SME
Classical Mechanics
Central forces
Use this for “A satellite passes 6800 km from Earth’s center at 8.8 km/s, its closest point. Find the orbit’s eccentricity, semi-major axis, farthest distance and period.”
Diagram
a
=
rₚ + rₐ
2
=
7,000 + 12,150
2
a = 9,574 km
v
=
√(GM(
2
r
−
1
a
))
v
=
√(398,600
×
(
2
7,000
−
1
9,574
))
v = 8.5 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.
e
=
rₚ (km)
(
vₚ (km/s)
)
2
GM (km³/s²)
−
1
=
At the closest point h = rₚ vₚ, and the orbit gives rₚ = h² ÷ (GM(1 + e)), solved here for e.
a
Semi-major axis
worked out (not typed)
rₐ
Farthest distance
worked out (not typed)
T
Period
worked out (not typed)
Formulas
e
=
rₚ
×
vₚ
2
GM
−
1
0.2688
=
7,000
×
8.5
2
398,600
−
1
1
a
=
2
rₚ
−
vₚ
2
GM
1
9,573.5118
=
2
7,000
−
8.5
2
398,600
rₐ
=
a
×
(1
+
e
)
12,147.0235 = 9,573.5118 × (1 + 0.2688)
T
=
2π
×
√(
a
3
GM
)
9,322.1939
=
2π
×
√(
9,
573.5118
3
398,600
)
Assumptions
•
The central body is far heavier, so it stays put (the one-body form; for two similar masses use the reduced mass m₁m₂ ÷ (m₁ + m₂)). GM: Earth 398,600 km³/s², Sun 1.327 × 10¹¹ km³/s².
•
Angular momentum and energy are conserved, and at the closest point the speed is square to r.
•
e < 1 is an ellipse; at e ≥ 1 the body escapes, and this page stops there.
Step-by-step
WE KNOW
GM = 398,600 km³/s², rₚ = 7,000 km, vₚ = 8.5 km/s
FIND
eccentricity (e), semi-major axis (a), farthest distance (rₐ), period (T)
Step 1 · Find eccentricity (e)
Use e = rₚ × vₚ² ÷ GM − 1
h
=
rₚvₚ
at
the
closest
point,
and
the
orbit
r
=
h
2
÷ GM
1 +
e
cos θ
at
θ
=
0
gives
1
+
e
=
rₚvₚ²
GM
.
e
=
7,000
×
8.5
2
398,600
−
1
e
=
7,000
×
72.25
398,600
−
1
e
=
505,750
398,600
−
1
e
=
1.2688
−
1
e
=
0.268816
Step 2 · Find semi-major axis (a)
Use 1 ÷ a = 2 ÷ rₚ − vₚ² ÷ GM
Energy
per
kilogram
at
rₚ,
1
2
vₚ
2
−
GM
rₚ
,
equals
−
GM
2
a
on
an
ellipse
(vis-viva).
a
=
1
2 ÷ rₚ −
vₚ
2
÷ GM
a
=
1
2 ÷ 7,000 −
8.5
2
÷ 398,600
a
=
1
2 ÷ 7,000 − 72.25 ÷ 398,600
a
=
1
0.00028571 − 0.00018126
a
=
1
0.00010445
a
=
9,573.51175
km
Step 3 · Find farthest distance (rₐ)
Use rₐ = a × (1 + e)
The farthest point is a past the center plus the focus’s offset ae.
rₐ = 9,573.51175 × (1 + 0.268816)
rₐ = 9,573.5118 × 1.2688
rₐ = 12,147.02 km
Step 4 · Find period (T)
Use T = 2π × √(a³ ÷ GM)
Kepler’s
third
law:
T
2
=
4π²
a
3
GM
,
so
the
period
depends
on
a
alone.
T
=
2π
×
√(
9,
573.51175
3
398,600
)
T
=
9,322.194
s
Check
Put the numbers back in. Both sides should match.
0.268816
=
7,000
×
8.5
2
398,600
−
1
✓
1
9,573.51175
=
2
7,000
−
8.5
2
398,600
✓
12,147.02 = 9,573.51175 × (1 + 0.268816) ✓
9,322.194
=
2π
×
√(
9,
573.51175
3
398,600
)
✓
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An orbit’s turning points from its energy
Topic check
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