IB Physics HL Topic 4 — Force Fields Paper 1 & 2 V = −GM/r ~16 min read

Gravitational Potential

Field lines told you which way gravity pushes. Now we ask a different question: how much energy does it cost to be here? Think of a valley. The arrows tell you which way a ball rolls; the contour lines tell you how deep you are. Potential is the depth — and because the top of the valley is infinitely far away, every point you can actually stand on turns out to be below zero.

📘 What you need to know

Gravitational potential at a point: the work done per unit mass in bringing a test mass from infinity to that point
Symbol V_g, unit J kg⁻¹
For a point mass or uniform sphere: V_g = −GM/r
V_g is a scalar — no direction. (Field strength g is a vector.)
Potential is always negative. It is defined as zero at infinity, and gravity is attractive, so everywhere else is below that
Move closer to a planet → V_g gets more negative (smaller). Move away → V_g gets less negative (larger), tending to 0
V_g depends only on the mass making the field and the distance from its centre
V_g falls off as 1/r, not 1/r²

The definition, taken slowly

Here is the sentence the examiner wants, word for word:

Gravitational potential — definition the work done per unit mass in bringing a test mass
from infinity to a defined point

Three phrases are doing all the work in there.

Work done — so this is an energy quantity, measured in joules.
Per unit mass — divide by the kilograms you carried, so the test mass drops out again. Units: J kg⁻¹.
From infinity — the journey has to start somewhere, and physicists chose the only place with no gravity at all. Infinity is the zero of potential.

You don’t have to push the mass in. Gravity drags it. Because gravity does the work rather than you, the potential at P comes out below zero.

Where the minus sign comes from

Students hate this minus sign. It is not a trick, and it is not arbitrary. It follows from two decisions:

Potential is defined as zero at infinity. That’s the reference point, chosen because it is the only spot free of the planet’s pull.
Gravity is attractive. Coming in from infinity, the planet pulls the test mass the way it is already going, so gravity does positive work on it and the work you do is negative.

Start at zero, go negative, and you never come back up. Every point in the universe has a negative potential.

Gravitational potential of a point mass V_g = − GM / r V_g = potential (J kg⁻¹) • M = mass making the field (kg) • r = distance from its centre (m)

Look at what is not in that equation: the mass you carried. Just like g = GM/r², only the planet appears. So when a question hands you the mass of a meteor, a probe or an astronaut and asks for the potential, that number is a red herring. It’s bait. Leave it on the page and walk past.

And notice the denominator: a single r, not r². Potential is not an inverse square quantity. Double the distance and the potential halves — that is, it becomes half as negative.

The shape of the potential curve

Plot V_g against r and you get a curve that lives entirely below the axis, climbing steeply at first and then creeping towards zero, which it reaches only at infinity.

Read it as a valley seen in cross-section. The planet’s surface is the bottom of the well; the axis at the top is the rim, infinitely far out.

Which way is “bigger”?

This is where marks disappear. Say it in words, never from the sign alone.

Point B is further out, so its potential is higher. −2 is greater than −5. Read the number line, not the digits.

Here’s the trick that saves you every time. Don’t argue about “bigger” and “smaller”. Ask: is it closer to zero, or further from zero? Closer to zero means higher potential, and it always means further from the planet. Then translate into English — “the potential increases as the satellite moves away” — and let the sign look after itself.

Potential and field strength side by side

Both describe the same planet. One is about force, the other about energy. Keep them separate.

	Field strength g	Potential V_g
Definition	Force per unit mass	Work done per unit mass from infinity
Equation	g = GM/r²	V_g = −GM/r
Falls off as	1/r²	1/r
Unit	N kg⁻¹	J kg⁻¹
Vector or scalar?	Vector — has a direction	Scalar — just a number
Sign	Quoted as a positive magnitude	Always negative

Mass M

digs a
potential well

V_g = −GM/r

deepest at the surface,
zero at infinity

Energy cost
of being there

🕳️ Working out a potential

Identify M. It is the planet, star or moon making the field — never the object sitting in it.
Cross out any other mass the question gives you. If it asks for potential, that mass is a red herring.
Find r from the centre. Watch for a diameter given instead of a radius, and for “at a height h above the surface”.
Convert to metres, then substitute into V_g = −GM/r.
Write the minus sign. If your answer is positive, something has gone wrong.

WE 1

Mars has mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m. Calculate the gravitational potential at the surface of Mars.

Step 1 — on the surface, so r is the radius V = −GM / r Step 2 — substitute V = −(6.67 × 10⁻¹¹) × (6.42 × 10²³) / (3.39 × 10⁶) V = −4.28 × 10¹³ / 3.39 × 10⁶ V = −1.3 × 10⁷ J kg⁻¹ Read it as an energy debt: every kilogram sitting on Mars would need 1.3 × 10⁷ joules of energy to be dragged out to infinity. Note there is no square on that r.

WE 2

A planet has radius 2800 km and mass 1.8 × 10²³ kg. A probe of mass 750 kg hovers 1200 km above the planet’s surface. Determine the gravitational potential at the position of the probe.

Step 1 — spot the red herring The probe’s mass of 750 kg is not needed. Potential depends only on the planet. Step 2 — find r from the centre r = R + h = (2800 + 1200) km = 4000 km = 4.0 × 10⁶ m Step 3 — substitute V = −(6.67 × 10⁻¹¹) × (1.8 × 10²³) / (4.0 × 10⁶) V = −1.20 × 10¹³ / 4.0 × 10⁶ V = −3.0 × 10⁶ J kg⁻¹ If you multiplied by 750 you found the probe’s potential energy, in joules — a different quantity, and the wrong answer to this question. Check the units you have been asked for.

WE 3

Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m. Determine the distance from Earth’s centre at which the gravitational potential is −2.0 × 10⁶ J kg⁻¹, and comment on your answer.

Step 1 — rearrange for r V = −GM/r → r = −GM / V Step 2 — substitute, keeping both minus signs r = −(6.67 × 10⁻¹¹)(5.97 × 10²⁴) / (−2.0 × 10⁶) r = 3.98 × 10¹⁴ / 2.0 × 10⁶ r = 2.0 × 10⁸ m Step 3 — comment 2.0 × 10⁸ / 6.37 × 10⁶ ≈ 31 about 31 Earth radii out Two minus signs make a plus, so r comes out positive — as a distance must. And look how far you have to travel to climb from −6.3 × 10⁷ at the surface to only −2.0 × 10⁶. The well is deep.

💡 Top tips

Learn the definition verbatim: work done per unit mass in bringing a test mass from infinity to that point.
Always write the minus sign. A positive potential is always an error.
One r, not r². Potential is a 1/r quantity — that’s the difference between V_g and g.
V_g is a scalar. No arrows, no components, no directions.
If a question gives you the mass of the object in the field and asks for potential, ignore it.
Describe changes in words: “the potential increases as it moves away”. Then the sign takes care of itself.
Check whether you were given a diameter. Halve it before you use it.

⚠ Common mistakes

Dropping the minus sign, or “cancelling” it because the answer looks odd
Using r² in the denominator. That is the equation for g, not for V_g
Multiplying by the mass of the satellite or meteor — that gives potential energy, not potential
Saying the potential “gets bigger” as you approach a planet. It gets more negative, so it gets smaller
Treating V_g as a vector and trying to resolve it into components
Thinking potential is zero at the Earth’s surface. It is zero at infinity
Using the radius of the planet when the object is at altitude, or a diameter in place of a radius

Quick recap: Gravitational potential is the work done per unit mass in bringing a test mass from infinity to a point, measured in J kg⁻¹. For a point mass, V_g = −GM/r — a scalar, always negative, zero only at infinity, and falling off as 1/r rather than 1/r². Move away from the planet and V_g increases towards zero; move in and it becomes more negative. The mass placed in the field never appears.

So far every object has sat in a radial field, where g changes as you move. But the very first energy equation you ever met, E_p = mgΔh, quietly assumed g was constant — which is only true in a small patch near the ground. What happens to gravitational potential energy when the field is not uniform? That is the next page.

That minus sign still bothering you?

Book a free meeting and we’ll work through potential, potential wells and the sign conventions until they feel obvious.

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Gravitational Potential

📘 What you need to know

The definition, taken slowly

Where the minus sign comes from

The shape of the potential curve

Which way is “bigger”?

Potential and field strength side by side

🕳️ Working out a potential

💡 Top tips

⚠ Common mistakes

That minus sign still bothering you?

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