Vis-Viva Equation: Orbital Speed, Escape Velocity and Energy Explained

What does the vis-viva equation calculate?

For an ideal Newtonian two-body orbit, v² = μ(2/r − 1/a) gives the relative speed at separation r on an orbit with relative semimajor axis a. The gravitational parameter is μ = G(M₁ + M₂). For a tiny satellite around a much more massive planet, the satellite's contribution to μ is normally neglected.

Two useful limits are circular speed, v꜀ = √(μ/r), and local escape speed, vₑ = √(2μ/r). The first requires a circular orbit; the second is the zero-energy threshold in the ideal isolated model. Neither can be applied by replacing centre-to-centre distance with height above the surface.

This guide uses synthetic orbits around a nominal Earth parameter. It explains the physical quantities and arithmetic; it does not provide a launch trajectory, a prediction of a real satellite's position or a complete propulsion budget.

Keep speed, position and energy separate

Speed is the magnitude of a velocity vector. The scalar formula does not tell you the direction of travel, inclination, which side of the orbit the body occupies, or whether it is approaching or receding. Those require additional orbital information.

The energy expression is ε = v²/2 − μ/r = −μ/(2a) for an ellipse. In the general two-body problem it is the system's orbital energy divided by the reduced mass. In the negligible-satellite limit it is approximately energy per unit satellite mass. Distinguishing these meanings prevents an incorrect total-energy calculation for comparable-mass binaries.

Derive the speed relation from conserved energy

Take gravitational potential energy to be zero at infinite separation. In the relative problem, specific kinetic energy is v²/2 and specific potential energy is −μ/r. Their sum stays constant when only the two-body gravitational interaction acts.

For a bound ellipse, that constant is −μ/(2a). Equating the two forms and multiplying by two gives v² − 2μ/r = −μ/a, hence v² = μ(2/r − 1/a). The positive square root supplies speed; a negative result inside the square root signals incompatible inputs for that energy.

As r decreases on the same ellipse, potential energy becomes more negative while kinetic energy rises. The total remains unchanged. The satellite does not need to fire an engine merely to accelerate on its inward arc. A force can change speed while energy trades between kinetic and potential forms.

The declared Earth model and its units

Use the IAU nominal terrestrial gravitational parameter 3.986004 × 10¹⁴ m³/s², equivalent to 398600.4 km³/s². The conversion divides by a billion because the length unit is cubed. Combining the metre value with distances in kilometres would not give kilometres per second.

Here μ is a declared nominal conversion value, not a current precision estimate of every feature of Earth's gravitational field. The examples omit atmospheric drag, oblateness, thrust and perturbations from the Moon and Sun. Distances, speeds and energies are reported in km, km/s and km²/s² respectively.

For the main example choose a = 14000 km and e = 0.5. The closest radius is a(1 − e) = 7000 km and the farthest radius is a(1 + e) = 21000 km. These parameters define the reachable radius interval for this particular ellipse.

A complete speed and energy table along one ellipse

At the closest point, 2/7000 − 1/14000 = 3/14000 km⁻¹. Multiplication by μ and a square root give 9.241990 km/s. At the farthest point the speed is 3.080663 km/s. Their ratio is exactly three before display rounding.

The ratio also equals 21000/7000 at these two apsides, where the velocity is perpendicular to the radius. Do not extend that reciprocal-distance speed rule to every point on the ellipse: away from the apsides the velocity can have a radial component.

Every row has the same ε = -14.235729 km²/s². The circular and escape columns are reference speeds at the listed radius, not additional speeds that the same elliptical satellite simultaneously possesses.

Why the elliptical and circular-speed curves cross

At r = a the vis-viva expression becomes μ/a, so the elliptical speed equals the circular speed for that radius. This equality does not make the orbit circular. The elliptical body generally has a nonzero radial velocity there; a circular orbit's velocity is entirely transverse.

For r smaller than a, the ellipse's speed exceeds the circular reference. Beyond a, it falls below that reference. The local escape threshold lies above both throughout the plotted interval because the ellipse's specific energy is negative.

The green curve compares many circular orbits, each at its own radius. It is not the path of one circular satellite moving outward through all those radii. Keeping that interpretation clear avoids confusing a graph of possible speeds with a trajectory through space.

Altitude must be added to a reference radius

A height of 400 km is not r = 400 km in the gravitational equation. For an illustrative spherical reference with the IAU nominal equatorial radius R = 6378.1 km, use r = R + h. At h = 400 km, the calculation therefore uses 6778.1 km.

The chosen reference radius is stated because real Earth is not a perfect sphere. Precision work distinguishes the reference surface, coordinate system and local position. For this teaching example a nominal equatorial spherical radius is sufficient to demonstrate the altitude error without claiming a full geodetic model.

At the main example's closest radius of 7000 km, the corresponding height above this reference is 621.9 km. Reporting that height does not alter the 7000 km distance used in the formula.

Negative, zero and positive energy

With the potential zero at infinity, negative energy describes a bound two-body energy state, zero is the parabolic escape limit, and positive energy permits escape with nonzero speed at infinity. For nonradial point-body motion these correspond to elliptical, parabolic and hyperbolic conics.

Energy alone does not guarantee a usable orbit around an extended planet. A negative-energy path can intersect the surface, and an inward-moving object with positive energy can also collide before escaping. Position, velocity direction and the body's physical size matter in addition to this scalar classification.

The table's C₃ is twice ε. Confusing the two produces a factor-of-two error even when the units look correct. At the escape threshold the exact model energy is zero; tiny numerical residuals from floating-point square roots should not be interpreted as a physical discovery.

A real square root is necessary but not sufficient

For a = 14000 km, the radicand remains positive at r = 27000 km. Yet the declared e = 0.5 ellipse reaches only 21000 km. The larger radius is compatible with the same orbital energy only if a different angular momentum and eccentricity are chosen.

The lower example r = 5000 km illustrates the same issue on the near side. An energy equation does not encode all geometric constraints. Before interpreting a calculated speed as belonging to a particular orbit, check its periapsis and apoapsis limits as well.

Escape speed and hyperbolic excess speed are different

For an escaping trajectory, write v² = 2μ/r + v∞². The local speed v includes both the escape requirement at that radius and the excess energy that remains as r tends to infinity. Consequently, C₃ = v∞² = 2ε in this ideal hyperbolic case.

At r = 10000 km with v∞ = 3 km/s, local speed is 9.419134 km/s, while C₃ is 9 km²/s² and ε is 4.5 km²/s². The craft slows as it climbs outward, approaching 3 km/s rather than stopping.

The zero-excess row is the parabolic limiting case. For a hyperbola one can also use the signed-semimajor-axis convention a < 0, so −μ/(2a) is positive. Check a source's sign convention before copying a hyperbolic axis into software; some descriptions instead quote its positive magnitude.

Escape velocity is not the additional speed from a circular orbit

A satellite already moving at circular speed does not need that full escape speed added again. For an instantaneous prograde tangential increase at the same radius, the ideal increment is Δv = (√2 − 1)√(μ/r). At 7000 km it is 3.125677 km/s, taking the speed from 7.546053 to 10.671730 km/s.

This subtraction assumes the initial and final velocity vectors point in the same direction. For a change of direction, use the magnitude of their vector difference. A finite burn with gravity effects, propulsion constraints or atmosphere needs a fuller calculation, so the table is not a ground-launch requirement.

It can sound contradictory that a prograde burn raises an orbit while higher circular orbits have lower speeds. The burn raises orbital energy locally; the body then trades kinetic energy for potential energy as it moves outward. The speed at the later high point is not the speed immediately after the burn.

A reliable workflow for using the equation

1. Declare the two-body approximation and whether satellite mass is negligible.
2. Use centre-to-centre radius and a consistently defined relative semimajor axis.
3. Keep μ and distance units compatible before taking the square root.
4. Check the orbit's allowed radius range, not only the radicand.
5. Distinguish speed from the full velocity vector.
6. Verify that kinetic plus potential energy is constant along the model orbit.
7. Separate local escape speed, hyperbolic excess and any requested Δv.

The Kepler third-law guide answers the related question of the full orbital period. The perihelion and aphelion guide connects changing radius with planetary motion. Neither the period nor a speed alone determines the body's current orbital phase.

An observed Doppler radial velocity is a line-of-sight component and is not automatically this total relative speed. For ordinary observing plans, local sunrise and sunset supply a separate time reference. Keep the measured quantity attached to its definition before bringing it into an orbital model.

Frequently asked questions