IB Physics HLTopic 4 — Force FieldsPaper 1 & 2g = −ΔV/Δr~15 min read
Gravitational Potential Gradient
Potential is the depth of the valley. Field strength is the steepness of its sides. That’s the whole page, and it’s a genuinely lovely piece of physics: two quantities you learned separately turn out to be the same thing seen from different angles. Draw the potential against distance, take the slope, flip the sign — and out drops g.
📘 What you need to know
Potential gradient: the rate of change of gravitational potential with respect to displacement in the direction of the field
The gravitational field at a point equals the negative gradient of a potential–distance graph at that point
g = −ΔVg / Δr
Units check: J kg−1 per metre = N kg−1 — exactly the units of g
The Vg–r graph is negative, follows a −1/r shape, and rises shallowly towards zero
To find g at a point on a curved graph, draw a tangent and take its gradient
The minus sign means the field points in the direction of decreasing potential — downhill, towards the mass
A steeper potential curve means a stronger field
The potential gradient is defined by the equipotential lines, which are always drawn perpendicular to the field lines
Two descriptions of one field
You now have two ways to describe what a planet does to the space around it.
g, the field strength: how hard the planet pulls on each kilogram. A vector, in N kg−1.
Vg, the potential: how much energy each kilogram has at that spot. A scalar, in J kg−1.
They cannot be independent — they describe the same planet. The link between them is the gradient.
Potential gradient — definitionthe rate of change of gravitational potential with respect to displacement in the direction of the field
Field strength from potentialg = − ΔVg / Δrg = field strength (N kg−1) • ΔVg = change in potential (J kg−1) • Δr = change in distance (m)
The units prove it
If you ever doubt the equation, divide the units and watch them fall into place. A joule is a newton-metre, so:
A one-line sanity checkJ kg−1 ÷ m = (N m) kg−1 ÷ m = N kg−1the metres cancel, and you are left with the units of field strength
Reading g off the graph
Here is the practical skill. Given a graph of Vg against r, the field strength at any point is the gradient of the tangent there, with a minus sign in front.
The curve rises as you move away, so ΔV/Δr is positive. The minus sign then aims g back at the planet, which is exactly where gravity points.
Why is the graph curved rather than straight? Because Vg goes as 1/r and its gradient goes as 1/r². Differentiating a 1/r curve hands you back the inverse square law. Potential and field strength were never two separate ideas — one is the slope of the other, and the extra power of r appears the moment you take the gradient.
Steep means strong
Because gis the gradient, the shape of the potential curve tells you the strength of the field at a glance. Near the surface the curve plunges — a steep slope, a strong field. Far away it flattens off — a gentle slope, a feeble field.
Both tangents touch the same curve. The steep one is at the surface, the flat one four radii out, where g is sixteen times weaker.
A shortcut worth knowing
For a point mass, the two equations are Vg = −GM/r and g = GM/r². Divide one by the other and the GM vanishes:
Handy relation at a single pointg = |Vg| / rvalid for a point mass or sphere — a fast way to get one from the other
Which way does the field point?
That minus sign is not decoration. It says the field points from high potential to low potential — from the shallow, less-negative regions far away, down into the deep negative well at the planet. Water runs downhill; so does gravity.
The dashed circles are equipotentials — lines of equal Vg. They get further apart as you go out, because the potential is changing more slowly there.
On the Vg–r graph
What it means physically
The curve lies below the axis
Potential is negative everywhere
Gradient at a point
Minus the field strength there
Steep section
Strong field — you are close to the mass
Shallow section
Weak field — you are far away
Gradient is positive
So g is negative: it points inwards
Curve approaches the axis
g → 0, but only at infinity
Potential Vg = −GM/r
take the gradient and flip the sign
Field strength g = −ΔV/Δr
which comes out as
g = GM/r²
📐 Getting g from a potential graph
Find the point on the curve at the distance you want.
Draw a tangent there, long and with a ruler. A short tangent gives a bad gradient.
Build a big triangle on the tangent. Read ΔV and Δr off the axes, not off the paper.
Divide: gradient = ΔV/Δr. Watch the powers of ten.
Flip the sign.g = −gradient. Quote the magnitude and say “directed towards the centre”.
WE 1
On a potential–distance graph for a planet, the potential is −3.0 × 10⁷ J kg⁻¹ at r = 8.0 × 10⁶ m and −2.0 × 10⁷ J kg⁻¹ at r = 1.2 × 10⁷ m. Estimate the gravitational field strength between these two points.
Step 1 — find the changesΔV = (−2.0 × 10⁷) − (−3.0 × 10⁷) = +1.0 × 10⁷ J kg⁻¹Δr = (1.2 × 10⁷) − (8.0 × 10⁶) = 4.0 × 10⁶ mStep 2 — apply the gradient equationg = −ΔV/Δr = −(1.0 × 10⁷) / (4.0 × 10⁶)g = −2.5 N kg⁻¹Step 3 — say what the sign means2.5 N kg⁻¹, directed towards the planetThis is an average over the interval, because the curve bends between the two points. The true value halfway out, at r = 1.0 × 10⁷ m, is 2.4 N kg⁻¹ — close, but not identical. That gap is why examiners ask for a tangent, not a chord.
WE 2
The gravitational potential at the Earth’s surface is −6.25 × 10⁷ J kg⁻¹ and the Earth’s radius is 6.37 × 10⁶ m. Use the relation between potential and field strength to determine g at the surface, and comment on your answer.
Step 1 — use the point-mass shortcutV = −GM/r and g = GM/r² → g = |V| / rStep 2 — substituteg = (6.25 × 10⁷) / (6.37 × 10⁶)g = 9.81 N kg⁻¹Step 3 — comment
This is exactly the familiar surface value. The two descriptions agree.Notice we never needed G or the Earth’s mass. Potential and field strength carry the same information about a planet — give me one and I can hand you the other.
WE 3
(a) Explain the significance of the minus sign in g = −ΔVg/Δr. (b) Explain why the potential–distance graph becomes shallower as r increases.
(a) the minus sign
As r increases, V increases (becomes less negative), so the gradient ΔV/Δr is positive.
But the field points inwards, opposite to increasing r.
the minus sign reverses the direction
The field points from high potential to low potential — downhill.
(b) why it flattens
The gradient of the graph is the field strength.
g = GM/r², so as r grows, g falls as an inverse square.
a weaker field means a shallower slopeBoth parts hang on one sentence: the gradient of the potential graph is the field strength. If you can say that, you can answer almost anything on this page.
💡 Top tips
Say it out loud: the field is minus the gradient of the potential. Everything else follows.
Quote the gradient definition in full: rate of change of potential with respect to displacement in the direction of the field.
For a curved graph you must draw a tangent. Using two points on the curve gives an average, not the value at a point.
Make the tangent and the triangle large. Small triangles multiply your reading errors.
Divide the units to check: J kg−1 m−1 = N kg−1. That is a free mark if asked to justify.
For a sphere, g = |Vg|/r lets you jump between the two without G or M.
Finish by stating a direction: “towards the centre of the planet”.
⚠ Common mistakes
Dropping the minus sign, then claiming the field points outwards
Using a chord between two points on the curve when the question asks for gat a point — that gives an average
Reading ΔV and Δr off the tangent’s pixel length rather than off the axis scales
Forgetting the powers of ten when the axes are labelled in 10⁷ J kg−1 and 10⁶ m
Thinking a steeper curve means a higher potential. It means a stronger field
Confusing this with the force–distance graph, where the area (not the gradient) is the quantity of interest
Saying the gradient is zero far away because the graph “reaches” the axis. It only approaches it
Quick recap: The potential gradient is the rate of change of Vg with displacement in the direction of the field, and the field strength is minus that gradient: g = −ΔVg/Δr. Check it with units — J kg−1 per metre is N kg−1. On a Vg–r graph, find g at a point by drawing a tangent: steep means strong. The minus sign points the field downhill, from high potential to low, always towards the mass. And for a sphere, g = |Vg|/r.
We have just met the equipotentials in passing — those dashed circles of equal potential, sitting at right angles to every field line. They deserve better than a cameo. Why are they always perpendicular? Why is no work done moving along one? And why do they spread apart as you climb out of the well? That’s the next page: Gravitational Equipotential Surfaces.
Tangents and gradients tripping you up?
Book a free meeting and we’ll practise reading field strength off potential graphs until it’s second nature.
