OxfordAQA International AS and A-level Physics: Newton's law of gravitation and gravitational field strength (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.7.1 (Newton’s gravitational law) and 3.7.2 (Gravitational field strength) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, for exams from May/June 2020 onwards. Section 3.7 is International A-level only. Start with the study guide and the revision notes. The course hub and printable checklist show where this unit sits.

Use G = 6.67 × 10⁻¹¹ N m² kg⁻². Other data is given in each question. Give answers to two or three significant figures, as the data allows.

Questions

1. State Newton’s law of gravitation. [3]

2. Two ships of masses 3.0 × 10⁴ kg and 4.5 × 10⁴ kg are moored with their centres 5.0 m apart. Treat them as point masses.

(a) Calculate the gravitational force between them. [2] (b) The ships drift apart until their centres are 15 m apart. Calculate the new force. [1]

3. (a) Define gravitational field strength, and state its unit. [2] (b) State the direction of the gravitational field strength at a point near a planet. [1]

4. (a) Describe the pattern of gravitational field lines around an isolated spherical planet. [2] (b) State what the spacing of the lines shows. [1] (c) Explain why the field lines in a small region near the planet’s surface are drawn parallel and equally spaced. [1]

5. Planet X has a surface gravitational field strength of 2.5 N kg⁻¹ and a radius of 3.0 × 10⁶ m.

(a) Calculate the mass of planet X. [2] (b) Calculate the gravitational field strength at a point 4.5 × 10⁶ m from the centre of planet X. [2]

6. The Earth has mass 5.97 × 10²⁴ kg and the Moon has mass 7.35 × 10²² kg. Their centres are 3.84 × 10⁸ m apart.

(a) Calculate the gravitational force that the Earth exerts on the Moon. [2] (b) State the size of the force that the Moon exerts on the Earth. [1] (c) Calculate the gravitational field strength of the Earth at the Moon’s distance. [2] (d) State the direction of the force on the Moon. [1]

7. A planet is a uniform sphere of radius R and density ρ.

(a) Show that the gravitational field strength at its surface is g = (4/3)πGρR. [3] (b) Earth has radius 6.37 × 10⁶ m and mean density 5510 kg m⁻³. Use the equation to calculate g at Earth’s surface. [1]

8. An astronaut has a mass of 65 kg. The gravitational field strength at the Moon’s surface is 1.62 N kg⁻¹.

(a) State the astronaut’s mass on the Moon. [1] (b) Calculate the astronaut’s weight on the Moon. [1] (c) Calculate the astronaut’s weight at Earth’s surface and the factor by which the weight is greater. [2]

9. Earth has radius 6.37 × 10⁶ m and surface field strength 9.81 N kg⁻¹. A satellite is at a point where the field strength is 4.9 N kg⁻¹.

(a) Calculate the distance of the satellite from Earth’s centre. [3] (b) Calculate its height above Earth’s surface. [1]

10. A uniform sphere A of mass 5.0 kg and a uniform sphere B of mass 0.50 kg are placed with their centres 0.12 m apart. Ignore all other forces.

(a) Calculate the gravitational force on B due to A. [2] (b) The weight of B at Earth’s surface is 4.9 N. Calculate the ratio of this gravitational force to B’s weight. [2] (c) The centres are moved to 0.24 m apart. Calculate the new force. [1] (d) Explain why B would accelerate more than A if both were free to move, although the forces on them are the same size. [2]

11. A gas giant has mass 5.7 × 10²⁶ kg and radius 6.0 × 10⁷ m. A probe of mass 350 kg hovers at the top of its atmosphere, so that its distance from the centre equals the radius.

(a) Calculate the gravitational field strength at the probe’s position. [2] (b) Calculate the gravitational force on the probe. [1] (c) The probe moves to a distance of three times the planet’s radius from the centre. Calculate the gravitational field strength there. [2] (d) Calculate how many times greater the field strength in part (a) is than the value at Earth’s surface. [1] (e) Compare the field lines at the two positions in parts (a) and (c), in terms of direction and spacing. [2]

Answers

1. Two point masses attract with a force proportional to the product of their masses [1] and falling off with the square of the separation, F ∝ 1/r² [1]. The force is attractive [1]. [3] Examiner insight: Answers usually gain credit for each of the three ideas stated separately; “proportional to the masses and the distance” misses the inverse square and the product.

2. (a) F = Gm₁m₂/r² = 6.67 × 10⁻¹¹ × 3.0 × 10⁴ × 4.5 × 10⁴ / 5.0² [1] = 3.6 × 10⁻³ N [1] [2] (b) The distance is three times larger, so the force is 1/9 as large: 3.60 × 10⁻³ / 9 = 4.0 × 10⁻⁴ N [1] [1] Examiner insight: Show the substitution before the answer. In part (b), saying “one ninth” and then giving the new value makes the reasoning clear.

3. (a) The force per unit mass [1] acting on a small mass placed at that point; unit N kg⁻¹ [1] [2] (b) Towards the centre of the planet, the direction of the force [1] [1] Examiner insight: A definition earns credit when it says force per unit mass; “the force of gravity” alone is too vague.

4. (a) The lines are straight and radial [1], with arrows pointing inwards towards the centre of the planet [1] [2] (b) Field strength: stronger where the lines are packed more closely [1] [1] (c) The region is tiny compared with the planet’s radius, so the field strength and direction hardly change over it [1] [1] Examiner insight: A sketch or description earns credit only with arrowheads and a clear statement that lines point towards the centre. Linking spacing to field strength is a separate point.

5. (a) M = gR²/G [1] = 2.5 × (3.0 × 10⁶)² / 6.67 × 10⁻¹¹ = 3.4 × 10²³ kg [1] [2] (b) g ∝ 1/r², so g = 2.5 × (3.0 / 4.5)² [1] = 1.1 N kg⁻¹ [1] [2] Examiner insight: A ratio method using the given data avoids carrying a rounded mass into part (b). Using the rounded mass gives the same answer to two significant figures.

6. (a) F = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 7.35 × 10²² / (3.84 × 10⁸)² [1] = 1.98 × 10²⁰ N [1] [2] (b) The same size, 1.98 × 10²⁰ N [1], because they are a Newton’s third law pair [1] (c) g = GM/r² = 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ / (3.84 × 10⁸)² [1] = 2.70 × 10⁻³ N kg⁻¹ [1] [2] (d) Towards the Earth [1] [1] Examiner insight: For “state the size of the force on the other body”, examiners usually expect the equal-and-opposite pair idea, not a recalculation with a different mass. Mention that the forces act on different bodies.

7. (a) Mass M = density × volume = ρ × (4/3)πR³ [1]. Surface field g = GM/R² [1]. Substituting: g = G × ρ(4/3)πR³ / R² = (4/3)πGρR [1] [3] (b) g = (4/3) × π × 6.67 × 10⁻¹¹ × 5510 × 6.37 × 10⁶ = 9.81 N kg⁻¹ [1] [1] Examiner insight: In “show that” questions, write every algebra step, including where R³ divided by R² leaves R. A final line that only restates the given result gains little.

8. (a) Mass is unchanged: 65 kg [1] [1] (b) W = mg = 65 × 1.62 = 105 N [1] [1] (c) W = 65 × 9.81 = 638 N [1]. The factor is 638 / 105 = 6.1 [1] [2] Examiner insight: Keep the roles clear: kg for mass, N for weight. A weight given in kg, or a mass said to change, is easy to avoid.

9. (a) g ∝ 1/r², so (R/r)² = 4.9 / 9.81 [1]. r = R × √(9.81 / 4.9) [1] = 9.0 × 10⁶ m [1] [3] (b) h = r − R = 9.01 × 10⁶ − 6.37 × 10⁶ = 2.6 × 10⁶ m [1] [1] Examiner insight: Part (a) gives the distance from the centre, and the question asks for height. Subtracting the radius at the end is the step most often forgotten.

10. (a) F = 6.67 × 10⁻¹¹ × 5.0 × 0.50 / 0.12² [1] = 1.2 × 10⁻⁸ N [1] [2] (b) Ratio = 1.16 × 10⁻⁸ / 4.9 [1] = 2.4 × 10⁻⁹ [1] [2] (c) The distance doubles, so the force falls by a factor of 4: 1.16 × 10⁻⁸ / 4 = 2.9 × 10⁻⁹ N [1] [1] (d) The forces form an equal and opposite pair [1]. a = F/m, so the smaller mass B has the larger acceleration [1] [2] Examiner insight: Saying the force is “very small” is stronger with a number, such as the ratio in part (b). In part (d), give both ideas: equal forces, then a = F/m.

11. (a) g = GM/R² = 6.67 × 10⁻¹¹ × 5.7 × 10²⁶ / (6.0 × 10⁷)² [1] = 10.6 N kg⁻¹ [1] [2] (b) F = mg = 350 × 10.6 = 3.7 × 10³ N [1] [1] (c) The distance is three times larger, so g falls to one ninth [1]: 10.56 / 9 = 1.17 N kg⁻¹ [1] [2] (d) 10.6 / 9.81 = 1.08 times greater [1] [1] (e) Both sets of lines are radial and point inwards towards the centre [1]. At the greater distance the lines are further apart, showing the weaker field [1] [2] Examiner insight: Using the rounded 10.6 in part (c) can change the third figure compared with a one-step calculation. Show your working and avoid rounding in the middle of a long calculation.

Where marks are usually lost

• Using altitude, not the distance from the centre, as r.
• Squaring only the number in r and not its power of ten.
• Swapping G and g in a formula.
• Giving the unit of g as N, kg or m, or leaving it out.
• Saying that a heavier object feels a larger g, instead of explaining that m cancels.
• Treating the two forces in a gravitational pair as unequal because the masses are unequal.
• Drawing field lines without arrows, or with arrows pointing away from the mass.
• Giving a mass in newtons or a weight in kilograms.
• Rounding mid-calculation, then reporting too many figures.
• Skipping steps in a “show that” argument.

Next steps

• Re-read the revision notes and the study guide for anything you missed.
• Check the course hub and tick off the printable checklist.
• Take one of the free 10-minute diagnostics to see what to revise next.
• Book a free trial class if you want help with any of these questions.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), Version 4.4, for International AS and A-level exams May/June 2020 onwards, published by OxfordAQA. Sections 3.7.1 Newton’s gravitational law and 3.7.2 Gravitational field strength (International A-level only).