What happens if you give a planet exactly circular orbit speed but point it slightly off?

Still an ellipse — slight angle change preserves energy but changes shape.

Earth orbits at 30 km/s. What is Earth's escape velocity from the Sun?

V_escape = √2 × v_circular = √2 × 30 ≈ 42.4 km/s.

Verify Kepler's Third Law: compare orbital periods for planets at 1 AU and 4 AU.

T² ∝ r³ → (T₂/T₁)² = (4/1)³ → T₂ = 8 years.

Pro 🔒~25 min

Gravity and Orbits

Model planetary motion and orbital mechanics

How it works

Gravity provides the centripetal force needed for orbital motion. For circular orbits, v = √(GM/r) — the orbital speed depends only on the central mass and radius. Kepler's Third Law states T² ∝ r³ for all planets around the same star. Elliptical orbits arise when the initial velocity differs from the circular orbit speed. Escape velocity is √2 times the circular orbit speed.

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Step-by-step

Click to place a planet, then set its initial velocity with the arrow.
Observe the orbit shape — circular, elliptical, or hyperbolic.
Enable the path tracer to see the full orbit.
Check the orbit period to verify Kepler's Third Law.

Key formulas

$v_{orbital} = \sqrt{\frac{GM}{r}}$ Circular orbit speed
$T^2 \propto r^3$ Kepler's Third Law
$v_{escape} = \sqrt{\frac{2GM}{r}}$ Escape velocity

Frequently asked questions

What happens if you give a planet exactly circular orbit speed but point it slightly off?: Still an ellipse — slight angle change preserves energy but changes shape.
Earth orbits at 30 km/s. What is Earth's escape velocity from the Sun?: V_escape = √2 × v_circular = √2 × 30 ≈ 42.4 km/s.
Verify Kepler's Third Law: compare orbital periods for planets at 1 AU and 4 AU.: T² ∝ r³ → (T₂/T₁)² = (4/1)³ → T₂ = 8 years.