IB DP Physics: Gravitational fields: Kepler, Newton's law and field strength (DP Physics) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise the standard level and higher level part of sub-topic D.1, Gravitational fields, from the IB Diploma Programme Physics guide, first assessment 2025: Kepler’s laws, Newton’s law of gravitation, field strength and field lines. All questions are suitable for SL and HL. Use G = 6.67 × 10⁻¹¹ N m² kg⁻² and give numerical answers to three significant figures unless told otherwise.

Read the study guide first, then use the revision notes for a quick recap. See the IB Physics hub and the printable checklist. To find gaps first, try the free 10-minute diagnostics.

Questions

1. State Kepler’s first law and Kepler’s third law of planetary motion. [2]

2. A pair of uniform spheres, one of 5.0 × 10³ kg and one of 2.0 × 10³ kg, sit with centres 15 m from each other.

(a) Find the gravitational attraction between them. [2] (b) The larger mass is replaced by one of twice the mass, and the separation is increased to 45 m. Determine the new force. [1]

3. The table gives data for two of Jupiter’s moons, supplied for this question.

Moon	Orbital radius / m	Period / days
Europa	6.71 × 10⁸	3.55
Ganymede	1.07 × 10⁹	to find

Use Kepler’s third law to determine the orbital period of Ganymede in days. [4]

4. An asteroid orbits the Sun on an ellipse. At its closest point it is 1.4 × 10¹¹ m from the Sun and moving at 3.6 × 10⁴ m s⁻¹. Use Kepler’s second law (state it first) to find the speed at its furthest point, 4.2 × 10¹¹ m from the Sun. At both points the velocity is perpendicular to the line to the Sun. [4]

5. For Venus, take a mass of 4.87 × 10²⁴ kg and a radius of 6.05 × 10⁶ m.

(a) Define gravitational field strength. [1] (b) Calculate the field strength at the surface of Venus. [2] (c) Determine the weight of a 450 kg probe at the surface. [1]

6. A 950 kg spacecraft is 2.4 × 10⁶ m above the surface of Venus (data from question 5). Calculate the gravitational force on it, and state why the distance used is not 2.4 × 10⁶ m. [4]

7. Two bodies, A and B, have masses 1.8 × 10²³ kg and 2.0 × 10²² kg. Their centres are 6.0 × 10⁸ m apart.

(a) Determine the distance from A of the point on the line joining them where the resultant gravitational field strength is zero. [3] (b) Calculate the resultant field strength at the point on the line joining them that is 1.0 × 10⁸ m from B, and state its direction. [3]

8. Take the Earth’s mass as 5.97 × 10²⁴ kg and its radius as 6.37 × 10⁶ m.

(a) Calculate the percentage decrease in g between the surface and a point 2.0 km above it. [3] (b) At a height of 2000 km the decrease is 42 %. Explain why the field can be treated as uniform near the surface but not at 2000 km. [2]

9. Gravitational field lines are drawn around a planet.

(a) Describe the field lines of the uniform field near the surface of the planet. [2] (b) In a diagram of the radial field, 6 lines pass through a small area at point P and 2 lines pass through an area of the same size at point Q. State the ratio of field strengths gP : gQ and the ratio rQ : rP of their distances from the centre. [3]

10. The Earth has surface field strength 9.81 N kg⁻¹ and radius 6.37 × 10⁶ m.

(a) Show that the mass of a uniform spherical planet of density ρ and radius R gives g proportional to R. [3] (b) A planet has the same average density as the Earth and a radius 1.5 times larger. Calculate its surface field strength. [2] (c) Determine the mass of the Earth from the data given. [2]

11. A moon orbits a planet in a circular orbit of radius 4.2 × 10⁸ m with a period of 1.5 × 10⁵ s. For a circular orbit, T² = 4π²r³/(GM), where M is the mass of the planet (given for this question).

(a) Calculate the mass of the planet. [3] (b) Calculate the gravitational field strength of the planet at the moon’s orbit. [2] (c) A second moon orbits at a radius of 1.05 × 10⁹ m. Use Kepler’s third law to find its period. [2] (d) The planet has radius 7.0 × 10⁷ m. Calculate the field strength at its surface. [2]

Answers

1. Planets move in elliptical orbits with the Sun at one focus [1]. The square of the orbital period is proportional to the cube of the mean orbital radius (T² ∝ r³) [1]. [2] Examiner insight: Naming the Sun at one focus is the point that separates a first-law answer from a vague “orbits are ellipses”. For the third law, saying that the period is proportional to the radius cubed, without the squaring, loses the mark.

2. (a) F = Gm₁m₂/r² [1] = 6.67 × 10⁻¹¹ × 5.0 × 10³ × 2.0 × 10³ / 15² = 2.96 × 10⁻⁶ N [1] (b) One mass doubled gives a factor of 2, and r tripled gives a factor of 1/9, so F = 2/9 × 2.96 × 10⁻⁶ = 6.59 × 10⁻⁷ N [1] [3] Examiner insight: A scaling argument shows full understanding, but state the factor for each change.

3. Kepler’s third law: T² ∝ r³, so T₂ = T₁(r₂/r₁)^(3/2) [1]. r₂/r₁ = 1.07 × 10⁹ / 6.71 × 10⁸ = 1.595 [1]. (1.595)^(3/2) = 2.014 [1]. T₂ = 3.55 × 2.014 = 7.15 days [1] [4] Examiner insight: The ratio form means no unit conversion is needed. Converting to seconds and back is allowed but gives more chances for slips.

4. The Sun-asteroid line covers equal areas in equal times [1]. At these two points, r₁v₁ = r₂v₂ [1]. v₂ = 1.4 × 10¹¹ × 3.6 × 10⁴ / 4.2 × 10¹¹ [1] = 1.20 × 10⁴ m s⁻¹ [1] [4] Examiner insight: The shortcut r₁v₁ = r₂v₂ only works where the velocity is perpendicular to the line to the Sun. Saying that is part of justifying the method.

5. (a) The gravitational force exerted on a small point mass at that point, divided by that mass [1]. (b) g = GM/R² [1] = 6.67 × 10⁻¹¹ × 4.87 × 10²⁴ / (6.05 × 10⁶)² = 8.87 N kg⁻¹ [1] (c) W = mg = 450 × 8.87 = 3.99 × 10³ N [1] [4] Examiner insight: A definition needs the words “per unit mass” and the idea of a small test mass. “The force of gravity at a point” is not enough.

6. r = 6.05 × 10⁶ + 2.4 × 10⁶ = 8.45 × 10⁶ m. F = GMm/r² [1] = 6.67 × 10⁻¹¹ × 4.87 × 10²⁴ × 950 / (8.45 × 10⁶)² [1] = 4.32 × 10³ N [1] The distance must be measured from the centre of Venus, since the planet acts as a point mass at its centre [1]. [4] Examiner insight: Using the height above the surface as r gives a wrong final value. Show the addition of the radius so the working is clear.

7. (a) At the null point, M_A/x² = M_B/(d − x)² [1]. So (d − x)/x = √(2.0 × 10²² / 1.8 × 10²³) = 1/3 [1]. x = d/(1 + 1/3) = 4.5 × 10⁸ m from A [1] (b) The point is 5.0 × 10⁸ m from A. g_A = 6.67 × 10⁻¹¹ × 1.8 × 10²³ / (5.0 × 10⁸)² = 4.80 × 10⁻⁵ N kg⁻¹ towards A, and g_B = 6.67 × 10⁻¹¹ × 2.0 × 10²² / (1.0 × 10⁸)² = 1.33 × 10⁻⁴ N kg⁻¹ towards B [1]. The fields oppose, so the resultant = 1.334 × 10⁻⁴ − 4.80 × 10⁻⁵ [1] = 8.54 × 10⁻⁵ N kg⁻¹ towards B [1] [6] Examiner insight: Marks are earned by showing both individual field strengths before combining them. The direction is part of the answer. A resultant with no direction is incomplete for a vector quantity.

8. (a) g₀ = GM/R² = 9.813 N kg⁻¹ [1]. g at 2.0 km = GM/(R + 2000)² = 9.807 N kg⁻¹ [1]. Decrease = (9.813 − 9.807)/9.813 = 0.063 % [1] (b) At 2.0 km, g changes by under 0.1 %, so the lines are effectively parallel and equally spaced [1]. At 2000 km, g changes by a large fraction, so the lines converge and the field is radial, not uniform [1] [5] Examiner insight: An explanation that mentions the lines converging towards the centre is stronger than “g changes”.

9. (a) The lines are parallel [1], equally spaced, and directed towards the planet’s surface (downward) [1]. (b) The field strength is proportional to line density, so gP : gQ = 6 : 2 = 3 : 1 [1]. Since g ∝ 1/r², rQ²/rP² = 3 [1], so rQ : rP = √3 : 1 = 1.73 : 1 [1] [5] Examiner insight: A “describe” question wants each feature of the pattern named: direction, spacing and straightness.

10. (a) M = ρ × (4/3)πR³ [1]. g = GM/R² = G × ρ(4/3)πR³/R² [1] = (4/3)πGρR, so g ∝ R for constant ρ [1] (b) g = 1.5 × 9.81 [1] = 14.7 N kg⁻¹ [1] (c) M = gR²/G [1] = 9.81 × (6.37 × 10⁶)² / 6.67 × 10⁻¹¹ = 5.97 × 10²⁴ kg [1] [7] Examiner insight: On “show that” questions, every algebraic step must appear. In (a), stating the volume formula and cancelling the powers of R is what earns the marks.

11. (a) M = 4π²r³/(GT²) [1] = 4π² × (4.2 × 10⁸)³ / (6.67 × 10⁻¹¹ × (1.5 × 10⁵)²) [1] = 1.95 × 10²⁷ kg [1] (b) g = GM/r² = 6.67 × 10⁻¹¹ × 1.95 × 10²⁷ / (4.2 × 10⁸)² [1] = 0.737 N kg⁻¹ [1] (c) T₂ = T₁(r₂/r₁)^(3/2) = 1.5 × 10⁵ × (2.5)^(3/2) [1] = 5.93 × 10⁵ s [1] (d) g = 6.67 × 10⁻¹¹ × 1.95 × 10²⁷ / (7.0 × 10⁷)² [1] = 26.5 N kg⁻¹ [1] [9] Examiner insight: Carry unrounded values forward between parts. If (b) uses a rounded mass, the third significant figure may drift. In (d) the radius is the planet’s radius, not the orbital radius.

Where marks are usually lost

• Taking r as the height rather than adding the planet’s radius to it.
• Leaving out “per unit mass” and “small” from the definition of field strength.
• Giving a resultant field strength with no direction.
• Adding fields between two bodies where they oppose, or subtracting them outside where they add.
• Using r₁v₁ = r₂v₂ without saying it applies where the velocity is perpendicular to the line to the Sun.
• Writing T ∝ r³ instead of T² ∝ r³.

Next steps

• Recap with the revision notes.
• Re-read the study guide.
• Visit the IB Physics hub and the printable checklist.
• Take the free 10-minute diagnostics to find your weakest topics.
• Book a free trial class.

Official syllabus

This page follows sub-topic D.1 (Gravitational fields), standard level and higher level content, of the IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization.