# Gravitational field strength

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u17-gravitational-field-strength/

This sub-topic covers the definition of gravitational field strength, its calculation in radial and uniform fields, and its relation to gravitational force. It is a core foundation for orbital motion and gravitational potential topics.

**Prerequisites:** [Newton's law of universal gravitation](https://www.owlsprep.com/study/cie-9702-u17-newtons-law-gravitation/); [Force, mass and acceleration](https://www.owlsprep.com/study/cie-9702-u02-forces-acceleration/)

## Learning objectives

- Define gravitational field strength and state its units
- Calculate gravitational field strength in uniform and radial fields
- Derive the expression for g in a radial field from Newton's law of gravitation
- Explain the equivalence of units N kg⁻¹ and m s⁻² for g

## Definition of gravitational field strength

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

*Notation:* g

*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 $g$: $F = mg$.

**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.

## Gravitational field strength in radial fields

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:** 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 $F = \frac{GMm}{r^2}$

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:

*Conclusion:* 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 = $5.97 \times 10^{24}$ kg, radius = 6370 km, $G = 6.67 \times 10^{-11}$ 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. $$g = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(7.37 \times 10^6)^2} \approx 7.3 \text{ N kg}^{-1}$$

> **tip**
>
> In a radial field, $g \propto \frac{1}{r^2}$. If distance from the centre doubles, g decreases to one quarter of its original value.

## Uniform gravitational fields

**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.

> **info**
>
> The acceleration of free fall in a uniform gravitational field is equal to g. This proves that the units N kg⁻¹ and m s⁻² are equivalent: $1$ N kg⁻¹ = $1$ (kg m s⁻²) kg⁻¹ = $1$ m s⁻².

**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 $F = mg$:
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.

## Common pitfalls

- **Wrong:** Using height above the planet's surface as $r$ in $g = \frac{GM}{r^2}$
  - Why it fails: $r$ is defined as the distance from the centre of the mass creating the field, not the surface
  - Correct: Always add the planet's radius to the height above the surface to get the correct value of $r$
- **Wrong:** Treating g as a scalar when adding field strengths from two masses
  - Why it fails: g is a vector, so direction must be accounted for when combining
  - Correct: Draw a vector diagram and add components, subtract magnitudes if field strengths point in opposite directions
- **Wrong:** Using only m s⁻² as units for gravitational field strength
  - Why it fails: CIE examiners expect you to use the definition-based unit N kg⁻¹ for field strength
  - Correct: Use N kg⁻¹ when answering questions about gravitational field strength, even though it is equivalent to m s⁻²
- **Wrong:** Applying $g = \frac{GM}{r^2}$ for points inside a planet
  - Why it fails: The inverse square law only applies to points outside the mass creating the field
  - Correct: For uniform density planets, g decreases linearly from the surface to the centre, and is zero at the centre

## 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 $g$ | $g = \text{constant}$ | Near planet surface, ~9.81 N kg⁻¹ on Earth |
| Units | $1 \text{ N kg}^{-1} \equiv 1 \text{ m s}^{-2}$ | N kg⁻¹ is definition-based unit for field strength |

## 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:

- [Gravitational potential](https://www.owlsprep.com/study/cie-9702-u17-gravitational-potential/)
- [Orbital motion of satellites](https://www.owlsprep.com/study/cie-9702-u17-orbital-motion/)
- [Oscillations](https://www.owlsprep.com/study/cie-9702-u18-overview/)

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