ISEGORIA / MATH ENCYCLOPEDIA
Classical mechanics: motion and forces
Projectiles, oscillators, and orbital motion.
Before you begin: Vectors and calculus
Predict, manipulate, then check your reasoning against the example and question. Graphs illustrate the mathematics; they do not replace a proof.
1. Projectile motion
Drag the tip of the orange launch arrow, or use the sliders. The trajectory is integrated from Newton’s second law, and the hollow dots mark ten equal time steps: they are evenly spaced sideways because the horizontal velocity never changes, while gravity steadily changes the vertical velocity. The dashed curve is the complementary angle 90° − θ, which lands at the same range, and the dotted curve bounds every path at this speed. Press Launch to follow the velocity components, and switch on air drag to see what the vacuum model leaves out.
Worked example. At 45° without drag the range is maximal for a fixed speed, R = v₀²/g; at 18 m/s that is about 33 m.
Watch out. The vacuum model neglects air resistance and Earth’s curvature. The drag option uses a quadratic law a = −k|v|v with k = 0.006 m⁻¹, a rough value for a light ball.
What remains constant?
Horizontal velocity, while vertical velocity changes by gravity.
2. Harmonic oscillator
Tune the damping c and stiffness k of a mass on a spring (m = 1). The motion is integrated from the equation of motion and drawn beside the roots of the characteristic equation ms² + cs + k = 0. Complex roots mean oscillation, a double root is critical damping, and two real roots mean a slow, overdamped return.
Worked example. With k = 4 and m = 1, critical damping is c = 2√(mk) = 4. It returns fastest without overshoot.
Watch out. The spring is linear and the mass is point-like.
What sets the regime?
The discriminant c²−4mk.
3. Orbital motion
A body is launched tangentially at distance 1 from an attracting centre. Its path is the exact conic that follows from the inverse-square law: a circle at v₀ = √μ, an ellipse below the escape speed √(2μ), a parabola at it and a hyperbola above it. The shaded sectors are swept in equal times, which is Kepler’s second law: they all have the same area.
Worked example. At circular speed v₀ = √(μ/r₀), e = 0 and the radius and speed stay constant.
Watch out. The two-body model omits perturbations and relativistic corrections.
What is conserved?
In the ideal model, energy and angular momentum.