AQA A-Level Physics Paper 2, June 2023: Question 3

10 marks · Medium difficulty · Short Answer

Describe properties of a radial gravitational field, interpret a force-distance graph, calculate gravitational field strength ratios, and determine the mass of an astronomical body using energy conservation.

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Question

Exam question with five parts: 03.1 asks to describe two properties of a radial gravitational field (2 marks). 03.2 shows Figure 3, a graph of gravitational force on a space probe versus height above Earth's surface from 0 to 10 × 10^6 m, with the area under the curve shaded from 0 to 8 × 10^6 m, asking for the physical significance of the shaded area (1 mark). 03.3 asks to calculate the ratio g_S / g_E at Earth's surface given the Earth-Sun distance of 1.50 × 10^11 m (2 marks). 03.4 asks why g_S is more important than g_E as the probe escapes the Solar System (1 mark). 03.5 provides Table 2 with probe distances from centre of mass of body X and speeds at points A (6.0 × 10^6 m, 1.1 × 10^3 m/s) and B (0.17 × 10^6 m, 1.3 × 10^3 m/s) with probe mass 4.9 × 10^4 kg, asking to calculate the mass of X (4 marks).
Question text

03.1 Describe two properties of a radial gravitational field.

[2 marks]

A space probe is launched from the Earth’s surface.

Figure 3 shows how the gravitational force acting on the space probe varies with

height above the Earth’s surface.

Figure 3

03.2 State the physical significance of the shaded area in Figure 3.

[1 mark]

At the Earth’s surface,

• the gravitational field strength of the Sun is gS

• the gravitational field strength of the Earth is gE.

03.3 gS

Calculate .

gE

distance from the Earth to the Sun = 1.50 × 1011 m

[2 marks]

gS

gE =

03.4 Explain why gS is more important than gE in predicting the motion of the space probe

as it escapes from the Solar System.

[1 mark]

03.5 The space probe eventually reaches a point where the gravitational influence of

the Solar System is negligible.

The probe is unpowered as it approaches an isolated interstellar body X.

The gravitational field of X changes the kinetic energy of the space probe.

Table 2 shows the distance of the space probe from the centre of mass of X and the

speed for two positions A and B of the space probe.

Table 2

Distance of space probe from 3 −1

6 Speed of space probe / 10 m s

centre of mass of X / 10 m

A 6.0 1.1

B 0.17 1.3

*07* The space probe has a mass of 4.9 × 104 kg.

Calculate the mass of X.

[4 marks]

mass of X = kg

Mark scheme

Show the mark scheme Mark scheme for question 03: 03.1 awards 1 mark for a general property (e.g. attractive force, field lines point towards centre) and 1 mark for inverse-square variation with distance. 03.2 awards 1 mark for work done/energy required to launch the probe to 8 × 10^6 m or change in GPE. 03.3 awards 1 mark for calculating g_S = 5.90 × 10^-3 N/kg or showing valid substitution, and 1 mark for g_S / g_E = 6.0 × 10^-4. 03.4 awards 1 mark for recognizing Earth's force is much smaller than the Sun's at similar distances due to smaller mass. 03.5 awards 4 marks: equating change in kinetic energy (1.18 × 10^10 J) to change in gravitational potential energy, solving for mass of X = 6.3 × 10^20 kg.

Question Answers Additional comments/Guidelines Mark AO

03.1 Any one from: 1 1 is a general point about gravitational fields 2 AO1

×2

• region in which a mass experiences force due to another

2 is specific to a radial gravitational field

mass

• (the field is conservative so) any change in potential

The ‘force is attractive’ is insufficient

energy only depends on the initial and final positions and

not on path taken OWTTE

When required the term “force” must be used

• the force is always attractive OR field lines point to the

rather than words like “effect”, “gravity” etc.

(centre of) mass OR the equipotential surfaces are

spherical about the (centre of ) mass

When required the word ‘mass’ must be used

• gravitational force is a non-contact force

rather than object or similar.terms

The field strength/force has an inverse-square variation

Condone reference to “circular” for “spherical“

(with distance). 2

in describing an equipotential surface

03.2 The shaded area corresponds to: 1 AO2

the (minimum) energy needed/work done to launch the

space probe to a height of 8 × 106 m

OR

the change in GPE of the probe when moved from Earth's

surface to a height of 8 × 106 m

03.3 Evaluates the gravitational field strength due to the Sun = GM 6.67 × 10–11 × 1.99 × 1030 2 AO2

g = S =

−3 −1 1 S 2 2

5.90 × 10 (N kg ) r �1.50 × 1011� ×2

OR shows substitution into a valid equation 1 1 ALTERNATIVE

The Universal gravitational equation for both

bodies may be used:

gSun MSun rEarth

= ×

gEarth MEarth rSun

2 Answer needed to at least 2 sf. If 3+ sf are

g 5.90 × 10–3

Evaluates S( = = ) 6.0×10-4 or 0.060% seen, the answer must round to (6.01 or

g 9.81 6.02)×10-4.

E

Condone reference to “edge of Solar System”

03.4 The force from Earth because of its smaller mass, is less 1 AO2

as “a similar distance”.

than the force from the Sun at a similar distance.

OR

The total work done in moving a long way from the Sun is

much greater than that in moving a long way from the Earth

because 𝑚𝑚𝐸𝐸 ≪ 𝑚𝑚𝑆𝑆.

03.5 change in kinetic energy or kinetic energy per unit mass 1 2 2 10 4 AO1

formulated 1 allow 2 sf 1 m v( B – v A ) OR 1.18 × 10 (J) OR

2 AO1

∆Ek 1 ( 2 2 ) 5 AO2

= v B – v A OR m × 2.40 × 10 J

m 2 ×2

2 ∆Ep = GMm 5 – 6

change in gravitational potential energy / potential 1.7×10 6.0×10

formulated 2

= GMm × 5.72 × 10-6

∆Ep 1 1

OR = GM 5 – 6

m 1.7×10 6.0×10

Loss and gain may be replaced with change

or Δ’s for this mark.

evidence of intention to equate their ∆Ek and ∆Ep (even if

not carried through; allow ecf) 3 1 ( 2 2 ) –GM –GM

3 v B – v A = – and

2 rA rB

substitution of data into working equation. 13

This can be seen anywhere in the solution.

mass of X = 6.3 × 1020 kg

Total 10

How to answer it

Gravitational Fields, Potential Energy & Escape Dynamics

📋 Specification Focus

What this question tests

This multi-step question assesses your understanding of gravitational field properties, graphical interpretation of force-distance graphs, comparative field strengths, and conservation of mechanical energy in astronomical contexts:

  • Field Characteristics: Defining properties of general and radial gravitational fields (inverse-square law, field line directions).
  • Work Done & Force-Distance Graphs: Understanding that the area under a force-displacement graph represents work done or change in GPE.
  • Comparative Field Strengths: Calculating field strength using Newton's Law of Gravitation ( g = GM/r² ) and understanding why massive distant bodies dominate over large distances.
  • Energy Conservation: Applying ΔEk = ΔEp to calculate an unknown body's mass using orbital/trajectory data.
Part 03.1 • 2 Marks

Properties of a Radial Gravitational Field

Describe two properties of a radial gravitational field.

✅ Mark Scheme Requirements

First Mark (General property): Any one from:

  • Region in which a mass experiences a force due to another mass.
  • Conservative field: work done / change in potential energy depends only on initial and final positions (not on the path taken).
  • Field lines point towards the centre of mass / equipotential surfaces are spherical.
  • Force is a non-contact force (always attractive).

Second Mark (Radial-specific property):

  • The field strength / gravitational force obeys an inverse-square variation with distance ( g ∝ 1/r² or F ∝ 1/r² ).

❌ Common Errors & Examiner Traps

  • Vague terminology: Simply writing "the force is attractive" or "there is gravity" is insufficient for mark 1. You must specify that a mass experiences a force.
  • Missing the radial aspect: Giving two general definitions without stating the inverse-square law or spherical geometry loses the second mark.
  • Using "object" or "body" instead of mass.
Mark Breakdown: [Mark 1] General property of gravitational fields • [Mark 2] Specific inverse-square variation for a radial field.
Part 03.2 • 1 Mark

Significance of Area Under Force–Distance Graph

State the physical significance of the shaded area in Figure 3.

✅ Accepted Answers

  • The (minimum) work done / energy needed to launch the space probe from the Earth's surface to a height of 8 × 10⁶ m .
  • OR: The increase / change in gravitational potential energy (GPE) of the probe when moved from Earth's surface to a height of 8 × 10⁶ m .

🧠 Exam Technique

Always state the boundaries when interpreting an area over a specific interval:

  • Notice the shaded region ends at exactly 8 × 10⁶ m .
  • Do not just write "work done" in general; specify that it is the work done to move the probe to that specific height ( 8 × 10⁶ m ).
Mark Breakdown: [1 Mark] Explicitly mentions work done / ΔGPE and references the height to 8 × 10⁶ m .
Part 03.3 • 2 Marks

Ratio of Gravitational Field Strengths (gS / gE)

Calculate gS / gE at Earth's surface.

📐 Step-by-Step Calculation

Step 1: Identify constants from the Data Booklet

  • G = 6.67 × 10⁻¹¹ N m² kg⁻²
  • Mass of Sun MS = 1.99 × 10³⁰ kg
  • Distance Sun–Earth r = 1.50 × 10¹¹ m
  • gE = 9.81 N kg⁻¹

Step 2: Calculate gS at Earth's orbit

gS = (G × MS) / r²

gS = (6.67 × 10⁻¹¹ × 1.99 × 10³⁰) / (1.50 × 10¹¹)²

gS = 1.327 × 10²⁰ / 2.25 × 10²² = 5.90 × 10⁻³ N kg⁻¹

Step 3: Calculate the ratio gS / gE

gS / gE = (5.90 × 10⁻³) / 9.81 = 6.0 × 10⁻⁴ (or 0.060% )

💡 Alternative Method & Precision

Alternative ratio formula:

gS / gE = (MSun / MEarth) × (rEarth / rSun)²

Significant Figures:

  • Answer must be given to at least 2 s.f. ( 6.0 × 10⁻⁴ ).
  • If giving 3 s.f., it must round correctly to 6.01 × 10⁻⁴ or 6.02 × 10⁻⁴ depending on intermediate rounding.
Mark Breakdown: [Mark 1] Correct substitution to find gS = 5.90 × 10⁻³ N kg⁻¹ • [Mark 2] Correct ratio evaluated to 6.0 × 10⁻⁴ .
Part 03.4 • 1 Mark

Dominance of the Sun at Large Distances

Explain why gS is more important than gE in predicting the motion of the probe as it escapes from the Solar System.

✅ Key Reason

Because the Sun has a vastly greater mass ( MS ≫ ME ), the gravitational force exerted by the Earth becomes negligible compared to that of the Sun at large distances from Earth.

OR: The total work required to escape the Sun's gravitational well is far greater than that required to escape Earth's field because ME ≪ MS .

💡 Why Distance Changes Everything

At the Earth's surface, gE dominates because the probe is extremely close to Earth's centre ( r ≈ 6.4 × 10⁶ m ). However, once the probe travels away from Earth into interplanetary space, the distances to both bodies become comparable, meaning the mass difference ( ~330,000× ) completely dominates.

Mark Breakdown: [1 Mark] Identifies that Earth's force is far smaller at large/comparable distances due to its smaller mass (or compares total work required).
Part 03.5 • 4 Marks

Conservation of Energy to Find Mass of Interstellar Body X

Calculate the mass of X given distances and speeds at positions A and B.

📐 Step-by-Step Calculation

Step 1: Calculate change in kinetic energy (ΔEk)

At A: vA = 1.1 × 10³ m s⁻¹ , rA = 6.0 × 10⁶ m

At B: vB = 1.3 × 10³ m s⁻¹ , rB = 0.17 × 10⁶ m

mprobe = 4.9 × 10⁴ kg

ΔEk = ½ m (vB² − vA²)

ΔEk = ½ × (4.9 × 10⁴) × [(1.3 × 10³)² − (1.1 × 10³)²]

ΔEk = 0.5 × (4.9 × 10⁴) × (1.69 × 10⁶ − 1.21 × 10⁶)

ΔEk = 0.5 × (4.9 × 10⁴) × (0.48 × 10⁶) = 1.176 × 10¹⁰ J (or ΔEk/m = 2.40 × 10⁵ J kg⁻¹ ) [Mark 1]

Step 2: Formulate change in gravitational potential energy (ΔEp)

ΔEp = G M m (1/rB − 1/rA)

(1/rB − 1/rA) = (1 / (0.17 × 10⁶)) − (1 / (6.0 × 10⁶))

= 5.882 × 10⁻⁶ − 0.167 × 10⁻⁶ = 5.715 × 10⁻⁶ m⁻¹ [Mark 2]

Step 3: Equate ΔEk = ΔEp

½ (vB² − vA²) = G M (1/rB − 1/rA) [Mark 3]

2.40 × 10⁵ = (6.67 × 10⁻¹¹) × M × (5.715 × 10⁻⁶)

Step 4: Solve for Mass M

M = (2.40 × 10⁵) / [(6.67 × 10⁻¹¹) × (5.715 × 10⁻⁶)]

M = (2.40 × 10⁵) / (3.812 × 10⁻¹⁶) ≈ 6.3 × 10²⁰ kg [Mark 4]

❌ Common Calculation Traps

  • Powers of 10 in Table 2: Note that distance is in 10⁶ m and speed is in 10³ m s⁻¹ . Missing the table headers causes an error of several orders of magnitude.
  • Squaring bracket terms: Writing (vB − vA)² instead of (vB² − vA²) . (1.3 − 1.1)² = 0.04 , whereas 1.3² − 1.1² = 0.48 !
  • Radius subtraction error: Incorrectly doing 1/(rB − rA) instead of (1/rB − 1/rA) . Potential depends on reciprocal distance.
  • Not converting 0.17: 0.17 × 10⁶ m = 1.7 × 10⁵ m .
Mark Breakdown:
• Mark 1: Correct calculation of ΔEk = 1.18 × 10¹⁰ J (or ΔEk/m = 2.40 × 10⁵ J kg⁻¹ ).
• Mark 2: Correct formulation of ΔEp = GMm(1/rB − 1/rA) with values substituted.
• Mark 3: Clear evidence of equating ΔEk = ΔEp .
• Mark 4: Final evaluated answer of 6.3 × 10²⁰ kg (accept 2 s.f., 6.28 × 10²⁰ kg to 6.30 × 10²⁰ kg ).

Topics

Physics · 3.7 Fields and their consequences (A-level only) · 3.4 Mechanics and materials

Question and mark scheme from the AQA A-Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.