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Specific Orbital Energy

Total mechanical energy per unit mass of an orbiting body, constant along a given orbit.

Orbital MechanicsAstrodynamicsFundamental

Specific Orbital Energy Calculator

ε = v²/2 − μ/r = −μ/(2a)
Solve for ε, v, μ, r, or a
ε v, μ, r, a
km²/s²
km/s
km³/s²
km
km
Solve for:
Result
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Specific Energy vs. Radius ε(r) = v²/2 − μ/r
ε(r) for fixed v, μ Computed point (r, ε)
r > 0, μ > 0 • ε = v²/2 − μ/r = −μ/(2a)

Interpretation

Specific orbital energy: ε = v²/2 − μ/r = −μ/(2a), where μ is gravitational parameter, r radial distance, a semi‑major axis. It is the total energy per unit mass. Example: For a circular orbit at r=7000 km, v≈7.55 km/s, ε = (7.55e3)²/2 − 3.986e14/7e6 ≈ 2.85e7 − 5.69e7 = −2.84e7 J/kg.

ε = v^2/2 - μ/r = -μ/(2a)
Specific Orbital Energy

Variables

SymbolQuantityUnit
εSpecific orbital energyJ/kg
vOrbital speedm/s
μGravitational parameterm3/s2
rRadial distancem
aSemi-major axism

What it means

Specific orbital energy is the sum of kinetic and potential energy per unit mass. For a bound orbit, it is negative. It is a conserved quantity in the two‑body problem. The vis‑viva equation is derived from this. This energy determines the orbital parameters and is used in trajectory design and in calculating Δv requirements. Understanding specific energy is essential for orbital mechanics and for mission analysis.

Worked example

Specific Orbital Energy – Two Examples

Real‑World
Scenario: LEO with semi‑major axis a = 6.678×10⁶ m, μ = 3.986×10¹⁴. Find specific energy.
ParameterValue
μ3.986×10¹⁴
a6.678×10⁶
1ε = -μ/(2a) = -3.986e14/(2×6.678e6) = -3.986e14/1.3356e7 = -2.984×10⁷ J/kg
Result -2.98×10⁷ J/kg ✓ LEO
Scenario: GEO with a = 4.216×10⁷. Find ε.
ParameterValue
a4.216×10⁷
1ε = -3.986e14/(2×4.216e7) = -3.986e14/8.432e7 = -4.727×10⁶ J/kg
Result -4.73×10⁶ J/kg ✓ GEO
Key insight: Specific orbital energy is negative for bound orbits – less negative means higher orbit.

Common mistakes

  • Specific orbital energy: ε = v²/2 − μ/r = −μ/(2a).
  • v: Speed (m/s).
  • μ: Gravitational parameter.
  • r: Radial distance (m).
  • a: Semi‑major axis (m).
  • Units: J/kg (m²/s²).
  • Negative for bound orbits, zero for parabolic, positive for hyperbolic.

Applications

Specific orbital energy, ε = v²/2 − μ/r = −μ/(2a), is the total energy per unit mass of an orbiting body. It determines the size of the orbit (semi‑major axis). Engineers use this to calculate the energy required for orbital transfers and to assess the orbit type (elliptical, parabolic, hyperbolic). By understanding specific energy, aerospace engineers can plan interplanetary missions, compute launch energy, and design orbit insertion manoeuvres. It is a fundamental parameter in astrodynamics.

  • Orbit determination and tracking
  • Orbital transfer design (Hohmann, bi‑elliptic)
  • Launch vehicle energy requirements
  • Interplanetary trajectory design (energy balance)
  • Computation of orbital elements from state vectors

Frequently Asked Questions

Q01What is the Specific Orbital Energy used for?
A01

It is the total mechanical energy per unit mass of an orbiting body, constant along a given orbit. It is used to classify orbits and compute orbital parameters.

Q02What do the variables ε, v, μ, r, and a represent?
A02

ε = specific orbital energy (J/kg)
v = orbital speed (m/s)
μ = gravitational parameter (m³/s²)
r = orbital radius (m)
a = semi‑major axis (m)

Q03Why is the specific energy important?
A03

It determines the type of orbit: ε < 0 for bound orbits (elliptical), ε = 0 for parabolic, ε > 0 for hyperbolic.

Q04What are common mistakes when using this formula?
A04

  • Forgetting that specific orbital energy is negative for bound (elliptical/circular) orbits and only zero or positive for parabolic/hyperbolic trajectories.
  • Using the wrong sign for μ.
  • Confusing specific energy with total energy (which includes mass).

Q05Give a worked example.
A05

For a circular orbit at r = 7,000 km, v = 7.5 km/s, μ = 3.986e14. ε = 7500²/2 − 3.986e14/7e6 = 28,125,000 − 56,942,857 = −28,817,857 J/kg. Also, −μ/(2a) = −3.986e14/(2×7e6) = −28,471,429 J/kg (small difference due to rounding).

Q06How does the specific energy relate to the vis‑viva equation?
A06

The vis‑viva equation is derived from the specific energy: v²/2 − μ/r = ε.

Q07What is the significance of ε = 0?
A07

It corresponds to the escape trajectory (parabolic), where the object has just enough kinetic energy to escape the gravitational field.

Q08How does the specific energy change with altitude?
A08

For a circular orbit, ε = −μ/(2r); as r increases, ε becomes less negative (approaches zero).

Q09What is the specific energy of a geostationary orbit?
A09

At r = 42,164 km, ε = −3.986e14/(2×4.2164e7) ≈ −4.727e6 J/kg.

Q10How do you use specific energy to find the semi‑major axis?
A10

a = −μ/(2ε). For bound orbits, ε is negative, so a is positive.