IB DP Physics: Gravitational potential, escape speed and orbital energy (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 cover the additional higher level (HL only) part of D.1 Gravitational fields in the IB Physics guide, first assessment 2025. Read the study guide or the revision notes first. The course pages are the IB DP Physics hub and the checklist. Use G = 6.67 × 10⁻¹¹ N m² kg⁻². Treat all bodies as non-rotating, ignore drag unless stated, and measure r from the centre.

Data: Callisto mass 1.08 × 10²³ kg, radius 2.41 × 10⁶ m. Triton mass 2.14 × 10²² kg, radius 1.35 × 10⁶ m. Earth mass 5.97 × 10²⁴ kg.

Questions

1. State what is meant by the gravitational potential at a point. [2]

2. A 1500 kg probe is at a distance of 4.82 × 10⁶ m from the centre of Callisto. Calculate the gravitational potential energy of the probe-Callisto system. [3]

3. A 280 kg lander rises from the surface of Callisto to a point 2.0 × 10⁶ m above the surface.

(a) Calculate the work that must be done on the lander. [3] (b) The surface field strength is 1.24 N kg⁻¹. Using mgh with this value gives 6.95 × 10⁸ J. Explain why this is larger than your answer. [2]

4. The gravitational potential of Callisto is -1.44 × 10⁶ J kg⁻¹ at r = 5.0 × 10⁶ m and -1.20 × 10⁶ J kg⁻¹ at r = 6.0 × 10⁶ m.

(a) Estimate the field strength between these points. [2] (b) Calculate GM/r² at r = 5.5 × 10⁶ m and comment on the agreement. [2]

5. Describe the equipotential surfaces around an isolated spherical planet, state how they relate to the field lines, and explain why surfaces drawn at equal steps in potential become further apart with increasing distance. [4]

6. The surface field strength of Io is 1.80 N kg⁻¹ and its radius is 1.82 × 10⁶ m.

(a) Show that the escape speed at the surface is given by √(2gR). [2] (b) Calculate the escape speed. [2]

7. A projectile is launched vertically from the surface of Callisto.

(a) It leaves at 3.0 × 10³ m s⁻¹. Calculate its speed at a very large distance. [3] (b) A second projectile leaves at 1.5 × 10³ m s⁻¹. Calculate the greatest height it reaches above the surface. [3]

8. A satellite orbits 5.0 × 10⁵ m above the surface of Callisto in a circular orbit.

(a) Calculate its orbital speed. [3] (b) Calculate its orbital period. [2]

9. A 900 kg satellite moves from a circular orbit of radius 2.0 × 10⁶ m around Triton to one of radius 3.5 × 10⁶ m.

(a) Calculate its kinetic, potential and total energy in the first orbit. [3] (b) Calculate the energy that must be supplied. [2] (c) State, with the new speed, how the speed changes. [1]

10. A 500 kg probe starts at rest on the surface of Triton and enters a circular orbit 4.0 × 10⁵ m above the surface. Ignore fuel mass and Triton’s rotation.

(a) Calculate the energy that must be supplied. [4] (b) Calculate the fraction that appears as kinetic energy of the probe in orbit. [2] (c) Calculate the additional energy needed to send it from this orbit to infinity, and relate it to its kinetic energy. [2]

11. A 350 kg satellite orbiting the Earth has orbital radius 6.90 × 10⁶ m. Drag reduces the radius to 6.82 × 10⁶ m. Use GM = 3.982 × 10¹⁴ N m² kg⁻¹ for the Earth.

(a) Calculate the orbital speed before and after. [2] (b) Calculate the total energy lost. [2] (c) Explain why the speed increases although drag opposes the motion. [2]

12. Two small asteroids of mass 3.0 × 10¹⁵ kg and 1.2 × 10¹⁶ kg are 1.2 × 10⁴ m apart.

(a) Show that the field strength is zero 4.0 × 10³ m from the lighter asteroid, on the line joining them. [2] (b) Calculate the gravitational potential at that point. [2] (c) Calculate the work needed to move a 40 kg object from that point to a very large distance. [2]

Answers

1. The work done on each kilogram of a small test mass [1] as it is carried from infinity to that point [1]. [2] Examiner insight: Both “per unit mass” and “from infinity” are separate ideas, and a definition missing either one is incomplete.

2. GM = 6.67e-11 × 1.08e23 = 7.204e12 [1]. Ep = -GMm/r = -7.204e12 × 1500 / 4.82e6 [1] = -2.24 × 10⁹ J [1]. [3] Examiner insight: A positive answer shows the sign convention is not understood, so check that your result is negative.

3. (a) Vg at surface = -7.204e12 / 2.41e6 = -2.989e6 J kg⁻¹ [1]. Vg at r = 4.41e6 m = -1.633e6 J kg⁻¹ [1]. W = mΔVg = 280 × 1.356e6 = 3.80 × 10⁸ J [1] (b) The field strength decreases as the lander rises [1], so mgh with the surface value overestimates the work [1]. [5] Examiner insight: r is the distance from the centre (surface radius plus 2.0 × 10⁶ m), and using 2.0 × 10⁶ m directly is a typical slip.

4. (a) ΔVg = +2.4e5 J kg⁻¹ over Δr = 1.0e6 m [1]. |g| = 0.24 N kg⁻¹ [1] (b) GM/r² = 7.204e12 / (5.5e6)² = 0.238 N kg⁻¹ [1]. They agree closely because a gradient over a short interval gives the field strength near the midpoint [1]. [4] Examiner insight: A gradient is only an estimate of the field at one point, so compare it at the middle of the interval rather than at an end.

5. Concentric spheres centred on the planet [1]. Field lines are radial, meet the surfaces at right angles and point towards lower potential [1]. Equal steps in Vg are further apart at large r [1] because g is the gradient of Vg and is smaller there, so a bigger Δr is needed for the same ΔVg [1]. [4] Examiner insight: Link the spacing to the gradient. Saying only “the field is weaker” does not explain the spacing.

6. (a) At the surface ½mv² = GMm/R [1], and GM = gR² so v = √(2gR) [1] (b) v = √(2 × 1.80 × 1.82e6) = √(6.552e6) [1] = 2.56 × 10³ m s⁻¹ [1]. [4] Examiner insight: A “show that” needs the equation GM = gR² to appear, since this is where the new form comes from.

7. (a) Energy per kg: ½v² - GM/R = 4.5e6 - 2.989e6 = 1.511e6 J kg⁻¹ [1]. This equals ½v∞² [1], so v∞ = 1.74 × 10³ m s⁻¹ [1] (b) Energy per kg = 1.125e6 - 2.989e6 = -1.864e6 [1]. At the highest point -GM/r = -1.864e6, so r = 3.86e6 m [1]. Height = 1.45 × 10⁶ m [1]. [6] Examiner insight: The question asks for the height above the surface, and an answer left as the distance from the centre misses the final subtraction.

8. (a) r = 2.91e6 m [1]. GMm/r² = mv²/r gives v = √(GM/r) [1] = √(7.204e12 / 2.91e6) = 1.57 × 10³ m s⁻¹ [1] (b) T = 2πr/v [1] = 2π × 2.91e6 / 1573 = 1.16 × 10⁴ s (3.23 hours) [1]. [5] Examiner insight: Use the unrounded speed in part (b) to avoid a rounding drift in the third figure.

9. (a) GM = 1.4274e12. Ek = GMm/2r = 3.21 × 10⁸ J [1]; Ep = -6.42 × 10⁸ J [1]; E = -3.21 × 10⁸ J [1] (b) E at 3.5e6 m = -1.835e8 J [1]. Energy supplied = 1.38 × 10⁸ J [1] (c) The speed falls from 8.45 × 10² to 6.39 × 10² m s⁻¹ [1]. [6] Examiner insight: Energy and speed move in opposite directions in this change, so state both clearly.

10. (a) E(surface) = -GMm/R = -1.4274e12 × 500 / 1.35e6 = -5.287e8 J [1]. r = 1.75e6 m, E(orbit) = -GMm/2r = -2.039e8 J [1]. Difference [1] = 3.25 × 10⁸ J [1] (b) Ek in orbit = 2.039e8 J [1]. Fraction = 2.039e8 / 3.248e8 = 0.63 (63%) [1] (c) Extra energy = |E| = 2.04 × 10⁸ J [1], which equals the kinetic energy in orbit [1]. [8] Examiner insight: State your assumptions (non-rotating planet, no drag, fuel mass ignored), since the energy is a minimum.

11. (a) v = √(GM/r): before √(3.982e14/6.90e6) = 7.60 × 10³ m s⁻¹ [1]; after 7.64 × 10³ m s⁻¹ [1] (b) E = -GMm/2r. Loss = (GMm/2)(1/6.82e6 - 1/6.90e6) [1] = 1.18 × 10⁸ J [1] (c) Drag removes energy, so the total energy becomes more negative and the radius decreases [1]. The potential energy falls by twice the energy lost, so gravity’s work on the satellite exceeds drag’s and the kinetic energy rises [1]. [6] Examiner insight: Show that Ek rises as Ep falls and compare the two changes. A bare claim that gravity speeds it up is weaker.

12. (a) GM1/x² = GM2/(d - x)² gives x/(d - x) = √(M1/M2) = 1/2 [1], so x = d/3 = 4.0 × 10³ m [1] (b) Vg = -G M1/x - G M2/(d - x) = -(2.001e5/4.0e3) - (8.004e5/8.0e3) [1] = -150 J kg⁻¹ [1] (c) W = mΔVg = 40 × 150.1 [1] = 6.0 × 10³ J [1]. [6] Examiner insight: Potentials add as scalars, so each term keeps its minus sign. Do not add magnitudes of field strengths here.

Where marks are usually lost

• Putting the altitude into a formula that needs r from the planet centre.
• Writing a positive potential or potential energy.
• Using mgh for a large change in height.
• Quoting the escape speed formula without the factor 2.
• Giving only the new radius and not converting it to a height.
• Claiming that drag slows a satellite down.
• Leaving off units, especially J kg⁻¹ for potential.
• Mixing up energy to change orbit with energy to escape.

Next steps

• Revise with the revision notes and the study guide.
• Course hub: IB DP Physics. Checklist: IB DP Physics checklist.
• Take all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. Section D.1, additional higher level part.