Gravitational Potential in Radial Fields Revision Notes | A Level AQA Physics

1. Why Gravitational Potential Matters

You already know that gravitational field strength $g$ tells you the force per unit mass at a point. But force is a vector — it has direction, and working with vectors across large distances gets complicated fast. Gravitational potential $V$ gives you a scalar alternative that captures the energy landscape of the field instead.

Think of it this way: field strength tells you how steeply the landscape slopes at one spot; potential tells you the height of the landscape at every point. Once you know the potential at two locations, you can calculate the energy needed to move between them without worrying about direction at every step.

This section covers the formula for $V$ in a radial field, the crucial meaning of its negative sign, and how both $g$ and $V$ vary graphically with distance $r$. These three ideas are tightly linked — understanding one deepens your grasp of the others.

2. Gravitational Potential in a Radial Field

For a point mass $M$ (or any body that behaves as one, such as a uniform sphere), the gravitational potential at a distance $r$ from the centre is given by a clean, compact formula. The derivation comes from integrating the gravitational field strength from infinity to $r$, but at A-level the key requirement is to know and apply the result.

Notice that $V$ depends on $\frac{1}{r}$, not $\frac{1}{r^2}$ as field strength $g$ does. As $r$ increases, $V$ becomes less negative, approaching zero as $r \to \infty$. As $r$ decreases towards the surface, $V$ becomes more negative — the potential well deepens.

2a. The Significance of the Negative Sign

The negative sign is not a quirk of the maths — it carries real physical meaning. Gravitational fields are always attractive, so moving a mass from infinity (where $V = 0$ by definition) towards a massive body releases energy. The system loses potential energy as the mass falls inward, so $V$ must be negative at every finite distance.

The reference point is chosen at infinity precisely because it is the only location where the gravitational influence of the body is truly zero. Every other point is closer to the mass, so every other point has a lower (more negative) potential than infinity.

A useful way to picture this: imagine a funnel-shaped potential well. Infinity is the flat rim at the top, where $V = 0$. Moving towards the central mass is like sliding down into the funnel — the potential decreases (becomes more negative) the deeper you go. You would need to do positive work on the mass to pull it back up to the rim.

A common source of confusion is mixing up the sign of $V$ with the direction of $g$. Field strength $g$ points inward (towards the mass), but that is a vector direction. Potential $V$ is a scalar — its negative value reflects the energy stored in the system, not a direction.

3. Calculating Gravitational Potential: Worked Example

The formula $V = -\frac{GM}{r}$ is straightforward to apply, but exam questions often test whether you handle the negative sign correctly and use consistent SI units. The worked example below models the full method you should show in an exam.

Worked example: Calculating gravitational potential at the Earth's surface

Calculate the gravitational potential at the surface of the Earth. Use: mass of Earth $M = 5.97 \times 10^{24}$ kg, radius of Earth $R = 6.37 \times 10^{6}$ m, $G = 6.67 \times 10^{-11}$ N m$^2$ kg$^{-2}$.

1. $V = -\frac{GM}{r}$ — State the formula first. This earns the method mark even if arithmetic errors follow.
2. $V = -\frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{6.37 \times 10^{6}}$ — Substitute with units consistent in SI. $r$ is the radius of the Earth here because we want the potential at the surface.
3. $GM = 6.67 \times 10^{-11} \times 5.97 \times 10^{24}$ $= 6.67 \times 5.97 \times 10^{-11+24}$ $= 39.82 \times 10^{13}$ $= 3.982 \times 10^{14}$ N m$^2$ kg$^{-1}$ — Multiply the coefficients: $6.67 \times 5.97 = 6.67 \times 6 - 6.67 \times 0.03 = 40.02 - 0.20 = 39.82$. Then combine the powers of ten: $10^{-11} \times 10^{24} = 10^{13}$.
4. $V = -\frac{3.982 \times 10^{14}}{6.37 \times 10^{6}}$ $= -\frac{3.982}{6.37} \times 10^{14-6}$ $= -0.6251 \times 10^{8}$ $= -6.25 \times 10^{7}$ J kg$^{-1}$ — $3.982 \div 6.37 \approx 0.6251$. Powers: $10^{14} \div 10^{6} = 10^{8}$. The negative sign is carried through from the formula.
5. $V = -6.25 \times 10^{7}$ J kg$^{-1}$ — Always include the negative sign and the unit J kg$^{-1}$. A positive answer here would be physically wrong and would lose the accuracy mark.

Answer: $V = -6.25 \times 10^{7}$ J kg$^{-1}$

4. Graphical Variation of $g$ and $V$ with $r$

Two of the most frequently examined graphs in this topic show how $g$ and $V$ each change with distance $r$ from the centre of a massive body. They look different, and understanding why is essential for both description and analysis questions.

Recall from the gravitational fields section that $g = \frac{GM}{r^2}$, which follows an inverse-square law. The potential formula $V = -\frac{GM}{r}$ follows an inverse law — a weaker dependence on $r$. This difference in the power of $r$ produces two distinct curve shapes on a graph.

The $g$-$r$ graph shows a steep curve that falls rapidly from the surface value and flattens asymptotically towards zero. The curve never reaches zero at any finite $r$ — it only reaches zero at infinity. This is the classic inverse-square shape.

The $V$-$r$ graph is also a curve approaching zero at infinity, but it rises more gradually than the $g$-$r$ graph because $V$ depends on $\frac{1}{r}$ rather than $\frac{1}{r^2}$. Crucially, all values on the $V$-$r$ graph are negative — the curve lies entirely below the $r$-axis. At $r = R$ (the surface), $V$ has its most negative value.

The relationship $g = -\frac{dV}{dr}$ is the key link between the two graphs. The gradient of the $V$-$r$ graph at any point, taken with a negative sign, gives the gravitational field strength at that point. Where the $V$-$r$ curve is steepest (near the surface), $g$ is largest. Where the curve is nearly flat (far from the body), $g$ is small.

4a. Comparing the Two Graphs in Detail

Exam questions often ask you to compare the $g$-$r$ and $V$-$r$ graphs. The word 'compare' requires both similarities and differences — a common mark-losing error is to describe only one graph.

Both graphs show curves that decrease in magnitude as $r$ increases and both approach zero asymptotically at infinity. Neither graph ever reaches zero at a finite distance. These are the similarities.

The differences are equally important. The $g$-$r$ graph shows positive values (field strength is a magnitude here) and follows a $\frac{1}{r^2}$ relationship, making it steeper. The $V$-$r$ graph shows negative values throughout and follows a $\frac{1}{r}$ relationship, making it shallower. At any given $r$, the gradient of the $V$-$r$ graph equals $-g$ at that point.

5. Deeper Analysis: Linking $g$ and $V$

Earlier we established that $g = -\frac{dV}{dr}$. This is worth examining more carefully because it connects the two graphs in a precise, testable way. If you know the $V$-$r$ graph, you can deduce the $g$-$r$ graph by finding the negative gradient at each point — and vice versa.

You can verify this algebraically. Starting from $V = -\frac{GM}{r}$, differentiate with respect to $r$: $\frac{dV}{dr} = \frac{GM}{r^2}$. Applying the negative sign: $g = -\frac{dV}{dr} = -\frac{GM}{r^2}$. But wait — $g$ is a magnitude here, so $g = \frac{GM}{r^2}$, which is exactly Newton's law of gravitation per unit mass. The two formulae are fully consistent.

This also tells you something useful for graph questions: the $g$-$r$ graph falls off faster than the $V$-$r$ graph. At double the distance, $g$ drops to one quarter of its original value (inverse-square), whereas $V$ drops to only half its original magnitude (inverse). Examiners test this distinction directly.

6. Exam Technique and Common Mistakes

The three specification requirements for this topic — the formula, the negative sign, and the graphical variations — each carry predictable exam traps. Knowing them in advance is the difference between dropping marks and collecting them.

For extended-response questions asking you to explain the significance of the negative sign, structure your answer in three beats: (1) state that the zero of potential is defined at infinity, (2) explain that gravity is attractive so energy is released as a mass moves inward, and (3) conclude that potential at any finite distance must therefore be less than zero — i.e. negative. Each beat typically corresponds to a separate marking point.

Taken together, the formula $V = -\frac{GM}{r}$, the physical meaning of its negative sign, and the contrasting shapes of the $g$-$r$ and $V$-$r$ graphs form a coherent picture of how gravitational influence spreads through space — and the AQA specification tests all three aspects directly.