IB Physics HLTopic 4 — Force FieldsPaper 1 & 2F = Gm1m2/r²~15 min read
Newton’s Law of Gravitation
Drop a pen and it falls. That much is obvious. What is not obvious — and what took a genius to see — is that the same pull that drags the pen to the floor is the pull holding the Moon in its orbit, and the pull keeping the Earth swinging around the Sun. One equation, three lines long, covers all of it. Every mass in the universe attracts every other mass, and this page tells you exactly how hard.
📘 What you need to know
Newton’s law of gravitation: the 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
G = 6.67 × 10−11 N m² kg−2 — the same everywhere in the universe
r is measured centre to centre, never from the surface
Gravity is always attractive. There is no such thing as gravitational repulsion
The two forces are equal in size, opposite in direction — a Newton’s third law pair
Inverse square law: double the separation and the force drops to a quarter
Planets and stars are treated as point masses at their centres
The law in words
Newton’s law of gravitation says this:
Newton’s law of gravitationthe gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation
Read that slowly, because it is really two separate claims bolted together.
Proportional to the product of the masses. Double either mass and you double the force. Double both and the force goes up four times.
Inversely proportional to the square of the separation. Move them twice as far apart and the force does not halve — it drops to a quarter.
Put both claims into symbols and you get the equation you will use in every question on this topic:
Newton’s law of gravitation, in symbolsF = G m1m2 / r2F = gravitational force (N) • m1, m2 = the two masses (kg) • r = separation of centres (m)
The constant G is Newton’s gravitational constant. It is tiny, and it never changes:
The gravitational constantG = 6.67 × 10−11 N m2 kg−2
Notice how small G is — that little 10−11 is why you don’t feel yourself being tugged towards the person sitting next to you. Gravity is by far the weakest of the fundamental forces. It only ever looks strong because planets are enormous. Take one mass up to 1024 kg and suddenly that tiny G has something to work with.
Two masses, two forces
Whenever you draw this situation, draw both arrows. Mass 1 pulls on mass 2, and mass 2 pulls back on mass 1, with exactly the same size of force. It does not matter that one is a boulder and the other is a pebble — the forces are equal and opposite. That is Newton’s third law, quietly doing its job.
The big mass does not pull “harder”. The two forces are a Newton’s third law pair, so they match exactly.
The inverse square law
This is the part students get wrong, so let’s kill the mistake now. The r in the equation is squared. So the force does not fade in a straight line as you move away — it collapses.
The inverse square lawF ∝ 1/r2twice as far → a quarter of the force • three times as far → a ninth
New separation
Factor on r
Factor on F
Force becomes
r
×1
1/1²
F
2r
×2
1/2²
F/4
3r
×3
1/3²
F/9
4r
×4
1/4²
F/16
Not a straight line, and not a halving. Squaring the distance in the denominator makes the force fall away fast — but never quite to zero.
Here’s the trick for ratio questions. Don’t reach for G and a calculator. Just ask: by what factor did r change? Then square it and divide. Moved from r to 5r? The force is F/25. That’s the whole calculation, done in your head, and the exam is happy.
Where do you measure r from?
From the centre. Always from the centre. This single sentence rescues more marks than anything else on this page.
A satellite orbiting 350 km “above the surface” is not 350 km from the Earth. It is 350 km plus the whole radius of the Earth from the Earth’s centre — which is nearly 6 720 km. Get this wrong and your answer is out by a factor of hundreds.
The altitude h is the bit above the ground. The equation wants r, the whole way down to the centre. Add them.
Why can we treat a planet as a point?
Fair question. Newton’s law is written for point masses, and the Earth is very much not a point. But two things save us.
A uniform sphere behaves, from the outside, exactly as though all its mass were squashed into a single point at its centre. Newton proved it.
The distances involved are huge compared with the sizes. The Earth–Sun separation is about 12 000 Earth diameters. At that range, a planet is a dot.
So we point our arrow at the centre, measure r from there, and the equation works for oranges, asteroids and galaxies alike.
Two masses
separated by r (centre to centre)
F = Gm1m2/r²
acting on each of them
Equal and opposite pull
🪐 Attacking a gravitation question
Write down the two masses. Which is m1 and which is m2 never matters — they multiply.
Find r properly. Is it given centre to centre, or is it an altitude? If it’s an altitude, add the radius.
Convert to metres. Kilometres in a squared denominator are a disaster. 6400 km = 6.4 × 106 m.
Square the r. Say it out loud if you have to. This is the number one lost mark.
Sanity check. A satellite feels kilonewtons, not meganewtons. Two lab masses feel less than a microewton.
WE 1
Two small lead spheres of mass 5.0 kg and 12 kg are placed so that their centres are 0.25 m apart. Calculate the gravitational force between them. Take G = 6.67 × 10−11 N m² kg−2.
Step 1 — write the equationF = Gm1m2 / r²Step 2 — substitute, squaring the rF = (6.67 × 10⁻¹¹) × 5.0 × 12 / (0.25)²F = (6.67 × 10⁻¹¹) × 60 / 0.0625F = 6.4 × 10⁻⁸ NSixty-four billionths of a newton. That is the entire gravitational grip between two lead spheres a hand’s width apart — which is exactly why you have never noticed it.
WE 2
A satellite of mass 1200 kg orbits 350 km above the Earth’s surface. The Earth has mass 5.97 × 10²⁴ kg and radius 6.37 × 10⁶ m. Calculate the gravitational force on the satellite.
Step 1 — find r, from the centre of the Earthr = R + h = (6.37 × 10⁶) + (350 × 10³)r = 6.72 × 10⁶ mStep 2 — substitute into the lawF = (6.67 × 10⁻¹¹) × (5.97 × 10²⁴) × 1200 / (6.72 × 10⁶)²F = 4.78 × 10¹⁷ / 4.52 × 10¹³F = 1.1 × 10⁴ NHad you used 350 km as the whole of r, you would have got about 3.9 × 10⁶ N — nearly four hundred times too big. Always add the radius.
WE 3
A rock of mass 4.0 kg resting on the surface of a spherical asteroid of radius 480 m experiences a gravitational force of 1.2 × 10−3 N. Determine the mass of the asteroid.
Step 1 — the rock is on the surface, so r = 480 mF = Gm1m2 / r²Step 2 — rearrange for the asteroid’s massm1 = F r² / (G m2)Step 3 — substitutem1 = (1.2 × 10⁻³) × (480)² / [(6.67 × 10⁻¹¹) × 4.0]m1 = 276.5 / (2.67 × 10⁻¹⁰)m1 = 1.0 × 10¹² kgA billion tonnes of rock, and it can barely hold onto a 4 kg stone. Gravity really is feeble until you make things planet-sized.
💡 Top tips
Square the r. Write the bracket and the little 2 before you touch the calculator.
Centre to centre. If a question says “above the surface”, you are being tested on r = R + h.
Convert km to metres first, while the number is still simple.
For ratio questions, ignore G entirely. Change r by a factor n, and F changes by 1/n².
Both masses feel the same size of force. Draw both arrows, pointing at each other.
The word is attraction. Never write “gravitational repulsion” — there is no such thing.
⚠ Common mistakes
Forgetting to squarer — by far the most common error in this whole topic
Using the altitude as r instead of adding the planet’s radius
Leaving distances in kilometres, so the answer is out by 106
Confusing G (the universal constant, 6.67 × 10−11) with g (field strength, 9.81 on Earth). They are completely different quantities
Thinking the bigger mass pulls harder — the forces are equal and opposite
Halving the force when the distance doubles. It goes to a quarter
Saying gravity “runs out” far away. It gets very small, but it is never zero
Quick recap: Every pair of masses attracts with F = Gm1m2/r², where G = 6.67 × 10−11 N m² kg−2 and r is measured centre to centre. The force on each mass is the same size and points at the other one. Because r is squared, moving twice as far apart leaves only a quarter of the force — and spheres can be treated as point masses sitting at their centres.
You now have the force between any two masses. But carrying a second mass around in the equation is clumsy — what if we want to describe what the Earth does to any object placed near it, whatever its mass? That question turns a force into a field, and the answer is the next page: Gravitational Field Strength.
Losing marks on the squared r?
Book a free meeting and we’ll drill gravitation questions until centre-to-centre distances and inverse squares are automatic.
