IB Physics HLTopic 4 — Force FieldsPaper 1 & 2No work done along one~15 min read
Gravitational Equipotentials
A walking map has contour lines. Follow one round a hillside and you never climb, never descend, never break a sweat. Cut straight across them and you are gasping within minutes. Gravitational fields have contour lines too. They are called equipotentials, and the same rule holds with beautiful exactness: travel along one and gravity charges you nothing at all.
📘 What you need to know
Equipotential lines (in 2D) and surfaces (in 3D) join points of equal gravitational potential
They are always perpendicular to the field lines — in radial and uniform fields
They are drawn as dotted or dashed lines, with no arrows (they are not vectors and have no direction)
No work is done moving along an equipotential, because ΔV = 0. Work is only done moving between them
In a radial field: concentric circles around the planet, getting further apart further out
In a uniform field: horizontal, parallel, equally spaced straight lines
The potential gradient is defined by the equipotentials: closely packed → strong field
An object travelling along an equipotential neither gains nor loses gravitational potential energy
What an equipotential is
Pick a value of potential — say −3.0 × 10⁷ J kg−1. Now find every single point in space where the potential has exactly that value, and join them up. That’s an equipotential.
Equipotential — definitiona line (2D) or surface (3D) joining points that all have the same gravitational potential
Around a lone planet the answer is obvious once you say it out loud. Potential depends only on r, through Vg = −GM/r. So “all the points with the same potential” means “all the points at the same distance from the centre” — a sphere. Drawn on paper, a circle.
Radial and uniform fields
Two standard pictures. Learn to draw both, with a ruler, in about twenty seconds.
Dashed = equipotential, solid + arrowhead = field line. In both pictures they meet at exactly 90°, and in both the field points from high potential to low.
Why do the circles spread apart as you go out? Draw them at equal steps of potential — say every 1.0 × 107 J kg−1. Near the surface the potential is changing fast, so you only have to move a short way to drop by that step. Far out it changes lazily, so you must travel enormously further for the same drop. Crowded equipotentials mean a steep gradient, and a steep gradient means a strong field. They are a contour map, and the contours bunch up where the hill is steep.
No work done along an equipotential
This is the result the whole page exists for. Move a mass from one point to another on the same equipotential and the potential has not changed — so ΔV = 0, and the work done is zero.
Work done moving a massW = mΔVgalong an equipotential, ΔVg = 0, so W = 0 • work is only done moving between equipotentials
It does not matter how long the journey is, or how winding. A satellite in a perfectly circular orbit is sliding around a single equipotential the whole way, which is precisely why it needs no engine to keep going.
The green journey is free however far it winds. The orange one costs mΔV, and only the endpoints matter — not the route taken.
Why must they be perpendicular?
Not a coincidence, and not a drawing convention. It is forced on us by the last result. Suppose an equipotential met a field line at some angle other than 90°.
The argument in one line: no work along an equipotential means no force along it, which means the field must be at right angles to it.
Follow that logic once and you will never forget it. If the field had any component along the equipotential, that component would push a mass sideways and do work on it — but moving along an equipotential is defined to cost nothing. The only way to have a force and yet do no work is for the force to be at right angles to the motion. So perpendicular it is. The same reasoning, incidentally, is why the tension in a string does no work on a mass swinging in a circle.
Field lines
Equipotentials
Drawn as
Solid lines with arrows
Dashed lines, no arrows
Show
Direction of the force on a mass
Points of equal potential
Vector or scalar?
Vector quantity, g
Scalar quantity, Vg
Radial field
Straight, pointing at the centre
Concentric circles, spreading out
Uniform field
Parallel, equally spaced, downwards
Horizontal, parallel, equally spaced
Crowded together means
Strong field
Strong field (steep gradient)
Angle between them
Always 90°
Same potential everywhere on it
so ΔV = 0 along it
No work done W = mΔV = 0
so no force component along it
Field must be perpendicular
🗺️ Drawing equipotentials for full marks
Dashed or dotted lines only. Solid lines are field lines, and the examiner will read them as such.
No arrowheads. Potential is a scalar; there is nothing for an arrow to point at.
Radial field: concentric circles about the centre, with the gaps growing as you move out.
Uniform field: straight, horizontal, equally spaced. Use a ruler.
Check the angle. Every crossing with a field line must be a clean 90°.
WE 1
A satellite of mass 1200 kg moves in a circular orbit around a planet. (a) State the work done by the gravitational force during one complete orbit. (b) The satellite is then raised from an equipotential where Vg = −4.0 × 10⁷ J kg⁻¹ to one where Vg = −2.5 × 10⁷ J kg⁻¹. Calculate the work done.
(a) one complete orbit
A circular orbit lies entirely on one equipotential, so ΔV = 0.
W = 0 J(b) Step 1 — find the change in potentialΔV = (−2.5 × 10⁷) − (−4.0 × 10⁷) = +1.5 × 10⁷ J kg⁻¹Step 2 — multiply by the massW = mΔV = 1200 × (1.5 × 10⁷)W = 1.8 × 10¹⁰ JThe potential went up (less negative), so ΔV is positive and work had to be done on the satellite. If you got a negative answer, you subtracted the wrong way round.
WE 2
For a planet with GM = 4.0 × 10¹⁴ N m² kg⁻¹, equipotentials are drawn at −4.0, −3.0, −2.0 and −1.0 (× 10⁷ J kg⁻¹). Calculate the radius of each, and use your answers to explain why equipotentials spread out with distance.
Step 1 — rearrange the potential equationV = −GM/r → r = GM / |V|Step 2 — work through the four valuesr = 4.0×10¹⁴ / 4.0×10⁷ = 1.0 × 10⁷ mr = 4.0×10¹⁴ / 3.0×10⁷ = 1.3 × 10⁷ mr = 4.0×10¹⁴ / 2.0×10⁷ = 2.0 × 10⁷ mr = 4.0×10¹⁴ / 1.0×10⁷ = 4.0 × 10⁷ mStep 3 — look at the gaps3.3 × 10⁶ then 6.7 × 10⁶ then 2.0 × 10⁷ mthe gaps grow rapidly
Equal steps of potential need ever larger steps of distance, because the field out there is weak.
Six times further apart by the last step. Since g = −ΔV/Δr, a bigger Δr for the same ΔV simply is a smaller g. The picture and the equation agree.
WE 3
Explain why gravitational equipotential surfaces must always be perpendicular to gravitational field lines.
Step 1 — state the defining property
All points on an equipotential have the same potential, so ΔV = 0 between any two of them.
Step 2 — apply the work equationW = mΔV = 0, so no work is done moving a mass along the surface.
Step 3 — the contradiction
If the field had a component along the surface, that force would do work on a mass moving along it.
Step 4 — conclude
So the field can have no component along the surface.
the field must be at 90° to the equipotentialFour sentences, four marks. The exam wants the logic chain, not the phrase “because they just are”.
💡 Top tips
Dashed, and no arrows. Two of the easiest marks on any field diagram.
Use a ruler for uniform-field equipotentials and a steady hand for radial ones.
Say “no work is done because ΔV = 0“. The equation W = mΔV is the reason, so quote it.
Radial equipotentials get further apart; uniform ones stay equally spaced. Examiners check.
Work done between two equipotentials depends only on the start and end potentials — never on the path.
A circular orbit sits on one equipotential. That single sentence answers a lot of questions.
Crowded equipotentials mean a steep potential gradient, hence a strong field.
⚠ Common mistakes
Drawing equipotentials with arrows on them. Potential is a scalar — there is no direction
Drawing them solid, so they cannot be told apart from field lines
Spacing radial equipotentials equally. The gaps must widen with distance
Saying work is done when a satellite completes a circular orbit. ΔV = 0, so W = 0
Thinking equipotentials and field lines are parallel. They are always perpendicular
Believing the work done between two equipotentials depends on the route taken
Forgetting the mass: potential difference is in J kg−1, but work is mΔV, in joules
Quick recap:Equipotentials join points of equal potential. They are drawn dashed, without arrows, and are always perpendicular to the field lines — because no work is done along them, so the field can have no component along them. In a radial field they are concentric circles that spread apart with distance; in a uniform field they are parallel and equally spaced. Moving along one: W = 0. Moving between them: W = mΔV, and only the endpoints matter.
We have now mapped gravity every way there is: as a force, as a field, as a potential, and as a contour map. Time to point all of it at the sky. Planets do not wander at random — they obey three precise rules, worked out by Kepler from naked-eye observations decades before anyone knew why. Next page: Kepler’s Laws of Planetary Motion.
Equipotentials and field lines getting tangled?
Book a free meeting and we’ll practise the diagrams, the perpendicular argument and past-paper work-done questions.
