D.1 Gravitational Fields Questions

1D-1A-01

Gravitational field strength·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Planet X has the same average density as Earth but twice the radius.

What is the gravitational field strength at the surface of X in terms of g, the field strength at the surface of Earth?

Show mark scheme

Marking point	Mark	Notes
Step 1At the surface g = GM/R².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2With constant density, M = ρ × (4/3)πR³, so g = (4/3)πGρR ∝ R.	—
Step 3Doubling R at the same density doubles g: 2g.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis applies the inverse-square law with the mass held fixed — but the planet has 8 times the mass.
BThis would need the mass to double only; it increases by 2³ = 8.
CCorrect: mass ×8 and R² ×4 gives ×2.
DThis divides the eightfold mass by R instead of by R²: 8/2 = 4.

Syllabus understandingD.1 — that gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point as given by g = F/m = GM/r²; (guidance) Newton's universal law of gravitation extended to spherical masses of uniform density Command term: Deduce

2D-1A-02

Escape speed and orbital speed·D.1 Gravitational fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A small probe is launched vertically upwards from the surface of a non-rotating planet that has no atmosphere. The planet has mass M and radius R. The launch speed is equal to the orbital speed of a satellite in a circular orbit just above the surface, √(GM/R).

What is the maximum height reached by the probe above the surface of the planet?

Show mark scheme

Marking point	Mark	Notes
Step 1The field is not uniform over this range, so use energy conservation with E_p = −GMm/r: ½mv² − GMm/R = −GMm/r at the highest point, a distance r from the centre.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2With v² = GM/R: GMm/2R − GMm/R = −GMm/2R, so r = 2R.	—
Step 3The height above the surface is r − R = R.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis assumes a uniform field equal to the surface value g = GM/R² all the way up: h = v²/2g = R/2. The field weakens with height, so the probe rises further.
BCorrect: the probe reaches r = 2R, which is a height R above the surface.
CThis gives the distance from the centre of the planet (2R), not the height above the surface.
DThis omits the ½ in the kinetic energy: mv² = GMm/R is exactly the energy needed to escape, but the probe has only half of it (GMm/2R).

Syllabus understandingD.1 (HL) — the gravitational potential energy for a two-body system E_p = −Gm₁m₂/r; the orbital speed v_orbital = √(GM/r) and the escape speed v_esc = √(2GM/r) Command term: Deduce

3D-1A-03

Orbital energy·D.1 Gravitational fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A satellite of mass m is in a circular orbit of radius r around a planet of mass M. Its rockets move it to a circular orbit of radius 2r.

What is the work done by the rockets?

Show mark scheme

Marking point	Mark	Notes
Step 1In a circular orbit E_k = ½mv² = GMm/2r and E_p = −GMm/r, so the total energy is E = −GMm/2r.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2At radius 2r: E₂ = −GMm/4r.	—
Step 3Work done = E₂ − E₁ = −GMm/4r + GMm/2r = GMm/4r.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the total energy rises from −GMm/2r to −GMm/4r.
BThis is the change in potential energy alone, ignoring that the kinetic energy falls in the higher orbit.
CThis is the magnitude of the initial potential energy.
DThis adds the kinetic-energy change instead of subtracting it.

4D-1A-13

Geostationary orbits·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A geostationary satellite has an orbital period of 24 h. GM for the Earth is 4.0 × 10¹⁴ N m² kg⁻¹.

What is the radius of the orbit?

Show mark scheme

Marking point	Mark	Notes
Step 1Gravity provides the centripetal force: GMm/r² = m(2π/T)²r, so r³ = GMT²/4π².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2T = 24 × 3600 = 8.64 × 10⁴ s, so r³ = 4.0 × 10¹⁴ × (8.64 × 10⁴)²/39.5 = 7.6 × 10²² m³.	—
Step 3r = (7.6 × 10²²)^1/3 = 4.2 × 10⁷ m (about 36 000 km above the surface).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis uses T = 24 × 60 = 1440 s (minutes instead of hours) — and is smaller than the Earth's radius, so impossible.
BCorrect: r = (GMT²/4π²)^1/3 = 4.2 × 10⁷ m.
CThis omits the division by 4π²: (GMT²)^1/3.
DThis is r³ in m³ — the cube root has not been taken.

Syllabus understandingD.1 — Kepler's three laws of orbital motion; Newton's universal law of gravitation as given by F = Gm₁m₂/r² for bodies treated as point masses (guidance: the speed to maintain orbit; circular orbits) Command term: Determine

5D-1A-17

Kepler's second law·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A comet moves in an elliptical orbit around the Sun. Its greatest distance from the Sun is three times its least distance from the Sun.

What is (speed of the comet at its closest point to the Sun)/(speed of the comet at its furthest point from the Sun)?

Show mark scheme

Marking point	Mark	Notes
Step 1Kepler's second law: the line joining the comet to the Sun sweeps out equal areas in equal times.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the closest and furthest points the velocity is perpendicular to the radius, so in a short time Δt the area swept is ½rvΔt.	—
Step 3Equal areas: r₁v₁ = r₂v₂, so v₁/v₂ = r₂/r₁ = 3.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis inverts the ratio — by the second law the comet moves fastest when it is closest to the Sun.
BThis applies the circular-orbit result v = √(GM/r) to an elliptical orbit, where the speed is not set by the local radius alone.
CCorrect: equal areas in equal times give rv = constant at the two ends of the orbit, so the speed ratio is 3.
DThis treats the speed as following the inverse-square law of the gravitational force, v ∝ 1/r².

Syllabus understandingD.1 — Kepler's three laws of orbital motion (the second law: equal areas in equal times) Command term: Deduce

6D-1A-22

Variation of g with height·D.1 Gravitational fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The graph shows how the gravitational field strength g varies with the distance r from the centre of a planet of radius 4.0 × 10⁶ m, for points outside the planet.

What is the minimum energy needed to move a mass of 1.0 kg from the surface of the planet to a point where r = 8.0 × 10⁶ m?

Variation of gravitational field strength with distance from the centre of the planet (outside the planet).

Show mark scheme

Marking point	Mark	Notes
Step 1g = −ΔV_g/Δr, so the change in gravitational potential between two points is the area under the g–r graph between them.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The area between r = 4.0 × 10⁶ m and 8.0 × 10⁶ m is about 16 squares of 1 N kg⁻¹ × 10⁶ m, i.e. 1.6 × 10⁷ J kg⁻¹ (accept 1.5–1.7 × 10⁷). Check: V_g ∝ −1/r, so ΔV_g = 8.0 × 4.0 × 10⁶ × (1 − ½).	—
Step 3Minimum energy for 1.0 kg = 1.6 × 10⁷ J.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis multiplies the distance moved by the smallest field strength on the interval, 2.0 N kg⁻¹ at r = 8.0 × 10⁶ m.
BCorrect: the area under the curve is 1.6 × 10⁷ J kg⁻¹.
CThis joins the end points with a straight line: ½(8.0 + 2.0) × 4.0 × 10⁶. The curve lies below this chord, so the trapezium overestimates the area.
DThis assumes g keeps its surface value 8.0 N kg⁻¹ all the way: mgΔr = 1.0 × 8.0 × 4.0 × 10⁶ J.

Syllabus understandingD.1 (HL) — the gravitational field strength g as the gravitational potential gradient, g = −ΔV_g/Δr; the work done in moving a mass in a gravitational field W = mΔV_g Command term: Determine

7D-1A-29

Work from equipotentials·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The diagram shows equipotential surfaces around a planet. A spacecraft of mass 2000 kg moves from point A to point B.

What is the minimum work that must be done on the spacecraft?

Equipotential surfaces (dashed) around a planet. The values are the gravitational potential on each surface.

Show mark scheme

Marking point	Mark	Notes
Step 1A is on the −60 MJ kg⁻¹ surface and B on the −30 MJ kg⁻¹ surface: ΔV = −30 − (−60) = +30 MJ kg⁻¹.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2W = mΔV = 2000 × 30 × 10⁶.	—
Step 3W = 6.0 × 10¹⁰ J, independent of the path taken.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis is the potential difference alone (30 MJ), not multiplied by the mass.
BCorrect: W = mΔV = 2000 × 3.0 × 10⁷ J.
CThis uses the potential at A (−60) rather than the difference between A and B.
DThis uses the sum of the two potentials.

Syllabus understandingD.1 (HL) — the work done in moving a mass m in a gravitational field as given by W = mΔV_g; equipotential surfaces for gravitational fields Command term: Determine

8D-1A-30

Kepler's third law·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Planet X moves in a circular orbit of radius r around a star of mass M with period T. Planet Y moves in a circular orbit of radius 2r around a different star of mass 4M.

What is the orbital period of Y?

Show mark scheme

Marking point	Mark	Notes
Step 1Gravity provides the centripetal force: GMm/r² = m(4π²/T²)r, so T² = 4π²r³/(GM).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2For Y, r³ is multiplied by 8 and M by 4, so T² is multiplied by 8/4 = 2.	—
Step 3Period of Y = √2 T.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: T² ∝ r³/M gives a factor of √(8/4).
BThis uses T ∝ r³/M without taking the square root: 8/4 = 2.
CThis ignores the different mass of the star: T²/r³ is the same only for orbits around the same central body. √8 = 2√2.
DThis puts the mass in the numerator (T² ∝ Mr³): √(8 × 4) = 4√2.

Syllabus understandingD.1 — Kepler's three laws of orbital motion; Newton's universal law of gravitation as given by F = Gm₁m₂/r² for bodies treated as point masses; (guidance) for calculations involving orbital motion the orbits will be assumed to be circular Command term: Determine

9D-1A-31

Point masses·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Newton's universal law of gravitation, F = Gm₁m₂/r², is used to calculate the gravitational force between the bodies in each situation below, with r taken as the distance between their centres.

I. Two uniform solid lead spheres of radii 5.0 cm and 2.0 cm whose surfaces are touching, with r = 7.0 cm.

II. An irregularly shaped asteroid 3 km long and a planet whose centre is 2 × 10⁸ m away.

III. Two identical steel rods, each 1.0 m long, lying parallel and side by side with their centres 5.0 cm apart.

In which situations does the law give the force correctly?

Show mark scheme

Marking point	Mark	Notes
Step 1A uniform sphere (spherically symmetric mass distribution) acts on bodies outside it as if all its mass were at its centre, at any separation — so I is valid even with the spheres touching.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Any body acts as a point mass when the separation is much larger than its size: 3 km ≪ 2 × 10⁸ m, so II is valid.	—
Step 3The rods are neither spherical nor far apart (5.0 cm ≪ 1.0 m), so different parts are at very different distances and the formula with the centre separation fails: III is not valid.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis rejects II, assuming a non-spherical body can never be treated as a point mass — but it can when the separation is much larger than its size.
BThis rejects I, assuming spheres must be far apart compared with their radii — but uniform spheres act as point masses at any separation outside them.
CCorrect: I (uniform spheres) and II (separation ≫ size) only.
DThis assumes the law always works with the distance between centres; for long rods close together it does not.

Syllabus understandingD.1 — Newton's universal law of gravitation for bodies treated as point masses; conditions under which extended bodies can be treated as point masses Command term: Deduce

10D-1A-32

Equipotential surfaces·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Gravitational equipotential surfaces around an isolated spherical planet of radius R are drawn at equal intervals of potential. Close to the surface of the planet, adjacent surfaces are a distance s apart, where s is much smaller than R.

What is the distance between adjacent surfaces at a height 2R above the surface?

Show mark scheme

Marking point	Mark	Notes
Step 1g = −ΔV_g/Δr, so for a fixed interval ΔV_g the separation Δr = ΔV_g/g is inversely proportional to g.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2g = GM/r², so the separation is proportional to r².	—
Step 3At a height 2R the distance from the centre is 3R: separation = 3² × s = 9s.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis treats the field as uniform, so that the surfaces are equally spaced everywhere. That is true only close to the surface.
BThis takes the separation as proportional to r, as if g were proportional to 1/r.
CThis uses the height 2R as the distance from the centre: 2² = 4.
DCorrect: the separation ∝ 1/g ∝ r², and r = 3R.

Syllabus understandingD.1 (HL) — equipotential surfaces for gravitational fields and their relationship with field lines; the gravitational field strength as the potential gradient g = −ΔV_g/Δr; g = GM/r² Command term: Determine

11D-1A-33

Resultant field of two bodies·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two spherical asteroids of masses M and 4M have their centres a distance d apart. Point P lies on the line joining their centres, midway between them.

What is the magnitude of the resultant gravitational field strength at P?

Show mark scheme

Marking point	Mark	Notes
Step 1Each asteroid is d/2 from P, and the two fields at P point in opposite directions (each towards its own asteroid).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Field of the larger: G(4M)/(d/2)² = 16GM/d²; field of the smaller: GM/(d/2)² = 4GM/d².	—
Step 3Resultant = 16GM/d² − 4GM/d² = 12GM/d², towards the larger asteroid.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis uses d instead of d/2 as the distance from each asteroid: G(4M − M)/d².
BThis adds the two fields and also uses d as the distance: G(4M + M)/d².
CCorrect: G(4M − M)/(d/2)² = 12GM/d².
DThis adds the fields as if they pointed the same way: G(4M + M)/(d/2)² = 20GM/d².

Syllabus understandingD.1 — gravitational field strength g = F/m = GM/r²; determination of the resultant gravitational field strength at points along a line joining two bodies Command term: Determine

12D-1A-34

Potential gradient·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The graph shows the variation of the gravitational potential V_g with distance r from the centre of a planet. The tangent to the curve at r = 8.0 × 10⁶ m is drawn.

Which row gives the magnitude and the direction of the gravitational field strength at r = 8.0 × 10⁶ m?

Variation of gravitational potential with distance from the centre of a planet. The dashed line is the tangent to the curve at r = 8.0 × 10⁶ m; it passes through (0, −5.0) and (16, 0).

Magnitude / m s⁻²Direction

Show mark scheme

Marking point	Mark	Notes
Step 1Gradient of the tangent = ΔV_g/Δr = (0 − (−5.0) MJ kg⁻¹)/(16 × 10⁶ m − 0) = 5.0 × 10⁶/1.6 × 10⁷ = 0.31 J kg⁻¹ m⁻¹.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2g = −ΔV_g/Δr = −0.31 m s⁻².	—
Step 3The negative sign means the field points in the direction of decreasing r: towards the planet.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis ignores the prefix M on the potential axis (5.0 J kg⁻¹ instead of 5.0 × 10⁶ J kg⁻¹) while converting the distance: 5.0/1.6 × 10⁷.
BThe magnitude is right but the sign of g = −ΔV_g/Δr has been dropped: a positive potential gradient gives a field towards smaller r.
CCorrect: magnitude 0.31 m s⁻², directed towards the planet.
DThis is the gradient of the chord between r = 4.0 and 12.0 × 10⁶ m on the curve ((−1.67 − (−5.0))/8.0 = 0.42), not of the tangent at 8.0 × 10⁶ m.

Syllabus understandingD.1 (HL) — the gravitational field strength g as the gravitational potential gradient, g = −ΔV_g/Δr; V_g = −GM/r Command term: Determine

13D-1A-35

Gravitational potential energy·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two small moonlets, each of mass m, are at rest a distance a apart, far from any other mass.

What is the minimum work needed to increase their separation to 3a, leaving them at rest?

Show mark scheme

Marking point	Mark	Notes
Step 1The gravitational potential energy of the two-body system is E_p = −Gm²/r; the minimum work needed equals the increase in E_p (no kinetic energy is given to the moonlets).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Initially E_p = −Gm²/a; finally E_p = −Gm²/(3a).	—
Step 3W = −Gm²/(3a) − (−Gm²/a) = 2Gm²/(3a).	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis is the magnitude of the final potential energy only; the work done is the change in potential energy.
BCorrect: the potential energy rises from −Gm²/a to −Gm²/(3a), an increase of 2Gm²/(3a).
CThis treats the potential energy as varying as 1/r², like the force: (1 − 1/9)Gm²/a.
DThis counts the energy twice, once for each moonlet in the field of the other; the potential energy belongs to the pair, so there is only one term −Gm²/r.

Syllabus understandingD.1 (HL) — the gravitational potential energy E_p of a system as the work done to assemble the system from infinite separation of its components; E_p = −Gm₁m₂/r for a two-body system Command term: Determine

14D-1A-36

Launch and escape from orbit·D.1 Gravitational fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A probe starts from rest on the surface of a non-rotating moon of radius R that has no atmosphere. The minimum energy needed to place the probe in a circular orbit of radius 2R is E.

What is the minimum additional energy the probe needs, once in this orbit, to escape completely from the moon's gravitational field?

Show mark scheme

Marking point	Mark	Notes
Step 1At rest on the surface the total energy is −GMm/R. In a circular orbit of radius 2R the total energy is −GMm/2(2R) = −GMm/4R.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2So E = GMm/R − GMm/4R = 3GMm/4R.	—
Step 3To escape, the total energy must be raised to zero: additional energy = GMm/4R = E/3.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: the orbiting probe needs GMm/4R more, which is one third of E.
BThis supplies the magnitude of the potential energy in orbit, GMm/2R, forgetting that the probe already has kinetic energy GMm/4R.
CThis assumes the escape speed is twice the orbital speed, so the kinetic energy must be quadrupled: an extra 3 × GMm/4R = E.
DThis is the escape energy from rest on the surface, GMm/R, ignoring the energy E already supplied.

Syllabus understandingD.1 (HL) — energetics of a satellite going into orbit around a non-rotating planet starting from rest on its surface; energy conditions for an orbiting satellite to escape the gravitational influence of a planet; E_p = −GMm/r Command term: Deduce

15D-1A-37

Orbital speed·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Five moons move in circular orbits around the same planet. The graph shows the square of the orbital speed v² of each moon against the reciprocal of its orbital radius, 1/r, with the best-fit line.

What is the mass of the planet?

Square of orbital speed against the reciprocal of orbital radius for five moons of one planet.

Show mark scheme

Marking point	Mark	Notes
Step 1v_orbital = √(GM/r), so v² = GM × (1/r): the graph is a straight line through the origin with gradient GM.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Gradient of the line = 3.0 × 10⁸ m² s⁻² ÷ 8.0 × 10⁻⁹ m⁻¹ = 3.75 × 10¹⁶ m³ s⁻² (accept 3.7–3.8 × 10¹⁶).	—
Step 3M = gradient/G = 3.75 × 10¹⁶/6.67 × 10⁻¹¹ = 5.6 × 10²⁶ kg (accept 5.5–5.7 × 10²⁶ kg).	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis multiplies the gradient by G instead of dividing by it: 3.75 × 10¹⁶ × 6.67 × 10⁻¹¹ = 2.5 × 10⁶ kg.
BThis ignores the factor 10⁻⁹ in the label of the 1/r axis, so the gradient used (3.75 × 10⁷) is 10⁹ times too small.
CThis uses v² = 2GM/r, which is the escape-speed relation, so it halves the mass: 2.8 × 10²⁶ kg.
DCorrect: M = (3.75 × 10¹⁶)/(6.67 × 10⁻¹¹) = 5.6 × 10²⁶ kg.

Syllabus understandingD.1 (HL) — the orbital speed of a body orbiting a large mass, v_orbital = √(GM/r); determining a quantity from the gradient of a linear graph Command term: Determine

16D-1A-38

Newton's law of gravitation·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

A space probe is at a distance d from the centre of an asteroid and the gravitational force between them is F. The probe then releases a lander containing half of its mass, and moves away until it is at a distance 3d from the centre of the asteroid.

What is the gravitational force between the asteroid and the probe now?

Show mark scheme

Marking point	Mark	Notes
Step 1F ∝ m₁m₂/r²: the probe mass is multiplied by ½ and the distance by 3.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2New force = F × ½ × 1/3² = F/18.	✓ 1	Answer A

Answer: A · 2 stages of work, one mark

Every option, and why

ACorrect: ½ × 1/9 = 1/18.
BThis ignores the halving of the probe mass.
CThis halves the mass but does not square the distance: ½ × 1/3.
DThis doubles the mass instead of halving it: 2 × 1/9.

Syllabus understandingD.1 — Newton's universal law of gravitation F = Gm₁m₂/r² for bodies treated as point masses Command term: Determine

17D-1A-39

Escape speed·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark2 steps to full marksDetermine

The escape speed from the surface of a planet of radius R is v_esc.

What is the escape speed from a point at a height 3R above the surface?

Show mark scheme

Marking point	Mark	Notes
Step 1v_esc = √(2GM/r), where r is the distance from the centre of the planet.	—	All 2 steps must be completed — there is no mark for a part-answer.
Step 2At a height 3R, r = R + 3R = 4R, so the escape speed is multiplied by 1/√4 = 1/2.	✓ 1	Answer C

Answer: C · 2 stages of work, one mark

Every option, and why

AThis applies an inverse-square law to the speed: 1/4².
BThis takes the escape speed as inversely proportional to r.
CCorrect: v_esc ∝ 1/√r and r = 4R.
DThis uses the height 3R instead of the distance 4R from the centre.

Syllabus understandingD.1 (HL) — the escape speed v_esc at any point in a gravitational field, v_esc = √(2GM/r) Command term: Determine

18D-1A-60

Kepler's first law·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A planet moves around a star in an elliptical orbit. No other bodies affect its motion.

I. The star is at one focus of the ellipse.

II. The gravitational force on the planet is perpendicular to the velocity of the planet at every point of the orbit.

III. The kinetic energy of the planet is greatest at the point of the orbit closest to the star.

Which statements are correct?

Show mark scheme

Marking point	Mark	Notes
Step 1Kepler's first law: every planet moves in an ellipse with the star at one focus, so I is correct.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The gravitational force is directed towards the star. Except at the closest and furthest points, the velocity is not perpendicular to the line to the star, so the force has a component along (or against) the velocity and the speed changes. II is incorrect.	—
Step 3The total energy is constant. Closest to the star the gravitational potential energy is least (most negative), so the kinetic energy is greatest. III is correct: I and III only.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis accepts II, which is true only for a circular orbit (or at the two ends of the major axis), and misses that III follows from energy conservation.
BCorrect: the star is at a focus, and the kinetic energy is greatest where the potential energy is least.
CThis places the star at the centre of the ellipse instead of at a focus, and applies the circular-orbit result in II to an ellipse.
DThis treats all three statements as true, applying the circular-orbit property in II to an elliptical orbit.

Syllabus understandingD.1 — Kepler's three laws of orbital motion (the first law: elliptical orbits with the central body at one focus); A.3 — the conservation of energy Command term: Deduce

19D-1A-61

Uniform field near the surface·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

A small region just above the flat surface of a planet is considered. The equipotential surfaces in each diagram are drawn at equal intervals of gravitational potential.

Which diagram shows the gravitational field lines (solid) and the equipotential surfaces (dashed) in this region?

Field lines are drawn solid, with arrows; equipotential surfaces are dashed. Each diagram shows the same small region above the surface.

Show mark scheme

Marking point	Mark	Notes
Step 1Close to the surface, over a small region, the field is uniform: the field lines are parallel, equally spaced and directed towards the planet (the direction of the force on a mass).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Equipotential surfaces are perpendicular to the field lines, so they are horizontal.	—
Step 3Since g = −ΔV_g/Δr is constant, equal intervals of potential are equal distances apart: diagram D.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThe spacing grows with height as it would far from a planet, where g decreases; over a small region near the surface g is constant, so the spacing is equal.
BThe arrows point away from the planet; the field direction is the direction of the force on a mass, which is towards the planet.
CThe equipotentials are drawn parallel to the field lines; they must be perpendicular, because no work is done moving a mass along an equipotential.
DCorrect: parallel, equally spaced field lines towards the surface, with perpendicular equipotentials equally spaced.

Syllabus understandingD.1 — gravitational field lines; D.1 (HL) — equipotential surfaces for gravitational fields and the relationship between equipotential surfaces and gravitational field lines; the (assumed) uniform field close to the surface of massive bodies Command term: Identify

20D-1A-62

Kepler's third law·D.1 Gravitational fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The graph shows lg T against lg r for the moons of two planets X and Y, where T is the orbital period and r the radius of the circular orbit of a moon.

What is (mass of X)/(mass of Y)?

Graph drawn to scale. lg denotes the logarithm to base 10.

Show mark scheme

Marking point	Mark	Notes
Step 1T² = 4π²r³/(GM), so lg T = 1.5 lg r + lg(2π) − ½ lg(GM): both lines have gradient 1.5.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2At the same r, the line for X is 0.30 higher: lg T_X − lg T_Y = −½ lg(M_X/M_Y) = 0.30.	—
Step 3lg(M_X/M_Y) = −0.60, so M_X/M_Y = 10^−0.6 = 0.25.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: a longer period at the same radius means a smaller mass, and T ∝ M^−1/2.
BThis omits the factor ½, treating T as proportional to 1/M: 10^−0.3 = 0.50.
CThis omits the factor ½ and takes the longer period to mean the larger mass: 10^0.3 = 2.0.
DThis takes the longer period to mean the larger mass: 10^0.6 = 4.0.

Syllabus understandingD.1 — Kepler's three laws of orbital motion (the third law for circular orbits, T² ∝ r³); Newton's universal law of gravitation applied to circular orbits; Tools 3 — interpreting a logarithmic graph Command term: Determine

21D-1A-72

Atmospheric drag on a satellite·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A satellite moves in a low circular orbit around the Earth. The thin upper atmosphere exerts a very small drag force on the satellite, so that over many orbits its path remains almost circular.

Which row describes how the radius of the orbit and the speed of the satellite change?

Radius of orbitSpeed of satellite

Show mark scheme

Marking point	Mark	Notes
Step 1The drag force does negative work on the satellite, so its total energy E = −GMm/(2r) decreases (becomes more negative).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2A more negative total energy means a smaller r: the satellite moves into a lower orbit.	—
Step 3In a circular orbit v = √(GM/r), so the speed increases as r decreases; the kinetic energy GMm/(2r) = −E rises.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis assumes that a slower satellite needs a smaller centripetal force and so drifts outwards. The drag removes energy, which lowers the orbit, and in a lower orbit the satellite moves faster.
BThis lowers the orbit correctly but assumes that a force opposing the motion must slow the satellite down. As the satellite moves inwards the gravitational force does positive work, which more than replaces the energy removed by the drag.
CThis treats the radius as fixed and lets the drag simply reduce the speed. At a lower speed the gravitational force is larger than the centripetal force needed at that radius, so the satellite moves inwards.
DCorrect: the total energy falls, so the orbit becomes smaller, and the speed √(GM/r) increases.

22D-1A-73

Orbital decay·D.1 Gravitational fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A satellite in a low circular orbit around the Earth is slowly brought lower by a very small atmospheric drag force. At every instant its path is almost a circle.

Which statements are correct while the radius of the orbit decreases?

I. The magnitude of the work done by the drag force is equal to the increase in the kinetic energy of the satellite.

II. The orbital period of the satellite decreases.

III. The gravitational force on the satellite has a component in the direction of its velocity.

Show mark scheme

Marking point	Mark	Notes
Step 1In a circular orbit E_k = GMm/(2r) = −E, so ΔE_k = −ΔE. The work done by the drag force equals ΔE (a negative quantity), so its magnitude equals the increase in E_k: I is correct.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2T² = 4π²r³/(GM): the period decreases as r decreases, so II is correct.	—
Step 3The satellite spirals slowly inwards, so its velocity has a small component towards the Earth. The gravitational force, directed towards the centre, therefore has a component along the velocity and does positive work: III is correct.	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis rejects III by treating the gravitational force as exactly perpendicular to the velocity, as in a perfect circular orbit. On the slow inward spiral the velocity has a small inward component, so gravity does positive work, which is why the speed increases.
BThis rejects II, assuming that a satellite slowed by drag takes longer to go round. Its orbit becomes smaller and its speed greater, so the period decreases (T ∝ r^3/2).
CThis rejects I, assuming that the drag force removes kinetic energy, so the kinetic energy must fall. Since E_k = −E for a circular orbit, the kinetic energy rises by exactly the energy that the drag removes.
DCorrect: all three statements follow from E_k = −E in a circular orbit and from the slow inward spiral.

Syllabus understandingD.1 (HL) — the qualitative effect of a small viscous drag due to the atmosphere on the height and speed of an orbiting body; the orbital speed of a body orbiting a large mass as given by v_orbital = √(GM/r); (guidance) changes in energy when a satellite changes orbit; D.1 — Kepler's three laws of orbital motion; A.3 — work done by a force as a transfer of energy Command term: Deduce

23D-1A-74

Potential energy of a three-body system·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Three identical small spheres, each of mass m, are held at rest at the corners of an equilateral triangle of side a, far from all other masses.

What is the gravitational potential energy of the system of three spheres?

Show mark scheme

Marking point	Mark	Notes
Step 1The gravitational potential energy of a system is the work done to assemble it from infinite separation. For point masses it is the sum of −Gm₁m₂/r over every pair of bodies.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Three spheres form three pairs, and each pair is a distance a apart.	—
Step 3E_p = 3 × (−Gm²/a) = −3Gm²/a.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis counts every pair twice, once for each sphere in the field of the other. The potential energy belongs to the pair, so there is one term per pair: three terms in all.
BThis uses the distance a/√3 from each sphere to the centre of the triangle instead of the separation a of each pair: −3√3 Gm²/a.
CCorrect: three pairs, each contributing −Gm²/a.
DThis counts only the two pairs that include one particular sphere (the energy needed to remove that sphere to infinity) and leaves out the pair formed by the other two.

24D-1A-75

Potential at the point of zero field·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

Two stars of masses 4M and M have their centres a distance d apart. Point P lies on the line joining their centres, between the stars, where the resultant gravitational field strength is zero.

What is the gravitational potential at P?

Show mark scheme

Marking point	Mark	Notes
Step 1At P the two fields are equal and opposite: G(4M)/x² = GM/(d − x)², so 2(d − x) = x and x = 2d/3 from the larger star (d/3 from the smaller).	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Potential is a scalar, so the contributions add: V_g = −G(4M)/(2d/3) − GM/(d/3) = −6GM/d − 3GM/d.	—
Step 3V_g = −9GM/d.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis places P at the midpoint of the line: −G(4M)/(d/2) − GM/(d/2) = −10GM/d. The field is zero closer to the smaller star.
BCorrect: P is 2d/3 from the larger star, and the two negative potentials add to −9GM/d.
CThis subtracts the two potentials, as if they pointed in opposite directions like the fields: −6GM/d + 3GM/d. Potential has no direction; both contributions are negative.
DThis assumes that zero field strength means zero potential. The field strength is the potential gradient, g = −ΔV_g/Δr: at P the potential along the line is at a maximum, not zero.

Syllabus understandingD.1 (HL) — the gravitational potential V_g at a point as the work done per unit mass in bringing a mass from infinity to that point, as given by V_g = −GM/r; the gravitational field strength g as the gravitational potential gradient, as given by g = −ΔV_g/Δr; D.1 — (guidance) determination of the resultant gravitational field strength due to two bodies along the line joining them Command term: Determine

25D-1A-76

Field strength from equipotentials·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

The diagram shows equipotential lines in a plane through the centre of an elongated asteroid. The lines are drawn at equal intervals of gravitational potential; the values are in J kg⁻¹. Points P and Q are marked.

Which row gives the direction of the gravitational field at P, and compares the gravitational field strength at P with that at Q?

Equipotential lines (dashed) at equal intervals of potential around an elongated asteroid; values in J kg⁻¹. The asteroid is shaded. Drawn to scale.

Direction of field at PField strength at P compared with Q

Show mark scheme

Marking point	Mark	Notes
Step 1Field lines cross the equipotentials at right angles and point towards lower (more negative) potential, the direction of the force on a mass: at P the field points towards the asteroid.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2g = −ΔV_g/Δr: for equal steps of potential, the field is strongest where the equipotentials are closest together.	—
Step 3The lines are closer together at P than at Q, so the field strength at P is greater, although Q lies on a more negative potential (between the −8.0 and −7.0 lines).	✓ 1	Answer D

Answer: D · 3 stages of work, one mark

Every option, and why

AThis combines two errors: the field is taken to point towards higher potential, and the field strength is judged from how negative the potential is rather than from the spacing of the lines.
BThis compares the field strengths correctly but takes the field to point towards higher potential. A mass released at P accelerates towards lower potential, towards the asteroid.
CThis judges the field strength from the value of the potential: Q is on a more negative potential, but the field strength is the potential gradient, which is smaller where the lines are further apart.
DCorrect: the field points down the potential gradient, towards the asteroid, and the closer spacing at P means a stronger field.

Syllabus understandingD.1 (HL) — equipotential surfaces for gravitational fields; the relationship between equipotential surfaces and gravitational field lines; the gravitational field strength g as the gravitational potential gradient, as given by g = −ΔV_g/Δr Command term: Deduce

26D-1A-77

Orbital speed and period·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

Two moons, X and Y, move in circular orbits around the same planet. The orbital period of Y is 8 times the orbital period of X.

What is (orbital speed of Y)/(orbital speed of X)?

Show mark scheme

Marking point	Mark	Notes
Step 1Kepler's third law: T² ∝ r³, so r ∝ T^2/3 and the orbit of Y has 8^2/3 = 4 times the radius of the orbit of X.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2v = 2πr/T, so the speed ratio = 4/8.	—
Step 3= 1/2 (check: v = √(GM/r) gives 1/√4).	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis uses v = 2πr/T but keeps the radius the same, so the speed falls by the factor 8 of the period.
BThis finds the radius ratio of 4 correctly but takes the orbital speed as inversely proportional to r.
CCorrect: the radius is 4 times larger and the period 8 times longer, so the speed is halved.
DThis inverts the dependence on radius, taking v ∝ √r: √4 = 2. Outer moons move more slowly.

Syllabus understandingD.1 — Kepler's three laws of orbital motion; D.1 (HL) — the orbital speed of a body orbiting a large mass as given by v_orbital = √(GM/r) Command term: Deduce

27D-1A-78

Gravitational force and height·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The gravitational force on an astronaut standing on the surface of a planet of radius R is W. At a height h above the surface the gravitational force on the astronaut is 0.64W.

What is h?

Show mark scheme

Marking point	Mark	Notes
Step 1F ∝ 1/r², where r is the distance from the centre of the planet: (R/r)² = 0.64.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2R/r = 0.80, so r = 1.25R.	—
Step 3h = r − R = 0.25R.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: r = R/0.80 = 1.25R, so the height is 0.25R.
BThis assumes the force falls linearly with height: 1 − 0.64 = 0.36.
CThis omits the square root: r/R = 1/0.64 = 1.56, giving h = 0.56R.
DThis is the distance from the centre of the planet, not the height above its surface.

Syllabus understandingD.1 — Newton's universal law of gravitation as given by F = Gm₁m₂/r² for bodies treated as point masses; gravitational field strength g at a point as the force per unit mass experienced by a small point mass at that point, as given by g = F/m = GM/r² Command term: Determine

28D-1A-79

Escape from orbit·D.1 Gravitational fields

Paper 1AHard1 mark

Multiple choice · 1 mark3 steps to full marksDeduce

A satellite moves in a circular orbit around a planet with speed v₀. Its rocket engine fires briefly, so that its speed becomes 2v₀ without any change in its distance from the planet. The satellite then moves away and never returns.

What is the speed of the satellite when it is very far from the planet?

Show mark scheme

Marking point	Mark	Notes
Step 1In the circular orbit GM/r = v₀², so the gravitational potential energy per unit mass is −GM/r = −v₀².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Total energy per unit mass after the firing = ½(2v₀)² − v₀² = v₀².	—
Step 3Very far away the potential energy is zero, so ½v² = v₀² and v = √2 v₀.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis subtracts the escape speed √2 v₀ from the new speed 2v₀. Speeds do not subtract in this way; it is the energies that must be compared.
BCorrect: the energy per unit mass left over far away is 2v₀² − v₀² = v₀², so v = √2 v₀.
CThis uses −GM/(2r), the total energy per unit mass in the orbit, as the potential energy: ½v² = 2v₀² − ½v₀².
DThis ignores the work done against the gravitational force as the satellite moves away, as if its speed stayed at 2v₀.

Syllabus understandingD.1 (HL) — the gravitational potential energy for a two-body system as given by E_p = −Gm₁m₂/r; the orbital speed of a body orbiting a large mass as given by v_orbital = √(GM/r); the escape speed v_esc at any point in a gravitational field as given by v_esc = √(2GM/r); (guidance) energetics of a satellite going into orbit around a non-rotating planet starting from rest on its surface; energy conditions for a body to escape the gravitational influence of a planet Command term: Deduce

29D-1A-80

Synchronous orbit·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

A planet of uniform density ρ and radius R rotates once in a time T. A satellite moves in a circular orbit in the plane of the planet's equator, in the same direction as the planet rotates, so that it always stays above the same point on the surface.

What is the radius of the orbit of the satellite?

Show mark scheme

Marking point	Mark	Notes
Step 1The orbital period must equal T: GM/r² = (4π²/T²)r, so r³ = GMT²/4π².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2The mass of the planet is M = (4/3)πR³ρ, so r³ = (4/3)πGρR³T²/4π² = GρR³T²/3π.	—
Step 3r = R(GρT²/3π)^1/3.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: equating the gravitational field strength to the centripetal acceleration with M = (4/3)πR³ρ.
BThis uses ω = 1/T instead of 2π/T, so that r³ = GMT²: the factor 4π² is lost.
CThis takes the mass of the planet as ρR³, omitting the factor 4π/3 in the volume of a sphere.
DThis inverts the relationship, as if a longer rotation period required a smaller orbit. Kepler's third law gives r³ ∝ T².

Syllabus understandingD.1 — Kepler's three laws of orbital motion; gravitational field strength g at a point as the force per unit mass experienced by a small point mass at that point, as given by g = F/m = GM/r²; A.2 — centripetal acceleration a = 4π²r/T² Command term: Determine

30D-1A-81

Potential energy of a planet–moon system·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksCalculate

Pluto has a mass of 1.30 × 10²² kg and its largest moon, Charon, has a mass of 1.59 × 10²¹ kg. The distance between their centres is 1.96 × 10⁴ km.

What is the gravitational potential energy of the Pluto–Charon system?

Show mark scheme

Marking point	Mark	Notes
Step 1E_p = −Gm₁m₂/r, with r = 1.96 × 10⁷ m.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2E_p = −6.67 × 10⁻¹¹ × 1.30 × 10²² × 1.59 × 10²¹/1.96 × 10⁷.	—
Step 3E_p = −7.03 × 10²⁵ J.	✓ 1	Answer C

Answer: C · 3 stages of work, one mark

Every option, and why

AThis substitutes the distance in km (1.96 × 10⁴) instead of in m, making the energy 1000 times too large.
BThis counts the energy twice, once for Pluto in the field of Charon and once for Charon in the field of Pluto. The potential energy belongs to the system and is counted once.
CCorrect: E_p = −Gm₁m₂/r = −7.0 × 10²⁵ J.
DThis omits the minus sign. The potential energy is zero at infinite separation, and work must be done on the system to separate the bodies, so the potential energy of the bound system is negative.

31D-1A-82

Escape speed and orbital speed·D.1 Gravitational fields

Paper 1AEasy1 mark

Multiple choice · 1 mark3 steps to full marksDetermine

The graph shows how the escape speed v_esc from a planet varies with the distance r from its centre, for points outside the planet.

A satellite moves in a circular orbit of radius 3.2 × 10⁷ m around the planet. What is the orbital speed of the satellite?

Escape speed v_esc against distance r from the centre of the planet (drawn to scale).

Show mark scheme

Marking point	Mark	Notes
Step 1From the graph, at r = 3.2 × 10⁷ m the escape speed is 5.0 km s⁻¹.	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2v_orbital = √(GM/r) and v_esc = √(2GM/r), so v_orbital = v_esc/√2 at the same r.	—
Step 3v_orbital = 5.0/√2 = 3.5 km s⁻¹.	✓ 1	Answer B

Answer: B · 3 stages of work, one mark

Every option, and why

AThis halves the escape speed, as if the escape speed were twice the orbital speed.
BCorrect: the orbital speed is the escape speed divided by √2.
CThis takes the escape speed as the speed needed for a circular orbit. A satellite at the escape speed would leave the planet.
DThis multiplies by √2 instead of dividing: the orbital speed is smaller than the escape speed at the same point.

32D-1A-83

Resultant field between two stars·D.1 Gravitational fields

Paper 1AMedium1 mark

Multiple choice · 1 mark3 steps to full marksIdentify

Two identical stars, S₁ and S₂, are a distance d apart. x is the distance from the centre of S₁ along the line joining the centres. The resultant gravitational field strength g on this line is taken as positive when it points from S₁ towards S₂.

Which graph shows how g varies with x between the stars?

Graphs A to D of the resultant gravitational field strength g against x between the stars (S₁ at x = 0, S₂ at x = d).

Show mark scheme

Marking point	Mark	Notes
Step 1Each star produces a field directed towards its own centre, of magnitude GM/r².	—	All 3 steps must be completed — there is no mark for a part-answer.
Step 2Near S₁ its field dominates and points towards S₁ (negative); near S₂ the field points towards S₂ (positive).	—
Step 3At d/2 the two equal fields are opposite and cancel, so g rises steadily from large negative values through zero to large positive values: graph A.	✓ 1	Answer A

Answer: A · 3 stages of work, one mark

Every option, and why

ACorrect: negative near S₁, zero at the midpoint and positive near S₂.
BThis reverses the directions, as if each field pointed away from its star. The gravitational field is the force per unit mass on a small mass, which is attracted towards each star.
CThis plots the magnitude of the resultant field, ignoring the sign convention given: the field near S₁ points in the negative direction.
DThis adds the magnitudes of the two fields as if they pointed the same way. Between the stars they point in opposite directions, so there must be a point of zero field.

Syllabus understandingD.1 — gravitational field strength g at a point as the force per unit mass experienced by a small point mass at that point, as given by g = F/m = GM/r²; gravitational field lines; (guidance) determination of the resultant gravitational field strength due to two bodies along the line joining them Command term: Identify

33D-1B-01

Kepler's third law·D.1 Gravitational fields

Paper 1BEasy7 marks

Data-based question5 steps to full marksDetermine

A student uses published data for the four largest moons of Jupiter to test Kepler's third law. The mean orbital radius r of each moon was obtained from telescope measurements of its greatest angular distance from the planet, and the orbital period T from the times between successive eclipses of the moon by Jupiter. Each radius is known to within ±1 %; the uncertainties in the periods are negligible. The orbits may be treated as circular.

The student's hypothesis is that T²/r³ has the same value for every moon. 1 day = 8.64 × 10⁴ s.

Moon	r / 10⁸ m	T / days
Io	4.22	1.77
Europa	6.71	3.55
Ganymede	10.7	7.15
Callisto	18.8	16.7

(a)

The hypothesis.

(i)

Test the hypothesis, using the data for at least three of the moons.

(2)

(b)

The mass of Jupiter.

(i)

Determine the mass of Jupiter.

(2)

(ii)

Determine the absolute uncertainty in your answer to (b)(i).

(2)

(c)

A prediction.

(i)

Amalthea, a small inner moon of Jupiter, has an orbital radius of 1.81 × 10⁸ m. Predict its orbital period, in days.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
T²/r³ calculated with T in seconds for at least three moons: Io 3.11, Europa 3.11, Ganymede 3.12, Callisto 3.13 (× 10⁻¹⁶ s² m⁻³)	✓ 1	Award only if at least three moons are used. Accept 2 s.f. Consistent other units (e.g. days² m⁻³) are acceptable.
The values agree to within about 1 % (less than the 3 % uncertainty in r³), so the hypothesis is supported	✓ 1	The conclusion must be supported by the candidate's own values. Do not accept "they are roughly the same" without a comparison.
Part (b)(i)
Gravitational force provides the centripetal force: GMm/r² = m(2π/T)²r, so T²/r³ = 4π²/(GM) and M = 4π²/(G × T²/r³)	✓ 1
M = 4π²/(6.67 × 10⁻¹¹ × 3.12 × 10⁻¹⁶) = 1.90 × 10²⁷ kg	✓ 1	Allow ECF from (a). Accept 1.88–1.91 × 10²⁷ kg (any single moon's value of T²/r³ may be used).
Part (b)(ii)
M ∝ r³/T², so the percentage uncertainty in M is 3 × 1 % = 3 %	✓ 1	Using 1 % (not multiplying by the power 3) scores 0 for this mark.
ΔM = 0.03 × 1.90 × 10²⁷ ≈ 0.06 × 10²⁷ kg, so M = (1.90 ± 0.06) × 10²⁷ kg	✓ 1	Allow ECF from (b)(i). The absolute uncertainty must be given to 1 s.f., with the value to the same decimal place.
Part (c)(i)
T = √(3.12 × 10⁻¹⁶ × (1.81 × 10⁸)³) = 4.30 × 10⁴ s = 0.498 days	✓ 1	Allow ECF from (a) or (b)(i). Accept 0.49–0.51 days.

Answers: (a)(i) T²/r³ ≈ 3.12 × 10⁻¹⁶ s² m⁻³ for all four moons · (b)(i) 1.90 × 10²⁷ kg · (b)(ii) ± 0.06 × 10²⁷ kg · (c)(i) 0.50 days (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — Kepler's three laws of orbital motion; Newton's universal law of gravitation F = Gm₁m₂/r² applied to circular orbits; Tools 3 — evaluating a hypothesis with several data points; propagating uncertainties through a power Command term: Determine

34D-1B-05

Gravitational field strength·D.1 Gravitational fields

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

An astronaut measures the gravitational field strength g at the surface of the Moon. A steel ball is held by an electromagnet at a height s above a trapdoor switch. A timer starts when the current in the electromagnet is switched off and stops when the ball opens the trapdoor. Because the magnetic field takes a short time to decay, the ball is released slightly after the timer starts, by the same delay each time.

The fall time t is measured for six heights. The graph shows t against √s. The radius of the Moon is 1.74 × 10⁶ m.

s / m	t / s	√s / m^½
0.20	0.528	0.447
0.40	0.735	0.632
0.60	0.893	0.775
0.80	1.027	0.894
1.00	1.143	1.000
1.20	1.251	1.095

Graph drawn to scale.

(a)

The field strength.

(i)

Draw the line of best fit and determine its gradient, including its unit.

(2)

(ii)

Hence determine g at the surface of the Moon.

(1)

(b)

The delay and the Moon.

(i)

State the intercept of your line on the t axis and explain why the delay does not affect the value of g found in (a)(ii).

(2)

(ii)

Determine the mass of the Moon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Straight line of best fit and gradient from a large triangle, e.g. (0.40, 0.476) to (1.10, 1.256)	✓ 1
Gradient = 1.11 s m^−½	✓ 1	Accept 1.09–1.14. Unit required (accept s m⁻⁰·⁵).
Part (a)(ii)
s = ½gt² gives t = √(2/g) √s + delay, so gradient = √(2/g) and g = 2/1.11² = 1.61 m s⁻²	✓ 1	Allow ECF from (a)(i). Accept 1.54–1.68 m s⁻².
Part (b)(i)
Intercept ≈ 0.03 s (the release delay)	✓ 1	Accept 0.02–0.04 s. Allow ECF from the candidate's line.
The same delay is added to every time (a systematic error), so the line is shifted up without changing its gradient, and g depends only on the gradient	✓ 1
Part (b)(ii)
g = GM/R², so M = gR²/G = 1.61 × (1.74 × 10⁶)²/(6.67 × 10⁻¹¹)	✓ 1	Allow ECF from (a)(ii).
M = 7.3 × 10²² kg	✓ 1	Accept 7.0–7.6 × 10²² kg.

Answers: (a)(i) 1.11 s m^−½ · (a)(ii) 1.61 m s⁻² · (b)(i) ≈ 0.03 s · (b)(ii) 7.3 × 10²² kg (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — gravitational field strength g = F/m = GM/r² at the surface of a spherical body; A.1 — the equations of motion for uniformly accelerated motion; Tools 3 — linearising a relationship, gradient with its unit, interpreting an intercept as a systematic error Command term: Determine

35D-1B-06

Escape speed and orbital speed·D.1 Gravitational fields

Paper 1BHard7 marks

Data-based question6 steps to full marksDeduce

After its engine is shut down, a spacecraft moves directly away from the Earth. Radar tracking gives its speed v at several distances r from the centre of the Earth. The only force acting on the spacecraft is the gravitational force of the Earth.

The graph shows v² against 1/r.

r / 10⁷ m	v / km s⁻¹	(1/r) / 10⁻⁸ m⁻¹	v² / 10⁶ m² s⁻²
1.00	8.48	10.00	71.9
1.50	6.73	6.67	45.3
2.00	5.64	5.00	31.8
3.00	4.31	3.33	18.6
4.00	3.45	2.50	11.9
6.00	2.31	1.67	5.3

Graph drawn to scale. The dashed line marks v² = 0.

(a)

The gradient.

(i)

Draw the line of best fit and determine its gradient, including its unit.

(2)

(ii)

Use the conservation of energy to show that the gradient is equal to 2GM, where M is the mass of the Earth. Hence determine M.

(2)

(b)

Where the spacecraft goes.

(i)

Extrapolate your line to determine the value of v² when 1/r = 0. Deduce whether the spacecraft escapes from the Earth.

(2)

(ii)

Determine the greatest distance from the centre of the Earth that the spacecraft reaches.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Straight line of best fit and gradient from a large triangle, e.g. (71.9 − 5.3) × 10⁶/((10.0 − 1.67) × 10⁻⁸)	✓ 1
Gradient = 8.0 × 10¹⁴ m³ s⁻²	✓ 1	Accept 7.8–8.2 × 10¹⁴. Unit required (accept N m² kg⁻¹ or J m kg⁻¹).
Part (a)(ii)
½mv² − GMm/r = constant, so v² = 2GM × (1/r) + constant: the gradient is 2GM	✓ 1
M = 8.0 × 10¹⁴/(2 × 6.67 × 10⁻¹¹) = 6.0 × 10²⁴ kg	✓ 1	Allow ECF from (a)(i). Accept 5.8–6.2 × 10²⁴ kg.
Part (b)(i)
Intercept ≈ −8 × 10⁶ m² s⁻²	✓ 1	Accept −6 × 10⁶ to −10 × 10⁶. Allow ECF from the candidate's line.
v² cannot be negative, so the spacecraft never reaches 1/r = 0 (infinite distance): its total energy ½mv² − GMm/r is negative and it does not escape	✓ 1	The deduction must be consistent with the sign of the candidate's intercept.
Part (b)(ii)
v = 0 where the line cuts the 1/r axis: 1/r = 8.1 × 10⁶/8.0 × 10¹⁴ = 1.0 × 10⁻⁸ m⁻¹, so r = 1.0 × 10⁸ m	✓ 1	Allow ECF from (a)(i) and (b)(i). Accept 0.8–1.3 × 10⁸ m.

Answers: (a)(i) 8.0 × 10¹⁴ m³ s⁻² · (a)(ii) 6.0 × 10²⁴ kg · (b)(i) ≈ −8 × 10⁶ m² s⁻²; it does not escape · (b)(ii) 1.0 × 10⁸ m (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the gravitational potential energy E_p = −Gm₁m₂/r; the energy conditions for a body to escape the gravitational influence of a planet; A.3 — the conservation of energy; Tools 3 — linearising, gradient with its unit, extrapolation to an intercept Command term: Deduce

36D-1B-08

Determining the density of an asteroid·D.1 Gravitational fields

Paper 1BHard7 marks

Data-based question6 steps to full marksDeduce

Surveys measure the rotation periods of thousands of asteroids from the regular variation in their brightness. The table shows, for asteroids in five ranges of diameter, the shortest rotation period T_min observed in that range. Each value has an uncertainty of ±0.02 h.

Many asteroids are thought to be loose piles of rubble held together only by gravity. Such an asteroid is modelled as a uniform sphere of density ρ. A loose rock on its equator stays on the surface only if the gravitational field strength there is at least equal to the centripetal acceleration of the rock.

Diameter / km	0.3–1	1–3	3–10	10–30	30–100
T_min / h	2.22	2.20	2.26	2.31	2.45

(a)

The spin limit.

(i)

Show that a rubble-pile asteroid of density ρ cannot rotate with a period shorter than √(3π/(Gρ)).

(2)

(ii)

Use the data to deduce the minimum density of rubble-pile asteroids.

(2)

(b)

Uncertainty and a deduction.

(i)

Determine the absolute uncertainty in your answer to (a)(ii).

(2)

(ii)

Some asteroids smaller than 0.2 km rotate with periods of about 0.50 h. Deduce what this suggests about these asteroids.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
At the limit GM/R² = ω²R = (4π²/T²)R	✓ 1
With M = (4/3)πR³ρ: (4/3)πGρ = 4π²/T², so T = √(3π/(Gρ)); a shorter period would need a larger centripetal acceleration than g can provide	✓ 1
Part (a)(ii)
The fastest rotator sets the limit, so use the shortest period, 2.20 h = 7.92 × 10³ s	✓ 1	Using any other period or a mean scores [1 max] for the part if the rest is correct.
ρ = 3π/(GT²) = 3π/(6.67 × 10⁻¹¹ × (7.92 × 10³)²) = 2.25 × 10³ kg m⁻³	✓ 1	Allow ECF from (a)(i). Accept 2.2–2.3 × 10³ kg m⁻³.
Part (b)(i)
Percentage uncertainty in T = 0.02/2.20 = 0.9 %; ρ ∝ T⁻², so the percentage uncertainty in ρ is 2 × 0.9 % = 1.8 %	✓ 1	Using 0.9 % for ρ scores 0 for this mark.
Δρ = 0.018 × 2.25 × 10³ = 40 kg m⁻³, so ρ = (2.25 ± 0.04) × 10³ kg m⁻³	✓ 1	Allow ECF from (a)(ii). Uncertainty to 1 s.f., value to the same decimal place.
Part (b)(ii)
As rubble piles they would need ρ ≥ 2.25 × 10³ × (2.20/0.50)² ≈ 4 × 10⁴ kg m⁻³, far more than any rock, so they cannot be held together by gravity alone: they must be single solid bodies held together by the forces between their particles	✓ 1	Allow ECF from (a)(ii). A numerical comparison (density or period limit) is required for the mark.

Answers: (a)(ii) 2.25 × 10³ kg m⁻³ · (b)(i) ± 0.04 × 10³ kg m⁻³ (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — gravitational field strength g = GM/r² at the surface of a spherical body; A.2 — centripetal acceleration a = ω²r = 4π²r/T²; Tools 3 — choosing the relevant data, propagating uncertainty through a power, absolute uncertainty Command term: Deduce

37D-1B-11

Mapping a field using potential·D.1 Gravitational fields

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

An orbiter surveying a dwarf planet uses radio tracking to determine the gravitational potential V_g at several distances r from the centre of the dwarf planet, which may be treated as a point mass. The results are shown in the table and in the graph of V_g against 1/r.

r / 10⁵ m	6.0	7.5	9.0	11.0	14.0	18.0
V_g / 10⁴ J kg⁻¹	−10.00	−7.92	−6.52	−5.27	−4.06	−3.06
(1/r) / 10⁻⁶ m⁻¹	1.667	1.333	1.111	0.909	0.714	0.556

Graph drawn to scale. The dashed line marks V_g = 0.

(a)

The mass.

(i)

Draw the line of best fit and determine its gradient, including its unit.

(2)

(ii)

Hence determine the mass of the dwarf planet.

(1)

(b)

Checking the data.

(i)

Extrapolate your line to find its intercept on the V_g axis. Explain what this intercept shows about the values of V_g.

(2)

(ii)

Use the values of V_g at r = 9.0 × 10⁵ m and r = 11.0 × 10⁵ m to estimate the gravitational field strength at r = 10.0 × 10⁵ m. Compare your estimate with the value given by your answer to (a)(ii).

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Straight line of best fit; gradient from a large triangle, e.g. (−10.0 − (−3.06)) × 10⁴/((1.67 − 0.556) × 10⁻⁶)	✓ 1
Gradient = −6.2 × 10¹⁰ J m kg⁻¹	✓ 1	Accept −6.0 to −6.4 × 10¹⁰. Unit required (accept N m² kg⁻¹ or m³ s⁻²).
Part (a)(ii)
V_g = −GM/r, so M = −gradient/G = 6.24 × 10¹⁰/6.67 × 10⁻¹¹ = 9.4 × 10²⁰ kg	✓ 1	Allow ECF from (a)(i). Accept 9.0–9.6 × 10²⁰ kg.
Part (b)(i)
Intercept ≈ +0.4 × 10⁴ J kg⁻¹	✓ 1	Accept +0.3 to +0.5 × 10⁴. Allow ECF from the candidate's line.
V_g should be zero at infinity (1/r = 0), so every value has been shifted by the same constant amount: a systematic (offset) error; it changes the intercept but not the gradient, so the mass found in (a)(ii) is unaffected	✓ 1	Both the systematic nature and the unaffected mass are needed.
Part (b)(ii)
g = ΔV_g/Δr = (−5.27 − (−6.52)) × 10⁴/(2.0 × 10⁵) = 0.063 N kg⁻¹ (towards the centre); the offset cancels in the difference	✓ 1	Accept 0.062–0.063.
GM/r² = 6.24 × 10¹⁰/(1.00 × 10⁶)² = 0.062 N kg⁻¹, which agrees with the estimate to within about 1 %	✓ 1	Allow ECF from (a)(ii). A comparison must be stated.

Answers: (a)(i) −6.2 × 10¹⁰ J m kg⁻¹ · (a)(ii) 9.4 × 10²⁰ kg · (b)(i) ≈ +0.4 × 10⁴ J kg⁻¹ · (b)(ii) 0.063 N kg⁻¹; 0.062 N kg⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the gravitational potential V_g = −GM/r, defined as zero at infinity; g = −ΔV_g/Δr; an ability to map fields using potential; Tools 3 — gradient with its unit, extrapolation to an intercept, systematic error Command term: Determine

38D-1B-13

Newton's law of gravitation·D.1 Gravitational fields

Paper 1BHard7 marks

Data-based question6 steps to full marksDetermine

In a torsion-balance experiment the gravitational force F between a large lead sphere of mass 1.50 kg and a small lead sphere of mass 20.0 g is determined for several centre-to-centre separations r. The absolute uncertainty in each value of F is ±0.10 × 10⁻¹⁰ N; the uncertainty in r is negligible.

At every separation the small sphere is also attracted in the opposite direction by a second large sphere on the other side of the balance. This opposing force is 4.0 % of the force due to the near sphere, and the tabulated values of F are the measured (resultant) forces. The graph shows five of the six points.

r / cm	4.5	5.0	5.5	6.0	7.0	8.5
F / 10⁻¹⁰ N	9.47	7.70	6.37	5.36	3.90	2.64
(1/r²) / m⁻²	494	400	331	278	204

Graph drawn to scale. Error bars show ±0.10 × 10⁻¹⁰ N.

(a)

The value of G.

(i)

Complete the table.

(1)

(ii)

Plot the missing point, draw the line of best fit and determine a value for G.

(2)

(b)

Uncertainty and a systematic error.

(i)

Draw the lines of maximum and minimum gradient consistent with the error bars, and hence state G with its absolute uncertainty.

(2)

(ii)

Compare your result with the accepted value of G, and explain the difference with reference to the second large sphere.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1/0.085² = 138 m⁻²	✓ 1	Accept 138.4.
Part (a)(ii)
Point plotted with its error bar; gradient of the line of best fit = 1.93 × 10⁻¹² N m²	✓ 1	Accept 1.88–1.98 × 10⁻¹² N m². Allow ECF from (a)(i).
G = gradient/(Mm) = 1.93 × 10⁻¹²/(1.50 × 0.0200) = 6.4 × 10⁻¹¹ N m² kg⁻²	✓ 1	Allow ECF. Accept 6.3–6.6 × 10⁻¹¹.
Part (b)(i)
Maximum gradient ≈ 1.98 × 10⁻¹² and minimum gradient ≈ 1.87 × 10⁻¹² N m²	✓ 1	Accept each within ±0.03 × 10⁻¹². Both lines must pass through every error bar.
G_max ≈ 6.6 × 10⁻¹¹ and G_min ≈ 6.2 × 10⁻¹¹, so G = (6.4 ± 0.2) × 10⁻¹¹ N m² kg⁻²	✓ 1	Uncertainty = half the range, to 1 s.f. Allow ECF from (a)(ii).
Part (b)(ii)
The accepted value, 6.67 × 10⁻¹¹ N m² kg⁻², lies outside the range found in (b)(i) (about 6.2–6.6 × 10⁻¹¹), so the difference is not explained by the random uncertainties	✓ 1	The comparison must use the candidate's own range (ECF).
Every measured F is 4 % smaller than the force due to the near sphere, so the gradient and G are 4 % too small (a systematic error); corrected G = 6.42/0.96 = 6.7 × 10⁻¹¹ N m² kg⁻², in agreement with the accepted value	✓ 1	Accept the argument without the corrected value if the 4 % reduction of the gradient is clearly stated.

Answers: (a)(i) 138 m⁻² · (a)(ii) 6.4 × 10⁻¹¹ N m² kg⁻² · (b)(i) (6.4 ± 0.2) × 10⁻¹¹ N m² kg⁻² (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — Newton's universal law of gravitation F = Gm₁m₂/r² for bodies treated as point masses; Tools 3 — uncertainty bars, lines of maximum and minimum gradient, uncertainty in a gradient; Inquiry 3 — comparing with the accepted value; systematic error Command term: Determine

39D-1B-14

Uniform field near the surface·D.1 Gravitational fields

Paper 1BMedium7 marks

Data-based question6 steps to full marksDeduce

A gravity model of Mars, built from the tracking of orbiting spacecraft, gives the gravitational potential difference ΔV_g between the surface of Mars and a point at height h above it. Some values are shown in the table, together with values of h/ΔV_g.

It can be shown that h/ΔV_g = R²/(GM) + (R/(GM))h, where M and R are the mass and radius of Mars. The graph shows five of the six points.

h / km	100	250	500	750	1000	1500
ΔV_g / MJ kg⁻¹	0.362	0.868	1.62	2.29	2.88	3.87
(h/ΔV_g) / 10⁻² s² m⁻¹	27.62	28.80	30.86	32.75	34.72

Graph drawn to scale (the vertical axis starts at 26).

(a)

The graph.

(i)

Complete the table.

(1)

(ii)

Plot the missing point, draw the line of best fit and determine its gradient and its intercept on the vertical axis, with units.

(2)

(b)

Mars and the uniform-field model.

(i)

Hence determine the gravitational field strength g₀ at the surface of Mars and the radius R of Mars.

(2)

(ii)

A payload of mass 1200 kg is lifted from the surface to a height of 500 km. Determine the percentage by which the uniform-field estimate mg₀h exceeds the actual increase in its gravitational potential energy.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1500 × 10³/3.87 × 10⁶ = 0.3876 s² m⁻¹ = 38.76 × 10⁻² s² m⁻¹	✓ 1	Accept 38.8.
Part (a)(ii)
Gradient = (38.7 − 26.8) × 10⁻²/1.5 × 10⁶ m = 7.9 × 10⁻⁸ s² m⁻²	✓ 1	Accept 7.7–8.2 × 10⁻⁸ s² m⁻² (or per-km equivalent with consistent units). Allow ECF from (a)(i).
Intercept (line extended to h = 0) = 26.8 × 10⁻² = 0.268 s² m⁻¹	✓ 1	Accept 0.265–0.271 s² m⁻¹.
Part (b)(i)
Intercept = R²/(GM) = 1/g₀, so g₀ = 1/0.268 = 3.73 m s⁻²	✓ 1	Accept 3.69–3.77 m s⁻². Allow ECF from (a)(ii).
Intercept/gradient = R: R = 0.268/7.93 × 10⁻⁸ = 3.4 × 10⁶ m	✓ 1	Accept 3.2–3.5 × 10⁶ m. Allow ECF.
Part (b)(ii)
Actual: mΔV_g = 1200 × 1.62 × 10⁶ = 1.94 × 10⁹ J; uniform field: 1200 × 3.73 × 5.0 × 10⁵ = 2.24 × 10⁹ J	✓ 1	Allow ECF from (b)(i) for g₀.
Overestimate = (2.24 − 1.94)/1.94 × 100 ≈ 15 %	✓ 1	Accept 14–16 %.

Answers: (a)(i) 38.76 · (a)(ii) 7.9 × 10⁻⁸ s² m⁻²; 0.268 s² m⁻¹ · (b)(i) 3.73 m s⁻²; 3.4 × 10⁶ m · (b)(ii) ≈ 15 % (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — g = −ΔV_g/Δr; V_g = −GM/r; the work done in moving a mass in a gravitational field W = mΔV_g; the (assumed) uniform field close to the surface of massive bodies; Tools 3 — linearising a relationship, gradient and intercept with units Command term: Deduce

40D-1B-24

Orbital decay of a small satellite·D.1 Gravitational fields

Paper 1BMedium6 marks

Data-based question5 steps to full marksDetermine

A small satellite of mass 4.0 kg moves in a low, almost circular orbit around the Earth. Radar tracking gives the height h of the satellite above the surface of the Earth every 20 days. Each value of h has an uncertainty of ±0.5 km; the uncertainty in t is negligible. The table also gives the orbital speed v calculated from each height.

At every instant the orbit may be treated as circular. Mass of the Earth = 5.97 × 10²⁴ kg; radius of the Earth = 6.37 × 10⁶ m.

t / days	h / km	v / km s⁻¹
0	400.0	7.669
20	392.0	7.674
40	383.0	7.679
60	372.5	7.685
80	360.0	7.692
100	345.0

Height h of the satellite above the surface against time t (drawn to scale; the vertical axis starts at 340 km). The error bars of ±0.5 km are smaller than the points.

(a)

(i)

Calculate v at t = 100 days.

(1)

(ii)

Determine the work done by the drag force on the satellite between t = 0 and t = 100 days.

(2)

(b)

(i)

Determine the mean rate of decrease of h, with its absolute uncertainty, for the first 40 days and for the last 40 days. Hence deduce whether the rate of decrease of h changes.

(2)

(ii)

Suggest why the rate at which the height decreases changes in the way that you deduced in (b)(i).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴/(6.37 × 10⁶ + 3.45 × 10⁵)) = 7.701 km s⁻¹	✓ 1	Accept 7.700–7.702 km s⁻¹. Using r = 345 km instead of R_E + h: [0].
Part (a)(ii)
Total energy in a circular orbit E = −GMm/(2r); the work done by the drag force equals the change in total energy: W = (GMm/2)(1/r₁ − 1/r₂) with r₁ = 6.770 × 10⁶ m and r₂ = 6.715 × 10⁶ m	✓ 1	ALT: W = −ΔE_k = −½m(v₂² − v₁²) with speeds from the table and (a)(i). Allow ECF from (a)(i).
W = −9.6 × 10⁵ J (negative: the drag removes energy, while the kinetic energy rises by the same amount)	✓ 1	Accept −9.5 × 10⁵ to −9.9 × 10⁵ J (the ALT with 3-decimal speeds gives about −9.8 × 10⁵ J); a magnitude with the sign explained in words is accepted. Using the change in E_p only (1.9 × 10⁶ J): [1 max].
Part (b)(i)
First 40 days: (400.0 − 383.0)/40 = 0.425 km per day; last 40 days: (372.5 − 345.0)/40 = 0.688 km per day	✓ 1	Accept 0.42–0.43 and 0.69 km per day (0.688). Rates read from tangents to the graph are also accepted if consistent.
Uncertainty in each difference of heights = ±1.0 km, so each rate is ±0.03 km per day (±0.025); the ranges (0.40–0.45 and 0.66–0.71 km per day) do not overlap, so the rate of decrease of h increases	✓ 1	Allow ECF from the rates. The conclusion must be supported by the uncertainties.
Part (b)(ii)
As the satellite descends it moves into denser atmosphere (and it moves faster), so the drag force and the rate of loss of energy are larger, and the orbit decays more quickly	✓ 1	Reference to the greater density of the atmosphere at lower heights is needed.

Answers: (a)(i) 7.701 km s⁻¹ · (a)(ii) −9.6 × 10⁵ J · (b)(i) (0.43 ± 0.03) and (0.69 ± 0.03) km per day; the rate increases (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the qualitative effect of a small viscous drag force due to the atmosphere on the height and speed of an orbiting body; the orbital speed of a body orbiting a large mass as given by v_orbital = √(GM/r); (guidance) changes in energy when a satellite changes orbit; for calculations involving orbital motion the orbits will be assumed to be circular; Tools 3 — absolute uncertainty of a difference, comparing rates within their uncertainties Command term: Determine

41D-1B-25

The radius and mass of a dwarf planet·D.1 Gravitational fields

Paper 1BMedium7 marks

Data-based question6 steps to full marksDetermine

A space probe is placed in a series of circular orbits around the dwarf planet Ceres. For each orbit, the height h of the probe above the surface of Ceres and its orbital period T are measured. Ceres may be treated as a uniform sphere of mass M and radius R.

The table shows the data and the values of T^2/3. The graph shows T^2/3 against h, with the h axis extended to negative values.

h / km	T / 10³ s	T^2/3 / s^2/3
200	13.75	574
400	20.32	745
700	31.82	1004
1000	44.60	1258
1400	64.24	1604
1900	91.53	2031

T^2/3 against the height h of the orbit above the surface of Ceres (drawn to scale).

(a)

(i)

Show that T^2/3 = (4π²/GM)^1/3(R + h).

(2)

(b)

(i)

Draw the line of best fit and determine its gradient, including its unit.

(2)

(ii)

Hence determine M.

(1)

(c)

(i)

Use the graph to determine R.

(1)

(ii)

Determine the mean density of Ceres.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The gravitational force provides the centripetal force: GMm/r² = m(4π²/T²)r with r = R + h	✓ 1
T² = 4π²(R + h)³/GM; taking the cube root of both sides gives T^2/3 = (4π²/GM)^1/3(R + h)	✓ 1	Using h instead of R + h: [0] for this mark.
Part (b)(i)
Straight line of best fit through the points and gradient from a large triangle, e.g. (0, 403) to (2000, 2117)	✓ 1
Gradient = 0.857 s^2/3 km⁻¹ = 8.57 × 10⁻⁴ s^2/3 m⁻¹	✓ 1	Accept 0.84–0.87 s^2/3 km⁻¹. Unit required.
Part (b)(ii)
Gradient = (4π²/GM)^1/3, so M = 4π²/(G × gradient³) = 4π²/(6.67 × 10⁻¹¹ × (8.57 × 10⁻⁴)³) = 9.4 × 10²⁰ kg	✓ 1	Allow ECF from (b)(i), with the gradient in s^2/3 m⁻¹. Accept 9.0 × 10²⁰–1.0 × 10²¹ kg.
Part (c)(i)
The line meets the h axis where R + h = 0, i.e. at h = −R, so R = 4.7 × 10⁵ m	✓ 1	Allow ECF from the candidate's line. Accept 4.4–5.0 × 10⁵ m.
Part (c)(ii)
ρ = M/((4/3)πR³) = 9.4 × 10²⁰/((4/3)π × (4.7 × 10⁵)³) = 2.2 × 10³ kg m⁻³	✓ 1	Allow ECF from (b)(ii) and (c)(i). Accept 1.9–2.5 × 10³ kg m⁻³.

Answers: (b)(i) 8.57 × 10⁻⁴ s^2/3 m⁻¹ · (b)(ii) 9.4 × 10²⁰ kg · (c)(i) 4.7 × 10⁵ m · (c)(ii) 2.2 × 10³ kg m⁻³ (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — Kepler's three laws of orbital motion; Newton's universal law of gravitation as given by F = Gm₁m₂/r² for bodies treated as point masses; Tools 3 — linearising a relationship, gradient with its unit, extrapolating a line to an intercept; Tools 2 — density from mass and volume Command term: Determine

42D-1B-26

The Moon moving away·D.1 Gravitational fields

Paper 1BHard7 marks

Data-based question5 steps to full marksDetermine

Laser pulses fired from observatories on the Earth are reflected by mirrors placed on the Moon, and the round-trip time of each pulse gives the Earth–Moon distance d. The table gives, for six years, the yearly mean value of Δd, the amount by which d exceeds a fixed reference distance of 3.844 × 10⁸ m. The time t is measured from the start of 1970. Each value of Δd has an uncertainty of ±4 cm.

Model the Moon as moving in a circular orbit around a fixed Earth. Mass of the Earth = 5.97 × 10²⁴ kg; mass of the Moon = 7.35 × 10²² kg; radius of the orbit = 3.84 × 10⁸ m; 1 year = 3.16 × 10⁷ s.

t / years	2	10	20	30	40	50
Δd / cm	19	50	88	127	165	202

Δd against t, with error bars of ±4 cm (drawn to scale).

(a)

(i)

Draw the line of best fit and determine the rate at which the Earth–Moon distance increases, in cm per year.

(1)

(ii)

Draw the lines of maximum and minimum gradient consistent with the error bars, and hence state the rate with its absolute uncertainty.

(2)

(b)

(i)

The radius r of the circular orbit of a body of mass m around a mass M increases by a small amount Δr. Show that the total energy of the orbiting body increases by ΔE ≈ (GMm/2r²)Δr.

(2)

(ii)

Determine the mean rate at which the total energy of the Moon increases.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Gradient of the line of best fit = 3.8 cm per year	✓ 1	Accept 3.7–3.9 cm per year.
Part (a)(ii)
Maximum gradient ≈ 3.98 and minimum gradient ≈ 3.65 cm per year, both lines passing through every error bar	✓ 1	Accept each within ±0.05 cm per year.
Uncertainty = half the difference ≈ ±0.2 cm per year: rate = (3.8 ± 0.2) cm per year	✓ 1	Allow ECF from (a)(i). Uncertainty to 1 s.f., value to the same decimal place.
Part (b)(i)
E = −GMm/(2r), so ΔE = (GMm/2)(1/r − 1/(r + Δr)) = GMmΔr/(2r(r + Δr))	✓ 1	The total energy in a circular orbit must be used; starting from E_p = −GMm/r only: [0] for this mark.
Since Δr ≪ r, r(r + Δr) ≈ r², giving ΔE ≈ (GMm/2r²)Δr	✓ 1	The approximation must be stated.
Part (b)(ii)
ΔE per year = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.35 × 10²² × 0.038/(2 × (3.84 × 10⁸)²) = 3.8 × 10¹⁸ J	✓ 1	Allow ECF from (a)(i). Subtracting two values of −GMm/(2r) on a calculator is not reliable: Δr/r ≈ 10⁻¹⁰.
Rate = 3.8 × 10¹⁸/3.16 × 10⁷ = 1.2 × 10¹¹ W	✓ 1	Accept 1.1–1.3 × 10¹¹ W. Omitting the factor ½ (2 × 10¹¹ W): [1 max].

Answers: (a)(i) 3.8 cm per year · (a)(ii) (3.8 ± 0.2) cm per year · (b)(ii) 1.2 × 10¹¹ W (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the gravitational potential energy for a two-body system as given by E_p = −Gm₁m₂/r; (guidance) changes in energy when a satellite changes orbit; A.3 — power as the rate of transfer of energy; Tools 3 — lines of maximum and minimum gradient, uncertainty of a gradient Command term: Determine

43D-2-01

Orbital energy·D.1 Gravitational fields

Paper 2Hard13 marks

Short answer & extended response9 steps to full marksDetermine

A communications satellite of mass 1200 kg is released into a circular orbit of radius 7.00 × 10⁶ m around the Earth. An ion thruster then exerts a small constant force of 0.20 N on the satellite, always in the direction of its velocity, until the radius of the orbit has increased to 1.20 × 10⁷ m. The satellite spirals outwards very slowly, so at every instant its orbit may be treated as circular. Ignore the change in the mass of the satellite.

The graph shows how the orbital speed v of the satellite varies with the radius r of its orbit. Mass of the Earth = 5.97 × 10²⁴ kg.

Graph drawn to scale.

(a)

The orbit.

(i)

Show that the orbital speed of the satellite at the start of the transfer is about 7.5 km s⁻¹.

(1)

(ii)

Use the graph to determine the orbital period of the satellite when r = 1.00 × 10⁷ m.

(2)

(b)

Energy.

(i)

Determine the increase in the total energy of the satellite during the transfer.

(2)

(ii)

Explain why the speed of the satellite decreases during the transfer, although the force exerted by the thruster always acts in the direction of its velocity.

(3)

(c)

Duration of the transfer.

(i)

Determine the distance travelled by the satellite along its spiral path during the transfer.

(2)

(ii)

Hence, using the graph, estimate the time taken for the transfer.

(2)

(iii)

State one assumption made in your answer to (c)(i).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴/7.00 × 10⁶) = 7.54 × 10³ m s⁻¹	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
v read from the graph at r = 1.00 × 10⁷ m: 6.3 km s⁻¹	✓ 1	Accept 6.25–6.35 km s⁻¹. ALT: T = 2π√(r³/GM) without the graph [1 max].
T = 2πr/v = 2π × 1.00 × 10⁷/6.31 × 10³ = 9.96 × 10³ s (about 2.8 h)	✓ 1	Accept 9.8–10.1 × 10³ s, consistent with the reading.
Part (b)(i)
Total energy in a circular orbit E = −GMm/(2r), so ΔE = (GMm/2)(1/r₁ − 1/r₂)	✓ 1	ALT: ΔE = −ΔE_k = ½m(v₁² − v₂²) with speeds from the graph.
ΔE = ½ × 3.98 × 10¹⁴ × 1200 × (1/7.00 × 10⁶ − 1/1.20 × 10⁷) = 1.42 × 10¹⁰ J	✓ 1	Accept 1.4–1.5 × 10¹⁰ J. Award [1 max] for 2.8 × 10¹⁰ J (change in potential energy only).
Part (b)(ii)
The thruster does positive work, so the total energy −GMm/(2r) increases (becomes less negative) and the radius of the orbit increases	✓ 1
In a circular orbit E_k = GMm/(2r) = −E, so as the total energy increases the kinetic energy decreases: the speed falls	✓ 1	Allow ECF from (b)(i). Accept v = √(GM/r), which decreases as r increases, as part of this mark only if linked to energy.
On the outward spiral the velocity has a small component away from the Earth, so the gravitational force has a component opposite to the velocity. The work done against gravity (the increase in potential energy, twice the work done by the thruster) is larger than the work done by the thruster	✓ 1	Accept: the potential energy increases by 2 × 1.42 × 10¹⁰ J while the thruster supplies only 1.42 × 10¹⁰ J. Do not accept "gravity is weaker further out" on its own.
Part (c)(i)
The work done by the thruster equals the increase in total energy: Fs = ΔE	✓ 1	Allow ECF from (b)(i).
s = 1.42 × 10¹⁰/0.20 = 7.1 × 10¹⁰ m	✓ 1	Accept 7.0–7.5 × 10¹⁰ m.
Part (c)(ii)
Average speed from the graph ≈ ½(7.54 + 5.76) km s⁻¹ = 6.65 km s⁻¹	✓ 1	Accept any reasonable mean speed in the range 6.4–6.8 km s⁻¹ read from the graph.
t = 7.1 × 10¹⁰/6.65 × 10³ = 1.1 × 10⁷ s (about 4 months)	✓ 1	Allow ECF from (c)(i). Accept 1.0–1.15 × 10⁷ s.
Part (c)(iii)
All the work done by the thruster becomes orbital (kinetic + gravitational potential) energy of the satellite / the force acts exactly along the velocity / the orbit is circular at every instant / the mass of the satellite is constant	✓ 1	Any one.

Answers: (a)(i) 7.54 × 10³ m s⁻¹ · (a)(ii) 9.96 × 10³ s · (b)(i) 1.42 × 10¹⁰ J · (c)(i) 7.1 × 10¹⁰ m · (c)(ii) 1.1 × 10⁷ s (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the orbital speed of a body orbiting a large mass as given by v_orbital = √(GM/r); (guidance) changes in energy when a satellite changes orbit; A.3 — that work done by a force is equivalent to a transfer of energy; A.2 — circular motion, v = 2πr/T Command term: Determine

44D-2-06

Gravitational potential·D.1 Gravitational fields

Paper 2Hard12 marks

Short answer & extended response8 steps to full marksDetermine

The graph shows how the gravitational potential V_g varies with distance r from the centre of a planet of radius 3.0 × 10⁶ m, from its surface outwards. The planet has no atmosphere and its rotation can be ignored.

Graph drawn to scale.

(a)

The field of the planet.

(i)

Use the graph to determine the mass of the planet.

(2)

(ii)

By drawing a tangent to the graph, determine the gravitational field strength at r = 6.0 × 10⁶ m.

(2)

(iii)

Hence deduce the gravitational field strength at the surface of the planet.

(1)

(b)

A probe is launched vertically upwards from the surface with a speed of 3.5 km s⁻¹.

(i)

Determine the maximum distance from the centre of the planet reached by the probe.

(3)

(ii)

A student calculates the maximum height from mgΔh = ½mv² using the surface value of g. Explain why the student's answer is too small.

(2)

(c)

(i)

Use the graph to determine the minimum speed with which the probe would have to be launched from the surface to escape from the planet.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
V_g at the surface = −10.7 MJ kg⁻¹	✓ 1	Accept −10.5 to −10.9 MJ kg⁻¹.
M = −V_gR/G = 10.7 × 10⁶ × 3.0 × 10⁶/6.67 × 10⁻¹¹ = 4.8 × 10²³ kg	✓ 1	Accept 4.7–4.9 × 10²³ kg.
Part (a)(ii)
Tangent drawn at r = 6.0 × 10⁶ m and its gradient found from a large triangle	✓ 1
g = \|ΔV_g/Δr\| ≈ 0.89 N kg⁻¹	✓ 1	Accept 0.75–1.05 N kg⁻¹.
Part (a)(iii)
g ∝ 1/r², and the surface is at half this distance, so g_s = 4 × 0.89 = 3.6 N kg⁻¹	✓ 1	Allow ECF from (a)(ii).
Part (b)(i)
Kinetic energy per unit mass at launch = ½v² = ½ × 3500² = 6.1 MJ kg⁻¹	✓ 1
V_g at the highest point = −10.7 + 6.1 = −4.6 MJ kg⁻¹	✓ 1	Allow ECF from (a)(i).
From the graph (or r = GM/\|V_g\|), r ≈ 7.0 × 10⁶ m	✓ 1	Accept 6.8–7.3 × 10⁶ m.
Part (b)(ii)
Student: Δh = v²/(2g_s) = 3500²/(2 × 3.6) ≈ 1.7 × 10⁶ m, compared with 4.0 × 10⁶ m from (b)(i)	✓ 1	Allow ECF from (a)(iii) and (b)(i).
ΔE_p = mgΔh assumes a uniform field, but g falls with height (to about a fifth of the surface value at the top), so less energy is needed for each metre gained higher up and the probe rises further	✓ 1
Part (c)(i)
To escape, the total energy must be at least zero (V_g = 0 at infinity): ½mv² ≥ m\|V_g\| at the surface	✓ 1	Accept v = √(2GM/R) with GM/R = \|V_g\| from the graph.
v = √(2 × 10.7 × 10⁶) = 4.6 × 10³ m s⁻¹	✓ 1	Allow ECF from (a)(i). Accept 4.6–4.7 × 10³ m s⁻¹.

Answers: (a)(i) 4.8 × 10²³ kg · (a)(ii) 0.89 N kg⁻¹ · (a)(iii) 3.6 N kg⁻¹ · (b)(i) 7.0 × 10⁶ m · (c)(i) 4.6 × 10³ m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the gravitational potential V_g = −GM/r; the gravitational field strength as the potential gradient g = −ΔV_g/Δr; the escape speed v_esc = √(2GM/r); energetics of a launch from the surface of a non-rotating planet; A.3 — E_p = mgΔh only close to the surface Command term: Determine

45D-2-11

Kepler's third law·D.1 Gravitational fields

Paper 2Medium12 marks

Short answer & extended response8 steps to full marksDetermine

A star S orbits the massive black hole at the centre of our galaxy. Model the orbit as a circle of radius 1.0 × 10³ AU with a period of 16 years. The black hole is 8.0 kpc from Earth.

1 year = 3.16 × 10⁷ s; mass of the Sun = 2.0 × 10³⁰ kg.

(a)

The mass of the black hole.

(i)

Show that the radius of the orbit is about 1.5 × 10¹⁴ m.

(1)

(ii)

Show that, for a body in a circular orbit of radius r around a mass M, T² = (4π²/GM)r³.

(2)

(iii)

Determine the mass of the black hole, in solar masses.

(2)

(b)

(i)

Determine the angle, in arcseconds, subtended at Earth by the radius of the orbit.

(2)

(c)

Escape from the black hole.

(i)

Estimate the radius of the region around the black hole from which the escape speed would be greater than the speed of light.

(2)

(ii)

State one assumption made in your estimate.

(1)

(d)

(i)

Calculate the orbital speed of S and comment on whether Newtonian mechanics is adequate for describing its orbit.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
1.0 × 10³ × 1.50 × 10¹¹ m = 1.50 × 10¹⁴ m	✓ 1	The AU conversion from the data booklet must be seen.
Part (a)(ii)
GMm/r² = mv²/r with v = 2πr/T	✓ 1	Accept mω²r with ω = 2π/T.
GM/r² = 4π²r/T², so T² = (4π²/GM)r³	✓ 1
Part (a)(iii)
M = 4π²r³/(GT²) = 4π² × (1.50 × 10¹⁴)³/(6.67 × 10⁻¹¹ × (16 × 3.16 × 10⁷)²) = 7.8 × 10³⁶ kg	✓ 1	Allow ECF from (a)(i).
= 3.9 × 10⁶ solar masses	✓ 1	The answer must be expressed in solar masses for this mark. Accept 3.8–4.0 × 10⁶.
Part (b)(i)
1 AU at a distance of 1 pc subtends 1 arcsecond (the definition of the parsec), so θ = (1.0 × 10³ AU)/(8.0 × 10³ pc)	✓ 1	ALT: θ = 1.5 × 10¹⁴/(8.0 × 10³ × 3.09 × 10¹⁶) = 6.1 × 10⁻⁷ rad, converted to arcseconds.
θ = 0.125 arcsecond	✓ 1	Accept 0.12–0.13 arcsecond.
Part (c)(i)
v_esc = √(2GM/r) = c gives r = 2GM/c²	✓ 1
r = 2 × 6.67 × 10⁻¹¹ × 7.8 × 10³⁶/(3.00 × 10⁸)² = 1.2 × 10¹⁰ m	✓ 1	Allow ECF from (a)(iii).
Part (c)(ii)
Newtonian gravitation (and the escape-speed equation) still applies so close to the black hole / the black hole behaves as a point mass	✓ 1
Part (d)(i)
v = 2πr/T = 2π × 1.50 × 10¹⁴/(16 × 3.16 × 10⁷) = 1.9 × 10⁶ m s⁻¹	✓ 1	Allow ECF from (a)(i).
v/c ≈ 0.006, so γ ≈ 1.00002: relativistic effects such as time dilation are negligible and the Newtonian model is adequate	✓ 1	A reasoned comparison with c is needed.

Answers: (a)(i) 1.50 × 10¹⁴ m · (a)(iii) 3.9 × 10⁶ solar masses · (b)(i) 0.13″ · (c)(i) 1.2 × 10¹⁰ m · (d)(i) 1.9 × 10⁶ m s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — Kepler's three laws of orbital motion; Newton's universal law of gravitation for bodies treated as point masses; D.1 (HL) — the escape speed v_esc = √(2GM/r); E.5 — the parsec, the light year and the astronomical unit; A.5 — time dilation (γ) Command term: Determine

46D-2-14

A mission to Mars·D.1 Gravitational fields

Paper 2Medium12 marks

Short answer & extended response8 steps to full marksDetermine

The graph shows how the gravitational field strength g due to Mars varies with distance r from its centre, from the surface (r = 3.39 × 10⁶ m) outwards. Ignore the rotation and the thin atmosphere of Mars.

A sample-return vehicle of mass 300 kg is to be launched from rest on the surface into a circular orbit of radius 4.0 × 10⁶ m.

Graph drawn to scale. The dashed line marks the surface of Mars.

(a)

Mars and the orbit.

(i)

Use the graph to determine the mass of Mars.

(2)

(ii)

Show, using the graph, that the orbital speed of the vehicle will be about 3.3 km s⁻¹.

(2)

(b)

The launch.

(i)

Determine the minimum energy needed to place the vehicle in this orbit.

(3)

(ii)

The propellant releases 1.3 × 10⁷ J of energy per kilogram, and 12 % of this energy becomes mechanical energy of the vehicle. Estimate the minimum mass of propellant needed.

(2)

(iii)

Suggest why the actual mass of propellant needed is greater than your answer to (b)(ii).

(1)

(c)

(i)

Show that the minimum energy per unit mass needed to reach a circular orbit just above the surface of any planet is ½g₀R, where g₀ is the field strength at the surface and R is the radius of the planet.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
g at the surface = 3.7 N kg⁻¹	✓ 1	Accept 3.6–3.8 N kg⁻¹.
M = gR²/G = 3.7 × (3.39 × 10⁶)²/6.67 × 10⁻¹¹ = 6.4 × 10²³ kg	✓ 1	Accept 6.2–6.5 × 10²³ kg.
Part (a)(ii)
g = 2.7 N kg⁻¹ at r = 4.0 × 10⁶ m, and mg = mv²/r	✓ 1	The value of g must be read from the graph (accept 2.6–2.75 N kg⁻¹).
v = √(gr) = √(2.68 × 4.0 × 10⁶) = 3.27 × 10³ m s⁻¹	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
Total energy at rest on the surface = −GMm/R	✓ 1
Total energy in orbit = −GMm/(2r) (= −½mv²)	✓ 1	Allow ECF from (a)(i) and (a)(ii).
ΔE = GMm(1/R − 1/(2r)) = 2.2 × 10⁹ J	✓ 1	Accept 2.1–2.3 × 10⁹ J.
Part (b)(ii)
Energy from the propellant = 2.2 × 10⁹/0.12 = 1.8 × 10¹⁰ J	✓ 1	Allow ECF from (b)(i).
Mass = 1.8 × 10¹⁰/1.3 × 10⁷ = 1.4 × 10³ kg	✓ 1
Part (b)(iii)
The propellant itself must be lifted and accelerated (the vehicle is much heavier than 300 kg during the burn) / energy is carried away by the exhaust gases	✓ 1
Part (c)(i)
Energy per unit mass changes from −GM/R to −GM/(2R), an increase of GM/(2R)	✓ 1
g₀ = GM/R², so GM = g₀R² and the energy per unit mass is ½g₀R	✓ 1

Answers: (a)(i) 6.4 × 10²³ kg · (a)(ii) 3.27 × 10³ m s⁻¹ · (b)(i) 2.2 × 10⁹ J · (b)(ii) 1.4 × 10³ kg (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — g = GM/r²; D.1 (HL) — the orbital speed v = √(GM/r); energetics of a satellite going into orbit around a non-rotating planet starting from rest on its surface; A.3 — energy density of fuel sources and efficiency Command term: Determine

47D-2-17

Field lines and Kepler's laws·D.1 Gravitational fields

Paper 2Easy13 marks

Short answer & extended response7 steps to full marksCalculate

A planet of mass 4.8 × 10²⁴ kg orbits a red dwarf star of mass 8.0 × 10²⁹ kg in a circular orbit of radius 3.0 × 10⁹ m. The radius of the star is 2.8 × 10⁸ m and its luminosity is 1.0 × 10²⁵ W.

(a)

The gravitational field of the star.

(i)

Outline why the star and the planet can be treated as point masses.

(1)

(ii)

Calculate the gravitational field strength due to the star at the position of the planet.

(1)

(iii)

Hence show that the orbital speed of the planet is about 1.3 × 10⁵ m s⁻¹.

(2)

(b)

Kepler's third law.

(i)

Determine the orbital period of the planet, in days.

(2)

(ii)

A second planet of the same star has an orbital period of 13 days. Determine its orbital radius.

(2)

(c)

(i)

Sketch the gravitational field lines around the star, treated as an isolated sphere.

(2)

(d)

Radiation from the star.

(i)

Calculate the intensity of the star's radiation at the orbit of the first planet.

(1)

(ii)

Estimate the mean surface temperature of this planet. Treat the planet as a black body with no atmosphere.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Both are (approximately) spherically symmetric, so their external fields are those of point masses at their centres; and/or their separation is much larger than the radius of either	✓ 1
Part (a)(ii)
g = GM/r² = 6.67 × 10⁻¹¹ × 8.0 × 10²⁹/(3.0 × 10⁹)² = 5.9 N kg⁻¹	✓ 1
Part (a)(iii)
g is the centripetal acceleration: g = v²/r	✓ 1	Allow ECF from (a)(ii).
v = √(gr) = √(5.93 × 3.0 × 10⁹) = 1.33 × 10⁵ m s⁻¹	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
T = 2πr/v = 2π × 3.0 × 10⁹/1.33 × 10⁵ = 1.41 × 10⁵ s	✓ 1	Allow ECF from (a)(iii).
= 1.64 days	✓ 1
Part (b)(ii)
T² ∝ r³, so r₂ = r₁(T₂/T₁)^2/3	✓ 1	Allow ECF from (b)(i).
r₂ = 3.0 × 10⁹ × (13/1.64)^2/3 = 1.2 × 10¹⁰ m	✓ 1	Accept 1.19–1.22 × 10¹⁰ m.
Part (c)(i)
Radial straight lines, evenly distributed around the star and meeting its surface at right angles	✓ 1
Arrows on the lines pointing towards the centre of the star	✓ 1
Part (d)(i)
b = L/(4πd²) = 1.0 × 10²⁵/(4π × (3.0 × 10⁹)²) = 8.8 × 10⁴ W m⁻²	✓ 1
Part (d)(ii)
Power absorbed bπR_p² = power emitted 4πR_p²σT⁴, so T⁴ = b/(4σ)	✓ 1	Allow ECF from (d)(i).
T = (8.8 × 10⁴/(4 × 5.67 × 10⁻⁸))^1/4 ≈ 790 K	✓ 1	Accept 780–800 K.

Answers: (a)(ii) 5.9 N kg⁻¹ · (a)(iii) 1.33 × 10⁵ m s⁻¹ · (b)(i) 1.64 days · (b)(ii) 1.2 × 10¹⁰ m · (d)(i) 8.8 × 10⁴ W m⁻² · (d)(ii) ≈ 790 K (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — Kepler's three laws of orbital motion; Newton's universal law of gravitation for bodies treated as point masses; conditions under which extended bodies can be treated as point masses; g = GM/r²; gravitational field lines; B.1 — b = L/(4πd²); B.2 — energy balance of a planet Command term: Calculate

48D-2-18

Resultant field of two bodies·D.1 Gravitational fields

Paper 2Medium12 marks

Short answer & extended response9 steps to full marksDetermine

A binary asteroid consists of a primary of mass 5.2 × 10¹¹ kg and radius 390 m, and a small moon (the secondary) of mass 4.9 × 10⁹ kg and radius 80 m. The distance between their centres is 1.20 km. Treat both bodies as uniform spheres and ignore all other masses.

(a)

Field strengths.

(i)

Calculate the gravitational field strength at the surface of the primary due to the primary alone.

(1)

(ii)

A spacecraft is held stationary at point P, on the line joining the centres and 600 m from the centre of each body. Determine the magnitude and direction of the resultant gravitational field strength at P.

(2)

(iii)

Point S is on the surface of the secondary, on the line joining the centres, on the side facing the primary. Determine the resultant gravitational field strength at S.

(3)

(b)

Energy.

(i)

Calculate the gravitational potential energy of the two-body system.

(1)

(ii)

The spacecraft, of mass 550 kg, moves from P to the point on the surface of the primary that is nearest to the secondary. Determine the change in its gravitational potential energy.

(3)

(c)

(i)

The spacecraft hovers at P using an ion thruster that ejects xenon at a speed of 3.0 × 10⁴ m s⁻¹ relative to the spacecraft. Determine the mass of xenon ejected per hour.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
g = GM/R² = 6.67 × 10⁻¹¹ × 5.2 × 10¹¹/390² = 2.3 × 10⁻⁴ N kg⁻¹	✓ 1
Part (a)(ii)
Primary: 9.63 × 10⁻⁵ N kg⁻¹ towards the primary; secondary: 9.1 × 10⁻⁷ N kg⁻¹ towards the secondary	✓ 1
Resultant = 9.5 × 10⁻⁵ N kg⁻¹, towards the primary	✓ 1	Direction required.
Part (a)(iii)
S is 1200 − 80 = 1120 m from the centre of the primary	✓ 1	This step is required.
Secondary: GM₂/80² = 5.1 × 10⁻⁵ N kg⁻¹ towards its centre; primary: GM₁/1120² = 2.8 × 10⁻⁵ N kg⁻¹ towards the primary	✓ 1
Resultant = 2.3 × 10⁻⁵ N kg⁻¹ towards the centre of the secondary	✓ 1	Direction required.
Part (b)(i)
E_p = −GM₁M₂/r = −6.67 × 10⁻¹¹ × 5.2 × 10¹¹ × 4.9 × 10⁹/1200 = −1.4 × 10⁸ J	✓ 1	Negative sign required.
Part (b)(ii)
At P: V_g = −G(5.2 × 10¹¹ + 4.9 × 10⁹)/600 = −5.84 × 10⁻² J kg⁻¹	✓ 1
At the surface point (390 m from the primary, 810 m from the secondary): V_g = −8.93 × 10⁻² J kg⁻¹	✓ 1
ΔE_p = mΔV_g = 550 × (−0.0893 − −0.0584) = −17 J (a decrease)	✓ 1	Accept −16 to −18 J.
Part (c)(i)
Thrust needed = weight at P = 550 × 9.54 × 10⁻⁵ = 5.2 × 10⁻² N	✓ 1	Allow ECF from (a)(ii).
F = (Δm/Δt)v gives Δm/Δt = 5.2 × 10⁻²/3.0 × 10⁴ = 1.7 × 10⁻⁶ kg s⁻¹ = 6.3 × 10⁻³ kg per hour	✓ 1

Answers: (a)(i) 2.3 × 10⁻⁴ N kg⁻¹ · (a)(ii) 9.5 × 10⁻⁵ N kg⁻¹ towards the primary · (a)(iii) 2.3 × 10⁻⁵ N kg⁻¹ · (b)(i) −1.4 × 10⁸ J · (b)(ii) −17 J · (c)(i) 6.3 × 10⁻³ kg (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — g = GM/r²; the resultant gravitational field strength at points along a line joining two bodies; D.1 (HL) — E_p = −Gm₁m₂/r; V_g = −GM/r; W = mΔV_g; A.2 — F = Δp/Δt for a changing mass Command term: Determine

49D-2-19

Orbital energy·D.1 Gravitational fields

Paper 2Medium11 marks

Short answer & extended response7 steps to full marksDeduce

A space tug lowers a defunct satellite of mass 1500 kg from a circular orbit at an altitude of 800 km to a circular orbit at an altitude of 300 km, from which atmospheric drag will bring it down. The graph shows how the total energy E of the satellite in a circular orbit varies with the radius r of the orbit.

M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m.

Graph drawn to scale.

(a)

Lowering the orbit.

(i)

Show that the orbital speed of the satellite at an altitude of 800 km is about 7.5 km s⁻¹.

(1)

(ii)

Use the graph to determine the minimum energy that the tug must remove from the satellite.

(2)

(iii)

Confirm the value of E at 800 km by calculation, using your answer to (a)(i).

(1)

(b)

Kinetic and potential energy.

(i)

Show that, when the radius of a circular orbit changes, the change in kinetic energy of the satellite is equal and opposite to the change in its total energy.

(2)

(ii)

Hence deduce the change in kinetic energy and the change in gravitational potential energy of the satellite when it is lowered.

(2)

(c)

The tug.

(i)

After releasing the satellite, the tug, of mass 900 kg, is in the 300 km orbit. Calculate the minimum energy it needs to escape from the Earth from this orbit.

(1)

(ii)

Explain why this is less than half of the energy the tug would need to escape from rest on the surface of the Earth.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = √(GM/r) = √(6.67 × 10⁻¹¹ × 5.97 × 10²⁴/7.17 × 10⁶) = 7.45 × 10³ m s⁻¹	✓ 1	r = 7.17 × 10⁶ m must be used; full substitution or 3 s.f. required.
Part (a)(ii)
E = −4.17 × 10¹⁰ J at r = 7.17 × 10⁶ m and −4.48 × 10¹⁰ J at r = 6.67 × 10⁶ m	✓ 1	Accept readings within ±0.02 × 10¹⁰ J.
Energy removed = 3.1 × 10⁹ J	✓ 1	Accept 2.7–3.5 × 10⁹ J.
Part (a)(iii)
E = −½mv² = −½ × 1500 × (7.45 × 10³)² = −4.17 × 10¹⁰ J, in agreement with the graph	✓ 1	Allow ECF from (a)(i). E = −Ek for a circular orbit.
Part (b)(i)
E_k = GMm/(2r) and E = E_k + E_p = −GMm/(2r)	✓ 1
So E_k = −E at every radius, and ΔE_k = −ΔE	✓ 1
Part (b)(ii)
ΔE_k = +3.1 × 10⁹ J (the satellite speeds up)	✓ 1	Allow ECF from (a)(ii).
ΔE_p = ΔE − ΔE_k = −6.2 × 10⁹ J (twice the energy removed)	✓ 1
Part (c)(i)
Energy = GMm/(2r) = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 900/(2 × 6.67 × 10⁶) = 2.7 × 10¹⁰ J	✓ 1
Part (c)(ii)
From rest on the surface it needs GMm/R_E (= 5.6 × 10¹⁰ J) to raise its total energy from −GMm/R_E to zero	✓ 1
In orbit its total energy is already −GMm/(2r), and r > R_E, so it needs less than half of GMm/R_E	✓ 1	Accept: the orbiting tug already has kinetic energy and is higher in the potential well.

Answers: (a)(i) 7.45 × 10³ m s⁻¹ · (a)(ii) 3.1 × 10⁹ J · (a)(iii) −4.17 × 10¹⁰ J · (b)(ii) ΔE_k = +3.1 × 10⁹ J; ΔE_p = −6.2 × 10⁹ J · (c)(i) 2.7 × 10¹⁰ J (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — v = √(GM/r); changes in energy when a satellite changes orbit; energy conditions for an orbiting satellite to escape the gravitational influence of a planet; E_p = −GMm/r Command term: Deduce

50D-2-20

Mapping a field using potential·D.1 Gravitational fields

Paper 2Hard13 marks

Short answer & extended response7 steps to full marksEstimate

The diagram shows equipotential lines in the plane containing the centres of a planet and its moon, which are 4.0 × 10⁸ m apart. The values of the gravitational potential are in MJ kg⁻¹, and the scale gives the distance x from the centre of the planet along the line joining the centres. Point A lies off this line; point B lies on it.

Equipotential lines (values in MJ kg⁻¹) in the plane containing the centres of a planet and its moon. The scale gives the distance x from the centre of the planet in units of 10⁸ m; small ticks on the line joining the centres are 0.2 × 10⁸ m apart. The planet and moon are not drawn to scale.

(a)

Field lines.

(i)

Sketch, on the diagram, the gravitational field line that passes through point A. Indicate its direction.

(2)

(b)

Field strength from potential.

(i)

Estimate the magnitude of the gravitational field strength at B.

(3)

(ii)

Explain why the method used in (b)(i) gives only an estimate.

(1)

(c)

The point of zero field.

(i)

Identify the point N on the line joining the centres at which the resultant gravitational field strength is zero, and explain how the equipotential pattern shows this.

(2)

(ii)

Determine the ratio (mass of planet)/(mass of moon).

(2)

(d)

Energy.

(i)

A probe of mass 600 kg is at rest on the −4.0 MJ kg⁻¹ equipotential near the planet. Calculate the minimum energy needed to move it to N.

(1)

(ii)

Determine the minimum speed with which the probe must leave the −4.0 MJ kg⁻¹ equipotential, with its engines off, in order to reach the moon. Ignore the orbital motion of the moon.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
Line crosses every equipotential it meets at right angles, curving to end on the planet	✓ 1	A line ending on the moon scores 0: from A the potential decreases most steeply towards the planet.
Arrow pointing towards the planet (towards lower potential)	✓ 1
Part (b)(i)
Along the line joining the centres, the −4.0 line crosses at x ≈ 1.04 × 10⁸ m and the −3.0 line at x ≈ 1.41 × 10⁸ m	✓ 1	Accept readings ±0.05 × 10⁸ m.
ΔV_g = 1.0 × 10⁶ J kg⁻¹ over Δx ≈ 0.37 × 10⁸ m	✓ 1
g ≈ ΔV_g/Δx ≈ 2.7 × 10⁻² N kg⁻¹	✓ 1	Accept 2.2–3.2 × 10⁻² N kg⁻¹.
Part (b)(ii)
The field strength (potential gradient) changes across the interval, so ΔV_g/Δx gives only the average over the interval, not the value at B (and reading the positions from the diagram is imprecise)	✓ 1
Part (c)(i)
N is at x ≈ 3.0 × 10⁸ m, where the −1.78 MJ kg⁻¹ equipotential crosses itself (the two loops meet)	✓ 1	Accept 2.9–3.1 × 10⁸ m.
Along the line the potential rises to a maximum at N, so the potential gradient, and therefore g = −ΔV_g/Δr, is zero there	✓ 1
Part (c)(ii)
GM_P/x_N² = GM_M/(4.0 × 10⁸ − x_N)²	✓ 1
M_P/M_M = (3.0/1.0)² = 9	✓ 1	Accept 8–10 consistent with the reading in (c)(i); allow ECF.
Part (d)(i)
W = mΔV_g = 600 × (−1.78 − (−4.0)) × 10⁶ = 1.3 × 10⁹ J	✓ 1
Part (d)(ii)
The probe need only just reach N (beyond N the resultant field points towards the moon), so ½mv² = mΔV_g with ΔV_g = 2.22 × 10⁶ J kg⁻¹	✓ 1	Or ½mv² = 1.33 × 10⁹ J from (d)(i).
v = √(2 × 2.22 × 10⁶) = 2.1 × 10³ m s⁻¹	✓ 1	Allow ECF from (d)(i).

Answers: (b)(i) ≈ 2.7 × 10⁻² N kg⁻¹ · (c)(i) x ≈ 3.0 × 10⁸ m · (c)(ii) 9 · (d)(i) 1.3 × 10⁹ J · (d)(ii) 2.1 km s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — equipotential surfaces for gravitational fields; the relationship between equipotential surfaces and gravitational field lines; g = −ΔV_g/Δr; W = mΔV_g; an ability to map fields using potential; D.1 — resultant field strength along a line joining two bodies; gravitational field lines Command term: Estimate

51D-2-21

Escape speed and orbital speed·D.1 Gravitational fields

Paper 2Medium11 marks

Short answer & extended response7 steps to full marksDetermine

A proposed lunar mass driver is an electromagnetic track, 20 km long, laid horizontally on the Moon's surface. It accelerates payloads uniformly from rest. Treat the Moon as a non-rotating uniform sphere of mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m, with no atmosphere. M_E = 5.97 × 10²⁴ kg, R_E = 6.37 × 10⁶ m.

(a)

A very low orbit.

(i)

Calculate the gravitational field strength at the surface of the Moon.

(1)

(ii)

Show that a payload leaving the track horizontally at about 1.7 km s⁻¹ could orbit the Moon just above its surface.

(2)

(iii)

Determine the period of this orbit.

(2)

(b)

Escape.

(i)

Calculate the escape speed from the surface of the Moon.

(1)

(ii)

Determine the acceleration the track must provide for a payload to leave it at the escape speed.

(2)

(c)

Moon versus Earth.

(i)

Determine the ratio (minimum energy per kilogram needed to escape from the Earth's surface)/(minimum energy per kilogram needed to escape from the Moon's surface).

(2)

(ii)

Suggest one reason why the real energy advantage of launching deep-space payloads from the Moon differs from this ratio.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
g = GM/R² = 6.67 × 10⁻¹¹ × 7.35 × 10²²/(1.74 × 10⁶)² = 1.62 N kg⁻¹	✓ 1
Part (a)(ii)
For a circular orbit of radius R the gravitational force provides the centripetal force: GMm/R² = mv²/R (or mg = mv²/R)	✓ 1
v = √(GM/R) = √(1.62 × 1.74 × 10⁶) = 1.68 × 10³ m s⁻¹	✓ 1	Answer to at least 3 s.f.
Part (a)(iii)
T = 2πR/v = 2π × 1.74 × 10⁶/1.68 × 10³	✓ 1	Or T = 2π√(R³/GM).
T = 6.5 × 10³ s (≈ 109 min)	✓ 1	Accept 6.5–6.6 × 10³ s. Allow ECF from (a)(ii).
Part (b)(i)
v_esc = √(2GM/R) = √2 × 1.68 × 10³ = 2.37 × 10³ m s⁻¹	✓ 1	Allow ECF from (a)(ii).
Part (b)(ii)
v² = 2as with s = 2.0 × 10⁴ m	✓ 1
a = (2.37 × 10³)²/(2 × 2.0 × 10⁴) = 1.4 × 10² m s⁻²	✓ 1	About 14g; accept 140–141 m s⁻². Allow ECF from (b)(i).
Part (c)(i)
Energy per kg to escape = GM/R (= ½v_esc²): Earth 6.25 × 10⁷ J kg⁻¹, Moon 2.82 × 10⁶ J kg⁻¹	✓ 1
Ratio = 22	✓ 1	Accept 21–23.
Part (c)(ii)
The payload leaving the Moon is still in the Earth's gravitational field (the Moon orbits the Earth), so extra energy is needed to leave the Earth–Moon system; or: launching from the Earth also requires energy to overcome air resistance	✓ 1	Any one sensible reason.

Answers: (a)(i) 1.62 N kg⁻¹ · (a)(ii) 1.68 km s⁻¹ · (a)(iii) 6.5 × 10³ s · (b)(i) 2.37 km s⁻¹ · (b)(ii) 1.4 × 10² m s⁻² · (c)(i) 22 (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the escape speed v_esc = √(2GM/r); the orbital speed v_orbital = √(GM/r); energetics of a satellite going into orbit around a non-rotating planet starting from rest on its surface; D.1 — g = GM/r²; A.2 — circular motion (T = 2πr/v); A.1 — equations of motion for uniform acceleration Command term: Determine

52D-2-32

A lander on Europa·D.1 Gravitational fields

Paper 2Easy12 marks

Short answer & extended response8 steps to full marksDetermine

A robotic lander is to land on Europa, a moon of Jupiter that has no atmosphere. Europa may be treated as a uniform sphere of mass 4.80 × 10²² kg and radius 1.56 × 10⁶ m.

(a)

Europa's gravitational field.

(i)

Show that the gravitational field strength at the surface of Europa is about 1.3 N kg⁻¹.

(1)

(b)

The uniform-field model.

(i)

The descent begins at a height of 15 km above the surface. Determine the percentage by which the gravitational field strength at this height is less than the value at the surface.

(3)

(ii)

State and explain whether the gravitational field may be treated as uniform during the descent.

(1)

(c)

Landing.

(i)

The engine is switched off when the lander is momentarily at rest 2.0 m above the surface. Determine the speed with which the lander reaches the surface.

(2)

(ii)

The landing legs are tested on Earth by dropping the lander from rest onto the ground. Determine the height from which it must be dropped on Earth so that it reaches the ground with the speed found in (c)(i). Ignore air resistance.

(2)

(iii)

Europa moves around Jupiter in a circular orbit of radius 6.71 × 10⁸ m with a period of 3.55 days. Determine the gravitational field strength due to Jupiter at the position of Europa.

(2)

(iv)

Your answer to (c)(iii) is about one-fifth of your answer to (a)(i). Suggest why Jupiter's field does not affect the landing speed found in (c)(i).

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
g = GM/R² = 6.67 × 10⁻¹¹ × 4.80 × 10²²/(1.56 × 10⁶)² = 1.316 N kg⁻¹	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (b)(i)
r = 1.56 × 10⁶ + 1.5 × 10⁴ = 1.575 × 10⁶ m	✓ 1	The radius of Europa must be added to the height.
g = 6.67 × 10⁻¹¹ × 4.80 × 10²²/(1.575 × 10⁶)² = 1.291 N kg⁻¹	✓ 1
Percentage decrease = (1.316 − 1.291)/1.316 × 100 = 1.9 %	✓ 1	Allow ECF from (a)(i). Accept 1.8–2.0 %. ALT: 1 − (R/(R + h))² = 0.019.
Part (b)(ii)
Yes: the field strength changes by only about 2 % over the descent (the height is much less than the radius of Europa), so g may be taken as constant	✓ 1	Allow ECF from (b)(i). The decision must be consistent with the candidate's percentage.
Part (c)(i)
Uniform field: v² = 2gs (or mgh = ½mv²)	✓ 1
v = √(2 × 1.32 × 2.0) = 2.3 m s⁻¹	✓ 1	Allow ECF from (a)(i). Award [1 max] for 6.3 m s⁻¹, which uses g = 9.81 N kg⁻¹.
Part (c)(ii)
v² = 2gh with g = 9.81 N kg⁻¹ (equivalently g_Earthh_Earth = g_Europa × 2.0 m)	✓ 1	Allow ECF from (c)(i).
h = 2.3²/(2 × 9.81) = 0.27 m	✓ 1	Accept 0.26–0.28 m (1.316 × 2.0/9.81 = 0.268 m).
Part (c)(iii)
The field strength of Jupiter equals the centripetal acceleration of Europa: g_J = 4π²r/T², with T = 3.55 × 8.64 × 10⁴ = 3.07 × 10⁵ s	✓ 1	ALT: M_J = 4π²r³/(GT²) followed by g = GM_J/r².
g_J = 4π² × 6.71 × 10⁸/(3.07 × 10⁵)² = 0.28 N kg⁻¹	✓ 1	Accept 0.27–0.29 N kg⁻¹.
Part (c)(iv)
Europa and the lander are both in free fall towards Jupiter with the same acceleration, so Jupiter's field produces no acceleration of the lander relative to Europa's surface	✓ 1	Accept: Jupiter's field (almost) equally accelerates the lander and Europa. Do not accept "Jupiter is too far away".

Answers: (a)(i) 1.316 N kg⁻¹ · (b)(i) 1.9 % · (c)(i) 2.3 m s⁻¹ · (c)(ii) 0.27 m · (c)(iii) 0.28 N kg⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — gravitational field strength g = F/m = GM/r²; the (assumed) uniform field close to the surface of planetary bodies; centripetal acceleration a = 4π²r/T²; A.1 — the equations of motion for uniform acceleration; A.3 — the conservation of energy Command term: Determine

53D-2-35

Escape speed and atmospheres·D.1 Gravitational fields

Paper 2Medium13 marks

Short answer & extended response8 steps to full marksDeduce

Titan, the largest moon of Saturn, has a thick atmosphere composed mainly of nitrogen (N₂, molar mass 0.028 kg mol⁻¹). Titan may be treated as a uniform sphere of mass 1.35 × 10²³ kg and radius 2.57 × 10⁶ m. The temperature at its surface is 94 K.

(a)

Titan's gravitational field.

(i)

Show that the gravitational field strength at the surface of Titan is about 1.4 N kg⁻¹.

(1)

(ii)

Show that the escape speed from the surface of a body of radius R and surface gravitational field strength g₀ is √(2g₀R), and hence calculate the escape speed from Titan.

(2)

(b)

A planetary scientist's rule states that a body keeps a gas in its atmosphere for billions of years if its escape speed is more than six times the typical speed of the gas molecules. The typical speed is the speed of a molecule that has the average translational kinetic energy.

(i)

Show that the typical speed of the nitrogen molecules at the surface of Titan is about 290 m s⁻¹.

(2)

(ii)

Deduce whether, according to the rule, Titan keeps its nitrogen.

(2)

(iii)

Determine the maximum surface temperature at which Titan could keep its nitrogen according to the rule.

(3)

(c)

Comparison with the Moon.

(i)

The Moon has an escape speed of 2.4 km s⁻¹ and a daytime surface temperature of about 390 K. Use the rule to explain why the Moon cannot keep an atmosphere of nitrogen.

(2)

(ii)

The gravitational field strength at the surface of the Moon, 1.6 N kg⁻¹, is greater than at the surface of Titan. Suggest why Titan nevertheless keeps an atmosphere while the Moon does not.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
g₀ = GM/R² = 6.67 × 10⁻¹¹ × 1.35 × 10²³/(2.57 × 10⁶)² = 1.363 N kg⁻¹	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
½mv_esc² = GMm/R and GM = g₀R², so v_esc = √(2g₀R)	✓ 1
v_esc = √(2 × 1.363 × 2.57 × 10⁶) = 2.65 × 10³ m s⁻¹	✓ 1	Allow ECF from (a)(i). Accept 2.6–2.7 km s⁻¹.
Part (b)(i)
m = 0.028/6.02 × 10²³ = 4.65 × 10⁻²⁶ kg and E_k = (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 94 = 1.95 × 10⁻²¹ J	✓ 1
v = √(2E_k/m) = 289 m s⁻¹	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (b)(ii)
6 × 289 = 1.7 × 10³ m s⁻¹	✓ 1
This is less than the escape speed of 2.6 × 10³ m s⁻¹, so Titan keeps its nitrogen	✓ 1	Allow ECF from (a)(ii) and (b)(i).
Part (b)(iii)
Largest typical speed = v_esc/6 = 2647/6 = 441 m s⁻¹	✓ 1	Allow ECF from (a)(ii).
(3/2)k_BT = ½mv², so T = mv²/(3k_B)	✓ 1
T = 4.65 × 10⁻²⁶ × 441²/(3 × 1.38 × 10⁻²³) = 2.2 × 10² K	✓ 1	ALT: T ∝ v², so T = 94 × (441/289)² = 220 K. Accept 210–230 K.
Part (c)(i)
Typical speed = √(3k_BT/m) = 589 m s⁻¹	✓ 1	Allow ECF from (b)(i), e.g. 290 × √(390/94) = 590 m s⁻¹.
6 × 589 = 3.5 × 10³ m s⁻¹, which is greater than 2.4 km s⁻¹, so the nitrogen escapes	✓ 1
Part (c)(ii)
Titan is much colder, so its molecules move more slowly; also the escape speed depends on g₀R, and Titan's larger radius gives it the larger escape speed	✓ 1	Either point scores the mark.

Answers: (a)(i) 1.363 N kg⁻¹ · (a)(ii) 2.65 × 10³ m s⁻¹ · (b)(i) 289 m s⁻¹ · (b)(iii) 2.2 × 10² K (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the escape speed v_esc at any point in a gravitational field, v_esc = √(2GM/r); D.1 — g = GM/r²; B.3 — the average kinetic energy of the molecules of an ideal gas E_k = (3/2)k_BT Command term: Deduce

54D-2-37

A lander on a comet nucleus·D.1 Gravitational fields

Paper 2Easy13 marks

Short answer & extended response8 steps to full marksDetermine

A lander of mass 100 kg is sent to the nucleus of a comet. The nucleus has a mass of 1.0 × 10¹³ kg and an average radius of 2.0 km. In the calculations, model the nucleus as a uniform sphere with no atmosphere and ignore its rotation.

(a)

(i)

Calculate the gravitational field strength at the surface of the nucleus, according to the model.

(1)

(ii)

The real nucleus is irregular in shape, with two lobes. Outline why the model gives only an estimate of the field strength at its surface, but is reliable for the field strength at a distance of 20 km from its centre.

(2)

(b)

The lander is released from rest relative to the nucleus at a distance of 20 km from its centre, and falls freely to the surface.

(i)

Determine the speed with which the lander reaches the surface.

(2)

(ii)

The landing legs compress by 5.0 cm as they bring the lander to rest. Determine the average force exerted on the lander by the legs.

(2)

(c)

The lander then fires a harpoon of mass 0.50 kg vertically downwards into the surface, at a speed of 60 m s⁻¹ relative to the nucleus. The lander is not anchored, and it moves vertically upwards.

(i)

Show that the escape speed from the surface of the nucleus is about 0.8 m s⁻¹.

(1)

(ii)

Determine the speed with which the lander starts to move upwards.

(2)

(iii)

Deduce whether the lander escapes from the nucleus, and determine the maximum height above the surface that it reaches.

(3)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
g = GM/R² = 6.67 × 10⁻¹¹ × 1.0 × 10¹³/(2.0 × 10³)² = 1.7 × 10⁻⁴ N kg⁻¹	✓ 1
Part (a)(ii)
Close to an irregular (non-spherical) body, different parts of it are at very different distances and directions, so it does not act like a point mass at its centre	✓ 1	Reference to the shape or mass distribution not being spherically symmetric is needed.
At 20 km the distance is large (10 times) compared with the size of the nucleus, so the nucleus can be treated as a point mass	✓ 1	"It is far away" without comparison with the size: [0].
Part (b)(i)
Loss of potential energy = gain in kinetic energy: ½mv² = mΔV_g = GMm(1/R − 1/r)	✓ 1	ALT: ½v² = GM/R − GM/r per unit mass.
v = √(2 × 6.67 × 10⁻¹¹ × 1.0 × 10¹³ × (1/2000 − 1/20 000)) = 0.77 m s⁻¹	✓ 1	Accept 0.77–0.78 m s⁻¹. Using the surface value of g as a uniform field over 18 km (2.5 m s⁻¹): [1 max].
Part (b)(ii)
Kinetic energy = ½ × 100 × 0.775² = 30 J	✓ 1	Allow ECF from (b)(i).
Average force = 30/0.050 = 6.0 × 10² N	✓ 1	The weight of the lander (≈ 0.02 N) is negligible. Accept 590–610 N.
Part (c)(i)
v_esc = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 1.0 × 10¹³/2.0 × 10³) = 0.817 m s⁻¹	✓ 1	Full substitution or an answer to at least 2 s.f. is required.
Part (c)(ii)
The total momentum of the lander and the harpoon is zero before and after firing, so the upward momentum of the lander equals the downward momentum of the harpoon: m_Lv_L = m_hv_h	✓ 1	Any clear statement of momentum conservation.
v = 0.50 × 60/100 = 0.30 m s⁻¹	✓ 1	Accept 0.302 m s⁻¹ (using 99.5 kg).
Part (c)(iii)
0.30 m s⁻¹ is less than the escape speed (0.82 m s⁻¹), so the total energy is negative and the lander does not escape	✓ 1	Allow ECF from (c)(i) and (c)(ii).
½v² − GM/R = −GM/r_max: 0.045 − 0.3335 = −GM/r_max	✓ 1	Energy per unit mass; GM/R = 0.3335 J kg⁻¹.
r_max = 2312 m, so the height is 3.1 × 10² m	✓ 1	Accept 300–320 m. Using h = v²/2g with the surface value of g (270 m): [2 max].

Answers: (a)(i) 1.7 × 10⁻⁴ N kg⁻¹ · (b)(i) 0.77 m s⁻¹ · (b)(ii) 6.0 × 10² N · (c)(i) 0.817 m s⁻¹ · (c)(ii) 0.30 m s⁻¹ · (c)(iii) 3.1 × 10² m; it does not escape (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — gravitational field strength g at a point as the force per unit mass experienced by a small point mass at that point, as given by g = F/m = GM/r²; conditions under which extended bodies can be treated as point masses; D.1 (HL) — the escape speed v_esc at any point in a gravitational field as given by v_esc = √(2GM/r); the gravitational potential V_g at a point as the work done per unit mass in bringing a mass from infinity to that point, as given by V_g = −GM/r; A.2 — conservation of linear momentum; A.3 — work done as energy transferred Command term: Determine

55D-2-38

A super-Earth·D.1 Gravitational fields

Paper 2Easy9 marks

Short answer & extended response8 steps to full marksDetermine

Planet K orbits a nearby star. Observations give its mass and radius in terms of the mass M_E and the radius R_E of the Earth, listed in the table. Both planets may be treated as spheres in which the density depends only on the distance from the centre. Ignore the rotation and the atmosphere of each planet.

Gravitational field strength at the surface of the Earth = 9.81 N kg⁻¹; M_E = 5.97 × 10²⁴ kg; R_E = 6.37 × 10⁶ m.

	Mass	Radius
Earth	M_E	R_E
Planet K	6.0M_E	1.7R_E

(a)

(i)

Show that the gravitational field strength at the surface of K is about 20 N kg⁻¹.

(2)

(ii)

Calculate (mean density of K)/(mean density of the Earth).

(1)

(iii)

Outline why the field strength at the surface of K may be calculated as if all its mass were at its centre, even though its density is not uniform.

(1)

(b)

(i)

Calculate the escape speed from the surface of K.

(2)

(ii)

Hence determine the gravitational potential at the surface of K.

(1)

(c)

(i)

An astronaut can jump vertically to a height of 0.45 m on the Earth. Estimate the height that she would reach on K with the same take-off speed.

(1)

(ii)

Calculate the period of a simple pendulum of length 1.00 m on the surface of K.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
g = GM/R², so g_K = 9.81 × 6.0/1.7²	✓ 1	ALT: direct substitution of G, 6.0M_E and 1.7R_E.
g_K = 20.4 N kg⁻¹	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
ρ ∝ M/R³: 6.0/1.7³ = 1.22	✓ 1	Accept 1.2.
Part (a)(iii)
The mass is distributed with spherical symmetry (the density depends only on the distance from the centre), so outside the planet, and at its surface, it acts as a point mass at its centre	✓ 1	"Because it is a sphere" alone: [0].
Part (b)(i)
v_esc = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 6.0 × 5.97 × 10²⁴/(1.7 × 6.37 × 10⁶))	✓ 1	ALT: 11.2 × √(6.0/1.7) with 11.2 km s⁻¹ for the Earth.
v_esc = 2.1 × 10⁴ m s⁻¹	✓ 1	Award [1 max] for 3.9 × 10⁴ m s⁻¹ (ratio 6.0/1.7 not square-rooted).
Part (b)(ii)
V_g = −½v_esc² = −½ × (2.1 × 10⁴)² = −2.2 × 10⁸ J kg⁻¹	✓ 1	Allow ECF from (b)(i). ALT: V_g = −GM/R. Negative sign required.
Part (c)(i)
h = v²/2g ∝ 1/g: 0.45 × 9.81/20.4 = 0.22 m	✓ 1	Allow ECF from (a)(i).
Part (c)(ii)
T = 2π√(1.00/20.4) = 1.39 s	✓ 1	Allow ECF from (a)(i).

Answers: (a)(i) 20.4 N kg⁻¹ · (a)(ii) 1.22 · (b)(i) 2.1 × 10⁴ m s⁻¹ · (b)(ii) −2.2 × 10⁸ J kg⁻¹ · (c)(i) 0.22 m · (c)(ii) 1.39 s (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — gravitational field strength g at a point as the force per unit mass experienced by a small point mass at that point, as given by g = F/m = GM/r²; conditions under which extended bodies can be treated as point masses; D.1 (HL) — the escape speed v_esc at any point in a gravitational field as given by v_esc = √(2GM/r); the gravitational potential V_g at a point as the work done per unit mass in bringing a mass from infinity to that point, as given by V_g = −GM/r; A.1 — equations of motion for uniform acceleration; C.1 — the time period of a simple pendulum Command term: Determine

56D-2-39

Energy released when a moon forms·D.1 Gravitational fields

Paper 2Medium13 marks

Short answer & extended response8 steps to full marksExplain

An icy moon is thought to have formed when a very large number of small fragments, initially at rest and very far apart, were pulled together by their mutual gravitational attraction. The moon is now a uniform sphere of mass 6.0 × 10²⁰ kg and radius 5.0 × 10⁵ m.

The gravitational potential energy of a uniform sphere of mass M and radius R is E_p = −3GM²/(5R).

(a)

(i)

Explain, with reference to the work done in assembling the moon, why E_p is negative.

(2)

(ii)

Calculate E_p for the moon.

(1)

(iii)

Assume that all the energy released as the moon formed became internal energy of the moon. Estimate the resulting rise in its mean temperature. The specific heat capacity of the material of the moon is 1.5 × 10³ J kg⁻¹ K⁻¹.

(2)

(b)

The last fragment to arrive falls from rest, very far away, onto the surface of the completed moon.

(i)

Show that its speed at impact is about 400 m s⁻¹.

(2)

(ii)

Compare the energy released per kilogram by the last fragment with the average energy released per kilogram during the formation of the whole moon. Explain the difference.

(3)

(c)

(i)

Show that, for moons made of material of the same density ρ, the energy released per kilogram during formation is proportional to R².

(2)

(ii)

Another moon is made of the same material and has a radius 5 times larger. Estimate its rise in mean temperature during formation, and suggest one consequence for this moon.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
E_p of a system is the work done to assemble it from infinite separation of its parts, where the potential energy is defined as zero	✓ 1
The gravitational forces between the parts are attractive, so they pull the parts together: the work done by an external agent in assembling them is negative (energy is released), so E_p < 0	✓ 1	"Gravity is attractive" without reference to work or energy: [0] for this mark.
Part (a)(ii)
E_p = −3 × 6.67 × 10⁻¹¹ × (6.0 × 10²⁰)²/(5 × 5.0 × 10⁵) = −2.9 × 10²⁵ J	✓ 1	Negative sign required.
Part (a)(iii)
Energy released = \|E_p\| = 2.9 × 10²⁵ J = mcΔθ	✓ 1	Allow ECF from (a)(ii).
Δθ = 2.9 × 10²⁵/(6.0 × 10²⁰ × 1.5 × 10³) = 32 K	✓ 1	Accept 32 K (or °C).
Part (b)(i)
Kinetic energy per kilogram at impact = loss of potential energy per kilogram = GM/R	✓ 1	Accept: the impact speed equals the escape speed.
v = √(2 × 6.67 × 10⁻¹¹ × 6.0 × 10²⁰/5.0 × 10⁵) = 400.1 m s⁻¹	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (b)(ii)
Last fragment: GM/R = 8.0 × 10⁴ J kg⁻¹	✓ 1	Allow ECF from (b)(i): ½v².
Average: \|E_p\|/M = 3GM/(5R) = 4.8 × 10⁴ J kg⁻¹, which is smaller (60 % of the value for the last fragment)	✓ 1	Allow ECF from (a)(ii).
Earlier fragments fell onto a smaller moon of smaller mass: the potential at its surface, −GM/R, was less negative, so they lost less potential energy per kilogram	✓ 1	Accept: the potential well deepens as the moon grows.
Part (c)(i)
Energy per kilogram = \|E_p\|/M = 3GM/(5R) with M = (4/3)πR³ρ	✓ 1
= (4/5)πGρR², which is proportional to R² when ρ is fixed	✓ 1
Part (c)(ii)
Δθ = 5² × 32 = 8 × 10² K, which is enough to melt ice, so the larger moon could become partly liquid inside	✓ 1	Allow ECF from (a)(iii). Both the value and a consequence are needed.

Answers: (a)(ii) −2.9 × 10²⁵ J · (a)(iii) 32 K · (b)(i) 400.1 m s⁻¹ · (b)(ii) 8.0 × 10⁴ and 4.8 × 10⁴ J kg⁻¹ · (c)(ii) 8 × 10² K (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the gravitational potential energy E_p of a system as the work done to assemble the system from infinite separation of the components of the system; the gravitational potential V_g at a point as the work done per unit mass in bringing a mass from infinity to that point, as given by V_g = −GM/r; the escape speed v_esc at any point in a gravitational field as given by v_esc = √(2GM/r); B.1 — specific heat capacity, Q = mcΔθ Command term: Explain

57D-2-40

A shrinking binary star·D.1 Gravitational fields

Paper 2Hard14 marks

Short answer & extended response8 steps to full marksDeduce

Two white dwarf stars, each of mass M = 0.60 solar masses, move in circular orbits about the midpoint of the line joining their centres. The separation of their centres is d. Regular eclipses show that their orbital period is T = 1200 s.

The system loses energy by emitting gravitational waves, so the stars slowly spiral towards each other and the period decreases. The graph shows the change ΔT in the orbital period, measured from its value at t = 0, over 18 years.

Mass of the Sun = 1.99 × 10³⁰ kg; 1 year = 3.16 × 10⁷ s.

Change ΔT in the orbital period against time t, with error bars (drawn to scale).

(a)

(i)

Show that T = π√(2d³/GM).

(2)

(ii)

Determine d.

(2)

(b)

(i)

Show that the total energy of the system is E = −GM²/(2d).

(2)

(ii)

Calculate E.

(1)

(c)

(i)

Use the graph to determine the rate of change of the orbital period, in s s⁻¹.

(2)

(ii)

For a small change in the period, the fractional change in the separation is Δd/d = (2/3)ΔT/T. Deduce the rate at which the system loses energy.

(3)

(d)

(i)

Each white dwarf has a radius of 8.7 × 10⁶ m and a surface temperature of 1.2 × 10⁴ K. Compare the rate at which the system loses energy with the total luminosity of the two stars.

(2)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
The gravitational force between the stars provides the centripetal force on each star, whose orbit has radius d/2: GM²/d² = M(4π²/T²)(d/2)	✓ 1	Using d instead of d/2 for the radius of the orbit: [0] for this mark.
T² = 2π²d³/GM, so T = π√(2d³/GM)	✓ 1
Part (a)(ii)
d³ = GMT²/(2π²) with M = 0.60 × 1.99 × 10³⁰ = 1.19 × 10³⁰ kg	✓ 1	Allow ECF from (a)(i).
d = 1.8 × 10⁸ m	✓ 1	Accept 1.79–1.81 × 10⁸ m.
Part (b)(i)
For each star Mv²/(d/2) = GM²/d², so its kinetic energy is GM²/(4d); total kinetic energy = GM²/(2d)	✓ 1
E_p = −GM²/d, so E = GM²/(2d) − GM²/d = −GM²/(2d)	✓ 1	The potential energy of the pair is counted once.
Part (b)(ii)
E = −6.67 × 10⁻¹¹ × (1.19 × 10³⁰)²/(2 × 1.8 × 10⁸) = −2.6 × 10⁴¹ J	✓ 1	Allow ECF from (a)(ii). Negative sign required.
Part (c)(i)
Gradient of the best-fit line = −0.30 ms per year	✓ 1	Accept −0.27 to −0.32 ms per year (magnitude only also accepted).
= −0.30 × 10⁻³/3.16 × 10⁷ = −9.5 × 10⁻¹² s s⁻¹	✓ 1	Accept 8.5–10 × 10⁻¹² s s⁻¹ in magnitude.
Part (c)(ii)
Fractional rate of change of d = (2/3) × 9.5 × 10⁻¹²/1200 = 5.3 × 10⁻¹⁵ s⁻¹	✓ 1	Allow ECF from (c)(i).
\|E\| ∝ 1/d, so \|E\| increases by the same small fraction as d decreases	✓ 1	Equivalent: P = (2/3)\|E\|\|ΔT/Δt\|/T.
Rate of energy loss = 5.3 × 10⁻¹⁵ × 2.6 × 10⁴¹ = 1.4 × 10²⁷ W	✓ 1	Allow ECF from (b)(ii). Accept 1.2–1.5 × 10²⁷ W. Omitting the factor 2/3 (2.1 × 10²⁷ W): [2 max].
Part (d)(i)
L = σ4πR²T⁴ = 5.67 × 10⁻⁸ × 4π × (8.7 × 10⁶)² × (1.2 × 10⁴)⁴ = 1.1 × 10²⁴ W for each star; total 2.2 × 10²⁴ W	✓ 1	Using one star only: accept for this mark.
The rate of energy loss is about 600 times the total luminosity: most of the energy leaves as gravitational waves, not as light	✓ 1	Allow ECF from (c)(ii). Accept 500–700 times.

Answers: (a)(ii) 1.8 × 10⁸ m · (b)(ii) −2.6 × 10⁴¹ J · (c)(i) −9.5 × 10⁻¹² s s⁻¹ · (c)(ii) 1.4 × 10²⁷ W · (d)(i) about 600 times larger (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 — Newton's universal law of gravitation as given by F = Gm₁m₂/r² for bodies treated as point masses; Kepler's three laws of orbital motion; D.1 (HL) — the gravitational potential energy for a two-body system as given by E_p = −Gm₁m₂/r; (guidance) changes in energy when a satellite changes orbit; B.1 — the Stefan–Boltzmann law L = σAT⁴; E.5 — white dwarfs; the determination of stellar radii; A.3 — power Command term: Deduce

58D-2-41

Escaping from the Solar System·D.1 Gravitational fields

Paper 2Medium12 marks

Short answer & extended response8 steps to full marksDetermine

A space probe of mass 720 kg is to be sent out of the Solar System. The Earth moves around the Sun in a circular orbit of radius 1.50 × 10¹¹ m. Ignore the gravitational effects of the Moon and of the other planets, and ignore the atmosphere and the rotation of the Earth unless stated.

Mass of the Sun = 1.99 × 10³⁰ kg; mass of the Earth = 5.97 × 10²⁴ kg; radius of the Earth = 6.37 × 10⁶ m.

(a)

(i)

Show that the orbital speed of the Earth is about 30 km s⁻¹.

(1)

(ii)

Show that, at any distance from the Sun, the escape speed is √2 times the speed of a body in a circular orbit at that distance. Hence calculate the escape speed from the Sun's field at the orbit of the Earth.

(2)

(b)

(i)

The probe leaves the neighbourhood of the Earth moving in the same direction as the orbital velocity of the Earth. Determine the minimum speed of the probe relative to the Earth at that point for it to escape from the Solar System.

(2)

(ii)

The probe is launched from the surface of the Earth. Determine the minimum launch speed relative to the Earth. Ignore the effect of the Sun's field while the probe moves away from the Earth.

(3)

(c)

(i)

Calculate the minimum kinetic energy that the probe must be given at launch.

(1)

(ii)

Launch sites close to the equator are often used, with the probe launched towards the east. Calculate the speed of a point on the equator due to the rotation of the Earth, and explain why this choice reduces the energy that the rocket must supply.

(2)

(d)

(i)

The probe, launched with the minimum speed, crosses the orbit of Jupiter, whose radius is 5.2 times that of the orbit of the Earth. Determine the speed of the probe relative to the Sun at that point.

(1)

Show mark scheme

Marking point	Mark	Notes
Part (a)(i)
v = √(GM/r) = √(6.67 × 10⁻¹¹ × 1.99 × 10³⁰/1.50 × 10¹¹) = 29.7 km s⁻¹	✓ 1	Full substitution or an answer to at least 3 s.f. is required.
Part (a)(ii)
½mv_esc² = GMm/r gives v_esc = √(2GM/r); mv_orbital²/r = GMm/r² gives v_orbital = √(GM/r); the ratio is √2	✓ 1	Both expressions are needed.
v_esc = √2 × 29.7 = 42.1 km s⁻¹	✓ 1	Allow ECF from (a)(i).
Part (b)(i)
Velocities in the same direction add: speed relative to the Sun = speed relative to the Earth + 29.7 km s⁻¹	✓ 1
Minimum speed = 42.1 − 29.7 = 12.3 km s⁻¹	✓ 1	Allow ECF from (a).
Part (b)(ii)
The probe must still have 12.3 km s⁻¹ relative to the Earth when it has escaped from the Earth's field	✓ 1	Allow ECF from (b)(i).
½v_L² − GM_E/R_E = ½(12.3 × 10³)², so v_L² = v_esc_E² + 12.3², with v_esc_E = √(2GM_E/R_E) = 11.2 km s⁻¹	✓ 1	Energy per unit mass.
v_L = √(11.2² + 12.3²) = 16.6 km s⁻¹	✓ 1	Adding the speeds (23.5 km s⁻¹): [1 max].
Part (c)(i)
½ × 720 × (16.6 × 10³)² = 9.9 × 10¹⁰ J	✓ 1	Allow ECF from (b)(ii). Accept 9.9 × 10¹⁰–1.0 × 10¹¹ J.
Part (c)(ii)
v = 2πR_E/T = 2π × 6.37 × 10⁶/(24 × 3600) = 463 m s⁻¹	✓ 1	Accept 460–470 m s⁻¹.
Before launch the probe already moves eastwards at this speed, in the direction in which it is launched, so part of the required launch speed (and kinetic energy) is already provided by the rotation of the Earth	✓ 1	Reference to the direction (eastwards, same as the rotation) is needed.
Part (d)(i)
With the minimum speed the total energy of the probe in the Sun's field is zero: ½v² = GM/r, so v ∝ 1/√r and v = 42.1/√5.2 = 18.4 km s⁻¹	✓ 1	Allow ECF from (a)(ii). Accept 18–19 km s⁻¹.

Answers: (a)(i) 29.7 km s⁻¹ · (a)(ii) 42.1 km s⁻¹ · (b)(i) 12.3 km s⁻¹ · (b)(ii) 16.6 km s⁻¹ · (c)(i) 9.9 × 10¹⁰ J · (c)(ii) 463 m s⁻¹ · (d)(i) 18.4 km s⁻¹ (the remaining parts are explanations — see the table above)

Syllabus understandingD.1 (HL) — the escape speed v_esc at any point in a gravitational field as given by v_esc = √(2GM/r); the orbital speed of a body orbiting a large mass as given by v_orbital = √(GM/r); (guidance) energetics of a satellite going into orbit around a non-rotating planet starting from rest on its surface; energy conditions for a body to escape the gravitational influence of a planet; A.5 — Galilean transformation of velocities; A.2 — circular motion, v = 2πr/T Command term: Determine

D.1 Gravitational fields: IB Physics HL exam-style questions

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