(a) (i) Briefly discuss the Kepler's laws of planetary motion. 5 marks (ii) Show that the escape velocity V_e on the surface of…

(a)(i) Kepler’s three laws are:

• Law of orbits: every planet moves in an ellipse with the Sun at one focus.
• Law of areas: the radius vector from the Sun to a planet sweeps equal areas in equal times; equivalently, dA/dt = L/(2m) is constant, so planetary angular momentum is conserved.
• Law of periods: T² ∝ a³, where a is the semi-major axis. For a circular orbit of radius r, T² = 4π²r³/(GM). These laws follow from the inverse-square gravitational force, with the Sun assumed much heavier than the planet.

(a)(ii) Let a body of mass m be projected from Earth’s surface with escape speed V_e. Using conservation of mechanical energy, the total energy at the surface is E = (1/2)mV_e² − GMm/R. At infinity, for the minimum escape speed, the body just reaches rest, so E = 0. Hence (1/2)mV_e² − GMm/R = 0, so V_e² = 2GM/R. At the Earth’s surface, g = GM/R², hence GM = gR². Therefore V_e² = 2gR²/R = 2gR, so V_e = √(2gR). Numerically, V_e = √(2 × 9.8 × 6.4 × 10⁶) ≈ 1.12 × 10⁴ m s⁻¹. This ignores air resistance, Earth’s rotation, and other bodies.

(a)(iii) For a circular orbit of radius r, gravity supplies the centripetal force: GMm/r² = mv²/r, so v² = GM/r. Thus kinetic energy K = (1/2)mv² = GMm/(2r), and gravitational potential energy U = −GMm/r. For satellite A, altitude R, so r_A = R + R = 2R. For satellite B, altitude 5R, so r_B = R + 5R = 6R. Therefore K_A/K_B = (GMm/(2 × 2R))/(GMm/(2 × 6R)) = 6R/2R = 3. Similarly, U_A/U_B = (−GMm/2R)/(−GMm/6R) = 6R/2R = 3. So K_A : K_B = 3 : 1 and U_A : U_B = 3 : 1; equivalently B : A = 1 : 3. For each satellite separately, K/U = −1/2.

(b) Let the rigid body consist of n particles of masses m_i at position vectors r_i from O. Its angular momentum about O is L = Σ m_i r_i × v_i. Since the body rotates with instantaneous angular velocity ω about an axis through O, the velocity of each particle is v_i = ω × r_i. Hence L = Σ m_i r_i × (ω × r_i). Using the vector identity a × (b × c) = b(a·c) − c(a·b), with a = r_i, b = ω, c = r_i, r_i × (ω × r_i) = ω(r_i·r_i) − r_i(r_i·ω). Therefore L = Σ m_i[r_i²ω − r_i(r_i·ω)]. Writing r_i = (x_i, y_i, z_i) and ω = (ω_x, ω_y, ω_z), the components are L_x = Σ m_i[(y_i² + z_i²)ω_x − x_i y_i ω_y − x_i z_i ω_z], L_y = Σ m_i[−y_i x_i ω_x + (x_i² + z_i²)ω_y − y_i z_i ω_z], L_z = Σ m_i[−z_i x_i ω_x − z_i y_i ω_y + (x_i² + y_i²)ω_z]. Define the inertia tensor I about O by I_xx = Σ m_i(y_i² + z_i²), I_yy = Σ m_i(x_i² + z_i²), I_zz = Σ m_i(x_i² + y_i²), I_xy = I_yx = −Σ m_i x_i y_i, I_xz = I_zx = −Σ m_i x_i z_i, I_yz = I_zy = −Σ m_i y_i z_i. Then L_x = I_xx ω_x + I_xy ω_y + I_xz ω_z, and similarly for L_y and L_z. Thus L_α = Σ_β I_αβ ω_β, i.e. L = I·ω. The tensor is symmetric. For a continuous body, the sums become integrals. If ω is along a principal axis, L = Iω.

(c)(i) The equation of motion is m d²x/dt² + γ dx/dt + kx = F₀ cosωt, F₀ = 5 N. Using the phasor method, the steady-state amplitude is A(ω) = F₀/√[(k − mω²)² + (γω)²]. Here ω₀ = √(k/m) = √(100/0.25) = 20 rad s⁻¹. At the standard resonance condition for a weakly damped oscillator, ω = ω₀ = 20 rad s⁻¹: A_res = F₀/(γω₀) = 5/(1 × 20) = 0.25 m. If by resonance one means the exact displacement maximum, then ω_r = √(ω₀² − γ²/(2m²)) = √392 = 14√2 rad s⁻¹, and A_max = F₀/[γ√(ω₀² − γ²/(4m²))] = 5/(6√11) m ≈ 0.251 m. Thus A_res = 0.25 m for the standard resonance condition, with exact maximum 5/(6√11) m.

(c)(ii) The Q-value is Q = ω₀m/γ = (20 × 0.25)/1 = 5. Equivalently, Q = √(mk)/γ = √(0.25 × 100)/1 = 5. Therefore Q = 5, dimensionless.