IB Physics HLTopic 4 — Force FieldsPaper 1 & 2g = GM/r²~16 min read
Gravitational Field Strength
Last page we found the force between two masses. But that is a clumsy way to describe the Earth. The Earth doesn’t care what you drop on it — a feather, a hammer, a satellite — it treats them all the same way. So let’s stop talking about the object and describe what the Earth alone does to the space around it. Divide the force by the mass, and the mass vanishes. What’s left is the field.
📘 What you need to know
A gravitational field is a region of space where a test mass experiences a force due to the gravitational attraction of another mass
Field strength is the force per unit mass on a test mass at that point: g = F/m, measured in N kg−1
For a point mass (or a uniform sphere): g = GM/r²
g depends only on the mass M creating the field and the distance r — not on the mass placed in it
g is a vector, always pointing towards the centre of the mass making the field. Gravity is always attractive
Gravity has infinite range — g gets small but never reaches zero
Mass is the same everywhere. Weight (W = mg) changes from planet to planet
g at a surface depends on the planet’s mass (or density) and its radius
g (field strength) and G (the universal constant) are completely different things
What is a gravitational field?
A field is not a thing you can touch. It’s a map of what would happen. Put a mass here, and it would feel this force. Put it there instead, and it would feel that one.
Gravitational field — definitiona region of space where a test mass experiences a force due to the gravitational attraction of another mass
Two things follow immediately, and both come up in exams:
The field always points towards the centre of the mass producing it, because gravity is only ever attractive. There is no repulsive gravity.
The field has infinite range. Every mass in the universe is, in principle, tugging on you right now. The tugs are just unimaginably small.
Field strength: force per kilogram
Here is the whole idea. Take a small test massm, put it at a point in the field, and measure the force F on it. Then divide.
Gravitational field strengthg = F / mg = field strength (N kg−1) • F = force on the test mass, i.e. its weight (N) • m = test mass (kg)
The test mass is just a probe. Change it for a heavier one and F grows in exactly the same proportion, so g comes out the same.
Notice what those units are telling you. N kg−1 means “newtons of pull for every kilogram you put there”. On Earth’s surface that’s 9.81 — every kilogram gets 9.81 newtons of pull. And since a = F/m too, field strength and free-fall acceleration are the same number. That’s why 9.81 shows up wearing two different hats: N kg−1 and m s−2.
The field of a planet
Now let’s get an equation for g that doesn’t need us to actually go and measure a force. Start with Newton’s law of gravitation for a planet of mass M and a test mass m a distance r from its centre:
Step 1 — the force on the test massF = GMm / r2
But that force is the weight of the test mass, so it also equals mg:
Step 2 — set the two expressions equalmg = GMm / r2the little m appears on both sides — cancel it
Step 3 — field strength of a point massg = GM / r2M = mass making the field (kg) • r = distance from its centre (m)
Look hard at that result. The test mass has gone. The field strength at a point depends only on the planet and how far out you are. That is exactly why a feather and a hammer, dropped together on the Moon, hit the ground at the same instant — something an Apollo astronaut famously proved on camera.
g falls off as an inverse square too
Since r is squared on the bottom, g behaves just like the force did: go twice as far from the centre and the field strength drops to a quarter.
The steepest part is right at the surface. Climb one extra Earth-radius and you have already lost three quarters of your weight.
Mass and weight are not the same thing
Your mass is how much stuff you are made of. Carry it to Jupiter and it is unchanged — not a single atom has left. Your weight is the force the field pulls on that stuff with, and Jupiter’s field is fierce.
WeightW = mgmass in kg • weight in N • only g changes when you change planet
Same block, same 60 kg, same number of atoms. Only the arrow changed — because only g changed.
What sets g at a planet’s surface?
At the surface, r is just the planet’s radius R, so g = GM/R². Two things, then: how much mass and how big. And if you’d rather think in terms of density, substitute M = ρ × (4/3)πR³:
Surface field strength, via densityg = G × ρ(4/3)πR3 / R2 → g = (4/3)πGρRso for a given density, g ∝ R — a bigger planet of the same stuff pulls harder
Careful with that last result — it looks like it contradicts the inverse square law, but it doesn’t. g ∝ 1/r² is about moving away from one fixed planet. g ∝ ρR is about comparing different planets at their own surfaces, where growing R also piles on extra mass. Different questions, different answers. Always ask yourself: am I moving, or am I swapping planets?
Symbol
Name
Value
Changes?
G
Newton’s gravitational constant
6.67 × 10−11 N m² kg−2
Never. Same in every galaxy
g
Gravitational field strength
9.81 N kg−1 on Earth’s surface
Yes — with planet and with r
m
Mass
e.g. 60 kg
Never, wherever you take it
W
Weight
mg, in newtons
Yes — it follows g
A mass M
warps the space around it
Field g = GM/r²
acts on anything you place in it
Force F = mg
🌍 Which equation do I use?
Given a force and a mass? Use g = F/m. That’s the definition — it always works, even in a weird field.
Given a planet’s mass and a distance? Use g = GM/r².
Check whose mass is which.M is the planet. m is the thing sitting in the field. They are never swapped.
Is r measured from the centre? “At a height h” means r = R + h. Metres, please.
Comparing two planets? Don’t substitute. Write the ratio g1/g2 and cancel G.
WE 1
A lander of mass 850 kg rests on the surface of Mars, where the gravitational field strength is 3.72 N kg⁻¹. Calculate the weight of the lander on Mars, and state its mass if it were brought back to Earth.
Step 1 — rearrange the definition of gg = F/m → F = mgStep 2 — substituteW = 850 × 3.72 = 3162 NW = 3.2 × 10³ N on MarsStep 3 — the mass back on Earthstill 850 kgMass is a count of matter, so it travels unchanged. Only the weight would change — on Earth it would be 850 × 9.81 ≈ 8.3 kN, over two and a half times heavier.
WE 2
The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m. Calculate the gravitational field strength at its surface, and hence show that it is roughly one sixth of the value on Earth (9.81 N kg⁻¹).
Step 1 — on the surface, so r is the radiusg = GM / r²Step 2 — substitute, remembering to square rg = (6.67 × 10⁻¹¹) × (7.35 × 10²²) / (1.74 × 10⁶)²g = 4.90 × 10¹² / 3.03 × 10¹²g = 1.6 N kg⁻¹Step 3 — compare with Earth9.81 / 1.62 = 6.06about 1/6 of Earth’sThe Moon is much less massive, but it is also much smaller — and the smaller radius fights back through that 1/r². The two effects together leave you at one sixth, not one eightieth.
WE 3
Planet Z is made of rock with exactly the same mean density as Earth, but its radius is twice Earth’s radius. Determine the ratio gZ / gE of the surface field strengths.
Step 1 — write g at a surfaceg = GM / R²Step 2 — write M in terms of densityM = ρV = ρ (4/3) π R³Step 3 — substitute and cancelg = G ρ (4/3) π R³ / R² = (4/3) π G ρ R
So for equal densities, g ∝ RStep 4 — take the ratiogZ/gE = RZ/RE = 2g on Z is twice Earth’sDoubling R divides g by 4 through the 1/R² — but it also multiplies the mass by 8, because volume goes as R³. Eight over four is two. Never guess this one; do the algebra.
💡 Top tips
Learn the definition in words: force per unit mass on a test mass at that point. Examiners want “per unit mass”.
g is a vector — pointing at the centre. Say so if asked for a direction.
Units of g are N kg−1, numerically the same as free-fall acceleration in m s−2.
In g = GM/r², M is the planet. The mass in the field never appears — if it shows up, it’s a red herring.
Ratio questions: divide the two equations and watch G cancel. It saves a minute and a mistake.
Density questions: remember ρ = M/V and V = (4/3)πr³. The volume formula is in your data booklet.
⚠ Common mistakes
Mixing up g (field strength, changes) with G (universal constant, never changes). “Little g” and “big G” are unrelated quantities
Putting the test mass into g = GM/r². Only the field-producing mass belongs there
Saying an object’s mass is smaller on the Moon. Its weight is smaller; its mass is identical
Using an altitude as r rather than R + h
Forgetting to squarer, or halving g when the distance doubles — it goes to a quarter
Claiming g “becomes zero” a long way out. It becomes small. It never becomes zero
Quoting g in newtons. It is newtons per kilogram
Quick recap: A gravitational field is a region where a test mass feels a force, and its strength is the force per unit mass: g = F/m, in N kg−1. For a planet of mass M, g = GM/r² — independent of whatever you put in the field, pointing always at the centre, and falling as an inverse square. Weight W = mg follows the field from planet to planet; mass never does.
You can now work out the strength of the field at any single point. But physics likes pictures, not just numbers — and there is a beautiful way to draw a whole field at once, so that its direction and its strength are both visible at a glance. Arrows, spacing, and one crucial rule about what they may never do. That’s the next page: Gravitational Field Lines.
Still mixing up little g and big G?
Book a free meeting and we’ll untangle field strength, weight and the inverse square law with past-paper questions.
