D.1 Gravitational fields

IB Physics HL· 25 min read

1. Gravitational Fields and Field Strength★★☆☆☆⏱ 5 min

📘 Definition

Gravitational field strength

$g = \frac{F}{m}$

The gravitational force per unit mass acting on an infinitesimally small test mass placed at a point in the field. Units are $N kg^{- 1}$ .

Example:

At Earth's surface, $g \approx 9.8 N kg^{- 1}$

Gravitational fields are always attractive, unlike electric fields which can be attractive or repulsive. The direction of the gravitational field vector at any point points directly towards the source mass that creates the field.

📐 Worked Example

A 3.0 kg test mass experiences a 29.4 N gravitational force at a point near Earth's surface. What is the gravitational field strength at this point?

1
Recall the definition of gravitational field strength:
2
$g = \frac{F}{m}$
3
Substitute the given values for force and mass:
4
$g = \frac{29.4 \text{ N}}{3.0 \text{ kg}} = 9.8 \text{ N kg}^{-1}$
5
This matches the expected value of gravitational field strength at Earth's surface.

Exam tip:

Always remember $g$ is a property of the field, independent of the test mass you place in it.

2. Newton's Law of Universal Gravitation★★★☆☆⏱ 7 min

📘 Definition

Newton's Law of Universal Gravitation

$F = G \frac{M m}{r ^{2}}$

Any two point masses attract each other with a force proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.

Example:

This law accurately describes the force between the Sun and Earth, when both are treated as point masses.

$G$ is the universal gravitational constant, with value $G \approx 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$ . This is an inverse square law: force decreases as $\frac{1}{r ^{2}}$ when distance increases. For uniform spherical masses, we can treat all mass as concentrated at the center, so the law applies even for large astronomical objects.

📐 Worked Example

Calculate the gravitational force between two 10 kg point masses separated by 0.5 m.

1
Write the formula for Newton's law:
2
$F = G\frac{Mm}{r^2}$
3
Substitute the known values:
4
$F = (6.67 \times 10^{-11}) \frac{(10)(10)}{(0.5)^2}$
5
Calculate denominator and simplify:
6
$F = \frac{6.67 \times 10^{-9}}{0.25} = 2.67 \times 10^{-8} \text{ N}$
7
This very small force explains why we do not observe gravitational attraction between everyday objects.

Exam tip:

Distance $r$ is always measured from center to center of the masses, not from their surfaces.

3. Gravitational Field Strength of Spherical Masses★★★☆☆⏱ 8 min

🔬 Derivation

Goal:

Derive the gravitational field strength due to a point source mass $M$

Starting from:

Newton's law of gravitation and the definition of $g$

1
Start with force on test mass $m$ : $F = G \frac{M m}{r ^{2}}$
2
Substitute into definition $g = \frac{F}{m}$ :
3
$g = \frac{1}{m} \left(G\frac{Mm}{r^2}\right)$
4
Cancel the test mass $m$ from numerator and denominator

Result:

$g = \frac{GM}{r ^{2}}$ , directed towards the source mass $M$

This formula applies to any point outside a uniform spherical mass, just like the force law. At the surface of a planet of radius $R$ , the field strength simplifies to $g_{s} = \frac{GM}{R ^{2}}$ , which is the constant value we use for problems near the surface.

📐 Worked Example

Earth has mass $6.0 \times 1 0^{24} kg$ and radius $6.4 \times 1 0^{6} m$ . Calculate the gravitational field strength at Earth's surface.

1
Use the formula for field strength at the surface of a sphere:
2
$g = \frac{GM}{R^2}$
3
Substitute values:
4
$g = \frac{(6.67 \times 10^{-11})(6.0 \times 10^{24})}{(6.4 \times 10^6)^2}$
5
Calculate numerator and denominator:
6
$g = \frac{4.00 \times 10^{14}}{4.10 \times 10^{13}} \approx 9.8 \text{ N kg}^{-1}$

4. Gravitational vs Inertial Mass★★☆☆☆⏱ 5 min

IB exams frequently test the distinction between two different definitions of mass. While the definitions are conceptually distinct, all experimental evidence confirms their values are identical.

📘 Definition

Gravitational mass vs inertial mass

Gravitational mass measures how strongly a body interacts with a gravitational field (via gravitational attraction to other masses). Inertial mass measures how strongly a body resists acceleration when a force is applied (from Newton's second law $F = ma$ ).

Example:

Two different masses fall at the same acceleration because gravitational mass is proportional to inertial mass.

📐 Worked Example

A block pulled by 12 N force accelerates at $2 m s^{- 2}$ . What is the inertial mass of the block, and what is its gravitational mass?

1
Calculate inertial mass from Newton's second law:
2
$m_{inertial} = \frac{F}{a} = \frac{12 \text{ N}}{2 \text{ m s}^{-2}} = 6.0 \text{ kg}$
3
By the principle of equivalence, gravitational mass equals inertial mass:
4
$m_{g r a v i t a t i o na l} = 6.0 kg$

Exam tip:

For a 2 mark question, always define both terms separately before stating they are equivalent.

5. Common Pitfalls

Wrong move:

Measuring distance $r$ from the surface of a planet instead of center to center

Why:

Newton's law assumes mass is concentrated at the center of the sphere, so $r$ must include the full planet radius

Correct move:

Calculate $r$ as the sum of the planet radius and any height above the surface

Wrong move:

Confusing gravitational force $F$ with gravitational field strength $g$

Why:

$F$ depends on the mass of the test object, while $g$ is a property of the field independent of the test mass

Correct move:

Use $F = G \frac{M m}{r ^{2}}$ for force, $g = \frac{GM}{r ^{2}}$ for field strength

Wrong move:

Using $g = 9.8 N kg^{- 1}$ for all points in a gravitational field

Why:

$g$ follows an inverse square law and decreases as you move away from the planet center

Correct move:

Only use $g = 9.8 N kg^{- 1}$ for points near Earth's surface; use $g = \frac{GM}{r ^{2}}$ otherwise

Wrong move:

Claiming gravitational and inertial mass are fundamentally different values

Why:

While their definitions are distinct, all experiments confirm they are equal

Correct move:

Define the two concepts separately, then state their measured values are identical

6. Quick Reference Cheatsheet

Concept	Formula/Definition	Key Notes
Gravitational field strength	$g = \frac{F}{m}$	Force per unit test mass, $N kg^{- 1}$
Newton's gravitation law	$F = G \frac{M m}{r ^{2}}$	Always attractive, $r$ = center-to-center distance
Field from point mass $M$	$g = \frac{GM}{r ^{2}}$	Inverse square dependence on $r$
Field at planet surface	$g_{s} = \frac{GM}{R ^{2}}$	R = planet radius
Inertial mass	From $F = m_{in er t ia l} a$	Measures resistance to acceleration
Gravitational mass	Measures gravitational interaction	Equal in value to inertial mass

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.

2025 · 1
Calculate g at height above planet surface
2023 · 2
Compare g for two planets of different radii
2022 · 1
Distinguish gravitational vs inertial mass

Going deeper

syllabusIB Physics HL Theme D SyllabusOfficial IB syllabus reference

What's Next

Now that you master the basics of gravitational fields, you can extend this knowledge to gravitational potential and equipotential surfaces, which are required to calculate work done moving masses in gravitational fields and solve orbital motion problems. Gravitational fields are the first example of field theory you learn in IB Physics, and the concepts you master here will directly apply to electric fields and magnetic fields later in Theme D. Understanding gravitational fields also builds the foundation for astrophysics problems, a common optional topic for IB Physics HL.