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

Potential and escape

Predict the sign of the gravitational potential, what happens to it as you move away from a planet, whether a heavier rocket needs a bigger escape speed, and what air resistance does to a satellite. Then test each prediction with a simulation.

1 Predict · 7
2 Potential · 18
3 V–r graphs · 16
4 Escape · 16
5 Orbit energy · 13
6 Show + reflect · 10

Learning goals: by the end you can…

state the definitions of gravitational potential and gravitational potential energy word for word.
explain why gravitational potential and potential energy are always negative.
use W = mΔV for the work done against the field, and g = −ΔV/Δr to read the field from a V–r graph (zero field where the gradient is zero).
derive and use the escape speed v = √(2GM/r) from energy conservation.
use E_T = −GMm/2r and describe what a small drag force does to an orbit.

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)

E_p = −GMm/rV_g = −GM/rg = −ΔV_g/ΔrW = mΔV_gv_esc = √(2GM/r)v_orbit = √(GM/r)

Earth: GM = 4.0 × 10¹⁴ N m² kg⁻¹, R = 6.4 × 10⁶ m. V = 0 at infinity.

2. Potential is negative: test predictions 1.1 and 1.2

18 minPairs → individual

ATL Research: learn two definitions in exact words. Communication: explain a sign with the markscheme points.

Simulation A · The potential well of a planet Earth: GM = 4.0 × 10¹⁴ N m² kg⁻¹, R = 6.4 × 10⁶ m

distance from the centre r probe mass m

potential V = −GM/r—

potential energy E_p = mV—

work needed to take it to infinity—

Learn it word for word

Gravitational potential V_g at a point is the work done per unit mass in bringing a small mass from infinity to that point. V_g = −GM/r, unit J kg⁻¹, a scalar.

Gravitational potential energy E_p of a system is the work done to assemble the system from infinite separation of the components of the system. E_p = −GMm/r = mV_g.

Important note: why the potential is always negative. (1) The potential is defined as zero at infinity. (2) Gravity is always attractive, so to take a mass from any point out to infinity, positive work must be done on it (energy supplied). (3) So the mass has less energy at that point than at infinity: less than zero, which is negative. (4) As r increases, V rises towards zero (it becomes less negative) but never becomes positive.

2.1Potential at the Earth’s surface, V =

J kg⁻¹

2.2Move the probe to r = 2R. V =

J kg⁻¹

2.3Potential energy of the 1000 kg probe at the surface, E_p =

Work in a field: two ways to say itWork done against the field (by an external force, at constant speed): W = mΔV_g = m(V_final − V_initial). Moving away from a planet, ΔV > 0, so this work is positive: we supply energy. Work done by the field (by gravity) is the opposite: −mΔV_g. Gravity does negative work on a mass that moves away, and positive work on a mass that falls in.

2.4Work done against the field to lift the probe from R to 2R, W = mΔV =

Back check

2.bCheck: read E_p at 2R and at R from the simulation. E_p(2R) − E_p(R) =

Find the mistake

Find the energy needed to lift a 500 kg probe from the surface to 2R. A student writes:

W = mgh = 500 × 9.8 × 6.4 × 10⁶ = 3.1 × 10¹⁰ J

What is wrong?

g is not constant over such a height. Use W = mΔV = 500 × (GM/R − GM/2R) = 1.6 × 10¹⁰ J. h should be 2R. Nothing is wrong.

Hinge question

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

Which statement explains why the gravitational potential near a planet is negative?

A. Zero potential is at infinity, and positive work must be done to move a mass from the point to infinity, because gravity attracts. B. The field points towards the planet, so the potential points inwards too. C. Potential is negative because energy is lost as heat.

2.5Were predictions 1.1 and 1.2 right? Explain with the simulation.

What a good answer looks like

1.1: C, negative: −6.25 × 10⁷ J kg⁻¹ at the surface. Zero is at infinity, and work must be done on a mass to take it there against the attraction, so the potential everywhere else is below zero. 1.2: A, it increases: −6.25 × 10⁷ → −3.1 × 10⁷ J kg⁻¹, closer to zero.

3. Reading V–r graphs: the field is minus the gradient

16 minPairs → individual

ATL Thinking: read a gradient from a graph. Communication: link two methods (graph and balance of pulls).

Simulation B · The V–r graph between the Earth and the Moon d = 3.8 × 10⁸ m; the green line is the tangent at the probe

probe position x (from the Earth)

potential V—

gradient ΔV/Δx—

field g = −gradient—

Field and potential: g = −ΔV_g/Δr The field strength is minus the gradient (the slope of the tangent) of the V–r graph. Steep graph → strong field. Flat graph (zero gradient) → zero net field.

3.1Move the probe until the tangent is horizontal. Where is it?

at the top of the hump, x = 0.90 d half way, x = 0.50 d next to the Earth

3.2Potential at the top of the hump (x = 3.42 × 10⁸ m) =

J kg⁻¹

3.3At x = 0.50 d, in which direction is the net field, and why?

towards the Earth: V rises towards the Moon, so g = −gradient points back towards the Earth towards the Moon zero

3.4For a single planet, V = −GM/r, so the gradient at r = R is GM/R². Find g at the Earth’s surface from the gradient =

N kg⁻¹

Equipotential surfacesAn equipotential surface joins points with the same potential: around a sphere, spheres. Field lines cross equipotentials at 90°. Moving along an equipotential, ΔV = 0, so no work is done (W = mΔV = 0). Drawn with equal steps of V, equipotentials are closer together where the field is stronger.

3.5A satellite moves along a circular orbit, which is an equipotential. Work done by gravity in one orbit =

Find the mistake

A student uses g = −ΔV/Δr to “show that” g = GM/r²:

g = −ΔV/Δr = −(−GM/r)/r = GM/r²

What is wrong?

The deltas cannot just be cancelled: Δr is a small change, not r. Markschemes do not allow this. The minus sign is wrong. Nothing is wrong.

3.6Exam style (2 marks): explain how the V–r graph between the Earth and the Moon shows where the net field is zero.

What a good answer looks like

The field strength is minus the gradient of the V–r graph (g = −ΔV/Δr) ✔; the net field is zero where the gradient is zero, at the maximum (top of the hump) of the curve, 0.90 d from the Earth ✔.

4. Escape speed: test prediction 1.3

16 minPairs → individual

ATL Thinking: use energy conservation instead of forces. Communication: write a derivation that earns the marks.

Simulation C · Launch a probe straight up no air; the bars show energy per kg: E_k (green), E_p (red), total (blue)

launch speed

total energy per kg—

highest point (from the centre)—

escape speed √(2GM/R)—

Learn it word for word

Escape speed is the minimum speed a body must have at the surface to reach infinity (to escape the planet’s gravitational field), with no further energy supplied. Not “to leave the orbit” and not “to escape the atmosphere”.

Show that v_esc = √(2GM/r)

To just reach infinity, the body arrives there with E_k = 0, and at infinity E_p = 0: its total energy is zero.
Energy is conserved, so the total energy at launch is also zero: ½mv² − GMm/r = 0.
m cancels: v² = 2GM/r, so v_esc = √(2GM/r). In terms of the potential: v = √(−2V).
Do not equate forces (GMm/r² = mv²/r) and then add a factor of 2: that gives no marks. Escape is an energy argument.

4.1Escape speed from the Earth’s surface =

m s⁻¹

4.2Launch at 7.9 km s⁻¹. The probe stops at r = (in units of R)

Back check

4.bCheck with algebra: launched at v_esc/√2, r_max =

4.3The Moon has mass 7.4 × 10²² kg and radius 1.74 × 10⁶ m. Its escape speed (ignore the Earth) =

m s⁻¹

4.4Why does the escape speed not depend on the mass of the rocket?

Both E_k = ½mv² and E_p = −GMm/r are proportional to m, so m cancels. Heavy rockets have more fuel. Gravity does not act on rockets.

4.5Was prediction 1.3 right? Explain with energy.

What a good answer looks like

B, the same. To escape, the total energy must be at least zero: ½mv² − GMm/R = 0. The mass m appears in both terms and cancels, so v_esc = √(2GM/R) = 11 km s⁻¹ for any rocket. B needs twice the energy, but not a bigger speed.

5. Energy of an orbit and drag: test prediction 1.4

13 minPairs → individual

ATL Thinking: follow energy through a process. Communication: use the three markscheme points for drag.

Energy of a circular orbitFrom v² = GM/r: E_k = ½mv² = GMm/2r. E_p = −GMm/r. E_T = E_k + E_p = −GMm/2r. So E_p = −2E_k and E_T = −E_k. The total is negative: the satellite is bound to the planet.

5.1An 800 kg satellite orbits at r = 7.0 × 10⁶ m. Total energy E_T =

5.2Its kinetic energy E_k =

A small drag forceAir resistance does negative work, so the total energy decreases (becomes more negative). E_T = −GMm/2r more negative means r decreases: the orbit slowly spirals in. Then v = √(GM/r) increases: the satellite speeds up. E_p falls by twice as much as E_k rises.

5.3A satellite loses 1.0 × 10⁸ J to air resistance and moves to a slightly lower orbit. Its kinetic energy…

increases by 1.0 × 10⁸ J decreases by 1.0 × 10⁸ J does not change

Find the mistake

A student explains drag on a satellite:

“Air resistance slows the satellite down, so its kinetic energy decreases and it falls to a lower orbit.”

What is wrong?

The TOTAL energy decreases; the orbit radius decreases and the speed INCREASES, since v = √(GM/r). Air resistance speeds satellites up directly. Nothing is wrong.

5.4Work done against the field to lift a 500 kg probe from the surface to r = 3R =

5.5Was prediction 1.4 right? Explain with E_T = −GMm/2r.

What a good answer looks like

C, it increases. Drag reduces the total energy, so E_T = −GMm/2r becomes more negative, which needs a smaller r. In a smaller orbit v = √(GM/r) is larger. The lost potential energy (twice as much) pays for the extra kinetic energy and the work done by drag.

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…

state the definition of gravitational potential word for word;
explain the negative sign with “zero at infinity” and “positive work must be done”;
derive the escape speed from energy, not forces;
use one number from a simulation today.

Option A · Written

One paragraph: “Why the gravitational potential near the Earth is negative, and what that means for escaping.”

Option B · Sketch

The V–r graph of the Earth and the Moon, with the hump, the horizontal tangent, and arrows showing the field direction on each side.

Option C · Spoken

Explain to your partner in 1 minute why a satellite speeds up when air resistance acts on it. Your partner ticks the success criteria.

6.1Write your paragraph or your notes.

What a good answer looks like

Model answer (Option A): Gravitational potential is the work done per unit mass in bringing a small mass from infinity to that point. It is defined as zero at infinity. Gravity is always attractive, so positive work must be done on a mass to take it from near the Earth to infinity; the mass therefore has less energy near the Earth than at infinity, so the potential is negative: −6.25 × 10⁷ J kg⁻¹ at the surface. To escape, a body needs a total energy of at least zero: ½mv² − GMm/R = 0, so v = √(2GM/R) = 11 km s⁻¹. In the simulation, launching at 7.9 km s⁻¹ (total −3.1 × 10⁷ J kg⁻¹) the probe stopped at 2R and fell back.

Exit ticket (no notes)

1A planet has GM = 2.0 × 10¹³ N m² kg⁻¹ and radius 2.0 × 10⁶ m. The potential at its surface, in J kg⁻¹ =

2The escape speed from that planet, in m s⁻¹ =

3The potential changes from −8.0 × 10⁶ to −5.0 × 10⁶ J kg⁻¹. Work done against the field on a 20 kg mass, in J =

4On a V–r graph, the net gravitational field is zero where…

V is zero the gradient of the graph is zero V is most negative

5A small drag force acts on a satellite. Its orbit radius and speed…

decrease, increases decrease, decreases increase, decreases

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 how the potential can rise towards zero but never become positive.”

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

What a good answer looks like

A good feedforward names an action: “Next time I will check the sign of V before I use W = mΔV.”

TOK · 2 minutes

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

TOKPhysicists choose zero potential at infinity, and this choice makes gravitational potential negative everywhere. If a choice by humans decides the sign, is the potential a property of nature or of our description?

Points you could make

Only differences in potential (ΔV) can be measured, and they do not depend on where zero is put.
The choice of zero is a convention that makes equations simple (V = −GM/r); a different choice gives the same physics.
Many quantities in physics contain conventions; knowing which parts are conventions and which are measured is part of understanding the knowledge.

Read the notes Back to D.1