Gravitational field strength

CIE A-Level Physics· Unit 17: Gravitational fields· 25 min read

1. Definition of gravitational field strength★★☆☆☆⏱ 8 min

📘 Definition

Gravitational field strength

g

The gravitational force per unit mass acting on an infinitesimally small test mass placed at that point in the field.

Example:

At Earth's surface, $g \approx 9.81$ N kg⁻¹.

Gravitational field strength is a vector quantity: its direction is always towards the mass that creates the field, matching the direction of the attractive gravitational force. Rearranging the definition gives the gravitational force on mass $m$ at a point where field strength is $g$ : $F = m g$ .

📐 Worked Example

A 3.0 kg test mass experiences a gravitational force of 10.5 N at a point above the Moon's surface. Calculate the gravitational field strength at this point.

1
Start with the definition of gravitational field strength:
2
$g = \frac{F}{m}$
3
Substitute the given values for force $F = 10.5$ N and mass $m = 3.0$ kg:
4
$g = \frac{10.5}{3.0} = 3.5 \text{ N kg}^{-1}$
5
The direction of $g$ is towards the centre of the Moon.

2. Gravitational field strength in radial fields★★★☆☆⏱ 10 min

A point mass or uniform spherical mass (like a planet) produces a radial gravitational field, where field strength depends on distance from the centre of the mass. We can derive the formula for $g$ directly from Newton's law of universal gravitation:

🔬 Derivation

Goal:

Derive $g = \frac{GM}{r ^{2}}$ for a radial field

Starting from:

Newton's law: force between mass $M$ (source) and test mass $m$ at distance $r$ from M's centre is $F = \frac{GM m}{r ^{2}}$

1
By definition, gravitational field strength is force per unit test mass:
2
$g = \frac{F}{m} = \frac{1}{m} \left( \frac{GMm}{r^2} \right)$
3
The test mass $m$ cancels out from numerator and denominator, leaving:

Result:

g = \frac{GM}{r^2} where $r \geq R$ (R = radius of the spherical source mass).

📐 Worked Example

Calculate g at a height of 1000 km above Earth's surface. Earth's mass = $5.97 \times 1 0^{24}$ kg, radius = 6370 km, $G = 6.67 \times 1 0^{- 11}$ N m² kg⁻².

1
Calculate total distance from Earth's centre (remember r is not height above surface):
2
$r = R_{\text{Earth}} + h = 6370 + 1000 = 7370 \text{ km} = 7.37 \times 10^6 \text{ m}$
3
Substitute into the radial field formula:
4
$g = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(7.37 \times 10^6)^2} \approx 7.3 \text{ N kg}^{-1}$

3. Uniform gravitational fields★★☆☆☆⏱ 7 min

📘 Definition

Uniform gravitational field

A region of space where gravitational field strength has the same magnitude and direction at all points.

Example:

The gravitational field close to the surface of a large planet is approximately uniform.

When we are close to the surface of a large planet like Earth, the change in $r$ (distance from Earth's centre) is tiny compared to $r$ itself. This means $g = \frac{GM}{r ^{2}}$ is approximately constant, and the radial field lines are approximately parallel and equally spaced, forming a uniform field.

📐 Worked Example

A 75 kg astronaut stands on the surface of Mars, where $g = 3.7$ N kg⁻¹. Calculate the astronaut's weight.

1
Weight is gravitational force, so use $F = m g$ :
2
$F = 75 \times 3.7 = 277.5 \approx 280 \text{ N}$
3
The astronaut's weight is ~280 N, compared to ~740 N on Earth.

4. Common Pitfalls

Wrong move:

Using height above the planet's surface as $r$ in $g = \frac{GM}{r ^{2}}$

Why:

$r$ is defined as the distance from the centre of the mass creating the field, not the surface

Correct move:

Always add the planet's radius to the height above the surface to get the correct value of $r$

Wrong move:

Treating g as a scalar when adding field strengths from two masses

Why:

g is a vector, so direction must be accounted for when combining

Correct move:

Draw a vector diagram and add components, subtract magnitudes if field strengths point in opposite directions

Wrong move:

Using only m s⁻² as units for gravitational field strength

Why:

CIE examiners expect you to use the definition-based unit N kg⁻¹ for field strength

Correct move:

Use N kg⁻¹ when answering questions about gravitational field strength, even though it is equivalent to m s⁻²

Wrong move:

Applying $g = \frac{GM}{r ^{2}}$ for points inside a planet

Why:

The inverse square law only applies to points outside the mass creating the field

Correct move:

For uniform density planets, g decreases linearly from the surface to the centre, and is zero at the centre

5. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Definition of $g$	$g = \frac{F}{m}$	Force per unit test mass, vector towards source
Radial field $g$	$g = \frac{GM}{r ^{2}}$	$r$ = distance from centre of $M$ , for $r \geq R$
Inverse square law	$g \propto \frac{1}{r ^{2}}$	Doubling $r$ quarters $g$
Uniform field $g$	$g = constant$	Near planet surface, ~9.81 N kg⁻¹ on Earth
Units	$1 N kg^{- 1} \equiv 1 m s^{- 2}$	N kg⁻¹ is definition-based unit for field strength

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 2
Calculate g at height above Earth
2022 · 1
Compare g on two planets
2021 · 2
Derive g = GM/r²

Going deeper

syllabusCIE 9702 SyllabusOfficial syllabus reference

What's Next

Gravitational field strength is the core foundation for all further topics in gravitational fields. Understanding how g varies with distance allows you to calculate gravitational potential and potential energy, which are used to solve problems involving satellite orbits, planetary motion, and escape velocity. These topics are heavily weighted in CIE A-Level Physics exams, so mastering field strength first is critical for exam success. Build on your knowledge with the following sub-topics: