Generated ERQ

✓ passed D.1 Gravitational fields × A.1 Kinematics 14 marks HL 3 passes 123.93s $0.8092
``` ## ERQ · 14 marks · Topics: D.1 Gravitational fields + A.1 Kinematics · Archetype: theory_application **Integration:** primary=D.1 Gravitational fields, secondary=A.1 Kinematics (strength: supporting) **Stem.** The International Space Station (ISS) orbits Earth in a near-circular orbit at an altitude of 4.00 × 10² km above Earth's surface. Take Earth's mass as M = 5.97 × 10²⁴ kg, Earth's radius as R = 6.37 × 10⁶ m, and the universal gravitational constant as G = 6.67 × 10⁻¹¹ N m² kg⁻². An astronaut performing a spacewalk gently releases a small tool from rest, as measured in the astronaut's (freely falling) frame of reference. Observers on the ground describe the same situation using an inertial frame in which both the ISS and the tool are subject to Earth's gravitational field. The students analysing this scenario are asked to compare predictions made in the two frames and to assess the limitations of treating Earth's gravitational field as that of a point mass with spherical symmetry. ### Part (a) Define [2 marks] · AO1 · Topic: D.1 Define *gravitational field strength* and state its SI unit. ### Part (b)(i) Show that [3 marks] · AO2 · Topic: D.1 Show that the gravitational field strength at Earth's surface is approximately 9.81 N kg⁻¹. ### Part (b)(ii) Calculate [3 marks] · AO2 · Topic: D.1 Calculate the orbital speed of the ISS at altitude 4.00 × 10² km. ### Part (c) Determine and explain [4 marks] · AO2+AO3 · Topic: D.1 + A.1 In the inertial frame of an observer on the ground, the tool is released from rest *relative to the ISS* but is in fact moving at the ISS orbital speed. Using equations of uniformly accelerated motion in the radial direction, determine the time taken for the tool to fall a radial distance of 10 m towards Earth as predicted by a *naive* kinematic model that ignores the tool's tangential motion. Then explain why the astronaut nonetheless observes the tool to remain (essentially) stationary beside them. ### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR · Topic: D.1 Treating Earth's gravitational field as that of a point mass with perfect spherical symmetry is a simplifying assumption used in elementary orbit calculations. Suggest **two** distinct physical limitations of this assumption, each with its consequence for the accuracy of predicted satellite orbits. --- ## Mark Scheme ### Part (a) [2 marks] — Define - M1: gravitational field strength = gravitational force per unit (test) mass at a point ✓ [no ECF] - M2: SI unit stated as N kg⁻¹ (accept m s⁻²) ✓ [no ECF] ### Part (b)(i) [3 marks] — Show that - M1: equates gravitational field strength with g = GM/R² (formula identified) ✓ [no ECF] - M2: correct substitution: g = (6.67 × 10⁻¹¹)(5.97 × 10²⁴) / (6.37 × 10⁶)² ✓ - M3: evaluates to 9.81 N kg⁻¹ (accept 9.80–9.82) — student value must be shown to at least 3 s.f. to earn the mark ✓ ### Part (b)(ii) [3 marks] — Calculate - M1: applies v = √(GM/r) with r = R + h = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m ✓ [ECF on r] - M2: correct substitution: v = √[(6.67 × 10⁻¹¹)(5.97 × 10²⁴) / (6.77 × 10⁶)] ✓ - M3: v = 7.67 × 10³ m s⁻¹ (accept 7.6–7.7 × 10³ m s⁻¹) ✓ [ECF from M1] ### Part (c) [4 marks] — Determine and explain (calculation + causal chain per §4.4.1) - M1: computes gravitational field strength at orbital altitude g′ = GM/r² = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)/(6.77 × 10⁶)² ≈ 8.69 m s⁻² ✓ [ECF from (b)(ii) M1] - M2: applies kinematics s = ½g′t² with s = 10 m, giving t = √(2s/g′) = √(20/8.69) ✓ - M3: t ≈ 1.5 s (accept 1.5–1.52 s) ✓ [ECF from M1] - M4: causal chain — the tool is released with the *same* orbital tangential speed as the ISS, therefore both experience the same centripetal acceleration g′ towards Earth, which means the tool co-orbits with the station and remains stationary relative to the astronaut despite the inertial-frame "fall" ✓ ### Part (d) [2 marks] — Suggest (proposal + physics-based justification per §4.4.1) - M1: first limitation stated AND linked to a quantitative orbital consequence ✓ - M2: second, *distinct* limitation stated AND linked to a quantitative orbital consequence ✓ ### Marker notes - (b)(i) Show-that target 9.81 N kg⁻¹ given to 3 s.f.; student-derived 9.80–9.82 acceptable. Award only if working is shown. - (b)(ii) Alternative method: from T = 2π√(r³/GM) then v = 2πr/T accepted, provided r is computed correctly. - (c) Alternative method for M1–M3: using g ≈ 8.7 m s⁻² (carried forward, or estimated as g(R/r)²) is acceptable; any answer in 1.45–1.55 s earns M3. - (c) M4: require explicit linking language ("therefore", "so", "which means", "because") tying the *equal tangential speed* OR *both freely falling* to the *zero relative acceleration*. A bare statement "they are weightless" without mechanism: do **not** award M4. - (d) Accept any two of the following limitation–consequence pairs (one mark each, only if BOTH halves are present): • Earth is oblate (equatorial bulge / J₂ term) → causes nodal regression / orbital plane precession not predicted by spherical model • Mass distribution is non-uniform (mascons, density variations) → produces local perturbations in satellite altitude/period • Atmospheric drag at low altitudes is neglected (field treatment ignores non-gravitational forces) → causes orbital decay over time • Gravitational influence of Moon and Sun ignored → causes long-period perturbations in orbital elements • Field treated as static while Earth rotates with its mass → fails to account for frame-dragging / tidal bulges affecting orbit - (d) Reject answers giving only a limitation with no stated orbital consequence (e.g. "Earth is not a sphere" alone earns 0). Reject two limitations that are physically the same idea reworded. ```