A Level Physics 13.1 Gravitational Field MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Section 13.1, Gravitational Field, from the Cambridge International AS & A Level Physics (9702) syllabus – an A Level (A2) topic that builds on the force and field ideas from AS Level. Work through 30 original multiple-choice questions on the field concept, field lines and how the field strength g varies from place to place, each with a fully worked answer.
Formula recap: g = F / m | W = m g. The field strength g at a point is the gravitational force per unit mass at that point, measured in N/kg. For the complete list of 9702 equations, see the A Level Physics Formula Sheet.
The Gravitational Field Concept
Q1. Which statement best describes a gravitational field?
- (A) A region of space in which a mass experiences a gravitational force
- (B) A region of space in which a charged particle experiences a force
- (C) A region of space in which a compass needle experiences a force
- (D) A region of space that exists only inside a planet
Show Answer
Answer: A – A gravitational field is any region in which a mass placed there experiences a gravitational force. Charged particles respond to electric fields and compass needles to magnetic fields.
Q2. The gravitational field strength at a point is defined as
- (A) the force acting on the object that creates the field
- (B) the gravitational force acting on any mass at that point
- (C) the gravitational force acting per unit mass placed at that point
- (D) the mass contained in each cubic metre at that point
Show Answer
Answer: C – Field strength g = F / m, the force experienced by each kilogram of mass placed at the point. The force itself depends on how much mass is placed there, so the field strength must be quoted per unit mass.
Q3. At a point in a gravitational field, a mass of 2.0 kg experiences a force of 16 N. The field strength at the point is
- (A) 32 N/kg
- (B) 8.0 N/kg
- (C) 0.25 N/kg
- (D) 18 N/kg
Show Answer
Answer: B – g = F / m = 16 / 2.0 = 8.0 N/kg.
Q4. The SI unit of gravitational field strength is the
- (A) newton
- (B) joule per kilogram
- (C) kilogram per metre
- (D) newton per kilogram
Show Answer
Answer: D – g = F / m, so its unit is N/kg. This is equivalent to m/s2, which is why g is also the acceleration of free fall at that point.
Q5. The direction of the gravitational field strength vector at a point near a planet is
- (A) towards the centre of the planet
- (B) away from the centre of the planet
- (C) parallel to the surface of the planet
- (D) in the direction of the planet’s rotation
Show Answer
Answer: A – g is a vector pointing in the direction of the force on a mass at that point, and gravity always pulls a mass towards the centre of the attracting body.
Q6. A gravitational field can never push two masses apart. This is because
- (A) field lines can never cross each other
- (B) there is only one kind of mass, and the gravitational force between any two masses is always attractive
- (C) every planet rotates on its axis
- (D) field strength is a scalar quantity
Show Answer
Answer: B – Unlike electric charge, mass comes in only one “type”, so the force between two masses is always attractive. (Note that g is in fact a vector, not a scalar.)
Q7. A small test mass is used to explore a gravitational field. Which statement about the test mass is correct?
- (A) The field exists only while the test mass is present
- (B) The test mass is what creates the field
- (C) The field exists at the point whether or not a test mass is placed there; the test mass simply experiences a force that reveals the field
- (D) The test mass must have a mass of exactly 1 kg
Show Answer
Answer: C – The field is a property of the space around the attracting mass and exists independently. Placing a test mass in it produces a force F = m g, which is how we detect and measure the field.
Q8. The gravitational field strength at the surface of Mars is 3.7 N/kg. The weight of a 60 kg astronaut standing on Mars is
- (A) 588 N
- (B) 222 N
- (C) 16 N
- (D) 64 N
Show Answer
Answer: B – W = m g = 60 × 3.7 = 222 N. (On Earth the same astronaut would weigh 60 × 9.81 = 588 N.)
Q9. An astronaut travels from the Earth to the Moon. Which statement is correct?
- (A) Her mass decreases but her weight stays the same
- (B) Her weight is measured in kilograms on both bodies
- (C) Her mass and her weight are exactly the same quantity
- (D) Her mass stays the same, but her weight changes because weight = mass × local field strength, and g is different on the Moon
Show Answer
Answer: D – Mass (in kg) is the amount of matter and does not change. Weight (in N) is the gravitational force on that mass, W = m g, so it changes wherever g changes.
Q10. At a point where the gravitational field strength is 25 N/kg, the force on a mass of 4.0 kg is
- (A) 100 N
- (B) 6.3 N
- (C) 29 N
- (D) 21 N
Show Answer
Answer: A – F = m g = 4.0 × 25 = 100 N.
Representing Gravitational Fields
Q11. The gravitational field of an isolated spherical planet is represented by field lines that
- (A) form circles around the planet
- (B) point radially outwards from the planet
- (C) point radially inwards towards the centre of the planet
- (D) run parallel to the surface of the planet
Show Answer
Answer: C – A spherical mass produces a radial field, with the lines pointing inwards because the force on any mass near the planet is towards its centre.
Q12. On a field-line diagram, the field is strongest in the region where the lines are
- (A) furthest apart
- (B) completely absent
- (C) parallel to each other
- (D) closest together
Show Answer
Answer: D – The spacing of field lines indicates the strength of the field: closely spaced lines mean a strong field, widely spaced lines mean a weak field.
Q13. Over a small area near the surface of the Earth, the field is treated as uniform. Its field lines are correctly drawn as
- (A) parallel and equally spaced
- (B) converging lines aimed at the centre of the Earth
- (C) concentric circles around the Earth
- (D) lines that get closer together with height
Show Answer
Answer: A – In a uniform field, g has the same magnitude and direction everywhere, which is shown by parallel, equally spaced lines. Over a small area this is an excellent approximation for the Earth’s field.
Q14. The direction of a gravitational field line at a point gives the direction of
- (A) the motion of every object at that point
- (B) the gravitational force on a mass placed at that point
- (C) the magnetic field of the Earth
- (D) the velocity an object needs in order to escape
Show Answer
Answer: B – A field line is drawn so that at every point it points along the force that a mass placed there would experience. An object does not have to move along the line – it moves according to its total velocity and all the forces on it.
Q15. In a diagram of the Earth’s radial gravitational field, the field lines are further apart at greater heights above the surface. This shows that the field strength
- (A) increases with height
- (B) is the same at all heights
- (C) decreases with height
- (D) becomes uniform at high altitude
Show Answer
Answer: C – Wider spacing means a weaker field, so g falls as you move away from the Earth. This is the graphical picture of the inverse-square decrease studied in Section 13.3.
Q16. Gravitational field lines point towards a mass, while the field lines of a positive electric charge point away from the charge. The reason for this difference is that
- (A) the force on a mass in a gravitational field is always attractive, so a test mass is pulled towards the mass creating the field
- (B) gravitational field lines are always curved
- (C) masses are always heavier than charges
- (D) electric field lines are never straight
Show Answer
Answer: A – Field lines show the direction of the force on a test object. A test mass is always attracted towards the source mass, so gravitational lines point inwards; a positive test charge is repelled by a positive charge, so its lines point outwards.
Q17. Between the Earth and the Moon there is a point where the resultant gravitational field strength is zero. This null point is located
- (A) exactly halfway between the Earth and the Moon
- (B) closer to the Earth, because the Earth has the larger mass
- (C) at the centre of the Moon
- (D) closer to the Moon, because the Moon has the smaller mass, so its weaker field balances the Earth’s field only at a point nearer to the Moon
Show Answer
Answer: D – At the null point the two fields are equal and opposite. The Moon’s field is weaker, so the balance point must be much nearer the Moon, where its field is still strong enough to match the Earth’s field at that distance.
Q18. A small mass placed exactly at the null point between two planets experiences
- (A) equal gravitational forces pulling it in the same direction
- (B) no resultant gravitational force, because the two fields cancel
- (C) a force that switches off the gravity of both planets
- (D) no mass
Show Answer
Answer: B – The pulls from the two planets are equal in size and opposite in direction, so they cancel and the resultant field strength (and force) is zero. The individual fields are still present.
Q19. Which comparison of gravitational and electric fields is correct?
- (A) Both gravitational and electric forces can only be attractive
- (B) Electric field lines point towards a positive charge
- (C) The field around a spherical object is radial in both cases; the gravitational lines point inwards, while the electric lines of a positive charge point outwards
- (D) Gravitational field lines point away from a mass
Show Answer
Answer: C – The patterns are geometrically similar (radial), but the directions differ because gravity only attracts, whereas a positive charge repels a positive test charge. Electric forces can be attractive or repulsive depending on the signs of the charges.
Q20. In a uniform gravitational field,
- (A) every object has the same weight
- (B) every mass has the same potential energy
- (C) the field lines converge on the centre of the Earth
- (D) the field strength has the same magnitude and the same direction at every point
Show Answer
Answer: D – “Uniform” means g is a constant vector throughout the region. Objects still have different weights because W = m g depends on each object’s mass.
Gravitational Field Strength in Practice
Q21. The gravitational field strength at the surface of the Earth is approximately
- (A) 1.6 N/kg
- (B) 9.81 N/kg
- (C) 3.7 N/kg
- (D) 25 N/kg
Show Answer
Answer: B – g = 9.81 N/kg at the Earth’s surface. For comparison, 1.6 N/kg is the Moon’s surface value, 3.7 N/kg is Mars, and about 25 N/kg is Jupiter.
Q22. An astronaut in a space station a few hundred kilometres above the Earth finds that the field strength there is
- (A) greater than at the surface, because she is closer to the Sun
- (B) exactly 9.81 N/kg, the same as everywhere
- (C) zero, because there is no air above the atmosphere
- (D) less than 9.81 N/kg, because she is further from the centre of the Earth
Show Answer
Answer: D – g decreases with distance from the centre of the Earth, so at orbital height it is noticeably less than 9.81 N/kg (about 8.7 N/kg at space-station height). Gravity certainly has not fallen to zero – it is what keeps the station in orbit.
Q23. A physicist travels deep below the surface of the Earth in a mine shaft. As she goes deeper, the field strength she measures
- (A) decreases, falling to zero at the centre of the Earth
- (B) increases steadily as she approaches the centre
- (C) is zero everywhere below the surface
- (D) remains 9.81 N/kg all the way to the centre
Show Answer
Answer: A – Below the surface, only the mass lying deeper than her position (the sphere between her and the centre) pulls her towards the centre, and that mass shrinks as she descends. At the centre the pulls from all sides cancel, so g = 0.
Q24. Planet X has twice the mass of the Earth but exactly the same radius. The field strength at its surface is
- (A) the same as on Earth
- (B) one half of the Earth value
- (C) twice the Earth value
- (D) four times the Earth value
Show Answer
Answer: C – At a fixed distance from the centre, the field strength is proportional to the mass creating the field. Doubling the mass at the same radius doubles g. The full relationship, g = G M / r2, is developed in Sections 13.2 and 13.3.
Q25. Planet Y has the same mass as the Earth but half the radius. The field strength at its surface is
- (A) four times the Earth value
- (B) twice the Earth value
- (C) the same as the Earth value
- (D) one quarter of the Earth value
Show Answer
Answer: A – The surface is now only half as far from the centre, and the field follows an inverse-square pattern with distance: halving the distance multiplies the field strength by 22 = 4.
Q26. The field strength at the poles of the Earth is slightly greater than at the equator. This is because
- (A) objects at the equator are thrown outwards so strongly that g nearly doubles there
- (B) at the equator, part of the gravitational force is needed to provide the centripetal force for the Earth’s rotation, reducing the measured g, and the equator is also slightly further from the centre of the Earth
- (C) g is exactly the same at every point on the surface of the Earth
- (D) the poles are closer to the Sun than the equator
Show Answer
Answer: B – The Earth is slightly flattened, so the poles are closer to the centre (stronger field), and at the equator some of the gravitational force is used up keeping objects moving in a circle as the Earth spins, which reduces the measured value of g there.
Q27. The field strength at the surface of Mars is 3.7 N/kg. A rock sample of mass 10 kg has a weight on Mars of
- (A) 98 N
- (B) 3.7 N
- (C) 370 N
- (D) 37 N
Show Answer
Answer: D – W = m g = 10 × 3.7 = 37 N. (98 N would be the answer if the Earth value 9.81 N/kg were used by mistake.)
Q28. An astronaut has a mass of 70 kg. On the Moon, where the field strength is 1.6 N/kg, the correct pair of values for the astronaut is
- (A) mass = 11 kg, weight = 112 N
- (B) mass = 70 kg, weight = 687 N
- (C) mass = 70 kg, weight = 112 N
- (D) mass = 0 kg, weight = 112 N
Show Answer
Answer: C – The mass is unchanged at 70 kg. The weight is W = m g = 70 × 1.6 = 112 N, much less than the 687 N she weighs on Earth.
Q29. A satellite is at a height above the surface of the Earth equal to one Earth radius, so it is twice as far from the centre of the Earth as a point on the surface. The field strength at the satellite is approximately
- (A) 2.5 N/kg
- (B) 4.9 N/kg
- (C) 9.81 N/kg
- (D) 0 N/kg
Show Answer
Answer: A – The field falls with the inverse square of the distance. Doubling the distance makes g one quarter of the surface value: 9.81 / 4 ≈ 2.5 N/kg. (Dividing by 2 to get 4.9 N/kg is the common mistake.)
Q30. At a point three Earth radii from the centre of the Earth, the field strength is approximately
- (A) 3.3 N/kg
- (B) 1.1 N/kg
- (C) 4.9 N/kg
- (D) 0.33 N/kg
Show Answer
Answer: B – Tripling the distance divides the field strength by 32 = 9: g = 9.81 / 9 ≈ 1.1 N/kg. The value 3.3 N/kg comes from dividing by 3 instead of by 9.
Keep practising: find more board-wise physics questions, organised by syllabus section, on the Physics MCQs hub, and keep the A Level Physics Formula Sheet open beside you while you revise Sections 13.2 to 13.4.
These are original practice questions written for revision purposes only. They are not Cambridge past-paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.