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## ERQ · 14 marks · Topics: A.2 Forces and momentum + D.1 Gravitational fields · Archetype: theory_application
**Integration:** primary=A.2 Forces and momentum, secondary=D.1 Gravitational fields (strength: supporting)
**Stem.** A spacecraft of mass 2500 kg is in a circular orbit around Earth at an altitude of 400 km. Mission control plans a Hohmann transfer to raise the spacecraft to a circular orbit at 800 km altitude. The manoeuvre uses two brief thruster burns: the first burn is applied tangentially at the 400 km orbit ("perigee burn") to place the spacecraft on an elliptical transfer orbit whose apogee lies at 800 km altitude; the second burn ("apogee burn") is applied tangentially at apogee to circularise the orbit. Treat both burns as instantaneous (impulsive) so that the spacecraft's position does not change during the burn, only its velocity. Take Earth's mass as 5.97 × 10²⁴ kg, Earth's radius as 6.37 × 10⁶ m, and G = 6.67 × 10⁻¹¹ N m² kg⁻².
### Part (a) State [2 marks] · AO1 · Topic: A.2
State Newton's second law in terms of momentum, and explain why this form is more appropriate than F = ma when analysing a thruster burn.
### Part (b)(i) Calculate [3 marks] · AO2 · Topic: A.2 + D.1
Show that the orbital speed of the spacecraft in the initial 400 km circular orbit is 7.67 × 10³ m s⁻¹.
### Part (b)(ii) Determine [3 marks] · AO2 · Topic: A.2
The perigee burn changes the spacecraft's speed from 7.67 × 10³ m s⁻¹ to 7.84 × 10³ m s⁻¹ (the speed required at perigee of the elliptical transfer orbit). The thruster expels exhaust gas at a speed of 3.10 × 10³ m s⁻¹ relative to the spacecraft. Determine the mass of propellant ejected during the perigee burn.
### Part (c) Derive [4 marks] · AO3 · Topic: A.2 + D.1
By equating the gravitational force to the centripetal force, derive an expression for the total mechanical energy E of a satellite of mass m in a circular orbit of radius r around a planet of mass M. Hence calculate the magnitude of the energy that must be supplied by the apogee burn to circularise the orbit at 800 km altitude, given that the speed of the spacecraft at apogee of the transfer orbit (before the burn) is 7.13 × 10³ m s⁻¹.
### Part (d) Explain [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR
Explain one way in which the assumption of an instantaneous (impulsive) burn leads to a systematic discrepancy between the predicted final orbit and the orbit actually achieved.
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## Mark Scheme
### Part (a) [2 marks]
- M1: Net force equals the rate of change of momentum, F = Δp/Δt (or dp/dt) [no ECF]
- M2: During a thruster burn the spacecraft's mass changes as propellant is expelled, so F = ma (which assumes constant mass) is not strictly valid; the momentum form correctly accounts for the changing mass [no ECF]
### Part (b)(i) [3 marks] — Show that
- M1: Equates gravitational force to centripetal force: GMm/r² = mv²/r, giving v = √(GM/r) [no ECF]
- M2: Substitutes r = 6.37 × 10⁶ + 0.400 × 10⁶ = 6.77 × 10⁶ m with M = 5.97 × 10²⁴ kg [no ECF]
- M3: v = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/(6.77 × 10⁶)] = 7.67 × 10³ m s⁻¹ (accept 7.66–7.68 × 10³) [no ECF]
### Part (b)(ii) [3 marks] — Determine (Calculate)
- M1: Identifies conservation of momentum / rocket equation approach. Using impulse–momentum equivalence: m_prop × v_exhaust = m_spacecraft × Δv (impulsive approximation), OR uses Tsiolkovsky: Δv = v_e ln(m_i/m_f) [no ECF]
- M2: Δv = 7.84 × 10³ − 7.67 × 10³ = 170 m s⁻¹; substitution: m_prop = (2500 × 170)/(3.10 × 10³) [ECF from (b)(i)]
- M3: m_prop = 1.4 × 10² kg (accept 137–140 kg). Tsiolkovsky route gives m_prop = 2500(1 − e^(−170/3100)) ≈ 133 kg — also accept. [ECF from (b)(i)]
### Part (c) [4 marks] — Derive
- M1: From GMm/r² = mv²/r obtains KE = ½mv² = GMm/(2r) [no ECF]
- M2: States gravitational PE = −GMm/r and hence E = KE + PE = −GMm/(2r) [no ECF]
- M3: At r₂ = 7.17 × 10⁶ m, circular orbital speed v₂ = √(GM/r₂) = 7.45 × 10³ m s⁻¹; computes ΔKE = ½ × 2500 × (7450² − 7130²) [ECF from M1–M2]
- M4: ΔE = ½ × 2500 × (7450² − 7130²) ≈ 5.8 × 10⁹ J (accept 5.7–5.9 × 10⁹ J). Equivalent method using E = −GMm/(2r) at both radii also accepted. [ECF from M1–M3]
### Part (d) [2 marks] — Explain (causal chain per §4.4.1)
- M1: A real thruster burn has finite duration, so during the burn the spacecraft moves along its orbit and the thrust vector is no longer applied at a single point with purely tangential orientation — part of the thrust acts against gravity ("gravity losses") rather than producing useful tangential Δv [premise / mechanism]
- M2: Therefore the impulsive-burn model overestimates the Δv delivered for a given propellant mass, so the actual apogee (and final orbital radius) is systematically **lower** than predicted [explicit "therefore" linking mechanism to orbital consequence]
### Marker notes
- (b)(i): Show-that target 7.67 × 10³ m s⁻¹ given to 3 sig figs; student working must show v = √(GM/r) substitution to earn full marks (do not award M3 if only the target value is quoted).
- (b)(ii): Accept either the impulsive momentum-conservation method or the Tsiolkovsky rocket equation. ECF applies to Δv if (b)(i) value differs.
- (c): Accept direct calculation via E = −GMm/(2r) at both radii: E₂ − E_apogee where E_apogee = ½mv² − GMm/r₂ at the transfer-orbit apogee, giving the same numerical answer to within rounding.
- (d) full credit requires BOTH (i) identification of a physical mechanism (gravity losses, finite burn arc, thrust direction rotating away from tangential, mass change during burn affecting acceleration profile) AND (ii) an explicit directional consequence linking that mechanism to the orbit (lower apogee / smaller final radius / orbit falls short of target). Award 1 mark only if the mechanism is stated without the directional orbital consequence, or vice versa.
- Accept for (d) M1 alternatives: finite burn means propellant is ejected over a range of positions where g varies; thrust applied while spacecraft rotates around Earth means impulse direction changes.
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