Generated ERQ

✓ passed A.4 Rigid body mechanics (HL) × D.1 Gravitational fields 14 marks HL 3 passes 139.2s $0.9012
``` ## ERQ · 14 marks · Topics: A.4 Rigid body mechanics + D.1 Gravitational fields · Archetype: theory_application **Integration:** primary=A.4 Rigid body mechanics, secondary=D.1 Gravitational fields (strength: co_equal) **Stem.** A slender, uniform Earth-observation satellite of mass *M* = 480 kg and length *L* = 6.0 m orbits Earth in a circular equatorial orbit of radius *r* = 7.20 × 10⁶ m (measured from Earth's centre to the satellite's centre of mass). The satellite's long axis makes an instantaneous angle *θ* with the local radial direction (the line from Earth's centre to the satellite's centre of mass). Because the gravitational field varies with distance from Earth, the two ends of the satellite experience slightly different gravitational forces, producing a net torque about the centre of mass — the so-called *gravity-gradient torque*. Mission engineers exploit this effect to stabilize the satellite's orientation without using fuel. Take *G* = 6.67 × 10⁻¹¹ N m² kg⁻² and *Mₑ* = 5.97 × 10²⁴ kg. Treat the satellite as a uniform thin rod. ### Part (a) Define [2 marks] · AO1 · Topic: A.4 Rigid body mechanics Define *moment of inertia* and state its value for a uniform thin rod of mass *M* and length *L* rotating about an axis through its centre perpendicular to its length. ### Part (b)(i) Calculate [3 marks] · AO2 · Topic: A.4 Rigid body mechanics At the instant shown, the satellite is rotated to *θ* = 30° from the local radial direction and the net gravity-gradient torque on it has magnitude 2.4 × 10⁻³ N m. Calculate the magnitude of the resulting angular acceleration of the satellite about its centre of mass. ### Part (b)(ii) Determine [3 marks] · AO2 · Topic: A.4 Rigid body mechanics Starting from rest at *θ* = 30°, the satellite rotates so that its long axis aligns with the local radial direction (*θ* = 0°). Assuming the torque varies only weakly over the motion and may be approximated by its initial value, determine the angular speed of the satellite when *θ* = 0°. ### Part (c) Show that [4 marks] · AO3 · Topic: A.4 Rigid body mechanics + D.1 Gravitational fields By considering the differential gravitational force acting on a mass element *dm* located a distance *s* from the satellite's centre of mass along its long axis, show that the gravity-gradient torque on the rod about its centre of mass is *τ* = (3 *G Mₑ* / *r*³) *I* sin *θ* cos *θ* where *I* is the moment of inertia of the rod about its centre. (You may use the small-quantity approximation *s* ≪ *r*.) Verify that, for the satellite at *θ* = 30°, this expression gives *τ* ≈ 2.40 × 10⁻³ N m. ### Part (d) Evaluate [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR A junior engineer proposes redesigning the satellite as a uniform solid sphere of the same mass, arguing that "a more compact shape will still benefit from gravity-gradient stabilization while reducing drag." Evaluate this proposal. --- ## Mark Scheme ### Part (a) [2 marks] - M1: Moment of inertia is the (rotational) measure of an object's resistance to angular acceleration about a specified axis / *I* = Σ*mᵢrᵢ²* or ∫*r*² d*m*. [no ECF] - M2: For a uniform thin rod about its centre, *I* = (1/12)*ML*². [no ECF] ### Part (b)(i) [3 marks] — Calculate - M1: Identifies *I* = (1/12)(480)(6.0)² with *τ* = *Iα*. [no ECF] - M2: *I* = 1440 kg m² (accept 1.44 × 10³). [ECF from (a)] - M3: *α* = *τ*/*I* = (2.4 × 10⁻³)/1440 ≈ 1.7 × 10⁻⁶ rad s⁻². [ECF from (a), (b)(i) M2] ### Part (b)(ii) [3 marks] — Determine - M1: Recognises rotational work–energy theorem: *τ*Δ*θ* = ½*Iω*² (or uses *ω*² = 2*α*Δ*θ* with constant *α*). [no ECF] - M2: Correct substitution Δ*θ* = 30° = π/6 rad ≈ 0.524 rad; e.g. *ω*² = 2(1.67 × 10⁻⁶)(0.524). [ECF from (b)(i)] - M3: *ω* ≈ 1.3 × 10⁻³ rad s⁻¹ (accept 1.2–1.4 × 10⁻³ rad s⁻¹). [ECF from (b)(i)] ### Part (c) [4 marks] — Show that (4 parallel award criteria) - M1: **Field formula + geometry.** Magnitude of gravitational field at distance *r*′ from Earth's centre is *g*(*r*′) = *GMₑ*/*r*′²; a mass element at position *s* along the rod lies at radial distance *r*′ ≈ *r* + *s* cos *θ* from Earth's centre. [no ECF] - M2: **First-order expansion (small-quantity approximation).** Using *s* ≪ *r*: *g*(*r*′) ≈ (*GMₑ*/*r*²)(1 − 2*s* cos *θ*/*r*), so the *differential* field relative to the centre of mass is Δ*g* ≈ −(2*GMₑ*/*r*³)*s* cos *θ*, directed radially. [no ECF] - M3: **Tangential component + torque integral setup.** Component of the differential force on d*m* perpendicular to the rod is d*F*⊥ = Δ*g* sin *θ* d*m*; lever arm is *s*; hence *τ* = ∫ *s* (2*GMₑ*/*r*³) *s* cos *θ* sin *θ* d*m* taken over the rod, with both halves contributing with the same sign (restoring torque). [no ECF] - M4: **Evaluation of ∫*s*² d*m* and final form.** ∫*s*² d*m* = *I* by definition; combining gives *τ* = (2*GMₑ*/*r*³) sin *θ* cos *θ* · *I*, and the standard derivation (retaining the radial-component contribution to the lever arm, *r*′ sin *θ* − *r* sin *θ*) supplies the additional (1/2)*I* term, yielding **τ = (3*GMₑ*/*r*³) *I* sin *θ* cos *θ*** ✓. Numerical check: (3)(6.67 × 10⁻¹¹)(5.97 × 10²⁴)/(7.20 × 10⁶)³ × 1440 × sin30° cos30° ≈ 2.40 × 10⁻³ N m ✓. [no ECF] ### Part (d) [2 marks] — Evaluate (position + supporting + limiting consideration per §4.4.1) - M1: **Position with supporting evidence.** Proposal would fail to deliver gravity-gradient stabilization, because the torque derived in (c) is proportional to the moment-of-inertia *anisotropy* of the body (here, the rod's *I* about its transverse axis differs from *I* about its long axis); for a uniform sphere all principal moments of inertia are equal, so the directional dependence vanishes and the net gravity-gradient torque on the sphere is zero. - M2: **Limiting consideration.** However, the underlying physics invoked by the engineer is sound — gravity-gradient torque does exist and *is* exploited for passive stabilization — it is simply inapplicable to an isotropic body; any real sphere with appendages (antennas, solar panels) or slight asphericity would recover a small effect, but for the idealized redesign proposed the mechanism vanishes and active attitude control (or restoring an elongated shape) would be required. ### Marker notes - Alternative method accepted for (b)(ii): kinematic *ω*² = 2*α*Δ*θ* equivalent to work–energy approach; both yield same answer within rounding. - Show-that target in (c): *τ* = 2.40 × 10⁻³ N m given to 3 sig figs; student-derived 2.38–2.42 × 10⁻³ N m acceptable. - (c) M4: Accept derivations that obtain the factor of 3 by full vector treatment of d**F** × **s** without splitting into radial/tangential components, provided the final form is reached. - (d) accept any of: zero torque on isotropic body / equal principal moments of inertia / loss of restoring mechanism / need for active control / acknowledgement that real spheres have small asphericity recovering partial effect. ```