Physics Simulation · gravitation

Orbital Mechanics

Kepler's laws and the vis-viva equation — watch an orbit and read its numbers.
v = √(μ(2/r − 1/a))
Earth · circular

Central body

Orbit

Real-world orbits

Orbit

Earth t = 0.0 min
speed
Orbit
Detail
Field notes

Reading an orbit

How it works

One number describes the whole orbit

Every two-body orbit is governed by a single gravitational parameter μ = GM of the central body — plug that and the orbit's geometry into a handful of exact formulas and everything else follows: period from Kepler's third law (T = 2π√(a³/μ)), speed at any point from the vis-viva equation (v² = μ(2/r − 1/a)), and the shape itself from just the periapsis and apoapsis distances. The animated orbit isn't just an ellipse traced at constant angular speed — position vs. time is solved from Kepler's equation (M = E − e·sin E), which is exactly what makes the orbiting point visibly speed up at periapsis and slow down at apoapsis: Kepler's second law, equal areas in equal time.

Worked example

The ISS orbits at ~400 km altitude, giving a period of about 92.4 minutes and an orbital speed of 7.67 km/s — numbers this tool reproduces from μ and altitude alone, matching the real station within rounding. Push the same orbit out to 35,786 km altitude and the period stretches to almost exactly one sidereal day: geostationary orbit.

Why do higher orbits move slower?

Gravity is weaker further out, so less centripetal force is available — and less force means less speed is needed to keep circling rather than falling in. It's the same reason outer planets crawl around the Sun far more slowly than Mercury does.

What's actually special about geostationary orbit?

It's the one altitude where the orbital period exactly matches Earth's rotation — the satellite traces its orbit at the same rate the ground turns beneath it, so from the surface it appears to hang motionless in the sky.

Why is escape velocity √2 times the circular orbital velocity?

Both come from the same vis-viva equation — circular orbit is v² = μ/r, escape is the limit as apoapsis goes to infinity, v² = 2μ/r. The ratio between them is exactly √2, at any altitude, for any body.

Does the animation really show Kepler's second law?

Yes — the orbiting point's position is solved from the true time-domain equations of orbital motion (via Kepler's equation), not just an ellipse traced at a fixed angular rate, so it genuinely speeds up approaching periapsis and slows down approaching apoapsis, sweeping out equal areas in equal time.

Two-body Keplerian orbits. No atmospheric drag, no other gravitating bodies, no orbital perturbations — the clean textbook problem, not a mission-planning tool.