# Gravitational potential

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

This subtopic explains gravitational potential, its key properties, its relation to work done and gravitational field strength, and how to use it to solve common exam problems about moving masses in gravitational fields.

**Prerequisites:** [Newton's law of gravitation](https://www.owlsprep.com/study/cie-9702-u17-gravitational-field-and-point-masses/); [Gravitational field strength](https://www.owlsprep.com/study/cie-9702-u17-gravitational-field-strength/)

## Learning objectives

- Define gravitational potential and gravitational potential difference
- Calculate gravitational potential for a point mass
- Relate gravitational potential to gravitational field strength
- Calculate work done to move masses between two points in a gravitational field

## Definition and Key Properties of Gravitational Potential

**Gravitational Potential** — The work done per unit mass to bring a small test mass from infinity (the reference point of zero potential) to the point in question. It is a scalar quantity, not a vector.

*Notation:* V

*Example:* For a point mass $M$, gravitational potential at distance $r$ from the center of mass is given by:

$$V = -\frac{GM}{r}$$

All gravitational potentials due to a mass are negative because gravity is always attractive. Zero potential is only achieved at infinite distance from any mass, where gravity has no effect.

**Worked example:** Calculate the gravitational potential at the surface of the Earth. Given $M = 5.97 \times 10^{24}$ kg, $R = 6.37 \times 10^6$ m, $G = 6.67 \times 10^{-11}$ N m² kg⁻².

1. Recall the formula for potential due to a point mass:
2. $$V = -\frac{GM}{r}$$
3. Substitute the given values:
4. $$V = -\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^6}$$
5. Calculate the final result:
6. $$V \approx -6.25 \times 10^7 \text{ J kg}^{-1}$$

> **Exam tip:** Always remember the negative sign in the potential formula. Losing the sign is the most common mistake in CIE exams for this topic.

## Potential Difference and Work Done

Gravity is a conservative force, meaning work done moving a mass between two points depends only on the potential difference between the points, not the path taken.

**Gravitational Potential Difference** — The difference between final and initial potential: $\Delta V = V_{final} - V_{initial}$. The total work done to move a mass $m$ between two points is $W = m\Delta V$.

*Notation:* \Delta V

*Example:* Moving a mass away from a planet takes it from a more negative potential to a less negative potential, so $\Delta V$ is positive, meaning work must be done on the mass against gravity.

**Worked example:** Calculate the work done to move a 1500 kg satellite from Earth's surface to an infinite distance away. Use $V_{surface} = -6.25 \times 10^7$ J kg⁻¹.

1. Potential at infinity is defined as $V_{\infty} = 0$, so calculate the potential difference:
2. $$\Delta V = V_{\infty} - V_{surface} = 0 - (-6.25 \times 10^7) = 6.25 \times 10^7 \text{ J kg}^{-1}$$
3. Calculate total work done using $W = m\Delta V$:
4. $$W = 1500 \times 6.25 \times 10^7 = 9.38 \times 10^{10} \text{ J}$$

> **info**
>
> This work is exactly the minimum energy needed for the satellite to escape Earth's gravity entirely.

## Relation Between Potential and Gravitational Field Strength

Gravitational field strength is the negative rate of change (gradient) of gravitational potential with distance. This relation holds for all gravitational fields, both uniform and radial.

$$g = -\frac{dV}{dr}$$

The negative sign indicates that gravitational field strength always points in the direction of decreasing potential, towards the mass creating the field. We can verify this for a radial point mass field:

$$V = -\frac{GM}{r} \implies \frac{dV}{dr} = \frac{GM}{r^2} \implies g = -\frac{GM}{r^2}$$

This matches the standard formula for gravitational field strength of a point mass, confirming the relation.

**Worked example:** The gravitational potential at distance $r$ from the center of a non-uniform planet is given by $V = -\frac{GM}{r^2}$ for $r > R$ (planet radius). Find an expression for gravitational field strength at distance $r$.

1. Use the relation $g = -\frac{dV}{dr}$. First differentiate $V$ with respect to $r$:
2. $$\frac{dV}{dr} = \frac{d}{dr}(-GM r^{-2}) = 2GM r^{-3} = \frac{2GM}{r^3}$$
3. Multiply by $-1$ to get $g$:
4. $$g = -\frac{dV}{dr} = -\frac{2GM}{r^3}$$
5. The negative sign confirms that field points towards the planet's center, where potential is lower.

> **Exam tip:** If you are given a graph of $V$ against $r$, the gravitational field strength at any point is equal to the negative gradient of the graph at that point.

## Common pitfalls

- **Wrong:** Forgetting the negative sign in gravitational potential values or formulas
  - Why it fails: Potential is defined as work done on the test mass, and attraction means this is always negative for finite distances from a mass
  - Correct: Always include the negative sign when writing potential, unless working with potential differences only
- **Wrong:** Calculating potential difference as $V_{initial} - V_{final}$ instead of $V_{final} - V_{initial}$
  - Why it fails: This flips the sign of work done, leading to wrong answers for energy requirements
  - Correct: Always use $\Delta V = V_{final} - V_{initial}$, then $W = m\Delta V$
- **Wrong:** Treating gravitational potential as a vector quantity
  - Why it fails: Unlike gravitational field strength, potential is scalar, so vector resolution is not needed
  - Correct: Add potentials from multiple masses algebraically (including their negative signs) directly
- **Wrong:** Assuming zero potential is at the Earth's surface
  - Why it fails: All CIE questions use the standard definition of zero potential at infinity unless explicitly stated otherwise
  - Correct: Always take $V = 0$ at infinity, so potential at Earth's surface is negative

## Cheatsheet

| Concept | Formula | Key Note |
| --- | --- | --- |
| Gravitational potential (point mass) | $V = -\frac{GM}{r}$ | Scalar, always negative, units J kg⁻¹ |
| Potential difference (A to B) | $\Delta V = V_B - V_A$ | Work done per unit mass moving from A to B |
| Work done moving mass m | $W = m\Delta V$ | Path independent for conservative gravity |
| Relation of g to V | $g = -\frac{dV}{dr}$ | g = negative gradient of potential vs distance |
| Reference potential | $V = 0$ at infinity | Standard definition for all CIE questions |

## What's next

Gravitational potential is a core scalar concept that underpins all further study of gravitational fields, including gravitational potential energy, satellite orbits, and escape velocity. Mastery of potential is essential for solving common CIE exam problems that ask for work done to move masses between orbits, or for calculating the total potential in systems with multiple masses like binary stars. Unlike vector field strength, potential simplifies calculations of work done because it adds algebraically, with no need to resolve vector components. This topic directly leads to the study of orbital motion and escape speed, which are heavily weighted in CIE A-level Physics exams.

- [Orbital Motion](https://www.owlsprep.com/study/cie-9702-u17-orbital-motion/)
- [Oscillations](https://www.owlsprep.com/study/cie-9702-u18-overview/)
- [Simple Harmonic Motion](https://www.owlsprep.com/study/cie-9702-u18-simple-harmonic-motion/)

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