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

(a)(i) Let a uniform spherical shell have radius R, mass M and surface density σ = M/(4πR²). By the shell/ring method, take a ring at angle θ from OP, where OP = r. Its mass is dm = σ(2πR sinθ)(R dθ) = (M/2) sinθ dθ. Distance of the ring from P is s = √(R² + r² − 2Rr cosθ). Its gravitational potential at P is dV = −G dm/s. Hence V(r) = −(GM/2) ∫₀^π sinθ dθ /√(R² + r² − 2Rr cosθ). Put u = cosθ, du = −sinθ dθ: V(r) = −(GM/2) ∫₋₁¹ du /√(R² + r² − 2Rr u). Evaluating the integral:

• for r > R, integral = 2/r;
• for r < R, integral = 2/R. Therefore, (I) outside the shell, r ≥ R: V(r) = −GM/r. (II) inside the shell, r ≤ R: V(r) = −GM/R. Inside the shell the field is zero and the potential is constant; at r = R both expressions give V = −GM/R. The convention V(∞) = 0 is used.

(a)(ii) Escape velocity is found from total energy = 0: (1/2)m vₑ² − GMm/R = 0, so vₑ = √(2GM/R). The mass 10 kg cancels.

Using G = 6.67 × 10⁻¹¹ N m² kg⁻², M = 7.3 × 10²² kg, R = 1.7 × 10⁶ m: vₑ = √(2 × 6.67 × 10⁻¹¹ × 7.3 × 10²² / 1.7 × 10⁶) = √(5.728 × 10⁶) m/s = 2.39 × 10³ m/s. Escape velocity ≈ 2.39 km/s.

(b) For two thin lenses of focal lengths f₁, f₂ separated by distance d in air, the equivalent power is P = 1/F = 1/f₁ + 1/f₂ − d/(f₁ f₂). Let the dispersive power of lens i be ωᵢ = δPᵢ/Pᵢ. For achromatism, δP = 0 for a small wavelength interval. Therefore, δP = ω₁P₁ + ω₂P₂ − d(ω₁P₁P₂ + ω₂P₁P₂) = 0. Substituting P₁ = 1/f₁, P₂ = 1/f₂: ω₁/f₁ + ω₂/f₂ = d(ω₁ + ω₂)/(f₁ f₂). Hence the achromatism condition is d = (ω₁ f₂ + ω₂ f₁)/(ω₁ + ω₂). Given ω₁ = 0.020, ω₂ = 0.028, f₁ = 10 cm, f₂ = 5 cm: d = (0.020 × 5 + 0.028 × 10)/(0.020 + 0.028) = (0.10 + 0.28)/0.048 = 0.38/0.048 = 7.9167 cm. Separation d ≈ 7.92 cm. Valid for thin lenses, paraxial rays, small dispersion, and d measured between optical centres.

(c)(i) Translatory motion ↔ Rotatory motion:

• mass m ↔ moment of inertia I
• velocity v ↔ angular velocity ω
• acceleration a ↔ angular acceleration α
• force F ↔ torque τ
• momentum p = mv ↔ angular momentum L = Iω
• impulse F Δt ↔ angular impulse τ Δt
• work W = ∫F dx ↔ W = ∫τ dθ
• kinetic energy (1/2)mv² ↔ (1/2)Iω²
• power P = Fv ↔ P = τω

Corresponding equations:

• F = ma ↔ τ = Iα
• v = u + at ↔ ω = ω₀ + αt
• s = ut + (1/2)at² ↔ θ = ω₀t + (1/2)αt²
• v² = u² + 2as ↔ ω² = ω₀² + 2αθ.

(c)(ii) Perpendicular axes theorem: For a plane lamina lying in the xy-plane, if x and y are two mutually perpendicular axes in the plane and z is the axis perpendicular to the lamina through their intersection, then I_z = I_x + I_y. Proof: take an element dm at (x, y). Then I_z = ∫(x² + y²)dm = ∫x²dm + ∫y²dm = I_y + I_x = I_x + I_y.

Parallel axes theorem: The moment of inertia about any axis is equal to the moment of inertia about a parallel axis through the centre of mass plus M d², where M is the mass and d is the perpendicular distance between the axes: I = I_cm + M d². Proof for a plane lamina: let the centre of mass be the origin and the old axis be the z-axis. Let the new parallel axis pass through (a, b, 0), so d² = a² + b². For an element mᵢ at (xᵢ, yᵢ, 0), distance squared to the old axis is xᵢ² + yᵢ², and to the new axis is (xᵢ − a)² + (yᵢ − b)². Therefore I = Σ mᵢ[(xᵢ − a)² + (yᵢ − b)²] = Σ mᵢ(xᵢ² + yᵢ²) + (a² + b²)Σ mᵢ − 2aΣ mᵢxᵢ − 2bΣ mᵢyᵢ. Since the origin is the centre of mass, Σ mᵢxᵢ = 0 and Σ mᵢyᵢ = 0. Also Σ mᵢ = M. Hence I = I_cm + M d². This theorem applies to any rigid body, including a plane lamina.