SME
Orbital Mechanics · Problem type
Radius at a true anomaly (the orbit equation)
Use this for “An orbit has a = 24,471 km and e = 0.723. How far from Earth’s centre is the satellite at true anomaly 90°?”
Diagram
r
=
a(1 −
e
2
)
1 + e cos ν
=
11,680
1 + 0.723 × 0
r = 11,680 km
v
=
√(μ(
2
r
−
1
a
))
v
=
√(398,600
×
(
2
11,680
−
1
24,471
))
v = 7.209 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.
r (km)
=
a (km)
(1
−
e
2
)
÷
(1
+
e
cos
ν
°)
Formulas
r
=
a
×
1 −
e
2
1 +
e
× cos(
ν
)
11,680
=
24,471
×
1 −
0.723
2
1 + 0.723 × cos(90°)
Assumptions
•
Two-body motion: the orbit is a fixed ellipse with Earth’s centre at one focus.
•
The true anomaly ν is the angle at Earth’s centre from perigee to the satellite.
•
ν runs 0° to 180° on the way out; the way back mirrors it, so 360° − ν gives the same r.
Step-by-step
WE KNOW
a = 24,471 km, e = 0.723, ν = 90°
FIND
radius (r)
Step 1 · Find radius (r)
Use r = a × (1 − e²) ÷ (1 + e × cos(ν))
The
orbit
equation:
the
semi-latus
rectum
a
(1
−
e
2
)
shrunk
or
stretched
by
1
+
e
cos
ν
.
r
=
24,471
×
1 −
0.723
2
1 + 0.723 × cos(90°)
r
=
24,471
×
1 − 0.5227
1 + 0.723 × 0
r
=
24,471
×
0.4773
r
=
11,680
km
Check
Put the numbers back in. Both sides should match.
11,680
=
24,471
×
1 −
0.723
2
1 + 0.723 × cos(90°)
✓
Topic check
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