(a) With x₀ = α/β we have β = α/x₀, so
U(x) = α/x² − (α/x₀)/x = (α/x₀²)·[(x₀/x)² − (x₀/x)].
Hence U(x₀) = (α/x₀²)(1 − 1) = 0 — the curve crosses zero exactly at x₀.
For x < x₀ the 1/x² term wins (U > 0, steeply repulsive); for x > x₀ the −1/x term wins (U < 0), and U → 0⁻ as x → ∞.
(Some printings print the prefactor as α/2x₀²; direct substitution gives α/x₀².)
Well minimum: dU/dx = −2α/x³ + β/x² = 0 ⇒ x = 2α/β = 2x₀, with
Umin = U(2x₀) = (α/x₀²)(¼ − ½) = −α/(4x₀²) = −β²/(4α).
(b) Released at x₀: E = U(x₀) = 0, so KE = −U and
v(x) = √(2(0 − U)/m) = √( (2α/(m x₀²))·[(x₀/x) − (x₀/x)²] ), valid for x ≥ x₀.
The proton starts at rest at x₀, is pushed to larger x (F > 0 there), speeds up until 2x₀, then slows down for ever: it escapes to infinity with v → 0. It never returns.
(c) v is maximum where U is minimum, i.e. x = 2x₀, and
vmax = √(2·(α/4x₀²)/m) = √(α/(2m x₀²)) = β/√(2mα) (= 0.707 in our units).
(d) F(x) = −dU/dx = 2α/x³ − β/x². At x = 2x₀: F = 2α/(8x₀³) − (α/x₀)/(4x₀²) = α/(4x₀³) − α/(4x₀³) = 0. Maximum speed happens exactly at the equilibrium point — that is why it is the speed maximum.
(e) Released at x₁ = 3x₀: E = U(3x₀) = (α/x₀²)(1/9 − 1/3) = −2α/(9x₀²) < 0, so the motion is bound:
v(x) = √( (2α/(m x₀²))·[(x₀/x) − (x₀/x)² − 2/9] ).
Turning points: put u = x₀/x in u² − u = −2/9 ⇒ 9u² − 9u + 2 = 0 ⇒ u = ⅔ or ⅓ ⇒ x = 1.5x₀ and 3x₀.
The proton oscillates back and forth between those two walls, fastest at 2x₀ where
vmax = √(2(E − Umin)/m) = √(2(α/4x₀² − 2α/9x₀²)/m) = √(α/(18 m x₀²)) (= 0.236 here).
The oscillation is periodic but not simple harmonic — U is not symmetric about 2x₀, so the outward swing is longer and lazier than the inward one.
(f) Range of the motion:
• release at x₀: xmin = x₀, xmax = ∞ (unbound).
• release at 3x₀: xmin = 1.5x₀, xmax = 3x₀.
General rule for release from rest at xr: E = U(xr); with u = x₀/x, the turning points solve u² − u − E x₀²/α = 0, i.e. u = ½(1 ± √(1 + 4Ex₀²/α)). A second (outer) root exists only when E < 0, i.e. only when xr > x₀.
On the model: everything is shown in dimensionless units α = m = x₀ = 1 (β = 1). The motion is integrated with velocity Verlet (h ≤ 0.004) and reflected exactly at the analytic turning points, so energy is conserved to ~10⁻⁸ over long runs.