SME
Orbital Mechanics · Problem type
Position after a time (Kepler’s equation)
Use this for “An orbit has a = 24,471 km and e = 0.723. Where is the satellite 1.684 h after perigee?”
Diagram
r
=
a(1 −
e
2
)
1 + e cos ν
r
=
11,680
1 + 0.723 × −0.7853
=
27,020
km
E = 1.715 rad, M = E − e sin E = 1.715 − 0.723 × 0.9896 = 0.9999 rad
Swept
area
πab
=
M
2π
=
t
T
=
0.1591:
equal
areas
in
equal
times.
v
=
√(μ(
2
r
−
1
a
))
v
=
√(398,600
×
(
2
27,020
−
1
24,471
))
v = 3.635 km/s
That is the vis-viva equation, energy per kilogram kept along the orbit.
This is an example. Type your own values to start a new problem.
T (h)
=
2π
(
a (km)
)
3
÷
μ
M (rad)
=
2π
t (h)
÷
T (h)
M (rad)
=
E (rad)
−
e
sin
E (rad)
ν
True anomaly
calculated
r
Radius
calculated · km
Formulas
T
=
2π
×
√(
a
3
398,600
)
38,100
=
2π
×
√(
24,
471
3
398,600
)
M
=
2π
×
t
T
0.9999
=
2π
×
6,062.4
38,100
M
=
E
−
e
×
sin(
E
)
0.9999 = 1.715 − 0.723 × sin(1.715)
ν
=
2
×
tan⁻¹(√(
1 +
e
1 −
e
)
×
tan(
E
2
))
2.474
=
2
×
tan⁻¹(√(
1 + 0.723
1 − 0.723
)
×
tan(
1.715
2
))
r
=
a
×
(1
−
e
×
cos(
E
))
27,020 = 24,471 × (1 − 0.723 × cos(1.715))
Formulas are worked in s, rad. The step-by-step shows the conversions.
Assumptions
•
Two-body motion round Earth, μ = 398,600 km³/s²; t is the time since perigee.
•
M, E and ν are in radians; the mean anomaly M grows evenly with time.
•
The eccentric anomaly E is measured on the circle round the ellipse, from its centre.
•
This page covers the way out, perigee to apogee (t up to T ÷ 2); the way back mirrors it.
Step-by-step
WE KNOW
a = 24,471 km, e = 0.723, t = 1.684 h
FIND
period (T), mean anomaly (M), eccentric anomaly (E), true anomaly (ν), radius (r)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to s, rad).
t = 1.684 h = 6,062.4 s (1 h = 3600 s)
Step 2 · 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 3 · Find mean anomaly (M)
Use M = 2π × t ÷ T
The mean anomaly grows evenly, 2π each period.
M
=
2π
×
6,062.4
38,096.83
M
=
0.999852
rad
Step 4 · Find eccentric anomaly (E)
Use M = E − e × sin(E)
No
rearrangement
puts
E
alone,
so
Newton’s
method
finds
it:
each
next
E
is
E
−
E
−
e
sin
E
−
M
1 −
e
cos
E
.
E
=
M
+
e
×
sin(
E
)
Start
with
E₀
=
M
=
0.9999
0.9999 − (0.9999 − 0.723 × sin(0.9999) − 0.999852) ÷ (1 − 0.723 × cos(0.9999)) = 1.9983
1.9983 − (1.9983 − 0.723 × sin(1.9983) − 0.999852) ÷ (1 − 0.723 × cos(1.9983)) = 1.7363
1.7363 − (1.7363 − 0.723 × sin(1.7363) − 0.999852) ÷ (1 − 0.723 × cos(1.7363)) = 1.7155
1.7155 − (1.7155 − 0.723 × sin(1.7155) − 0.999852) ÷ (1 − 0.723 × cos(1.7155)) = 1.7153
1.7153 − (1.7153 − 0.723 × sin(1.7153) − 0.999852) ÷ (1 − 0.723 × cos(1.7153)) = 1.7153
E
=
1.715
rad
Step 5 · Find true anomaly (ν)
Use ν = 2 × tan⁻¹(√((1 + e) ÷ (1 − e)) × tan(E ÷ 2))
The
half
angles
go
together:
tan(
ν
2
)
=
√(
1 +
e
1 −
e
)
×
tan(
E
2
).
ν
=
2
×
tan⁻¹(√(
1 + 0.723
1 − 0.723
)
×
tan(
1.715
2
))
ν
=
2.4739
rad
Step 6 · Find radius (r)
Use r = a × (1 − e × cos(E))
The
radius
from
the
eccentric
anomaly:
a
(1
−
e
)
at
perigee,
a
(1
+
e
)
at
apogee.
r
=
24,471
×
(1
−
0.723
×
cos(1.7153))
r
=
24,471
×
(1
−
0.723
×
(−0.144))
r
=
24,471
×
(1
−
(−0.1041))
r
=
24,471
×
1.1041
r
=
27,020
km
Step 7 · Convert the answers
Change each answer back to the unit you chose.
T = 38,100 s = 10.58245 h (1 h = 3600 s)
ν = 2.474 rad = 141.7° (180° = π rad)
Check
Put the numbers back in. Both sides should match.
38,096.83
=
2π
×
√(
24,
471
3
398,600
)
✓
0.999852
=
2π
×
6,062.4
38,096.83
✓
0.999852 = 1.715315 − 0.723 × sin(1.715315) ✓
2.474
=
2
×
tan⁻¹(√(
1 + 0.723
1 − 0.723
)
×
tan(
1.7153
2
))
✓
27,020 = 24,471 × (1 − 0.723 × cos(1.7153)) ✓
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Radius at a true anomaly (the orbit equation)
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What i, Ω, ω and ν describe
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