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D.1 Gravitational fields · SL + HL · Part 2 of 3 · student investigation · 80 minutes

Orbits and Kepler’s laws

Predict how fast a planet moves near the Sun, how the speed and period of a satellite change with the size of its orbit, and why astronauts float. Then test each prediction with a simulation and derive the most common “show that” in D.1: T² = 4π²r³/GM.

1 Predict · 7
2 Kepler 1 and 2 · 14
3 Circular orbits · 20
4 Weigh a planet · 16
5 Weightless · 13
6 Show + reflect · 10

Learning goals: by the end you can…

state Kepler’s three laws in the correct words (ellipse with the Sun at a focus; equal areas in equal times; T² ∝ r³).
explain that for a circular orbit gravity provides the centripetal force, and derive v = √(GM/r).
derive T² = 4π²r³/(GM) step by step, as in a “show that” question.
use T² ∝ r³ to compare two orbits, and find the mass of a planet from the period and radius of a moon.
explain why astronauts in orbit feel weightless.

How to work in this lesson

How to work. Pairs discuss and agree, but each of you writes. Pairs → individual discuss the method, then write your own answer. Individual no discussion until your teacher says so. Your typed answers are saved on this device only.

Everything you need is on this page. Each simulation uses the same numbers as the questions next to it, and stays locked until you have made your prediction. Type powers of ten like 3.2×10^-19 or 3.2e-19.

📘 From the data booklet (D.1)

F = GMm/r²v = √(GM/r)T² = 4π²r³/(GM)v = 2πr/T (A.2)a = v²/r = ω²r (A.2)

In the data booklet T² ∝ r³ appears as Kepler’s third law; for a circular orbit the semi-major axis is the radius r (from the centre).

2. Kepler’s first and second laws: test prediction 1.1

14 minPairs → individual

ATL Research: learn three laws in exact words. Thinking: turn a picture (equal areas) into an equation.

Simulation A · An elliptical orbit e = 0 is a circle; P = closest point (perihelion), A = furthest point (aphelion)

eccentricity e shade equal times

r(A) ÷ r(P)—

speed at P ÷ speed at A—

speed now—

Learn it word for word: Kepler’s three laws

First law: each planet moves on an ellipse with the Sun at one focus.
Second law: the line joining a planet to the Sun sweeps out equal areas in equal times.
Third law: the square of the orbital period is proportional to the cube of the semi-major axis of the orbit: T² ∝ a³ (for a circle, a = r).

2.1Set e = 0.60. r(A) ÷ r(P) =

Equal areas give the speedsIn a short time Δt near P the planet moves a distance v_PΔt; the swept area is a thin triangle of area ≈ ½ r_P v_PΔt. The same near A: ½ r_A v_AΔt. Equal areas: r_P v_P = r_A v_A, so v_P/v_A = r_A/r_P.

2.2A comet is 1.0 × 10¹¹ m from the Sun at P and 4.0 × 10¹¹ m at A. (speed at P) ÷ (speed at A) =

Back check

2.bCheck with the simulation (e = 0.60). speed at P ÷ speed at A =

Find the mistake

A student writes Kepler’s first two laws:

“A planet moves on an ellipse with the Sun at the centre. It moves through equal arc lengths in equal times.”

What is wrong?

The Sun is at a focus, not the centre, and it is equal AREAS (not arc lengths) in equal times. Planets move on circles. Nothing is wrong.

2.3Was prediction 1.1 right? Explain with Kepler’s second law.

What a good answer looks like

A, closest to the Sun. Equal areas in equal times: near the Sun the radius is short, so to sweep the same area the planet must cover a longer arc in the same time: it moves faster. With r_A = 4r_P, v_P = 4v_A.

3. Circular orbits and T² ∝ r³: test predictions 1.2 and 1.3

20 minPairs → individual

ATL Thinking: build the most common derivation of D.1 step by step. Communication: show every line in a “show that”.

Simulation B · A satellite in a circular orbit Earth: M = 6.0 × 10²⁴ kg, R = 6.4 × 10⁶ m; r is measured from the centre

orbit radius r

orbital speed v—

period T—

T² ÷ r³—

Show that T² = 4π²r³/(GM): learn these steps

For a circular orbit, gravity provides the centripetal force: GMm/r² = mv²/r.
Cancel m and one r: v² = GM/r, so v = √(GM/r).
One orbit is a distance 2πr in a time T: v = 2πr/T, so v² = 4π²r²/T².
Equate: 4π²r²/T² = GM/r. Rearrange: T² = 4π²r³/(GM). Since 4π²/GM is a constant for one central mass, T² ∝ r³ (Kepler’s third law).

How the marks go (2–3 marks): ✔ clear evidence that the centripetal force is equated to the gravitational force (GMm/r² = mv²/r or mω²r); ✔ v = 2πr/T (or ω = 2π/T) substituted and rearranged to the result. Do not start from the answer, and do not use equations of simple harmonic motion.

3.1r = 6.8 × 10⁶ m (400 km above the surface). Orbital speed v =

m s⁻¹

3.2Period T, in minutes =

min

3.3A geostationary satellite has T = 24 h = 86 400 s. Use T² = 4π²r³/(GM). r =

Back check

3.bCheck with the simulation: set r = 42.3 × 10⁶ m. T =

3.4Move r to three different values. T²/r³ stays the same. Its value =

s² m⁻³

What if? · stretch

3.wSatellite B has 4 times the orbit radius of A. T_B ÷ T_A =

Find the mistake

A satellite is 600 km above the Earth. A student writes:

v = √(GM/r) = √(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / 6.0 × 10⁵) = 2.6 × 10⁴ m s⁻¹

What is wrong?

r = R + h = 7.0 × 10⁶ m, so v = 7.6 × 10³ m s⁻¹. v = √(2GM/r) for orbits. Nothing is wrong.

Hinge question

Choose your answer. You will see at once if it is right, and why.

Two satellites orbit the Earth. X is in a higher orbit than Y. Which is correct?

A. X is slower and has a longer period. B. X is faster and has a longer period. C. X is slower and has a shorter period.

3.5Were predictions 1.2 and 1.3 right? Use the simulation and the equations.

What a good answer looks like

1.2: B, smaller. Gravity (the centripetal force) is weaker higher up, so v = √(GM/r) is smaller: 7.7 km s⁻¹ at 6.8 × 10⁶ m but 3.1 km s⁻¹ at 4.2 × 10⁷ m. 1.3: B, 8 times. T² ∝ r³, so T ∝ r^(3/2) and 4^(3/2) = 8.

4. Weigh a planet with Kepler’s third law

16 minPairs → individual

ATL Thinking: use an equation in reverse. Research: get a hidden quantity from measurements.

Simulation C · Weigh a mystery planet choose the radius of a moon’s orbit and time one orbit

orbit radius r

period T—

T² ÷ r³—

Use Kepler’s third law the other wayRearrange T² = 4π²r³/(GM): M = 4π²r³/(GT²). Time a moon (T) and measure its orbit (r), and you have “weighed” the planet. The moon’s own mass cancels, so it is not needed.

4.1Leave r = 4.2 × 10⁸ m. Period T =

4.2Mass of the planet, M =

4.3Set r = 8.4 × 10⁸ m (double). (new T) ÷ (old T) =

Back check

4.bCheck: T²/r³ = 4π²/(GM) =

s² m⁻³

4.4Our Moon orbits the Earth at 3.84 × 10⁸ m with a period of 27.3 days. Mass of the Earth =

4.5A graph of T² against r³ for the moons of a planet is a straight line through the origin. Its gradient is…

4π²/(GM) GM/(4π²) 4π²/G

4.6Exam style: explain how observations of a moon can be used to find the mass of its planet.

What a good answer looks like

Measure the period T and the orbit radius r of the moon ✔. Gravity provides the centripetal force: GMm/r² = m(2πr/T)²/r ✔, so M = 4π²r³/(GT²) ✔; the mass of the moon cancels.

5. Weightlessness: test prediction 1.4

13 minPairs → individual

ATL Communication: separate “has no weight” from “feels weightless”. Thinking: use free fall from A.1.

Weightless, not without gravityAt 400 km, g is still about 8.7 N kg⁻¹ (Part 1). The astronaut and the station both fall freely towards the Earth with the same acceleration g; their sideways speed keeps them going round instead of hitting the ground. The floor does not need to push on the astronaut: the normal (contact) force is zero, so the astronaut feels weightless. The weight (the gravitational force) is still there: it provides the centripetal force.

5.1In the space station, the force of gravity on a 70 kg astronaut is about…

610 N 0 N 690 N

5.2A scale under the astronaut’s feet reads…

zero, because the astronaut and the scale fall with the same acceleration 610 N less than 610 N but not zero

5.3The same feeling happens on Earth when…

a lift falls freely after its cable breaks a lift moves up at constant speed you stand still on the ground

5.4Speed of a geostationary satellite (r = 4.2 × 10⁷ m), v =

m s⁻¹

5.5Was prediction 1.4 right? Explain.

What a good answer looks like

B. Gravity is still there (g ≈ 8.7 N kg⁻¹ at 400 km), but the astronaut and the station fall together with the same acceleration, so there is no contact force between them and the astronaut feels weightless.

6. Show it, then reflect

10 minIndividual

ATL Communication + Self-management: choose how to show your understanding, retrieve without notes, plan your next step.

Success criteria: your answer must…

start a derivation with “gravity provides the centripetal force” and an equation;
show every line of T² = 4π²r³/(GM);
state at least one of Kepler’s laws with the exact key word (focus, equal areas, cube);
use one number from a simulation today.

Option A · Written

The full “show that” for T² = 4π²r³/(GM), then use it to find the radius of a geostationary orbit.

Option B · Sketch

An ellipse with the Sun at a focus and two equal shaded areas, labelled with Kepler’s three laws.

Option C · Spoken

Explain to your partner in 1 minute how to find the mass of Jupiter from one of its moons. Your partner ticks the success criteria.

6.1Write your derivation or your notes.

What a good answer looks like

Model answer (Option A): Gravity provides the centripetal force: GMm/r² = mv²/r, so v² = GM/r. One orbit takes T, so v = 2πr/T and 4π²r²/T² = GM/r, giving T² = 4π²r³/(GM). For T = 86 400 s: r³ = GMT²/(4π²) = 7.6 × 10²² m³, so r = 4.2 × 10⁷ m, about 36 000 km above the surface. The simulation gave T = 24.0 h at r = 42.3 × 10⁶ m.

Exit ticket (no notes)

1A satellite orbits at r = 8.0 × 10⁶ m around the Earth (GM = 4.0 × 10¹⁴ N m² kg⁻¹). Its speed, in m s⁻¹ =

2Moon Y orbits at 9 times the radius of moon X around the same planet. T_Y ÷ T_X =

3A comet is 2.0 × 10¹¹ m from the Sun at its closest and 6.0 × 10¹¹ m at its furthest. (speed closest) ÷ (speed furthest) =

4Kepler’s second law says that the line from the Sun to a planet sweeps out…

equal arc lengths in equal times equal areas in equal times equal angles in equal times

5The first step of the derivation of T² = 4π²r³/(GM) is…

gravitational force = centripetal force kinetic energy = potential energy a = −ω²x

6Which idea from today is still the least clear to you?

What a good answer looks like

There is no single right answer. A good answer names one idea exactly, for example: “I am not sure why the moon’s mass does not matter when we find the planet’s mass.”

7Feedforward: Next time I will check ________ before I ________.

What a good answer looks like

A good feedforward names an action: “Next time I will write “gravity provides the centripetal force” as my first line before I rearrange.”

TOK · 2 minutes

Theory of knowledge: how do we know? There is no single right answer. Write two or three sentences.

TOKKepler found his three laws from years of careful observations by Tycho Brahe, before anyone could explain them. Newton later derived them from gravity. Which gives stronger knowledge: a pattern found in data, or a law derived from a theory?

Points you could make

A pattern from data (induction) is reliable within the range of the data, but it does not say why, or whether it holds elsewhere.
A derivation from a theory links many facts (falling apples, the Moon, planets) and predicts new ones, such as the mass of a planet from its moon.
The two support each other: the theory is accepted because it reproduces the observed patterns.

Read the notes Back to D.1