellite’s
own
mass
cancels.
It orbits 35,790 km above the surface (r − R, with R = 6,378 km).
This is an example. Type your own values to start a new problem.
r (km)
=
3
μ
(
T (s)
÷
2π)
2
z (km)
=
r (km)
−
R
v (km/s)
=
μ
÷
r (km)
Formulas
r
=
∛(398,600
×
(
T
2π
)
2
)
42,164.1245
=
∛(398,600
×
(
86,164
2π
)
2
)
z
=
r
−
6,378
35,786.1245 = 42,164.1245 − 6,378
v
=
√(
398,600
r
)
3.0747
=
√(
398,600
42,164.1245
)
Assumptions
•
Earth is a point mass with μ = 398,600 km³/s²; its equatorial radius is 6378 km.
•
Geostationary: a circular, equatorial orbit turning with Earth once a sidereal day, 86,164 s.
•
Kepler’s third law for a circle: T² = 4π²r³ ÷ μ.
Step-by-step
WE KNOW
T = 86,164 s
FIND
orbit radius (r), altitude (z), orbital speed (v)
Step 1 · Find orbit radius (r)
Use r = ∛(398,600 × (T ÷ 2π)²)
Kepler’s
third
law,
T
2
=
4π²
r
3
μ
,
solved
for
r
:
the
period
fixes
the
radius.
r
=
∛(398,600
×
(
86,164
2π
)
2
)
r
=
42,164.12
km
Step 2 · Find altitude (z)
Use z = r − 6,378
Altitude
is
measured
from
the
surface:
take
Earth’s
radius
off
r
.
z
=
42,164.12
−
6,378
z
=
35,786.12
km
Step 3 · Find orbital speed (v)
Use v = √(398,600 ÷ r)
Gravity
is
the
centripetal
force:
μ
r
2
=
v
2
r
,
so
v
2
=
μ
r
.
v
=
√(
398,600
42,164.12
)
v
=
√
9.454
v
=
3.0747
km/s
Check
Put the numbers back in. Both sides should match.
42,164.12
=
∛(398,600
×
(
86,164
2π
)
2
)
✓
35,786.12 = 42,164.12 − 6,378 ✓
3.0747
=
√(
398,600
42,164.12
)
✓
Topic check
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