(a) (i) Derive the expressions for gravitational potentials at a point (I) outside the spherical shell, (II) inside the spherical…

(a)(i) Potential of a uniform spherical shell

Let the shell have radius R and mass M, with surface density σ = M/(4πR²), and let P be at distance r from the centre O. The potential at P is defined as the work done per unit mass in bringing a unit mass from infinity, with V(∞) = 0. Divide the shell into thin rings at polar angle θ about the line OP. The ring has area 2πR sinθ · R dθ, so its mass is dm = σ·2πR² sinθ dθ = (M/2) sinθ dθ. Every point of the ring is at the same distance s from P, where s² = R² + r² − 2Rr cosθ. dV = −G dm / s, so V = −(GM/2) ∫₀^π sinθ dθ / √(R² + r² − 2Rr cosθ). Put u = cosθ: V = −(GM/2) ∫₋₁¹ du / √(R² + r² − 2Rr u) = −(GM/2) · (1/Rr)[(R + r) − |R − r|]. (I) Outside the shell (r > R): the bracket is 2R, so V = −GM/r, as if the whole mass were at the centre. (II) Inside the shell (r < R): the bracket is 2r, so V = −GM/R, a constant independent of position. Since V is constant inside, the field g = −dV/dr = 0 there. At r = R both results agree, V = −GM/R, so the potential is continuous.

(a)(ii) Escape velocity from the Moon

Escape velocity: ½mv² = GMm/R, so v = √(2GM/R) = √(2gR), independent of the mass of the body (10 kg). With G = 6.67×10⁻¹¹ N m² kg⁻², M = 7.3×10²² kg and R = 1.7×10⁶ m: v = √(2 × 6.67×10⁻¹¹ × 7.3×10²² / 1.7×10⁶) = √(5.73×10⁶) = 2.39×10³ m/s. Using g_Moon = g/6 = 9.8/6 = 1.63 m/s² gives v = √(2 × 1.63 × 1.7×10⁶) = 2.36×10³ m/s. Escape velocity ≈ 2.4 km/s (the two routes agree within about 1.5%).

(b) Achromatic combination of two thin lenses separated by d

For two thin lenses in air, the combined power is P = P₁ + P₂ − dP₁P₂, where P = 1/f. For a small change of wavelength, δP₁ = ω₁P₁ and δP₂ = ω₂P₂, where ω is the dispersive power. Then δP = δP₁ + δP₂ − d(P₂δP₁ + P₁δP₂) = ω₁P₁ + ω₂P₂ − dP₁P₂(ω₁ + ω₂). For achromatism δP = 0: ω₁/f₁ + ω₂/f₂ = d(ω₁ + ω₂)/(f₁f₂), so d = (ω₁f₂ + ω₂f₁)/(ω₁ + ω₂). (When both lenses are of the same material, ω₁ = ω₂ and d = (f₁ + f₂)/2, the Huygens eyepiece condition.) Numbers: ω₁ = 0.020, ω₂ = 0.028, f₁ = 10 cm, f₂ = 5 cm: d = (0.020 × 5 + 0.028 × 10)/(0.020 + 0.028) = 0.38/0.048 = 7.92 cm. The equivalent focal length is F = f₁f₂/(f₁ + f₂ − d) = 50/(15 − 7.92) = 7.06 cm. This holds for thin lenses and paraxial rays.

(c)(i) Translational and rotational analogues

Mass m ↔ moment of inertia I; displacement x ↔ angle θ; velocity v ↔ angular velocity ω; acceleration a ↔ angular acceleration α; force F ↔ torque τ; momentum p = mv ↔ angular momentum L = Iω. Equations: F = ma ↔ τ = Iα F = dp/dt ↔ τ = dL/dt v = u + at ↔ ω = ω₀ + αt s = ut + ½at² ↔ θ = ω₀t + ½αt² v² = u² + 2as ↔ ω² = ω₀² + 2αθ Kinetic energy ½mv² ↔ ½Iω²; work F·x ↔ τθ; power Fv ↔ τω.

(c)(ii) Theorems of perpendicular and parallel axes for a plane lamina

Perpendicular axes: for a plane lamina in the xy-plane, the moment of inertia about the z-axis, perpendicular to the lamina through O, equals the sum of those about two mutually perpendicular axes x and y in its plane through O: I_z = I_x + I_y. Proof: for an element dm at (x, y), its distance from the z-axis is √(x² + y²). So I_z = ∫(x² + y²)dm = ∫y²dm + ∫x²dm = I_x + I_y (I_x uses the distance y from the x-axis). Parallel axes: the moment of inertia about any axis equals that about a parallel axis through the centre of mass plus Md², where d is the distance between the axes: I = I_cm + Md². Proof: take the centre of mass as origin, and the second axis through (a, b), so d² = a² + b². For an element mᵢ at (xᵢ, yᵢ): I = Σmᵢ[(xᵢ − a)² + (yᵢ − b)²] = Σmᵢ(xᵢ² + yᵢ²) + (a² + b²)Σmᵢ − 2aΣmᵢxᵢ − 2bΣmᵢyᵢ. About the centre of mass, Σmᵢxᵢ = Σmᵢyᵢ = 0, so I = I_cm + Md².