IB Physics HLTopic 4 — Force FieldsPaper 1 & 2T² ∝ r³~17 min read
Kepler’s Laws
Kepler had no telescope worth the name, no gravity, no calculus. What he had was twenty years of Tycho Brahe’s naked-eye observations and a stubborn refusal to make the numbers fit a circle. Out of that came three laws that describe every planet, every moon and every satellite ever launched — written down decades before Newton explained why they were true. Here they are, and here is how gravity delivers the third one on a plate.
📘 What you need to know
First law: the orbit of a planet is an ellipse, with the Sun at one of the two foci
Second law: a line joining the Sun to a planet sweeps out equal areas in equal times
Consequence of the second law: planets move faster nearer the Sun and slower further away
Third law: for bodies in circular orbit about the same central body, T² ∝ r³
In full: T² = 4π²r³ / GM, where M is the mass of the central body
It comes from equating the gravitational force to the centripetal force
A graph of log T against log r is a straight line of gradient 3/2, and it does not pass through the origin
An ellipse is a “squashed” circle. Earth’s orbit is nearly circular; Pluto’s is markedly elliptical
Kepler’s first law: orbits are ellipses
Kepler’s first lawthe orbit of a planet is an ellipse, with the Sun at one of the two foci
An ellipse has two foci. The Sun occupies one of them. The other is simply an empty point in space — nothing sits there, and nothing ever will. And the Sun is emphatically not at the centre.
Drawn here with a strong eccentricity so you can see the point. Earth’s real orbit is so close to circular that at this scale you could not tell it from a circle.
Kepler’s second law: equal areas in equal times
Kepler’s second lawa line segment joining the Sun to a planet sweeps out equal areas in equal time intervals
Imagine the planet trailing a rubber sheet back to the Sun. In any one month, that sheet covers the same amount of area — whether the planet is skimming past the Sun or crawling along at the far end of its orbit.
Near the Sun, the line is short. To sweep out the required area, the planet must travel a long way around. Far from the Sun the line is long, so a mere shuffle sideways sweeps the same area. Hence: fast near the Sun, slow far away.
These two wedges really are equal in area (they were computed, not sketched). But the arc near the Sun is about eight times longer, so the planet must race round it eight times faster.
Kepler found this by grinding through Mars data for years. We can see why in one line: the planet’s angular momentum about the Sun is conserved, because gravity always pulls straight at the Sun and so exerts no turning effect on it. The rate at which area is swept turns out to be exactly half the angular momentum per unit mass. Constant angular momentum, constant sweep rate, equal areas. Kepler’s second law is conservation of angular momentum, in disguise.
Kepler’s third law: T² ∝ r³
Kepler’s third lawfor planets or satellites in a circular orbit about the same central body, the square of the time period is proportional to the cube of the orbital radiusT2 ∝ r3
This one you can derive, and you should be able to on demand. Gravity supplies the centripetal force that holds a satellite in its circle. Nothing else is pulling it.
Step 1 — gravity provides the centripetal forceGMm / r2 = mv2 / r → v2 = GM / rthe orbiting mass m cancels — as it always does
Step 2 — write the speed in terms of the periodv = 2πr / T → v2 = 4π2r2 / T2one full circumference, 2πr, in one period T
Step 3 — set them equal and rearrange4π2r2 / T2 = GM / r → T2 = 4π2r3 / GMM = mass of the central body (kg) • r = orbital radius from its centre (m) • T = period (s)
Everything on the right except r3 is a constant for a given central body. So T² ∝ r³, exactly as Kepler found — except he found it by staring at tables of numbers, and we found it in three lines.
Read that final equation once more and notice what isn’t in it: the mass of the orbiting object. Swap the satellite for a bowling ball, or a fleck of paint, or the Moon — put any of them in the same orbit and they take exactly the same time to go round. The only mass that matters is the one at the centre.
Straightening the curve: the log graph
A plot of T against r is a curve, and curves are hard to test. Take logs of both sides of T² = 4π²r³/GM and the curve becomes a straight line:
Kepler’s third law, loggedlog T = (3/2) log r + ½ log(4π2/GM)a straight line of gradient 3/2, with a non-zero intercept set by the central mass
The gradient is 3/2 for every central body in the universe. Only the intercept changes, and it tells you the mass at the centre.
Law
What it says
What it gives you
First
Orbits are ellipses, Sun at one focus
The shape of the orbit
Second
Equal areas swept in equal times
The speed at each point: fast near, slow far
Third
T² ∝ r³ about the same central body
The period, and the central mass
Gravity is the centripetal force
so v² = GM/r
and v = 2πr/T
substitute and rearrange
T² = 4π²r³/GM
🪐 Using the third law
Identify the central body.M is its mass. The orbiting mass never appears.
Is r from the centre? Add the planet’s radius to any altitude given.
Convert the period to seconds. Days, years and hours are traps. 1 day = 86 400 s.
Comparing two orbits round the same body? Don’t substitute — write T1²/T2² = r1³/r2³ and watch GM cancel.
Finding a mass? Rearrange to M = 4π²r³/(GT²).
WE 1
A moon travels in a circular orbit of radius 2.4 × 10⁸ m around a planet, taking 5.6 days to complete one orbit. Determine the mass of the planet.
Step 1 — convert the period to secondsT = 5.6 × 86 400 = 4.84 × 10⁵ sStep 2 — rearrange the third law for MT² = 4π²r³/GM → M = 4π²r³ / (GT²)Step 3 — substituteM = 39.5 × (2.4 × 10⁸)³ / [(6.67 × 10⁻¹¹) × (4.84 × 10⁵)²]M = 5.46 × 10²⁶ / 15.6M = 3.5 × 10²⁵ kgAbout six times the mass of Jupiter. Notice we never needed the moon’s mass — if the question had given it, that would have been bait.
WE 2
Two satellites orbit the same star. Satellite X has an orbital radius four times that of satellite Y. Determine the ratio of their orbital periods, TX / TY.
Step 1 — same central body, so write the ratioT² ∝ r³ → TX²/TY² = rX³/rY³Step 2 — substitute the radius ratioTX²/TY² = 4³ = 64Step 3 — take the square rootTX/TY = √64TX / TY = 8Four times further out, eight times longer to go round. No G, no M, no calculator worth the name. Ratio questions are a gift — take them.
WE 3
A comet follows a highly elliptical orbit around the Sun. Use Kepler’s second law to explain why the comet moves fastest when it is closest to the Sun.
Step 1 — state the law
The line from the Sun to the comet sweeps out equal areas in equal times.
Step 2 — think about the length of that line
Close to the Sun the line is short. Far away it is long.
Step 3 — make the areas match
To sweep the same area with a short line, the comet must travel a long arc in that time.
Step 4 — concludea longer arc in the same time means a greater speedDon’t answer “because gravity is stronger there”. True, but it isn’t Kepler’s second law, and it isn’t what the question asked. Always answer with the law you were handed.
💡 Top tips
Learn all three laws word for word. IB asks you to state them, and marks the wording.
First law: “one of the two foci“. Not “the centre”, not “the middle”.
Second law: say “equal areas in equal times“, then explain the speed as a consequence.
Third law: it applies to bodies orbiting the same central body. That phrase is worth a mark.
Be able to deriveT² = 4π²r³/GM by equating gravitational and centripetal force. It’s a standard question.
Convert periods to seconds before doing anything else.
On a log–log plot, the gradient is 3/2 and the intercept is negative — the line misses the origin.
⚠ Common mistakes
Saying the Sun is at the centre of the ellipse. It is at a focus
Thinking something physical sits at the other focus. It is empty space
Stating the second law as “the planet moves faster near the Sun”. That is the consequence, not the law
Applying T² ∝ r³ across different central bodies — the constant 4π²/GM is different for each
Putting the orbiting body’s mass into M. M is always the central mass
Leaving T in days or years, or using an altitude as r
Expecting a plot of T against r to be straight, or a log–log plot to pass through the origin
Forgetting to square-root at the end of a ratio question — you found T², not T
Quick recap:First law — orbits are ellipses with the Sun at one focus. Second law — the Sun–planet line sweeps equal areas in equal times, so planets move fast near, slow far. Third law — for orbits about the same central body, T² ∝ r³, and equating gravity to the centripetal force gives T² = 4π²r³/GM. The orbiting mass cancels out. A log T against log r plot is straight with gradient 3/2 and a negative intercept.
Kepler’s third law tells you how fast a satellite must go to stay in a given orbit. Flip the question round: how fast must you go to leave altogether — to climb out of the potential well and never fall back? Set the kinetic energy you start with against the energy debt you must repay, and out drops one of the most quoted numbers in physics. Next page: Escape Speed.
Kepler’s laws not sticking?
Book a free meeting and we’ll practise the derivation, the ratio questions and the log-graph analysis together.
