Gravitational fields: worked solutions

Answer: Centre separation = 3.60 m; force magnitude = 1.29 × 10⁻⁴ N. The mutual forces remain equal in magnitude and opposite in direction.

1. r = 2.40 + 0.600 + 0.600 = 3.60 m. Spherical symmetry permits the external point-mass model.
2. F = GMm/r² = (6.67 × 10⁻¹¹)(5.00 × 10³)²/(3.60)² = 1.29 × 10⁻⁴ N.
3. Each force points along the line of centres towards the other sphere. They are a Newton’s third-law pair: equal magnitudes, opposite directions, acting on different spheres.
4. The support balances the force on the held sphere. It changes that sphere’s resultant force and acceleration, not the mutual gravitational force at the same masses and separation.

Check: G has units N m² kg⁻², so GMm/r² is in newtons. A tiny force is reasonable even for substantial laboratory masses.

Mark checklist · 5 marks

• 1 mark — Uses centre separation r = 3.60 m.
• 1 mark — Substitutes masses and distance into F = GMm/r².
• 1 mark — Obtains F ≈ 1.29 × 10⁻⁴ N.
• 1 mark — Describes equal, opposite attractive forces acting on different objects.
• 1 mark — Explains the support changes the resultant, not the interaction at fixed separation.

Common mistake: Using 2.40 m overestimates the force. Do not add the interaction pair to get a zero force on one sphere; its partner force acts on the other sphere.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Answer: M = 4.32 × 10²⁴ kg; uncertainty ≈ 4.0%, or 0.17 × 10²⁴ kg. Field lines point radially inward; locally they are approximately parallel and equally spaced.

1. Gravitational field strength is force per unit mass on a small test mass: g = F/m.
2. M = gR²/G = 8.00(6.00 × 10⁶)²/(6.67 × 10⁻¹¹) = 4.32 × 10²⁴ kg.
3. ΔM/M ≈ 0.16/8.00 + 2(0.06/6.00) = 0.040 = 4.0%.
   ΔM = 0.040M ≈ 0.17 × 10²⁴ kg, giving M ≈ (4.32 ± 0.17) × 10²⁴ kg.
4. For the full planet, draw radial lines with arrows pointing towards its centre. Lines spread out with increasing radius, indicating weaker field. In a patch whose size and height range are small compared with R, they are approximately parallel and equally spaced, so g is approximately uniform.

Check: The uncertainty is a conservative first-order estimate, as instructed, not a statistical quadrature result. More precise local g does not remove a radius uncertainty.

Mark checklist · 7 marks

• 1 mark — Defines g as force per unit mass on a test mass.
• 1 mark — Uses M = gR²/G with radius in metres.
• 1 mark — Obtains M ≈ 4.32 × 10²⁴ kg.
• 1 mark — Adds 2.0% from g and twice 1.0% from R to obtain 4.0%.
• 1 mark — Gives absolute uncertainty ≈ 0.17 × 10²⁴ kg with matching scale.
• 1 mark — Sketches radial inward arrows outside the sphere.
• 1 mark — Describes approximately parallel, equally spaced lines in a small surface region.

Common mistake: Mass is proportional to R² in this rearrangement, not inversely proportional. “Gravity acts down” is a local description; globally it points towards the centre.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, June 2018 — Q17(b)(ii)(3); report p. 23

Answer: r = 6.80 × 10⁶ m; g = 8.65 N/kg; v = 7.67 × 10³ m/s; T = 5.57 × 10³ s (92.8 min). Cabin and crew share free-fall acceleration, so the floating crew member has no support force.

1. r = 6.40 × 10⁶ + 400 × 10³ = 6.80 × 10⁶ m.
   g = GM/r² = 8.65 N kg⁻¹, towards the centre.
2. GMm/r² = mv²/r, so v = √(GM/r) = 7.67 × 10³ m s⁻¹.
3. T = 2πr/v = 5.57 × 10³ s = 92.8 min.
4. Both cabin and crew accelerate towards the planet at approximately the same g. The floating crew member needs no cabin support/contact force. This absence produces apparent weightlessness, even though the gravitational force remains substantial.

Check: Surface g = 9.77 N/kg, so the orbital field is about 89% of its surface value. The period is much shorter than a day, so this is not a stationary communications orbit.

Mark checklist · 7 marks

• 1 mark — Uses orbital radius R + h in metres.
• 1 mark — Obtains g ≈ 8.65 N/kg.
• 1 mark — Equates gravity to centripetal force or uses v² = GM/r.
• 1 mark — Obtains v ≈ 7.67 × 10³ m/s.
• 1 mark — Uses circumference to obtain T ≈ 5.57 × 10³ s.
• 1 mark — Explains cabin and crew share gravitational/free-fall acceleration.
• 1 mark — Links apparent weightlessness to absent support/contact force, not absent gravity.

Common mistake: A vacuum removes air resistance, not gravity. “No resultant force” would imply straight-line constant velocity and is incompatible with this orbit.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, June 2018 — Q17(a)(i)–(iii); report pp. 19–20 · OCR A H556/03, November 2020 — Q4(a)–(b)(i); report p. 5

Answer: T² = (4π²/GM)r³; B’s period = 40.5 days. Both orbits share the same central stellar mass M.

1. GMm/r² = mv²/r, with v = 2πr/T.
2. GM/r² = 4π²r/T², giving T² = (4π²/GM)r³.
3. For the same host star and negligible planet masses, 4π²/GM is constant. The circular orbit’s radius is measured from the star’s centre.
4. TB = TA(rB/rA)^(3/2) = 20.0(0.320/0.200)^(3/2) = 40.5 days. Consistent AU and day units cancel in the ratios.

Check: The radius ratio is 1.60, so period increases by 1.60^(3/2) ≈ 2.02, not by 1.60 or 1.60³. Dimensional check: r³/(GM) has unit s².

Mark checklist · 6 marks

• 1 mark — Equates gravitational force with the required centripetal force.
• 1 mark — Substitutes v = 2πr/T and cancels m.
• 1 mark — Rearranges to T² = (4π²/GM)r³.
• 1 mark — Forms the correct squared-period/cubed-radius ratio.
• 1 mark — Obtains TB ≈ 40.5 days.
• 1 mark — Explains the same central mass gives the same constant.

Common mistake: T² = r³ is only true for specially chosen units and normalization. In SI, keep the physical constant. A ratio from different stars cannot assume that it cancels.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, June 2022 — Q23(a); report p. 34 · OCR A H556/01, June 2025 — Q22(a)(i); report p. 41

Answer: T²/r³ ≈ (4.95, 4.93, 4.94) × 10⁻¹³ s²/m³. Physical gradient = 4.93 × 10⁻¹³ s²/m³; M = 1.20 × 10²⁴ kg.

1. TA = 4.42 × 3600 = 15 912 s.
   TA²/rA³ = 15 912²/(8.00 × 10⁶)³ = 4.95 × 10⁻¹³ s² m⁻³.
2. TB = 29 196 s gives 4.93 × 10⁻¹³; TC = 44 964 s gives 4.94 × 10⁻¹³ s² m⁻³. These are approximately constant at the stated precision, supporting T² ∝ r³.
3. S = 4.93(10⁸ s²)/(10²¹ m³) = 4.93 × 10⁻¹³ s² m⁻³.
4. M = 4π²/(GS) = 4π²/[(6.67 × 10⁻¹¹)(4.93 × 10⁻¹³)] = 1.20 × 10²⁴ kg.
5. One valid additional assumption is that other bodies provide negligible perturbations, so the planet supplies essentially all of the centripetal force. The supplied circular and negligible-moon-mass assumptions also matter; the inverse-square field must be appropriate at these exterior radii.

Check: The physical gradient has units s²/m³; GS has units kg⁻¹, so 4π²/(GS) is a mass. Moon C’s radius is twice A’s, so its period should be about √8 times A’s.

Mark checklist · 7 marks

• 1 mark — Converts hour periods into seconds.
• 1 mark — Correctly evaluates T²/r³ for at least two named rows.
• 1 mark — Evaluates the remaining row and justifies approximate constancy, rather than exact identity.
• 1 mark — Restores both graph axis scale factors to obtain S = 4.93 × 10⁻¹³ s²/m³.
• 1 mark — Uses M = 4π²/(GS).
• 1 mark — Obtains M ≈ 1.20 × 10²⁴ kg.
• 1 mark — States a relevant gravitational/circular-orbit/negligible-moon-mass assumption.

Common mistake: Unlabelled calculations make the comparison hard to assess. A plot of T against r is not straight: use T² and r³. Do not insert the unscaled number 4.93 into the SI formula.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/03, June 2023 — Q1(b)(i); report p. 6

Answer: Circular equatorial orbit, eastward with Earth, T = 86 164 s. r = 4.22 × 10⁷ m; h = 3.58 × 10⁷ m. A fixed antenna is useful; a lower orbit visiting many latitudes better serves close atmospheric sampling.

1. The orbit must be circular, in the equatorial plane, in the same eastward sense as Earth’s rotation, with T equal to Earth’s rotation period. Together these give a fixed position relative to the surface.
2. r = [GMT²/(4π²)]^(1/3) = 4.22 × 10⁷ m.
3. h = r − R = 4.2213 × 10⁷ − 6.40 × 10⁶ = 3.58 × 10⁷ m, about 35 800 km.
4. The apparent fixed direction permits a fixed receiving/transmitting ground antenna and continuous contact with the chosen region when it is in view.
5. For the atmospheric mission, this large altitude prevents close-up sampling of the upper atmosphere; the equatorial orbit also does not pass over many latitudes. A suitably inclined low orbit can address those goals. Either clearly linked limitation answers the requested one reason.

Check: 86 164 s is the supplied sidereal rotation period. Using the rounded 24 h value instead would alter r slightly; use the data given. Relative to Earth’s centre the satellite moves at about 3.08 km/s.

Mark checklist · 6 marks

• 1 mark — States circular orbit in the equatorial plane.
• 1 mark — States same rotational sense and period as Earth.
• 1 mark — Uses the supplied period to obtain r ≈ 4.22 × 10⁷ m.
• 1 mark — Subtracts Earth’s radius to obtain h ≈ 3.58 × 10⁷ m.
• 1 mark — Links fixed apparent direction/continuous regional coverage to broadcasting.
• 1 mark — Links altitude or latitude coverage to the stated atmospheric mission.

Common mistake: “It stays still” is incomplete: name the reference frame. A tilted orbit with a matching period is geosynchronous but would not remain fixed above one ground point.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, June 2024 — Q22(a)(i)–(ii); report p. 38 · OCR A H556/03, June 2023 — Q1(b)(ii); report p. 7

Answer: Gradient = −60.0 m³/s²; M = 9.00 × 10¹¹ kg. Potential falls from −0.0400 to −0.120 J/kg, giving impact speed 0.400 m/s.

1. Gravitational potential is the work done by an external agent per unit mass in bringing a small test mass slowly from infinity to the point, with zero potential at infinity. In an attractive field this external work is negative because the agent restrains the fall.
2. Gradient = (−0.120 − 0)/(2.00 × 10⁻³ − 0) = −60.0 J m kg⁻¹ = −60.0 m³ s⁻².
3. Vg = −GM(1/r), so GM = 60.0 m³ s⁻².
   M = 60.0/G = 9.00 × 10¹¹ kg.
4. Vstart = −60.0/1500 = −0.0400 J kg⁻¹; Vsurface = −0.120 J kg⁻¹.
5. Conservation: ½v² = Vstart − Vsurface = 0.0800 J kg⁻¹.
   v = √0.160 = 0.400 m s⁻¹. Probe mass cancels.
6. The fall spans a threefold radius change, so field strength changes by a factor of nine. Constant-g SUVAT across this interval is unsuitable.

Check: Surface escape speed is √(2GM/R) = 0.490 m/s. Falling from a finite distance gives a smaller speed than falling from rest at infinity.

Mark checklist · 6 marks

• 1 mark — Defines gravitational potential as work done per unit mass from infinity to the point, with zero at infinity.
• 1 mark — Calculates signed scaled gradient −60.0 m³/s² and identifies gradient = −GM.
• 1 mark — Obtains M ≈ 9.00 × 10¹¹ kg.
• 1 mark — Evaluates initial and final potentials or their difference correctly.
• 1 mark — Equates kinetic-energy gain per mass to the potential decrease.
• 1 mark — Obtains speed 0.400 m/s.

Common mistake: The horizontal axis is 1/r, not r. Subtract signed potentials in the correct order for the energy gain; do not take a square root of a negative kinetic energy.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, November 2020 — Q23(b)–(c); report p. 4

Answer: ΔEp = −2.40 × 10⁹ J; kinetic-energy gain = +2.40 × 10⁹ J; speed at B = 6.63 × 10³ m/s. Positive Vg–r slope gives inward gr; the largest displayed field magnitude is at the smallest radius.

1. ΔVg = VB − VA = −40.0 − (−20.0) = −20.0 MJ kg⁻¹.
   ΔEp = mΔVg = −2.40 × 10⁹ J.
2. ½vB² = ½vA² + (VA − VB) = 2.00 × 10⁶ + 20.0 × 10⁶ = 22.0 × 10⁶ J kg⁻¹.
   vB = √(44.0 × 10⁶) = 6.63 × 10³ m s⁻¹.
3. Vg increases as r increases, so dVg/dr is positive. The supplied relation makes gr negative: force/acceleration is inward. At the smallest displayed radius, the curve’s gradient magnitude is greatest, so the field magnitude is greatest.
4. The curve approaches zero from below at large r. This is inverse-distance variation, not exponential decay. The displayed model applies outside the spherical planet; it is not extended through the interior.

Check: Initial kinetic energy is 2.40 × 10⁸ J; final is 2.64 × 10⁹ J. Their difference matches −ΔEp. A negative potential is not a negative kinetic energy.

Mark checklist · 6 marks

• 1 mark — Finds signed ΔVg = −20.0 MJ/kg.
• 1 mark — Multiplies by 120 kg to obtain signed ΔEp = −2.40 × 10⁹ J.
• 1 mark — Adds initial kinetic energy per mass to the potential-energy loss per mass.
• 1 mark — Obtains vB ≈ 6.63 × 10³ m/s.
• 1 mark — Uses the positive gradient and supplied minus sign to identify inward field.
• 1 mark — Identifies steepest slope and strongest field at smallest displayed radius.

Common mistake: A graph reading in MJ/kg becomes an energy only after multiplying by mass and 10⁶. Starting with ½vA² = 0 would ignore the supplied initial speed.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, June 2025 — Q22(b)(i); report p. 42 · OCR A H556/01, June 2022 — Q23(b)(ii); report p. 36

Answer: Fext(r1) = 1.25 × 10³ N; Wext = +5.00 × 10⁹ J; Wgravity = −5.00 × 10⁹ J. The initial-force rectangle gives 1.00 × 10¹⁰ J, twice the correct work.

1. Fext(r1) = GMm/r1² = 1.25 × 10³ N; at r2 it is only 312.5 N.
2. The area under the positive outward balancing-force curve, between r1 and r2, is positive external work for outward displacement. It equals the increase in gravitational potential energy because ΔEk ≈ 0.
3. Wext = ΔEp = GMm(1/r1 − 1/r2) = +5.00 × 10⁹ J.
   Gravity is opposite to outward motion: Wgravity = −5.00 × 10⁹ J.
4. Wrectangle = F(r1)(r2 − r1) = 1250(8.00 × 10⁶) = 1.00 × 10¹⁰ J.
5. That rectangle stays above the decreasing curve, overestimating work by a factor of 2 (100%). FΔr is exact only for a constant force along the displacement; the local mgΔh approximation is unsuitable for doubling this radius.

Check: The tug and gravity nearly cancel at each point, so their works sum to approximately zero and the cargo’s kinetic energy stays almost constant. Positive potential-energy change is compatible with negative absolute potential energy.

Mark checklist · 6 marks

• 1 mark — Obtains the initial balancing force 1250 N.
• 1 mark — Identifies shaded area as external work/positive ΔEp for outward slow motion.
• 1 mark — Calculates Wext = +5.00 × 10⁹ J from the endpoint potential energies.
• 1 mark — Gives Wgravity = −5.00 × 10⁹ J.
• 1 mark — Evaluates initial-force rectangle as 1.00 × 10¹⁰ J.
• 1 mark — Explains overestimate using decreasing force/rectangle above curve; quantifies factor 2.

Common mistake: The plotted positive force is the external balancing force, not the signed inward gravitational force. The same positive shaded area cannot be labelled positive work by gravity for this outward move.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, November 2020 — Q23(b)–(c); report p. 4

Answer: x = 3.00 × 10⁸ m from A, or 1.00 × 10⁸ m from B. Each field magnitude = 0.00667 N/kg; resultant = 0. Total potential = −2.668 × 10⁶ J/kg. A small rightward displacement accelerates the test mass towards B.

1. gA = −GMA/x²; gB = +GMB/(d − x)².
   Zero resultant requires MA/x² = MB/(d − x)².
2. x/(d − x) = √(MA/MB) = 3.
   x = 3(d − x), so x = 3d/4 = 3.00 × 10⁸ m.
3. Vtotal = −G[MA/x + MB/(d − x)] = −2.668 × 10⁶ J kg⁻¹.
   Each field magnitude is 0.00667 N kg⁻¹, but the directions are opposite.
4. Fields are vectors and cancel at this point. Potential is a scalar: both contributions are negative and add, so it remains negative.
5. Displacing the mass slightly right increases B’s field magnitude and decreases A’s. The resultant is rightwards, further from the neutral point: it is unstable along this line in the fixed-source model. This is not an orbital Lagrange-point calculation.

Check: The neutral point lies closer to the smaller mass, which must be nearer to match the larger mass’s field. Using the mass ratio 9 directly as a distance ratio would fail the inverse-square equality.

Mark checklist · 7 marks

• 1 mark — Sets equal and opposite field magnitudes using x and d − x.
• 1 mark — Takes the square root of the mass ratio to obtain distance ratio 3.
• 1 mark — Finds x = 3.00 × 10⁸ m from A.
• 1 mark — Adds both signed point-mass potentials.
• 1 mark — Obtains Vtotal = −2.668 × 10⁶ J/kg.
• 1 mark — Explains vector cancellation versus scalar addition.
• 1 mark — Identifies acceleration towards B after a small displacement towards B and justifies it.

Common mistake: Zero gravitational force at one point does not make the entire region field-free. The fixed-source equilibrium discussed here must not be called a stable parking orbit.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, June 2018 — Q17(c)(i)–(ii); report pp. 24–25

Answer: Star at one focus; fastest at closest approach; vp/va = 3.00. a = 8.00 × 10¹⁰ m; T = 1.00 × 10⁷ s (116 days). From far to near: ΔEp = −1.00 × 10¹² J and ΔEk = +1.00 × 10¹² J.

1. Kepler’s first law places the star at one focus of the ellipse. The joining line sweeps equal areas in equal times. Near the star, the smaller radius requires a larger distance along the orbit in the same short time, so speed is greatest at closest approach.
2. At the two turning points, swept area in a short Δt is ½rvΔt.
   rpvp = rava, so vp/va = ra/rp = 3.00.
3. a = (4.00 × 10¹⁰ + 1.20 × 10¹¹)/2 = 8.00 × 10¹⁰ m.
   T = T0(a/a0)^(3/2) = 1.40 × 10⁷(0.800)^(3/2) = 1.00 × 10⁷ s, about 116 days.
4. ΔEp = Ep,near − Ep,far = −GMm(1/rp − 1/ra) = −1.00 × 10¹² J.
5. No thrust or drag: Ek + Ep is constant, so ΔEk = −ΔEp = +1.00 × 10¹² J.
6. Do not use v = √(GM/r) at each point of an elliptical orbit: that is the circular-orbit speed relation. This solution uses Kepler’s area law and energy conservation instead.

Check: The orbit period is between those of circular orbits at rp and ra. Its total energy is conserved while both kinetic and potential energy change. Rounded reference data explain tiny differences from the direct GM formula.

Mark checklist · 8 marks

• 1 mark — Places the star at one focus.
• 1 mark — Uses equal areas in equal times to locate maximum speed at closest approach.
• 1 mark — Uses perpendicular endpoint geometry to obtain vp/va = 3.00.
• 1 mark — Calculates semi-major axis a = 8.00 × 10¹⁰ m.
• 1 mark — Uses the same-star Kepler ratio to obtain T ≈ 1.00 × 10⁷ s.
• 1 mark — Evaluates endpoint potential energies or GMm(1/rp − 1/ra).
• 1 mark — Obtains signed ΔEp = −1.00 × 10¹² J.
• 1 mark — Applies conservation to obtain ΔEk = +1.00 × 10¹² J.

Common mistake: A focus is not the centre. Do not describe all three Kepler laws when a particular law is requested: apply that law to the named probe. Gravitational acceleration need not be perpendicular to velocity except at the turning points.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/03, June 2023 — Q1(a)(i)–(ii); report p. 5 · OCR A H556/01, June 2022 — Q23(b)(ii); report p. 36

Answer: Minimum gain = 34.7 MJ. Aircraft supplies about 0.148 MJ (0.428%), so the 1% claim fails. Fuel requirements also depend on propulsion efficiency, drag, gravity losses and the trajectory.

1. GMm/r² = mv²/r gives v² = GM/r.
   Ek = ½mv² = GMm/(2r).
2. Initial Etotal = Ep,surface = −GMm/R = −62.5 MJ.
   Final Ep = −55.6 MJ; final Ek = +27.8 MJ; final Etotal = −27.8 MJ.
3. ΔErequired = Efinal − Einitial = +34.7 MJ.
   Equivalently, ΔEp = +6.94 MJ and ΔEk = +27.8 MJ. Add them.
4. Aircraft credit ≈ mg0h + ½mv² = 1.00(9.77)(12 000) + ½(1.00)(250)² = 1.485 × 10⁵ J = 0.148 MJ.
5. Fraction = 0.14849/34.7222 × 100% = 0.428%, below 1%. In this ideal model the remaining required mechanical gain is about 34.6 MJ.
6. The ideal endpoint budget is not a fuel-energy prediction. For example, finite-time powered flight has gravity losses and drag; real engines have efficiency limits and expend energy on propellant. These change the fuel required. The altitude and speed credit is small but does not rule out operational advantages of an aircraft launch.

Check: The change is positive although both initial and final total energies are negative. Exact endpoint potential for 12 km gives an aircraft credit only about 0.00027 MJ smaller, which cannot rescue the 1% claim.

Mark checklist · 8 marks

• 1 mark — Equates gravitational and centripetal forces.
• 1 mark — Derives Ek = GMm/(2r).
• 1 mark — Finds initial energy −62.5 MJ and final potential −55.6 MJ.
• 1 mark — Includes final orbital kinetic energy +27.8 MJ.
• 1 mark — Finds total required gain ≈ 34.7 MJ.
• 1 mark — Adds aircraft altitude and kinetic credits to obtain ≈ 0.148 MJ.
• 1 mark — Quantifies ≈ 0.428% and rejects the stated 1% energy claim.
• 1 mark — Gives a physical limitation connecting the ideal endpoint budget to actual fuel/flight.

Common mistake: Using only ΔEp misses most of the launch energy. Do not mix ground-relative aircraft speed with planet-centred speed or introduce rotation into only one side of the stated non-rotating comparison.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, June 2024 — Q22(b)(i)–(ii); report pp. 39–40 · OCR A H556/01, June 2024 — Q22(c); report p. 40; MS pp. 23–24

Answer: vi = 7.07 km/s; vf = 5.00 km/s. ΔEk = −7.50 GJ; ΔEp = +15.0 GJ; ΔEtotal = +7.50 GJ. The potential increase is twice the kinetic decrease, so net energy must be supplied.

1. vi = √(GM/ri) = 7.07 × 10³ m s⁻¹; vf = √(GM/rf) = 5.00 × 10³ m s⁻¹.

2. Circular endpoint energies

   Endpoint	Ek / GJ	Ep / GJ	Etotal / GJ
   Initial	+15.0	−30.0	−15.0
   Final	+7.50	−15.0	−7.50

3. ΔEk = −7.50 GJ;
   ΔEp = +15.0 GJ;
   ΔEtotal = +7.50 GJ.
4. Gravitational potential becomes less negative as the satellite moves outwards. Its increase is larger than the kinetic-energy decrease, so the satellite’s total energy rises. In the ideal fixed-mass model, thrusters must supply net positive mechanical work.
5. A brief prograde burn initially increases speed and changes the path; later speed varies along the transfer. The lower final circular speed does not imply that the manoeuvre begins by simply braking into a higher circular orbit. No transfer-burn timing or fuel calculation is requested.

Check: At each circular endpoint Ek = −Ep/2 and Etotal = Ep/2. Doubling r halves the magnitudes of all three energies but reduces speed by √2.

Mark checklist · 7 marks

• 1 mark — Obtains vi ≈ 7.07 km/s.
• 1 mark — Obtains vf = 5.00 km/s.
• 1 mark — Obtains signed ΔEk = −7.50 GJ.
• 1 mark — Obtains signed ΔEp = +15.0 GJ.
• 1 mark — Adds them to obtain ΔEtotal = +7.50 GJ.
• 1 mark — Explains potential-energy increase exceeds the kinetic decrease, so work is positive.
• 1 mark — Distinguishes the two circular endpoints from the powered intermediate transfer path.

Common mistake: Do not assume total mechanical energy is conserved through firing thrusters. Conservation applies to unpowered gravity-only motion; here the satellite gains net mechanical energy.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.

Related examiner evidence: OCR A H556/01, June 2024 — Q22(b)(i)–(ii); report pp. 39–40

Answer: Surface vesc = 3.46 km/s. Helium crms = 1.37 km/s; nitrogen crms = 0.517 km/s. Helium is more vulnerable, but the high-speed tail matters. At r = 3.00 × 10⁶ m, vesc = 2.19 km/s, so the 2.50 km/s outward probe escapes.

1. For just escape, Etotal at infinity = 0. Energy conservation gives ½mvesc² − GMm/R = 0.
   vesc = √(2GM/R); launched mass cancels.
2. vesc(R) = √[2(7.20 × 10¹²)/(1.20 × 10⁶)] = 3.46 × 10³ m s⁻¹.
3. crms,He = √[3k(300)/(6.64 × 10⁻²⁷)] = 1.37 × 10³ m s⁻¹.
   crms,N₂ = √[3k(300)/(4.65 × 10⁻²⁶)] = 517 m s⁻¹.
4. At the same temperature, lighter helium particles have higher typical speeds and a larger high-speed tail above the escape threshold, so helium is more vulnerable to thermal escape. Both rms values are below vesc, but rms is not a maximum: some particles can be faster. Escape also depends on direction, collisions, altitude and heating. These estimates cannot establish complete long-term retention.
5. At the probe launch point, vesc = √[2GM/(3.00 × 10⁶)] = 2.19 × 10³ m s⁻¹.
   2.50 × 10³ > vesc, so the outward probe escapes in the stated isolated model.

Check: Surface g = GM/R² = 5.00 N/kg, so √(2gR) reproduces 3.46 km/s. Mass independence applies to a probe’s required escape speed; gas-particle mass affects the thermal speed, which is a different calculation.

Mark checklist · 7 marks

• 1 mark — Uses zero potential and minimum zero kinetic energy at infinity.
• 1 mark — Equates initial kinetic and binding-energy magnitude to derive vesc = √(2GM/R).
• 1 mark — Obtains surface vesc ≈ 3.46 km/s.
• 1 mark — Calculates both gas rms speeds ≈ 1.37 km/s and 517 m/s.
• 1 mark — Links higher helium speed at equal T to greater vulnerability to escape.
• 1 mark — Explains rms is not a maximum and a speed distribution/high-speed tail prevents a complete-retention conclusion.
• 1 mark — Uses local vesc ≈ 2.19 km/s to conclude the outward 2.50 km/s probe escapes.

Common mistake: A lower gas mass does not change the gravitational escape threshold at the same point. It changes the distribution of particle speeds at a given temperature. Escape speed also depends on launch radius; it is not one fixed number throughout space.

Exam technique: Show the chosen physical model, centre-to-centre distances and units before substituting. These original questions use PhysicsUK mark guidance; linked OCR lessons concern the named source parts.