What is the gravitational field of a point mass?

The gravitational field of a point mass is a region of space around the mass where another mass experiences a force of attraction. It is described by the gravitational field strength, which is the force per unit mass experienced by a small test mass placed in the field. The field strength is given by the formula g = G * M / r^2, where G is the gravitational constant, M is the mass of the point, and r is the distance from the point mass.

How do you calculate the gravitational field strength due to a point mass?

To calculate the gravitational field strength due to a point mass, use the formula g = G * M / r^2. Here, G is the universal gravitational constant (6.674 × 10^-11 N m²/kg²), M is the mass of the point mass, and r is the distance from the point mass to the location where you want to calculate the field strength. This formula shows that the field strength decreases with the square of the distance from the mass.

Why does the gravitational field strength decrease with distance from a point mass?

The gravitational field strength decreases with distance from a point mass because the field spreads out over a larger area as you move away from the mass. According to the inverse square law, the field strength is inversely proportional to the square of the distance from the mass. This means that if you double the distance, the field strength becomes one-fourth as strong.

What is the significance of the gravitational constant (G) in the gravitational field equation?

The gravitational constant (G) is a fundamental constant in physics that quantifies the strength of the gravitational force. In the equation for gravitational field strength, g = G * M / r^2, G ensures that the units are consistent and provides the proportionality factor needed to calculate the force between two masses. Its value is approximately 6.674 × 10^-11 N m²/kg², reflecting the relatively weak nature of gravitational interactions compared to other fundamental forces.

How does the concept of a gravitational field help in understanding gravitational interactions?

The concept of a gravitational field helps in understanding gravitational interactions by providing a way to describe how a mass influences the space around it. Instead of considering the force between two masses directly, the field concept allows us to consider how a mass creates a field that affects other masses. This approach simplifies the analysis of complex systems with multiple interacting masses and is fundamental in fields like astrophysics and cosmology, which are part of the Cambridge International A Levels Physics curriculum.

Why is "Measuring $r$ from Earth's surface" a common mistake in Gravitational field of a point mass?

Why it happens: Confusing radius and altitude. How to avoid it: r in gravity formulas is from CENTRE of source mass. Add Earth's radius to altitude. Source: 9702 Examiner Reports 2022-2024

Gravitational FieldsCambridge International A Levels Physics (9702)

Gravitational field of a point mass

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Gravitational Field of Point Mass Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

$g = GM/r^2$ . Orbit speed and period. Geostationary satellites.

What you’ll learn

Mapped to the Cambridge IGCSE 9702 syllabus (2025-2027).

13.3.1 — Use $g = GM/r^2$ .
13.3.2 — Apply to satellite orbits.
13.3.3 — Use Kepler's 3rd law.

$g$ from source mass

$GM/r^2$ .

Formula. $g = GM/r^2$ .

Earth surface. $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m: $g \approx 9.81$ N/kg.

Altitude $h$ . $r = R_E + h$ in the formula. $g$ decreases with height.

Cambridge tip. Don't forget Earth's radius — common error.

$g = GM/r^2$ .
$r$ from CENTRE.
$g$ decreases with $h$ .

See the full worked example for gravitational field of a point mass →

Satellite orbits

Gravity = centripetal.

Circular orbit. Gravitational force = centripetal: $\frac{GMm}{r^2} = \frac{mv^2}{r} \Rightarrow v = \sqrt{\frac{GM}{r}}$

Period. $T = 2\pi r/v$ . Combined with above: $T^2 = \frac{4\pi^2 r^3}{GM}$

This is Kepler's 3rd law: $T^2 \propto r^3$ .

Geostationary. Orbit period = Earth's rotation period (24 h). Solving gives $r \approx 4.2 \times 10^7$ m. Must orbit above equator.

Polar orbit (low altitude). Period ~90 min.

Cambridge tip. Geostationary altitude above SURFACE ≈ 36,000 km (subtract Earth's radius).

Gravity = centripetal.
$v = \sqrt{GM/r}$ .
$T^2 \propto r^3$ .

See the full worked example for gravitational field of a point mass →

How it’s examined

Orbits central to A2. Most-tested: orbit speed (6 marks), Kepler's 3rd (8 marks), geostationary (8 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/42 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — Gravitational field of a point mass

Step-by-step solutions to past-paper-style questions on gravitational field of a point mass, written exactly the way a tutor would explain them at the board.

1 $g$ at Earth surface (5 marks)
Extended• Adapted from 9702/42 May/Jun 2024• g
Question
Find $g$ at Earth's surface. $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m. (5 marks)
Step-by-step solution
1. Step 1
  $g = GM/r^2$ .
  $g = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.37 \times 10^6)^2} \approx 9.81 \text{ N/kg}$
Answer
$g \approx 9.81$ N/kg.
2Satellite speed (6 marks)
Extended• orbit
Question
A satellite orbits Earth at radius $r$ from centre. Find orbital speed in terms of $G$ , $M$ , $r$ . (6 marks)
Step-by-step solution
1. Step 1
  Gravity provides centripetal force. $GMm/r^2 = mv^2/r$ .
2. Step 2
  Solve.
  $v = \sqrt{\dfrac{GM}{r}}$
Answer
$v = \sqrt{GM/r}$ .
3Geostationary orbit (8 marks)
Extended• geostationary
Question
Find the radius of a geostationary orbit. $G = 6.67 \times 10^{-11}$ , $M_E = 5.97 \times 10^{24}$ kg, $T = 24$ h. (8 marks)
Step-by-step solution
1. Step 1
  Kepler's 3rd law. $T^2 = \frac{4\pi^2 r^3}{GM}$ .
2. Step 2
  $T = 24 \times 3600 = 86400$ s.
3. Step 3
  Solve for $r$ .
  $r = \left(\dfrac{GMT^2}{4\pi^2}\right)^{1/3} \approx 4.22 \times 10^7 \text{ m}$
Answer
$r \approx 4.22 \times 10^7$ m (~42,000 km from centre).

Key Formulae — Gravitational field of a point mass

The formulae you need to memorise for gravitational field of a point mass on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

$g$ from source mass
$g = \dfrac{GM}{r^2}$
When to use
Field strength at distance $r$ from point mass (or spherical) $M$ .
Orbital speed
$v = \sqrt{\dfrac{GM}{r}}$
When to use
Circular orbit speed at radius $r$ .
Kepler's 3rd law
$T^2 = \dfrac{4\pi^2 r^3}{GM}$
When to use
Period of orbit at radius $r$ .

Key Definitions and Keywords — Gravitational field of a point mass

Definitions to memorise and the exact keywords mark schemes credit for gravitational field of a point mass answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Orbit
Examiner keyword
Closed path of one body around another under gravity. Often circular or elliptical.
Geostationary orbit
Examiner keyword
Orbit above equator with period 24 h, so satellite appears stationary above ground. Radius ~42,000 km from centre.

Common Mistakes and Misconceptions — Gravitational field of a point mass

The traps other students keep falling into on gravitational field of a point mass questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Measuring $r$ from Earth's surface
9702 Examiner Reports 2022-2024
Why it happens
Confusing radius and altitude.
How to avoid it
$r$ in gravity formulas is from CENTRE of source mass. Add Earth's radius to altitude.

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