Documentation / @interstellar-tools/equations / specificMechanicalEnergy

Function: specificMechanicalEnergy()

function specificMechanicalEnergy(v: number, r: number, mu: number): number;

Defined in: orbits/specific-mechanical-energy.ts:68

Specific mechanical energy ($ \varepsilon $) of a point mass in a central gravity field.

Definition

$$\epsilon =\frac{v^2}{2}-\frac{\mu }{r}$$

where ($v$) is the speed, ($r$) the distance to the central body, and ($\mu$) the gravitational parameter.

Units

• Inputs: ($v$) in m/s, ($r$) in m, ($\mu$) in m³/s².
• Output: ($\epsilon$) in J/kg (i.e., m²/s²).

Orbit-type by sign

• ($\epsilon <0$) → elliptic (bound)
• ($\epsilon =0$) → parabolic
• ($\epsilon >0$) → hyperbolic

TIP

• For circular orbits ($v=\sqrt{\mu /r}$), so ($ \varepsilon = -\mu/(2r) $).
• If you have the semi-major axis ($a$), then ($ \varepsilon = -\mu/(2a) $).

In practiceEnergy of the ISS orbit

-29.35 MJ/kg

Specific orbital energy at 420 km altitude

Negative means the orbit is bound. Escaping Earth from there takes 29.35 MJ more for every kilogram, which brings the energy up to zero.

const mu = 3.986004418e14;
const r = 6.371e6 + 420e3;

const v = circularSpeed(r, mu);

specificMechanicalEnergy(v, r, mu); // ≈ -2.935e7 J/kg

Parameters

Parameter	Type	Description
v	number	Speed magnitude ($v$) (m/s).
r	number	Radius ($r=|\mathbf{r}|$) from the central body (m).
mu	number	Gravitational parameter ($\mu$) (m³/s²).

Returns

number

Specific mechanical energy ($\epsilon$) (J/kg).

Throws

Error If v is not finite or negative.

Throws

Error If r is not finite or non-positive.

Throws

Error If mu is not finite or negative.

Example

// LEO example around Earth (~400 km altitude)
const muEarth = 3.986004418e14; // m³/s²
const r = 6378e3 + 400e3; // Earth radius + 400 km
const v = Math.sqrt(muEarth / r); // near-circular speed
const eps = specificMechanicalEnergy(v, r, muEarth);
// eps ≈ -mu/(2r)