Newton's Cannon 3D – Falling Back, Orbiting or Escaping the Earth
PhysicsGravitation & AstronomyAges 16–17
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Sign in to playFire a cannonball horizontally from the top of a very high mountain on a 3D Earth. Its motion is computed numerically with gravity proportional to 1/r²: at low speed it falls back, at about 7.9 km/s (near the surface) it follows a circular orbit, faster still an ellipse, and from 11.2 km/s upwards it escapes the Earth.
Lesson: Orbital speed and escape velocity; path of a projectile launched horizontally around the Earth
What it shows
Newton imagined a cannon on a mountain so high that air resistance could be ignored. Fire slowly and the ball falls back; fire faster and the ground curves away beneath it as quickly as it falls, so it circles the Earth. Here the ball feels only the Earth's gravity, F = GMm/r², and its path is computed step by step with the Runge–Kutta method on a rotatable 3D globe. The circular orbital speed √(GM/r) and the escape speed √(2GM/r) are given for the chosen height; near the surface they are 7.9 and 11.2 km/s.
How to use
Set v₀ and Mountain height h, predict what will happen, then press Fire. The buttons v₀ = vI (circular) and v₀ = vII (escape) choose the exact speeds for that height. Tick Keep old orbits to compare shots by colour: red falls back, orange and green are ellipses, yellow is circular and blue escapes. Use Time warp, Pause and Clear orbits, drag to rotate the globe and press Reset view to return.
Parameters you can change
- Launch speed v₀ 1–14 km/s
- Mountain height h 100–2000 km
- Time warp (×100) 1–30
- Keep the orbits of earlier shots
Questions to explore
- Why does a ball fired at 7 km/s from 400 km fall back even though it travels thousands of kilometres?
- As v₀ rises from the circular speed towards the escape speed, how do the farthest point and the period change?
- What is the ratio of the escape speed to the circular speed, and does it depend on the height h?
