Linking g, V and ΔV Revision Notes | A Level AQA Physics

1. The Connection Between g and V

You already know two separate quantities: gravitational field strength $g$ and gravitational potential $V$. They describe the same field from two different angles — one tells you the force per unit mass, the other tells you the energy per unit mass. The question is: how are they mathematically linked?

Think about what happens when you move a small test mass a tiny distance $\Delta r$ outward from a planet. The field does work on the mass, and the potential changes by $\Delta V$. The field strength is essentially the potential energy gradient — the steeper the potential well, the stronger the field pulling you back in.

The negative sign is not cosmetic — it carries real physical meaning. Gravitational potential $V$ becomes less negative (increases toward zero) as you move away from a planet, so $\frac{\Delta V}{\Delta r}$ is positive. But $g$ points inward, toward the planet. The negative sign ensures $g$ has the correct inward direction, consistent with the field being attractive.

A useful analogy: imagine a valley. The steeper the slope, the harder gravity pulls you toward the bottom. The potential is like the height of the valley walls — it increases as you climb out. The field strength is the steepness of the slope at any point, and it always points downhill (toward lower potential), hence the negative sign.

2. Why the Negative Sign Matters — Direction and Sign Convention

This is where many students lose marks. Gravitational potential $V$ is always negative (it is zero at infinity and decreases as you approach a mass). As $r$ increases, $V$ becomes less negative, so $\Delta V$ is positive for an outward displacement $\Delta r > 0$. That makes $\frac{\Delta V}{\Delta r}$ positive — but $g$ must be negative (pointing inward). The negative sign in the equation corrects for this.

Conversely, if you treat $g$ as a magnitude (which AQA questions often do when asking for field strength at a point), you take the magnitude of the gradient: $g = \left|\frac{\Delta V}{\Delta r}\right|$. Always check whether the question asks for the vector field strength (include sign and direction) or the scalar magnitude.

3. Reading g from a V–r Graph

A graph of gravitational potential $V$ against radial distance $r$ from a planet's centre is a curve that starts very negative close to the surface and rises toward zero as $r$ increases. The shape follows $V = -\frac{GM}{r}$, so it is a reciprocal curve, not a straight line.

To find $g$ at any point on this graph, you calculate the gradient of the tangent to the curve at that point, then apply the negative sign. In practice, for a small interval you draw a chord between two nearby points and compute $\frac{\Delta V}{\Delta r}$. The steeper the curve (closer to the planet), the larger the magnitude of $g$.

Notice how the curve is steepest near the surface (small $r$) and flattens out at large $r$. A steep gradient means a large magnitude of $g$ — the field is strongest close to the planet. At large distances the curve is nearly flat, corresponding to a weak field. This is the graphical expression of the inverse-square law for $g$.

Worked example: Finding g from the gradient of a V–r graph

A V–r graph for a planet shows that at $r = 8.0 \times 10^6$ m, $V = -50.0 \times 10^6$ J kg⁻¹, and at $r = 10.0 \times 10^6$ m, $V = -40.0 \times 10^6$ J kg⁻¹. Calculate the magnitude of the gravitational field strength g at approximately $r = 9.0 \times 10^6$ m.

1. $\Delta V = V_2 - V_1 = (-40.0 \times 10^6) - (-50.0 \times 10^6) = +10.0 \times 10^6$ J kg⁻¹ $\Delta r = r_2 - r_1 = (10.0 \times 10^6) - (8.0 \times 10^6) = 2.0 \times 10^6$ m — Read the two coordinate pairs from the graph carefully. ΔV is positive because V becomes less negative as r increases. [M1]
2. $\frac{\Delta V}{\Delta r} = \frac{10.0 \times 10^6}{2.0 \times 10^6} = +5.0$ J kg⁻¹ m⁻¹ — Divide the change in potential by the change in distance. The result is positive, as expected for an outward displacement. [M1]
3. $g = -\frac{\Delta V}{\Delta r} = -(+5.0) = -5.0$ N kg⁻¹ — The negative value confirms g points inward (toward the planet). [M1]
4. $|g| = 5.0$ N kg⁻¹ — If the question asks for field strength as a magnitude, quote the positive value and state the direction separately. [A1]

Answer: The gravitational field strength at $r \approx 9.0 \times 10^6$ m is $5.0$ N kg⁻¹ directed toward the planet's centre.

4. Finding ΔV from the Area Under a g–r Graph

The relationship $g = -\frac{\Delta V}{\Delta r}$ can be rearranged to give $\Delta V = -g \, \Delta r$. For a small strip of width $\Delta r$ on a $g$ against $r$ graph, the area of that strip equals $g \, \Delta r$. Summing all such strips between two radii gives the total area under the curve — and that area equals $-\Delta V$, the negative of the change in potential.

In other words: the area under a $g$–$r$ graph between $r_1$ and $r_2$ gives the magnitude of the change in gravitational potential $|\Delta V|$ between those two points. This is the graphical integration of the field to recover the potential difference.

At A-level you will not be asked to perform calculus integration in the exam. Instead, you estimate the area using geometric methods: count grid squares, use the trapezoid rule, or approximate the region with triangles and rectangles. Always include units in your area calculation — the area has units of N kg⁻¹ × m = J kg⁻¹, which is exactly the unit of potential.

Worked example: Estimating ΔV from the area under a g–r graph

A g–r graph for a moon shows that g decreases approximately linearly from $1.80$ N kg⁻¹ at $r = 2.0 \times 10^6$ m to $0.80$ N kg⁻¹ at $r = 4.0 \times 10^6$ m. Estimate the change in gravitational potential $\Delta V$ between these two radii.

1. The region under a straight line between two points forms a trapezoid. — Since g decreases linearly over this interval, the area is a trapezoid with parallel sides equal to the two g values and width equal to Δr.
2. $\Delta r = (4.0 \times 10^6) - (2.0 \times 10^6) = 2.0 \times 10^6$ m — This is the width of the trapezoid. [M1]
3. Area $= \frac{1}{2}(g_1 + g_2) \times \Delta r = \frac{1}{2}(1.80 + 0.80) \times 2.0 \times 10^6$ $= \frac{1}{2} \times 2.60 \times 2.0 \times 10^6$ $= 1.30 \times 2.0 \times 10^6$ $= 2.6 \times 10^6$ J kg⁻¹ — The area under the graph has units N kg⁻¹ × m = J kg⁻¹, which is the unit of gravitational potential. [M1]
4. $\Delta V = -\text{Area} = -2.6 \times 10^6$ J kg⁻¹ — The area gives $-\Delta V$, so $\Delta V$ is negative. Moving outward from the moon, V increases (becomes less negative), so the change from $r_1$ to $r_2$ is actually positive. Check: $\Delta V = V(r_2) - V(r_1)$. Since V becomes less negative, $\Delta V > 0$. The area equals $|\Delta V|$. [A1]
5. $|\Delta V| = 2.6 \times 10^6$ J kg⁻¹ — The gravitational potential increases by $2.6 \times 10^6$ J kg⁻¹ as the distance from the moon's centre increases from $2.0 \times 10^6$ m to $4.0 \times 10^6$ m.

Answer: The change in gravitational potential $|\Delta V| = 2.6 \times 10^6$ J kg⁻¹. Potential increases (becomes less negative) moving outward.

5. Connecting Both Graphs — A Unified Picture

Earlier you saw how to extract $g$ from the gradient of a $V$–$r$ graph. Now you have the reverse: extracting $\Delta V$ from the area under a $g$–$r$ graph. These two operations are mathematical inverses — differentiation and integration — and they reflect the deep link between field and potential in all field theories.

A common exam question presents one graph and asks you to extract information about the other quantity. Practise moving fluently between both directions: gradient of $V$–$r$ gives $g$, and area under $g$–$r$ gives $\Delta V$.

6. Exam Technique and Common Mistakes

AQA examiners test this topic in two main ways: graph-reading questions where you extract a gradient or area, and calculation questions where you apply $g = -\frac{\Delta V}{\Delta r}$ directly. Both reward careful attention to sign, units, and the direction of the field.

When drawing a tangent to a $V$–$r$ curve to find the gradient, use a ruler and make your triangle as large as possible — a larger triangle reduces the percentage error in reading off $\Delta V$ and $\Delta r$. Show the triangle clearly on the graph and label the coordinates you used. Examiners award method marks for this even if your final numerical answer is slightly off.

For area calculations under a $g$–$r$ graph, state explicitly which geometric method you are using (trapezoid, counting squares, etc.) and show the calculation with units at every stage. A bare number with no units will not score the accuracy mark.

To consolidate both spec requirements, try the following exam-style question before checking the worked solution below.

Worked example: Exam-style multi-step: linking g, V and ΔV

A planet of mass $M = 4.0 \times 10^{24}$ kg has radius $R = 5.0 \times 10^6$ m. Use $G = 6.67 \times 10^{-11}$ N m² kg⁻². (a) Calculate the gravitational potential at the planet's surface. (b) Calculate the gravitational field strength at the surface using $g = -\frac{\Delta V}{\Delta r}$, treating the surface as a point where $V = -\frac{GM}{R}$ and using a small outward step $\Delta r = 1.0 \times 10^4$ m to estimate the gradient.

1. $V_1 = -\frac{GM}{R} = -\frac{6.67 \times 10^{-11} \times 4.0 \times 10^{24}}{5.0 \times 10^6}$ Numerator: $6.67 \times 10^{-11} \times 4.0 \times 10^{24} = 6.67 \times 4.0 \times 10^{-11+24} = 26.68 \times 10^{13} = 2.668 \times 10^{14}$ $V_1 = -\frac{2.668 \times 10^{14}}{5.0 \times 10^6} = -5.336 \times 10^7$ J kg⁻¹ — Use the formula for gravitational potential at distance R from the centre. [M1]
2. $r_2 = 5.0 \times 10^6 + 1.0 \times 10^4 = 5.01 \times 10^6$ m $V_2 = -\frac{2.668 \times 10^{14}}{5.01 \times 10^6} = -5.325 \times 10^7$ J kg⁻¹ — A small outward step allows us to estimate the local gradient. [M1]
3. $\Delta V = V_2 - V_1 = (-5.325 \times 10^7) - (-5.336 \times 10^7) = +1.1 \times 10^5$ J kg⁻¹ $\Delta r = 1.0 \times 10^4$ m — ΔV is positive (V becomes less negative moving outward), as expected. [M1]
4. $g = -\frac{\Delta V}{\Delta r} = -\frac{1.1 \times 10^5}{1.0 \times 10^4} = -11$ N kg⁻¹ — The negative sign confirms g points inward. The magnitude is 11 N kg⁻¹. [A1]
5. $g = \frac{GM}{R^2} = \frac{2.668 \times 10^{14}}{(5.0 \times 10^6)^2} = \frac{2.668 \times 10^{14}}{25 \times 10^{12}} = \frac{2.668 \times 10^{14}}{2.5 \times 10^{13}} = 10.67 \approx 10.7$ N kg⁻¹ — The gradient method gives approximately 11 N kg⁻¹, consistent with the direct formula result of 10.7 N kg⁻¹. The small discrepancy arises because the gradient method is an approximation over a finite Δr rather than an exact derivative. This confirms the method is physically sound.

Answer: (a) $V \approx -5.34 \times 10^7$ J kg⁻¹ at the surface. (b) $|g| \approx 11$ N kg⁻¹ directed toward the planet's centre (consistent with the direct calculation of $\approx 10.7$ N kg⁻¹).

Both spec requirements are now fully connected: the gradient of a $V$–$r$ graph gives $g$ via $g = -\frac{\Delta V}{\Delta r}$, and the area under a $g$–$r$ graph gives $|\Delta V|$. These are the two graphical tools AQA expects you to apply fluently in this topic.

The mathematical link between field strength and potential gradient is g = −ΔV/Δr. The negative sign matters because gravitational potential becomes less negative as you move away from the mass, while the gravitational field still points inward towards the mass.