Home · A-Level 9702 · Gravitational fields · Gravitational field and Newton's law · Practice
A2 · Practice questions · Gravitational field and Newton's law

Force per kilogram, falling as the inverse square.

Six original Cambridge-style questions on gravitational field strength, Newton's law of gravitation, treating spheres as point masses and g = GM / r².

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Field strength is a force per unit mass, and it is a vector.

01
Analysis
[2 marks]

A student writes that the gravitational field strength at a point is "the force on a mass placed at that point". State what is missing from this definition, then give the correct definition and its SI unit.

  • The definition must be per unit mass: the force on a mass depends on the mass placed there, whereas the field strength is a property of the point alone. ✓
  • Gravitational field strength is the gravitational force per unit mass acting on a small test mass placed at the point, g = F / m, with SI unit N kg⁻¹ (equivalent to m s⁻²). ✓
02
Calculation
[3 marks]

Two lead spheres, each of mass 1.5 kg, are placed with their centres 0.12 m apart. Calculate the gravitational force each exerts on the other, and compare it with the weight of one sphere on Earth (g = 9.81 N kg⁻¹).

  • F = Gm₁m₂ / r² = (6.67 × 10⁻¹¹ × 1.5 × 1.5) / 0.12². ✓
  • F = 1.50 × 10⁻¹⁰ / 0.0144 = 1.0 × 10⁻⁸ N. ✓
  • The weight of one sphere is 1.5 × 9.81 = 15 N, about 1.4 × 10⁹ times larger, which is why gravity between everyday objects is never noticed. ✓
03
Calculation
[3 marks]

The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Calculate the gravitational field strength at the Moon's surface, and hence the weight there of an astronaut of mass 85 kg.

  • g = GM / R² = (6.67 × 10⁻¹¹ × 7.35 × 10²²) / (1.74 × 10⁶)². ✓
  • g = 4.90 × 10¹² / 3.03 × 10¹² = 1.62 N kg⁻¹. ✓
  • W = mg = 85 × 1.62 = 1.4 × 10² N, about one sixth of the 834 N the astronaut weighs on Earth. ✓
04
Analysis
[3 marks]

A uniform sphere of mass 50 kg and radius 0.40 m is placed so that it just touches a uniform sphere of mass 2.0 kg and radius 0.10 m. State the separation r that must be used in Newton's law of gravitation for this pair, explaining why, and compare the gravitational force on the small sphere with that on the large sphere.

  • r = 0.40 + 0.10 = 0.50 m, the distance between the centres, not the gap between the surfaces (which is zero). ✓
  • A uniform sphere attracts, and is attracted, as if all its mass were concentrated at its centre, so Newton's law for point masses applies with the centre-to-centre distance. ✓
  • The two forces are equal in magnitude and opposite in direction (a Newton's third law pair), despite the different masses; the small sphere simply has the larger acceleration. ✓
05
Calculation
[3 marks]

The gravitational field strength at the Earth's surface is 9.81 N kg⁻¹ and the Earth's radius is 6.4 × 10⁶ m. Calculate the height above the surface at which the field strength has fallen to 2.45 N kg⁻¹.

  • g ∝ 1 / r² for a point (or spherical) mass, so g falls to one quarter when r doubles: 2.45 / 9.81 = 1 / 4. ✓
  • The required distance from the centre is therefore r = 2R = 1.28 × 10⁷ m. ✓
  • Height above the surface h = r − R = R = 6.4 × 10⁶ m. ✓
06
Calculation
[4 marks]

Mars has mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m.

(a) Calculate the gravitational field strength at the surface of Mars. (b) Calculate the gravitational force on a lander of mass 950 kg resting on the surface. (c) Calculate the field strength at a height of 2.0 × 10⁶ m above the surface.

  • (a) g = GM / R² = (6.67 × 10⁻¹¹ × 6.42 × 10²³) / (3.39 × 10⁶)² = 4.28 × 10¹³ / 1.15 × 10¹³ = 3.73 N kg⁻¹. ✓
  • (b) F = mg = 950 × 3.73 = 3.5 × 10³ N. ✓
  • (c) Distance from the centre r = 3.39 × 10⁶ + 2.0 × 10⁶ = 5.39 × 10⁶ m (the height is added to the radius, not used on its own). ✓
  • g = GM / r² = 4.28 × 10¹³ / (5.39 × 10⁶)² = 1.47 N kg⁻¹. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.