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

Gravity and its field

Predict what happens to the pull between two bodies when they move apart, which body feels the bigger force, how the field of a planet weakens with distance, and where the pulls of two bodies cancel. Then test each prediction with the simulation in each stage.

1 Predict · 7
2 Newton’s law · 16
3 Field strength · 16
4 Net field zero · 18
5 Field lines · 13
6 Show + reflect · 10

Learning goals: by the end you can…

use Newton’s law of gravitation F = GMm/r², with r measured from centre to centre, and state when a body can be treated as a point mass.
explain with Newton’s third law why two bodies pull each other with forces of the same size.
state the definition of gravitational field strength word for word and use g = GM/r².
find the point on the line between two bodies where the net gravitational field is zero.
sketch field lines for a sphere and close to a planet’s surface.

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 = G m₁m₂/r²g = F/mg = GM/r²G = 6.67 × 10⁻¹¹ N m² kg⁻²

r is the distance between the centres of the two bodies. Earth: M = 6.0 × 10²⁴ kg, R = 6.4 × 10⁶ m.

2. Newton’s law: test predictions 1.1 and 1.2

16 minPairs → individual

ATL Thinking: find a pattern by changing one variable at a time. Communication: use Newton’s third law in words.

Simulation A · Two bodies pull each other the starting values are the Earth and the Moon

mass M mass m distance r

force on each body, F—

acceleration of m, F/m—

acceleration of M, F/M—

Newton’s universal law of gravitation: every two point masses attract each other with a force F = G m₁m₂/r² where r is the distance between them (between their centres for spheres). The force is always attractive and acts along the line joining them.

2.1M = 6.0 × 10²⁴ kg, m = 7.4 × 10²² kg, r = 3.8 × 10⁸ m (the starting values). F =

2.2Set r = 7.6 × 10⁸ m (double). (new F) ÷ (old F) =

2.3Look at the two red arrows while you change M. What happens?

Both arrows change together and always stay the same length. Only the arrow on m changes. The arrow on the bigger mass is always longer.

2.4Back to the starting values. Acceleration of the Moon, a = F/m =

m s⁻²

Back check

2.bCheck: GM/r² =

N kg⁻¹

Find the mistake

Two spheres of radius 1.0 m and mass 500 kg each sit with a 2.0 m gap between their surfaces. A student writes:

F = G × 500 × 500 / 2.0² = 4.2 × 10⁻⁶ N

What is wrong?

r must be measured from centre to centre: r = 1.0 + 2.0 + 1.0 = 4.0 m, so F = 1.0 × 10⁻⁶ N. The masses must be added, not multiplied. Nothing is wrong.

Hinge question

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

A 60 kg student stands on the Earth. Which statement is correct?

A. The Earth pulls the student with 590 N and the student pulls the Earth with 590 N. B. The Earth pulls the student with 590 N; the student does not pull the Earth. C. The student pulls the Earth with a tiny force, much less than 590 N.

2.5Were predictions 1.1 and 1.2 right? Use numbers from the simulation.

What a good answer looks like

1.1: B, one quarter. F ∝ 1/r², so doubling r gives F/4: 2.1 × 10²⁰ N became 5.1 × 10¹⁹ N. 1.2: B, the same size. The forces are a Newton’s third law pair, both equal to GMm/r². The Moon accelerates much more (2.8 × 10⁻³ m s⁻²) only because its mass is much smaller.

3. Gravitational field strength: test prediction 1.3

16 minPairs → individual

ATL Research: learn a definition in the exact words. Thinking: use ratios instead of long calculations.

Simulation B · The field of a planet move the small test mass away; the graph shows g against r/R

planet mass M planet radius R distance from the centre r

distance from the centre, r—

field strength, g—

g ÷ g(surface) = (R/r)²—

Learn it word for word

Gravitational field strength g at a point is the force per unit mass experienced by a small point mass at that point. g = F/m, unit N kg⁻¹ (= m s⁻²). It is a vector, pointing towards the mass that makes the field.

Build g = GM/r² from Newton’s law

Put a small test mass m at distance r from the centre of a planet of mass M.
Newton: the force on it is F = GMm/r².
Field strength is force per unit mass: g = F/m = GMm/r² ÷ m.
m cancels: g = GM/r². It depends on the planet (M) and the distance (r), not on the test mass.

3.1Earth values: M = 6.0 × 10²⁴ kg, R = 6.4 × 10⁶ m, test mass at the surface (r = R). g =

N kg⁻¹

3.2Move the test mass to r = 2R. g =

N kg⁻¹

3.3A space station orbits 400 km above the surface. g there =

N kg⁻¹

Back check

3.bCheck with a ratio: g = 9.77 × (R/r)² =

N kg⁻¹

Find the mistake

Find g 400 km above the Earth. A student writes:

g = GM/h² = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ / (4.0 × 10⁵)² = 2500 N kg⁻¹

What is wrong?

r is measured from the centre: r = R + h = 6.8 × 10⁶ m, so g = 8.7 N kg⁻¹. G should be 9.8. h should be in km.

What if? · stretch

3.wA planet has twice the Earth’s mass and twice its radius. Use ratios with g = GM/R². g at its surface =

N kg⁻¹

3.4Was prediction 1.3 right? Explain with g = GM/r².

What a good answer looks like

C, about 2.5 N kg⁻¹. g = GM/r², so g ∝ 1/r². Doubling r from R to 2R divides g by 4: 9.8/4 = 2.4 N kg⁻¹. g is not a constant: 9.8 N kg⁻¹ is only its value at the Earth’s surface.

4. Where the net field is zero: test prediction 1.4

18 minPairs → individual

ATL Thinking: turn “the pulls balance” into an equation. Communication: explain with vectors.

Simulation C · Where the pulls cancel M = 6.0 × 10²⁴ kg, d = 3.8 × 10⁸ m; move the probe along the line

mass ratio M/m probe position x (from M)

distance from M, x—

field of M, GM/x²—

field of m, Gm/(d − x)²—

net field—

Important note. The net field is zero where the pull from one side equals the pull from the other: GM/x² = Gm/(d − x)². G cancels. Take the square root: x/(d − x) = √(M/m). The point is always closer to the smaller mass.

4.1Leave M/m = 81 (the Earth and the Moon). Move the probe until the net field is zero. x =

4.2Now calculate it. x/(d − x) = √(M/m). x =

Back check

4.bCheck that the two pulls are equal there. GM/x² =

N kg⁻¹

4.3Set M/m = 1 (two equal masses). Where is the net field zero?

exactly half way nowhere closer to M

Find the mistake

Mass M is 4 times mass m, and they are 12 m apart. A student writes:

x/(d − x) = M/m = 4, so x = 4(12 − x), x = 9.6 m from M

What is wrong?

The field goes as 1/r², so take the square root: x/(d − x) = √4 = 2, x = 8.0 m. The point must be half way: 6.0 m. Nothing is wrong.

Hinge question

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

At the point where the net field is zero, a student says “the two fields add up to zero, so I add GM/x² + Gm/(d − x)² = 0”. What is the right idea?

A. The fields point in opposite directions, so we set their sizes equal: GM/x² = Gm/(d − x)². B. The student is right. C. We set the masses equal.

4.4Was prediction 1.4 right? Explain why the point is not half way.

What a good answer looks like

C, closer to the Moon. The Moon’s mass is 81 times smaller, so its pull is weaker at the same distance. To balance the Earth’s pull, the probe must be much closer to the Moon: x/(d − x) = √81 = 9, so the point is 9 times further from the Earth than from the Moon (x = 0.90 d = 3.4 × 10⁸ m).

5. Field lines and point masses

13 minPairs → individual

ATL Communication: sketch and read field lines. Thinking: decide when a model (point mass) applies.

Field lines and point massesField lines show the direction of the force on a small mass. For gravity they always point towards the mass, meet the surface at 90° and never cross. Around a sphere they are radial: they spread out, so the field gets weaker. Close to the surface of a planet they are parallel and equally spaced: the field is nearly uniform. A sphere of uniform density acts as a point mass at its centre; a small body (a satellite) acts as a point mass when it is much smaller than the distance.

5.1Which statement about gravitational field lines is correct?

Where the lines are closer together the field is stronger. Field lines point away from a mass. Two field lines can cross where two fields meet.

5.2Why is the field close to the Earth’s surface treated as uniform in a classroom?

Over a few metres r hardly changes compared with 6.4 × 10⁶ m, so g hardly changes. The Earth is flat. g does not depend on r.

5.3A planet has 0.11 times the Earth’s mass and 0.53 times its radius. Use ratios: g = 9.8 × (M ratio)/(R ratio)² =

N kg⁻¹

5.4Weight of a 70 kg astronaut on that planet =

5.5A 3.0 × 10³ kg probe is 2.0 × 10⁶ m from the centre of a moon of mass 8.0 × 10²² kg. Force on the probe =

5.6Exam style (2 marks): state the two conditions that let us use F = GMm/r² for a satellite orbiting a planet.

What a good answer looks like

The planet is spherical (and of uniform density), so it acts as a point mass at its centre ✔; the satellite is much smaller than the planet / than the distance between them ✔.

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…

use the definition of gravitational field strength word for word;
measure r from the centre of the planet;
use 1/r² as a ratio at least once;
use one number from a simulation today.

Option A · Written

One paragraph: “Why the point where a spacecraft feels no net pull is much closer to the Moon than to the Earth.”

Option B · Sketch

The field lines of the Earth and the Moon, with the neutral point marked and the two pulls drawn there as equal and opposite arrows.

Option C · Spoken

Explain to your partner in 1 minute why g at 2R is a quarter of g at the surface. Your partner ticks the success criteria.

6.1Write your paragraph or your notes.

What a good answer looks like

Model answer (Option A): Gravitational field strength is the force per unit mass experienced by a small point mass at that point; for a sphere g = GM/r², with r from the centre. On the line between the Earth and the Moon the two fields point in opposite directions, so the net field is zero where their sizes are equal: GM/x² = Gm/(d − x)². The Moon’s mass is 81 times smaller, so x/(d − x) = √81 = 9 and the point is at 0.90 d = 3.4 × 10⁸ m from the Earth, much closer to the Moon. In the simulation the net field read ≈ 0 at x = 0.90 d.

Exit ticket (no notes)

1Two bodies attract with 36 N. The distance between their centres is tripled. The new force, in N =

2g at the surface of a planet is 12 N kg⁻¹. g at 2R from its centre, in N kg⁻¹ =

3Mass A is 9 times mass B. They are 20 m apart. Distance from A to the point where the net field is zero, in m =

4Gravitational field strength is defined as…

the force per unit mass experienced by a small point mass at that point the acceleration of any body near a planet the force on a mass in a field

5The Earth pulls a satellite with a force F. The satellite pulls the Earth with a force of…

F much less than F zero

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 we take the square root when finding the zero-field point.”

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

What a good answer looks like

A good feedforward names an action: “Next time I will add the planet’s radius to the height before I use g = GM/r².”

TOK · 2 minutes

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

TOKNewton never explained WHY masses attract; he gave a law that predicts how strongly they do. Is a law that describes without explaining real knowledge?

Points you could make

Yes in a practical sense: the law predicts orbits, tides and spacecraft paths very accurately, and it can be tested.
Newton himself said “I frame no hypotheses” about the cause; later, Einstein’s general relativity offered a deeper picture (curved space-time).
Science often starts with a good description; explanations can come later, and they are also tested.

Read the notes Back to D.1