Vis-Viva Equation

The vis-viva equation relates speed to position and orbit shape. Choose a central body and enter orbital parameters.

Worked example — Vis-Viva Equation

Orbital period	1.62 h

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Same semi-major axis, same energy, same period 🖖

The vis-viva equation is energy conservation in orbital form. At every point the total specific mechanical energy is ε = ½v² − GM/r = −GM/(2a) — a constant set only by the semi-major axis a. Two orbits with the same a but different eccentricities carry identical energy and have the same period; eccentricity shapes the trajectory without changing the energy budget. Speed is highest at periapsis (minimum potential energy) and lowest at apoapsis. The escape trajectory is the limiting case a → ∞, giving v_esc = √(2GM/r) = √2 · v_c — exactly √2 times the circular velocity at the same radius.

Only two numbers set the speed 🖖

The vis-viva equation says an orbiting object's speed depends on just two things: how far it currently is from the central body (r) and the overall size of its orbit, the semi-major axis a. Plug both into v = √(GM(2/r − 1/a)) and you get the speed. Notice the object's own mass never appears — a paperclip and a space station in the same orbit travel at exactly the same speed.

The name means 'living force' 🖖

Vis viva is Latin for 'living force', the term Leibniz coined in the 1680s for the quantity mv² — the ancestor of kinetic energy. It sparked a decades-long dispute with Newton's followers, who insisted momentum mv was the true 'measure of motion'. Both sides were partly right, and the v² that survives in this equation is a direct echo of Leibniz's living force.

Problem solved in full

Period and two extreme speeds of a Molniya orbit around Earth 5 steps
A Molniya orbit: semi-major axis 26 560 km, eccentricity 0.74, around Earth. Find its period and its two extreme speeds, and then the number the orbit was really designed around — how much of each pass is spent far out. This is the Molniya (HEO) state, with μ = GM = 3.986 × 10¹⁴ m³/s².
1. The semi-major axis is the average of the two extreme radii, so the eccentricity splits it apart symmetrically. This orbit skims 535 km above the ground at one end and reaches 39 843 km at the other, further out than geostationary.
  
  $r_{p} = a (1 - e) = 26560 \times 0.26 = 6905.6 km, r_{a} = a (1 + e) = 26560 \times 1.74 = 46 214.4 km$
2. The period ignores all of that. Kepler's third law sees only the semi-major axis, so this ellipse takes exactly as long as a circular orbit at 26 560 km would — and the answer is the design point, because 11.97 h is half a sidereal day. The satellite therefore retraces the same ground track twice a day and can be handed to a ground station on a fixed schedule.
  
  $T = 2 π \frac{a ^{3}}{μ} = 2 π \frac{( 2.656 \times 1 0 ^{7} ) ^{3}}{3.986 \times 1 0 ^{14}} = 43079 s = 11.97 h$
3. Vis-viva needs only the current radius and the size of the orbit, and periapsis supplies both. The result is 93% of the speed that would let it leave Earth from that height: a very eccentric orbit is a near-miss escape at the bottom.
  
  $v_{p} = μ (\frac{2}{r _{p}} - \frac{1}{a}) = 3.986 \times 1 0^{14} (\frac{2}{6.906 \times 1 0 ^{6}} - \frac{1}{2.656 \times 1 0 ^{7}}) = 10.021 km/s$
4. The same equation at the other end, with only the radius changed, gives a speed nearly seven times smaller. That factor is not new information — angular momentum rv is conserved, so the two speeds must be in inverse proportion to the two radii, and the radius ratio was already fixed by e in step 1.
  
  $v_{a} = μ (\frac{2}{r _{a}} - \frac{1}{a}) = 1.497 km/s, \frac{v _{p}}{v _{a}} = \frac{r _{a}}{r _{p}} = \frac{1 + e}{1 - e} = 6.69$
5. Slow at the top means long at the top, and Kepler's second law makes that exact: equal areas in equal times, so a fraction of the period is a fraction of the area. The minor axis cuts the ellipse into two equal halves and crosses the orbit precisely where r = a, so the area swept from Earth on the far side is that half plus the triangle between the focus and the axis. Both a and b cancel from the ratio, which leaves the eccentricity on its own.
  
  $\frac{t _{r > a}}{T} = \frac{\frac{1}{2} π ab + ab e}{π ab} = \frac{1}{2} + \frac{e}{π} = 0.5 + 0.2356 = 0.7355 ⟹ 0.7355 \times 11.97 = 8.80 h$
Answer

The tool prints a period of 11.97 h, 10.021 km/s at periapsis and 1.497 km/s at apoapsis. Step 5 is what those numbers are for: 8.80 hours of every 11.97-hour orbit is spent beyond r = a, and barely 3.17 hours inside it. Three satellites in staggered planes therefore cover a whole day with more than two hours of overlap, which is how the Soviet Molniya network gave high-latitude Russia continuous television — a geostationary satellite sits on the horizon at those latitudes and is useless. And ½ + e/π contains no μ and no a: an orbit of eccentricity 0.74 around any body whatever spends 73.6% of its life on the far side.

$\frac{t _{r > a}}{T} = \frac{1}{2} + \frac{e}{π} (a, μ cancel)$

Learning path

Orbits from two numbers

Leads to Orbital period the speed at any point of that ellipse from just two distances: where you are, and the semi-major axis.

References (1)

The vis-viva equation and the energy-from-a alone result: R. R. Bate, D. D. Mueller and J. E. White, Fundamentals of Astrodynamics, §1.4–1.6. Dover, 1971. ISBN 978-0-486-60061-1.

By the 42Math team · Updated 2026-08-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

LEO circular - LEO circular: v≈7.67 km/s
GEO (geostationary) - A geostationary orbit: 3.075 km/s at 42,164 km, and a period of 23.94 h — not 24. That is the sidereal day, the time Earth takes to turn once against the stars, and matching it is the whole requirement for staying over one spot on the ground.
Molniya (HEO) - At r = a the speed reads 3.874 km/s, but this orbit is worth reading at its ends: 10.021 km/s at periapsis against 1.497 km/s at apoapsis, a factor of 6.7. The satellite spends most of its 11.97-hour period crawling through the slow end, which is exactly what a Molniya orbit is for.
Earth-Sun orbit - Earth's own orbit — 29.788 km/s, period 365.22 d. An eccentricity of 0.017 sounds negligible and still moves the speed by a kilometre a second: 30.299 km/s at perihelion in early January against 29.286 km/s at aphelion in early July.