IB Physics HLTopic 4 — Force FieldsPaper 1 & 2When mgΔh stops working~15 min read
GPE in a Non-Uniform Field
You have used Ep = mgΔh since you were fourteen, and it has never let you down. It is also, strictly speaking, wrong — or at least, it is only right in one very small corner of the universe: the thin shell of air near the ground where g can be pretended to be constant. Lift something high enough and g starts falling away beneath you. This page is about what gravitational potential energy really means once that happens.
📘 What you need to know
Gravitational potential energy (GPE) is the energy an object has because of its position in a gravitational field
GPE of a system: the work done to assemble the system from infinite separation of its components
GPE of a point mass: the work done in bringing that mass from infinity to the point
Near a surface, Ep = mgΔh, and GPE is taken as zero at the surface
That equation is valid only where the field is uniform — a small region close to the surface
Far out, the field is radial, g changes with r, and mgΔh fails
In a radial field, GPE is measured from zero at infinity, so GPE is negative everywhere else
The area under a force–distance graph equals the work done, and therefore the change in GPE
Two ways to say the same thing
IB gives you two definitions of GPE. They look different. They aren’t.
GPE of a systemthe work done to assemble the system from infinite separation of its components
GPE of a point massthe work done in bringing the mass from infinity to that point
The first talks about a pair of objects — a planet and a satellite — dragged together from opposite ends of the universe. The second fixes the planet and moves just the satellite. Same journey, same energy bookkeeping, same answer.
Notice the family resemblance to potential. Potential was the work done per kilogram to bring a test mass in from infinity. Potential energy is the work done to bring your particular mass in from infinity. One is per-kilogram, the other is the total bill. If you can see that, you have already half-derived the equation on the next page.
Where mgΔh comes from — and what it assumes
Lift a mass m through a height Δh. You pull upwards with a force equal to its weight, mg, and you move it a distance Δh. Work done = force × distance:
GPE near a surfaceEp = mgΔhvalid only for objects close to a planet’s surface, where the field is uniform
Every step of that argument leaned on one hidden assumption: that the force mg stayed the same size all the way up. If g changes during the lift, “force × distance” is meaningless — which force would you use?
There is a second hidden choice, too. When we write Ep = mgΔh we quietly declare the GPE at the ground to be zero. That is a convenience, not a law. In a radial field the sensible zero is at infinity instead, and the two conventions give very different-looking numbers for the same physics.
Left: a local zero at your feet, and g pretending to be constant. Right: the honest zero, infinitely far away, with everything below it in energy debt.
Watching the assumption break
How high is “close to the surface”? Look at what g actually does as you climb.
A classroom is 3 metres tall out of 6 370 000. That is why mgΔh works so well down here, and why it collapses in orbit.
So mgΔh is not wrong so much as local. It is the flat-Earth approximation for energy, and like the flat-Earth map of your town, it is perfectly accurate right up until you try to use it for a journey across the world.
The honest way: area under the force–distance graph
If the force keeps changing, you cannot simply multiply. You have to add up force × distance over lots of tiny steps — and that is exactly what the area under a graph does for you.
Reading the areaforce × distance = work done so the area under a force–distance graph = work done = change in GPE
Steep near the planet, shallow far out. Most of the energy cost of leaving is paid in the first few thousand kilometres.
This graph is worth more marks than it looks. If an exam ever shows you a curve and asks what the area represents, look at the units of the axes and multiply them. Newtons × metres = joules. So the area is an energy — here, the change in gravitational potential energy. That trick works on every graph you will ever meet.
Uniform field (near a surface)
Radial field (everywhere)
Is g constant?
Yes, to a very good approximation
No — g = GM/r²
Zero of GPE
At the surface
At infinity
Sign of GPE
Positive above the ground
Negative everywhere
Equation
Ep = mgΔh
Needs the radial equation (next page)
Find the work by
Force × distance
Area under the force–distance graph
Valid when
Δh « planet radius
Always
Move a mass in a field
work is done against gravity
Energy stored as GPE
is g constant over the move?
mgΔh if yes area if no
🚀 Can I use mgΔh here?
Compare Δh with the planet’s radius. A few metres against 6400 km? Go ahead.
Kilometres or more? Stop. Check how much g has changed using g = GM/r².
If g has changed noticeably, mgΔh will overestimate the energy, because it charges you the surface value the whole way.
Use the area under the force–distance graph, or the radial GPE equation.
Check your zero. Surface-zero gives positive GPE; infinity-zero gives negative GPE. Say which you are using.
WE 1
A book of mass 2.5 kg is lifted from the floor onto a shelf 1.8 m above it. Taking g = 9.81 N kg⁻¹, calculate the gain in gravitational potential energy, and state the assumption you have made.
Step 1 — check the equation is allowed
1.8 m against an Earth radius of 6.37 × 10⁶ m. Utterly negligible.Step 2 — substituteEp = mgΔh = 2.5 × 9.81 × 1.8Ep = 44 JStep 3 — the assumption
That g is constant over the lift — i.e. the field is uniform in the room.
Also worth saying: this 44 J is measured from a zero at the floor. It is a change in GPE, not the book’s total GPE, which is a large negative number.
WE 2
A student calculates the energy needed to raise a 500 kg satellite from the Earth’s surface to an altitude of 3000 km using Ep = mgΔh. Calculate the value the student obtains, then calculate g at that altitude and explain why the student’s method is invalid. (ME = 5.97 × 10²⁴ kg, RE = 6.37 × 10⁶ m)
Step 1 — the student’s answerEp = 500 × 9.81 × (3000 × 10³)Ep = 1.5 × 10¹⁰ JStep 2 — find r at that altituder = 6.37 × 10⁶ + 3.00 × 10⁶ = 9.37 × 10⁶ mStep 3 — find g up thereg = GM/r² = 3.98 × 10¹⁴ / (9.37 × 10⁶)²g = 4.5 N kg⁻¹Step 4 — the verdict
g has fallen to less than half its surface value, so the field is not uniform over the climb.
mgΔh is invalid; it overestimatesThe student charged the full 9.81 for the entire 3000 km, when gravity was quietly getting cheaper the whole way up. The correct method is the area under the force–distance curve.
WE 3
In a radial field, gravitational potential energy is measured from a zero at infinity rather than at a planet’s surface. Explain why this choice is made, and state what it implies about the sign of the GPE of any object near a planet.
Step 1 — why not the surface?
Every planet has a different surface. A zero there would make GPE meaningless when comparing two bodies.
Step 2 — why infinity?
At infinite separation the objects no longer interact, so zero energy is the natural, universal choice.
Step 3 — the consequence
Gravity is attractive, so bringing a mass in from infinity means gravity does positive work and the mass’s energy falls below zero.
GPE is negative everywhere except at infinityAnd that is why “the satellite gains GPE as it climbs” and “its GPE is negative” are both true at once. It becomes less negative — it rises, but it never gets above zero.
💡 Top tips
Quote the definition with both key phrases: work done and from infinite separation.
Before using mgΔh, ask whether Δh is tiny compared with the planet’s radius. If not, say so — that is the mark.
mgΔh gives a change in GPE with a zero at the ground; it is not an object’s total GPE.
Area under a force–distance graph = work done = change in GPE. Multiply the axis units to check: N × m = J.
Notice that g appears in mgΔh but not in the radial equation — because g is no longer a single number.
State your zero. “Taking GPE as zero at infinity” costs one line and earns marks.
⚠ Common mistakes
Using Ep = mgΔh for satellites, rockets or orbits, where the field is not uniform
Not realising mgΔhoverestimates the energy, because it uses the surface value of g throughout
Thinking GPE is zero at the Earth’s surface in a radial field. It is zero at infinity
Treating a change in GPE as if it were the total GPE of the object
Saying an object “loses GPE” when it moves away from a planet. It gains GPE — it becomes less negative
Reading the area under a force–distance graph as a force, or under a force–time graph as energy (that one is impulse)
Forgetting that Δh is measured from the surface but r is measured from the centre
Quick recap:GPE is the work done to assemble a system from infinite separation, or equivalently the work done bringing a mass from infinity to a point. Close to a surface the field is uniform, GPE is taken as zero at the ground, and Ep = mgΔh. Further out the field is radial, g falls as 1/r², and that equation breaks — it overcharges you. The honest measure of the work done is then the area under the force–distance graph, and with a zero at infinity the GPE of anything near a planet is negative.
We have said what GPE means in a radial field, and how to find it from a graph. What we have not done is write it down. Take the definition — work done bringing m in from infinity — and remember that potential was the same thing per kilogram. Multiply by m and you have it. The next page makes it official: the Gravitational Potential Energy Equation.
Not sure when mgΔh is allowed?
Book a free meeting and we’ll work through uniform and radial fields, energy zeros and graph-area questions together.
