Gravitational Potential Fundamentals Revision Notes | A Level AQA Physics

1. Why Gravitational Potential Matters

You already know that masses attract each other and that moving against gravity requires energy. But how do physicists quantify the energy stored at every point in a gravitational field — not just near Earth's surface? That question is exactly what gravitational potential answers.

At A-level, gravitational field strength $g$ tells you the force per unit mass at a point. Gravitational potential goes one step further: it tells you the energy per unit mass at that point, relative to a universal reference. Understanding this unlocks calculations about orbits, satellite energy, and escape from any gravitational field.

2. Gravitational Potential — Definition and the Zero at Infinity

Picture a mass sitting on the surface of a planet. To remove it completely — to take it so far away that the planet's gravity has no further effect — you must do work against the gravitational attraction. The reference point physicists choose for zero potential is infinity: the point where the gravitational field has fallen to zero and no further work is needed.

Because you must do work on the field to move a mass away to infinity (the field pulls the mass back), the potential at every point in a gravitational field is negative. At infinity, no work has been done, so $V = 0$. Closer to the mass creating the field, the potential becomes more negative — you would need to supply more energy to escape.

For a spherical mass $M$, the gravitational potential at distance $r$ from its centre is $V = -\frac{GM}{r}$, where $G = 6.67 \times 10^{-11}$ N m² kg⁻². Notice the negative sign is built into the formula — it is not optional. As $r \to \infty$, $V \to 0$, confirming the reference convention.

The graph above shows the key features: $V$ is always negative, it becomes less negative (increases towards zero) as distance increases, and it approaches zero asymptotically at infinity. The curve is not linear — it follows a $1/r$ relationship.

3. Gravitational Potential Difference

Knowing the potential at a single point is useful, but most exam questions ask about moving a mass between two points. That is where gravitational potential difference comes in. Think of it like the difference in height between two floors of a building — what matters for energy calculations is the change, not the absolute height above sea level.

When a mass moves from the surface of a planet (where $V$ is large and negative) outward to a higher orbit (where $V$ is smaller in magnitude and less negative), $\Delta V$ is positive. The mass has moved to a region of higher potential. This is directly analogous to lifting an object to a greater height — energy must be supplied.

4. Work Done Moving a Mass: $\Delta W = m\Delta V$

Gravitational potential difference tells you the energy change per unit mass. To find the actual work done when moving a real mass $m$, you scale up by $m$. This gives the core equation for this topic.

Notice that $\Delta W = m\Delta V$ is dimensionally consistent: kg × J kg⁻¹ = J. The sign of $\Delta W$ carries physical meaning. If $\Delta V$ is positive (moving to higher potential, further from the source mass), $\Delta W$ is positive — you must supply energy. If $\Delta V$ is negative (falling towards the source mass), $\Delta W$ is negative — the field does work on the mass, releasing energy.

Worked example: Calculating Work Done Moving a Satellite

A satellite of mass $800$ kg moves from a point where the gravitational potential is $-6.4 \times 10^7$ J kg⁻¹ to a point where the gravitational potential is $-4.8 \times 10^7$ J kg⁻¹. Calculate the work done on the satellite. [3 marks]

1. $m = 800$ kg, $V_A = -6.4 \times 10^7$ J kg⁻¹, $V_B = -4.8 \times 10^7$ J kg⁻¹ — Always list what you are given before substituting. Check units are consistent — both potentials are in J kg⁻¹. [M1]
2. $\Delta V = V_B - V_A = (-4.8 \times 10^7) - (-6.4 \times 10^7)$ $\Delta V = -4.8 \times 10^7 + 6.4 \times 10^7 = +1.6 \times 10^7$ J kg⁻¹ — Subtracting a negative number adds it. The positive result confirms the satellite moves to higher (less negative) potential. [M1]
3. $\Delta W = 800 \times 1.6 \times 10^7$ $\Delta W = 800 \times 1.6 \times 10^7$ $800 \times 1.6 = 1280$ $\Delta W = 1280 \times 10^7 = 1.28 \times 10^{10}$ J — Multiply mass by potential difference. Carry the power of ten separately to avoid arithmetic errors. [A1]
4. $\Delta W = +1.28 \times 10^{10}$ J — Positive work done means energy must be supplied to the satellite to move it to the higher potential. This is consistent with moving away from the planet against gravity.

Answer: $\Delta W = +1.28 \times 10^{10}$ J (work done on the satellite)

5. Equipotential Surfaces

Now that you can calculate work done using potential difference, a natural question arises: what happens when $\Delta V = 0$? That is the key to understanding equipotential surfaces.

Picture the gravitational field around Earth. The equipotential surfaces are spherical shells at different distances from Earth's centre. Every point on a shell at radius $r$ has the same value of $V = -GM/r$. If you move a satellite along one of these shells — keeping $r$ constant — the potential does not change, so $\Delta V = 0$ and therefore $\Delta W = m\Delta V = 0$.

Equipotential surfaces are always drawn perpendicular to gravitational field lines. This is not a coincidence — field lines point in the direction of steepest potential change (the direction in which potential decreases most rapidly). A surface perpendicular to that direction has no component of potential change along it, which is precisely why no work is done moving along it.

• For a uniform gravitational field (e.g. near Earth's surface over small distances), equipotential surfaces are parallel horizontal planes.
• For a radial field (e.g. around a planet), equipotential surfaces are concentric spheres.
• Equipotential surfaces are always perpendicular to gravitational field lines.
• Closely spaced equipotential surfaces indicate a stronger field (larger potential gradient).
• No work is done moving along an equipotential surface, regardless of the path taken.

6. Visualising Equipotentials and Field Lines Together

A diagram showing both field lines and equipotential surfaces together is one of the most common features of A-level exam questions on this topic. The two sets of lines are always perpendicular — this is a reliable way to check whether a diagram is physically correct.

In the diagram, the concentric circles represent equipotential surfaces. The radial lines are gravitational field lines pointing towards the central mass $M$. Every field line crosses every equipotential at a right angle. The inner equipotential $V_1$ is more negative than the outer $V_2$, reflecting the $-GM/r$ dependence — closer to the mass means lower (more negative) potential.

For a uniform field — such as the approximation used near Earth's surface over small vertical distances — the field lines are parallel and vertical, and the equipotential surfaces are horizontal planes. A ball rolling horizontally across a flat surface moves along an equipotential: no work is done by gravity, only by any horizontal force applied. This is the everyday version of the same principle.

7. Exam Technique and Common Mistakes

The concepts in this topic are tightly linked, and exam questions often test whether you can move fluently between them. The most reliable approach is to identify what quantity the question is asking for, write the relevant equation, and pay careful attention to signs throughout.

Exam questions on this topic frequently use the command word 'explain'. A full-mark explanation must include: the relevant equation, the condition that makes a term zero or non-zero, and the physical consequence. For equipotential surfaces, that chain is: same potential at all points $\Rightarrow$ $\Delta V = 0$ $\Rightarrow$ $\Delta W = m\Delta V = 0$ $\Rightarrow$ no work done.

Taken together, the five specification requirements for this topic form a coherent picture: gravitational potential quantifies energy per unit mass at any point in a field (zero at infinity, negative everywhere else); potential difference drives energy transfer via $\Delta W = m\Delta V$; and equipotential surfaces are the geometric consequence of points sharing the same potential, always perpendicular to field lines, always requiring zero work to traverse.