Why does a slightly faster satellite end up on an ellipse instead of a circle?

At distance r an orbit is circular only if the speed is exactly √(μ/r) and perpendicular to the radius. Any faster and that point becomes the periapsis, so the satellite climbs onto an ellipse. At √(2μ/r), the escape speed and √2 times the circular speed, the specific orbital energy is zero and the path is a parabola. Faster still gives a hyperbola and the satellite never returns. Slower turns the point into the apoapsis, and if the periapsis falls below the surface the satellite crashes.

Why does Kepler's second law hold, and how are the sector areas measured?

Gravity points at the centre, so it exerts no torque and the angular momentum h = r × v is conserved. The rate at which the radius vector sweeps area is exactly dA/dt = h/2, so equal times sweep equal areas. The tool cuts the path into twelfths of the period, measures each sector with the shoelace formula and divides by h·Δt/2. The ratios should come out very close to 1, even though sectors near periapsis are short and wide and those near apoapsis long and thin.

Why not just use Euler's method? What is a symplectic integrator?

Euler's method adds a little energy every step, so the satellite spirals outward and after a few dozen orbits the result is meaningless. Leapfrog (velocity Verlet) is symplectic: it preserves phase-space volume, so the energy error oscillates within a small band instead of growing and orbits stay closed over long runs. Here the step also shrinks as √(r³/μ), which keeps high-eccentricity orbits such as Halley's accurate at perihelion. The relative energy drift is shown on screen.

Is the three-body problem really unsolvable? What is the figure-eight orbit?

There is no general closed-form solution, and most starting conditions give chaotic motion in which tiny differences grow exponentially. The Pythagorean problem, with masses 3, 4 and 5 released from rest, is the textbook example and ends by ejecting one star. Special periodic solutions do exist, though: Lagrange's equilateral triangle, and the figure-eight found by Moore in 1993 and proved by Chenciner and Montgomery in 2000, where three equal masses chase each other around one figure-eight curve. Šuvakov and Dmitrašinović later found more, such as the butterfly.

How are the synchronous orbit altitude and the Hohmann Δv calculated?

Set the orbital period equal to the body's rotation period (one sidereal day, 23 h 56 min 4 s, for Earth) and solve T = 2π√(r³/μ) for r = (μT²/4π²)^(1/3). Subtracting the radius gives about 35,786 km for Earth; the tool uses the mean radius, so it differs slightly. A Hohmann transfer uses an ellipse tangent to both circles: burn Δv₁ to enter it and Δv₂ at apoapsis to circularise. LEO to GEO takes about 3.9 km/s in total and roughly 5.3 hours.

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🪐Orbit Simulator - Kepler's Laws, Orbital Elements & 3-Body Problem

Place a satellite anywhere, drag the yellow arrow to give it a velocity, and see whether it follows a circle, an ellipse, a parabola or a hyperbola. The semi-major axis, eccentricity, periapsis, apoapsis and period are computed live from the state vectors, and equal-time sectors make Kepler's second law visible. The integrator is a symplectic leapfrog with steps that shrink as √(r³/μ), so energy drift stays tiny; there are also classic N-body scenes (figure-eight, butterfly, Pythagorean) and a calculator for orbital and escape speeds, synchronous orbits and Hohmann transfers.

Free · No sign-up · Runs in your browserLast updated: 2026-10-26

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How to use Orbit & Gravity Simulator

1Pick a preset (ISS, geostationary, Molniya, Moon, Halley's Comet, escape velocity…) or a different central body
2Press play; drag the arrow tip to change the velocity, drag the satellite to move it, or type an altitude, speed and flight-path angle
3Read the orbit type, orbital elements, energy and sector areas, then try the N-body and calculator tabs

FAQ

Why does a slightly faster satellite end up on an ellipse instead of a circle?: At distance r an orbit is circular only if the speed is exactly √(μ/r) and perpendicular to the radius. Any faster and that point becomes the periapsis, so the satellite climbs onto an ellipse. At √(2μ/r), the escape speed and √2 times the circular speed, the specific orbital energy is zero and the path is a parabola. Faster still gives a hyperbola and the satellite never returns. Slower turns the point into the apoapsis, and if the periapsis falls below the surface the satellite crashes.
Why does Kepler's second law hold, and how are the sector areas measured?: Gravity points at the centre, so it exerts no torque and the angular momentum h = r × v is conserved. The rate at which the radius vector sweeps area is exactly dA/dt = h/2, so equal times sweep equal areas. The tool cuts the path into twelfths of the period, measures each sector with the shoelace formula and divides by h·Δt/2. The ratios should come out very close to 1, even though sectors near periapsis are short and wide and those near apoapsis long and thin.
Why not just use Euler's method? What is a symplectic integrator?: Euler's method adds a little energy every step, so the satellite spirals outward and after a few dozen orbits the result is meaningless. Leapfrog (velocity Verlet) is symplectic: it preserves phase-space volume, so the energy error oscillates within a small band instead of growing and orbits stay closed over long runs. Here the step also shrinks as √(r³/μ), which keeps high-eccentricity orbits such as Halley's accurate at perihelion. The relative energy drift is shown on screen.
Is the three-body problem really unsolvable? What is the figure-eight orbit?: There is no general closed-form solution, and most starting conditions give chaotic motion in which tiny differences grow exponentially. The Pythagorean problem, with masses 3, 4 and 5 released from rest, is the textbook example and ends by ejecting one star. Special periodic solutions do exist, though: Lagrange's equilateral triangle, and the figure-eight found by Moore in 1993 and proved by Chenciner and Montgomery in 2000, where three equal masses chase each other around one figure-eight curve. Šuvakov and Dmitrašinović later found more, such as the butterfly.
How are the synchronous orbit altitude and the Hohmann Δv calculated?: Set the orbital period equal to the body's rotation period (one sidereal day, 23 h 56 min 4 s, for Earth) and solve T = 2π√(r³/μ) for r = (μT²/4π²)^(1/3). Subtracting the radius gives about 35,786 km for Earth; the tool uses the mean radius, so it differs slightly. A Hohmann transfer uses an ellipse tangent to both circles: burn Δv₁ to enter it and Δv₂ at apoapsis to circularise. LEO to GEO takes about 3.9 km/s in total and roughly 5.3 hours.