AQA A-Level Physics Paper 2, June 2025: Question 4

📐 Step-by-Step Calculation

1. Find orbital radii from the centre of Saturn:
   Orbital radius = radius of planet + height above surface
   r_Mimas = 5.8 × 10⁷ + 1.28 × 10⁸ = 1.86 × 10⁸ m
   r_Enceladus = 5.8 × 10⁷ + 1.80 × 10⁸ = 2.38 × 10⁸ m
2. Apply Kepler's Third Law:
   T² / r³ = constant  or  T_E / T_M = (r_E / r_M)^(3/2)
   T_E = 0.94 × (2.38 × 10⁸ / 1.86 × 10⁸)^(3/2)
   T_E = 0.94 × (1.2796)^(1.5) = 0.94 × 1.4474
3. Calculate final period:
   T_E = 1.36 Earth days (to 2 or 3 s.f.)

✅ Mark Scheme Breakdown

• Mark 1: Adding the radius of Saturn to orbital height for at least one moon ( r_Mimas = 1.86 × 10⁸ m or r_Enceladus = 2.38 × 10⁸ m ).
• Mark 2: Correct use of T₁² / r₁³ = T₂² / r₂³ OR finding proportionality constant k = T / r^(1.5) .
• Mark 3: Final answer of 1.36 Earth days (allow ecf to 1.57 days if radius of Saturn was omitted in both radii).

❌ Common Errors

• Forgetting planet radius: Using height directly as orbital radius ( r = h ). In gravity calculations, r is ALWAYS measured from the centre of mass of the central body.
• Converting time units unnecessarily: Converting 0.94 days to seconds and back again. While not incorrect, it adds unnecessary arithmetic steps where calculator errors can occur.

🧠 Exam Technique

Notice the question states "at a height ... above the surface". Train your eyes to spot "height" vs "orbital radius". Whenever you see "height", immediately write down: r = R_planet + h .

✅ Correct Answer

The variation of gravitational potential with distance is approximately a straight line / linear / has a constant gradient

AND gravitational field strength equals the potential gradient ( g = -ΔV/Δr ).

💡 Key Knowledge

• The relationship between field strength and potential is:
  g = - dV/dr (field strength is the negative gradient of the V-r curve).
• A constant gradient directly implies a constant field strength.

❌ Common Errors & Examiner Warnings

• Saying "potential is directly proportional to distance": A straight line with a non-zero intercept is linear, NOT directly proportional. Saying "directly proportional" loses the mark.
• Incomplete reasoning: Only mentioning that the line is straight, without stating that field strength is equal to the gradient. Both elements are required for the single mark.

✅ Correct Answer

distance = 12.025 × 10⁸ m to 12.045 × 10⁸ m

🧠 Exam Technique: Reading the Maximum

• Resultant gravitational field strength g = 0 where the gradient of the total potential curve (solid line) is zero: dV/dr = 0 .
• Locate the crest/turning point of the solid line on Figure 2.
• Look at grid divisions: 1 major square = 0.1 × 10⁸ m, subdivided into 10 small squares. Therefore, 1 small square = 0.01 × 10⁸ m .
• The peak lies between 3 and 4 small squares past 12.0, giving 12.03 × 10⁸ m to 12.04 × 10⁸ m .

📐 Step-by-Step Calculation

1. Isolate potential due to Titan at 12.08 × 10⁸ m:
   Read both curves at distance = 12.08 × 10⁸ m (8 small squares past 12.0):
   • Dashed line (Saturn alone): V_Saturn ≈ -3.135 × 10⁷ J kg⁻¹ (allow -3.135 to -3.138)
   • Solid line (Total potential): V_total ≈ -3.200 × 10⁷ J kg⁻¹ (allow -3.200 to -3.203)
   V_Titan = V_total - V_Saturn
   V_Titan = -3.200 × 10⁷ - (-3.135 × 10⁷) = -0.065 × 10⁷ = -6.5 × 10⁵ J kg⁻¹
   (Matches the "show that ... is about -7 × 10⁵ J kg⁻¹")
2. Find distance from the centre of Titan:
   Centre of Titan is at 12.22 × 10⁸ m from Saturn.
   r_Titan = 12.22 × 10⁸ - 12.08 × 10⁸ = 0.14 × 10⁸ m = 1.4 × 10⁷ m
3. Calculate Titan's Mass using V = -GM / r:
   M = |V| × r / G
   Using calculated V = -6.5 × 10⁵ J kg⁻¹ :
   M = (6.5 × 10⁵ × 1.4 × 10⁷) / (6.67 × 10⁻¹¹) = 1.36 × 10²³ kg
   (Or using given value -7 × 10⁵ J kg⁻¹: M = 1.47 × 10²³ kg)

✅ Mark Scheme Breakdown

• Mark 1: Gravitational potential from Titan = solid line - dashed line read at 12.08 × 10⁸ m:
  Shows calculation giving a value in range -6.2 × 10⁵ to -6.8 × 10⁵ J kg⁻¹ .
• Mark 2: Correct distance from Titan ( 1.4 × 10⁷ m ) OR correct substitution into V = -GM/r .
• Mark 3: Final answer consistent with their V :
  • If using -6.5 × 10⁵ : 1.36 × 10²³ kg (range 1.3 × 10²³ to 1.4 × 10²³)
  • If using "show that" value -7 × 10⁵ : 1.47 × 10²³ kg

❌ Common Errors

• Using distance from Saturn instead of distance from Titan: Using r = 12.08 × 10⁸ m instead of the distance to Titan's centre ( 1.4 × 10⁷ m ). Gravitational potential of a body depends on distance from that body's centre.
• Reading off the wrong distance on the graph: Not checking grid square scaling carefully (each small square = 0.01 × 10⁸ m).
• Sign confusion in subtraction: Subtracting negative numbers incorrectly, e.g. -3.20 - 3.14 = -6.34 × 10⁷ instead of -3.20 - (-3.14) .

🧠 Exam Technique: "Show that... go on to determine"

When an exam paper says "Show that [X] is about -7 × 10⁵ J kg⁻¹; go on to determine...":

• You must show your own full calculation for the first mark to get a value close to -7 × 10⁵ .
• If you get stuck on the first step, you can still gain the subsequent marks by using the provided value ( -7 × 10⁵ J kg⁻¹ ) in the second part!