IB Physics SLTopic 4 — Force FieldsPaper 1 & 2F = Gm1m2/r²~7 min read
Newton’s Law of Gravitation
Let go of your phone and it drops. The Moon is dropping too — it just keeps missing, swinging round the Earth instead of hitting it. Newton’s great realisation was that both are the very same pull, captured in one compact rule: every mass in the universe attracts every other mass.
📘 What you need to know
Newton’s law of gravitation: the gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation
In symbols: F = Gm1m2 ÷ r², where G is the gravitational constant (it’s in the data booklet)
r is measured centre to centre — for a satellite, that’s the planet’s radius plus the height above the surface
Planets and stars are treated as point masses, because their separations are enormous compared with their sizes
The inverse square law: double the separation → the force drops to a quarter; triple it → a ninth
The two masses pull on each other with equal-sized, opposite forces — a Newton’s third law pair, however different the masses are
Every Mass Pulls on Every Other
Gravity isn’t something special that planets do — it’s what all masses do. Your phone pulls on the Earth, the Earth pulls on your phone, and the Sun pulls on both. For any two separate bodies — a star and a planet, a planet and its moon, two asteroids drifting past each other — the force between them is given by Newton’s law of gravitation. The statement is worth learning word for word:
The gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation.
Two dials control the force. Bigger masses pull harder — double either mass and the force doubles. And distance weakens it fast — double the gap and the force falls to a quarter, because the separation is squared. More on that shortly.
What counts as a “point mass”?
A point mass is a body whose size doesn’t matter for the problem — all of its mass can be treated as sitting at a single point, its centre. Planets and stars are obviously not points, but the Sun–Earth separation is more than twenty thousand Earth-radii, so from that far away the Earth might as well be a dot. That’s why the law works so well for planets orbiting the Sun.
From far enough away, everything looks like a dot. That single approximation is what lets one small formula run the entire solar system.
The Equation
Newton’s law of gravitationF = Gm1m2 ÷ r²
Where:
F = the gravitational force between the two masses (N)
G = the gravitational constant, 6.67 × 10⁻¹¹ N m² kg⁻² — it lives in the data booklet, so look it up rather than memorising it
m1 and m2 = the two masses (kg)
r = the distance between their centres (m)
That last line is where most marks are lost: r is measured centre to centre. A satellite “400 km above the surface” is not 400 km from the centre — you must add the planet’s radius first.
Newton’s law in one picture: the force acts along the line joining the centres, r is measured centre to centre, and the two pulls form an equal-and-opposite pair — the small mass tugs the big one exactly as hard as the big one tugs it.
Notice the two arrows are the same length. The Earth pulls the Moon and the Moon pulls the Earth back with a force of exactly the same size, in opposite directions — a Newton’s third law pair. A bigger mass doesn’t pull “harder” than it is pulled; it just responds less, because the same force barely accelerates something so heavy. You can see the symmetry in the formula: swap m1 and m2 and F doesn’t change.
The Inverse Square Law
The r² on the bottom line has its own name — the inverse square law — and it’s the part examiners test again and again. Because the separation gets squared before it divides, distance punishes gravity quickly:
Inverse square lawF ∝ 1 ÷ r²
separation × 2
square it: 2² = 4
r² grows × 4
F ∝ 1 ÷ r²
force ÷ 4 — a quarter
separation × 3
square it: 3² = 9
r² grows × 9
F ∝ 1 ÷ r²
force ÷ 9 — a ninth
Plotted against distance, this gives the classic falling curve — steep at first, then a long shallow tail. The force fades fast, but it never quite reaches zero:
The inverse square curve: at separation 2r the force is down to F/4, at 3r just F/9 — steep at first, then a long fade that never quite reaches zero.
The inverse square law is a gift in ratio questions: if the distance changes by a factor k, the force changes by 1 ÷ k² — no G, no masses, no calculator gymnastics needed.
🧭 Using F = Gm1m2 ÷ r² in the exam
Sketch the two bodies and mark their centres — the force acts along the line joining them
Build r centre to centre — for anything orbiting, r = planet radius + height above the surface
Convert to metres before squaring — kilometres slipped into r² is the classic power-of-ten disaster
Rearrange first, then substitute — keep r² together as a single block when it moves across, and take G from the data booklet
Sanity-check the size — planet masses land around 10²³–10²⁷ kg, and forces on satellites are kilonewtons, not meganewtons
Quick recap: every mass attracts every other; F = Gm₁m₂ ÷ r² with r measured centre to centre; double the distance and the force drops to a quarter.
WE 1
An 850 kg orbiter circles Mars at a height of 400 km above the surface. The gravitational force between Mars and the orbiter is 2.5 kN. Calculate the mass of Mars. (Radius of Mars = 3400 km)
Build r centre to centre (radius + height), in metresr = 3400 + 400 = 3800 km = 3.8 × 10⁶ mRearrange Newton’s law for the planet’s massF = GMm/r² → M = Fr²/(Gm)Substitute (G from the data booklet)M = (2500 × (3.8 × 10⁶)²) ÷ (6.67 × 10⁻¹¹ × 850) = 6.37 × 10²³ kgmass of Mars ≈ 6.4 × 10²³ kgForgetting the radius (using r = 400 km) would shrink r² by a factor of about 90 — and the mass with it. Centre to centre, always.
WE 2
Two asteroids attract each other with a gravitational force of 8.0 × 10⁴ N when their centres are 60 km apart. They slowly drift until their centres are 180 km apart. Calculate the new gravitational force between them.
Compare the separations180 ÷ 60 = 3 — the separation has tripledApply the inverse square lawF ∝ 1/r², so the force drops by 3² = 9: F = 8.0 × 10⁴ ÷ 9F ≈ 8.9 × 10³ NNo G, no masses, no unit conversions — ratio questions are exactly what F ∝ 1/r² is for.
💡 Top tips
r is centre to centre — add the planet’s radius to the orbital height before you do anything else
The separation must be squared, and converted to metres before squaring — check both every single time
Big G (6.67 × 10⁻¹¹ N m² kg⁻²) is in the data booklet — don’t confuse it with little g, the field strength, which arrives on the next page
Distance-change questions rarely need G at all — write F ∝ 1/r² and compare (×3 the distance → ÷9 the force)
⚠ Common mistakes
Using the height above the surface as r and forgetting the planet’s radius underneath it
Substituting kilometres straight into the formula — r must be in metres, and any slip gets squared
Thinking the more massive body pulls harder — the two forces are an equal-and-opposite pair; only the responses differ
Halving the force when the distance doubles — the square means it drops to a quarter
That’s the engine of the whole topic: one formula, two masses, one centre-to-centre distance. Up next: Gravitational Field Strength — where we stop tracking two particular masses and ask a slicker question instead: how strong is the pull a planet offers at a point in space? That’s g = F ÷ m.
Want this to actually click before the exam?
Book a free meeting and let’s work through the tricky bits together.
