SME
Classical Mechanics · Problem type
An orbit’s turning points from its energy
Use this for “A satellite has energy per mass ε = −25 km²/s² and angular momentum per mass h = 55,000 km²/s round Earth. How close and how far does it go?”
Chart
Level ε = −20.82 meets the curve at rₘᵢₙ = 7,001, rₘₐₓ = 12,140
This is an example. Type your own values to start a new problem.
ε
r²
+
GM
r
−
½
(
)
2
=
0
The turning points are where the curve meets ε: −GM ÷ r + h² ÷ (2r²) = ε. Multiply by r² and gather: they are the two roots of this quadratic.
rₘᵢₙ
Closest distance
worked out (not typed) · km
rₘₐₓ
Farthest distance
worked out (not typed) · km
2a
Major axis
worked out (not typed) · km
Formulas
rₘᵢₙ
=
h
2
GM +
√
GM² + 2 ×
ε
×
h
2
7,000.9804
=
59,
500
2
÷
(398,600
+
√(398,
600
2
+
2
×
(−20.82)
×
59,
500
2
))
rₘₐₓ
=
−GM −
√
GM² + 2 ×
ε
×
h
2
2 ×
ε
12,144.0724
=
(−398,600
−
√(398,
600
2
+
2
×
(−20.82)
×
59,
500
2
))
÷
(2
×
(−20.82))
2a
=
−
GM
ε
19,145.0528
=
−
398,600
−20.82
Assumptions
•
Per unit mass, Ueff = −GM ÷ r + h² ÷ (2r²): gravity plus the centrifugal term; the radial speed is 0 where Ueff = ε.
•
A bound orbit: ε < 0 and above the floor of the well, so there are two turning points.
•
Distances in km, ε in km²/s² (1 km²/s² = 1 MJ/kg), h in km²/s; GM for Earth is 398,600 km³/s².
Step-by-step
WE KNOW
GM = 398,600 km³/s², ε = −20.82 km²/s², h = 59,500 km²/s
FIND
closest distance (rₘᵢₙ), farthest distance (rₘₐₓ), major axis (2a)
Step 1 · Find closest distance (rₘᵢₙ)
Use rₘᵢₙ = h² ÷ (GM + √(GM² + 2 × ε × h²))
Set
Ueff
=
ε
and
multiply
by
r
2
:
εr²
+
GMr
−
h
2
2
=
0.
The
nearer
root,
written
so
nothing
cancels.
rₘᵢₙ
=
59,
500
2
÷
(398,600
+
√(398,
600
2
+
2
×
(−20.82)
×
59,
500
2
))
rₘᵢₙ
=
59,
500
2
÷
(398,600
+
√(158,881,960,000
+
2
×
(−20.82)
×
3,540,250,000))
rₘᵢₙ
=
59,
500
2
÷
(398,600
+
√(158,881,960,000
+
(−41.64)
×
3,540,250,000))
rₘᵢₙ
=
59,
500
2
÷
(398,600
+
√(158,881,960,000
+
(−147,416,010,000)))
rₘᵢₙ
=
59,
500
2
398,600 +
√
11
,465,950,000
rₘᵢₙ
=
59,
500
2
398,600 + 107,100
rₘᵢₙ
=
59,
500
2
505,700
rₘᵢₙ
=
3,540,250,000
505,700
rₘᵢₙ = 7,000.98 km
Step 2 · Find farthest distance (rₘₐₓ)
Use rₘₐₓ = (−GM − √(GM² + 2 × ε × h²)) ÷ (2 × ε)
The other root of the same quadratic, by the quadratic formula: the farther turning point.
rₘₐₓ
=
(−398,600
−
√(398,
600
2
+
2
×
(−20.82)
×
59,
500
2
))
÷
(2
×
(−20.82))
rₘₐₓ = (−398,600 − √(158,881,960,000 + 2 × (−20.82) × 3,540,250,000)) ÷ (2 × (−20.82))
rₘₐₓ = (−398,600 − √(158,881,960,000 + (−41.64) × 3,540,250,000)) ÷ (2 × (−20.82))
rₘₐₓ = (−398,600 − √(158,881,960,000 + (−147,416,010,000))) ÷ (2 × (−20.82))
rₘₐₓ
=
−398,600 −
√
11
,465,950,000
2 × (−20.82)
rₘₐₓ
=
−398,600 − 107,080
−41.64
rₘₐₓ
=
−
505,680
−41.64
rₘₐₓ = 12,144.07 km
Step 3 · Find major axis (2a)
Use 2a = −GM ÷ ε
The
two
roots
of
εr²
+
GMr
−
h
2
2
=
0
add
up
to
−
GM
ε
:
rₘᵢₙ
+
rₘₐₓ,
the
major
axis
2a.
2a
=
−
398,600
−20.82
2a = 19,145.05 km
Check
Put the numbers back in. Both sides should match.
7,000.98
=
59,
500
2
÷
(398,600
+
√(398,
600
2
+
2
×
(−20.82)
×
59,
500
2
))
✓
12,144.07
=
(−398,600
−
√(398,
600
2
+
2
×
(−20.82)
×
59,
500
2
))
÷
(2
×
(−20.82))
✓
19,145.05
=
−
398,600
−20.82
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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