Lesson 22: Gravitational and Electric Fields (HL) HL ONLY

Lesson 9 built fields at SL depth; this HL lesson (Theme D.1–D.2 completed) unifies them: gravitational and electric potential, potential energy, field–potential relations, orbital energy and escape — with the mathematical link E = −ΔV/Δx that makes potential the master quantity of field physics.

Potential: the scalar heart of a field

Gravitational potential at a point is the work done per unit mass bringing a test mass from infinity: Vg = −GM/r. It is negative because gravity is attractive and the zero is set at infinity; a mass in a potential well needs energy to escape. Electric potential Ve = kQ/r carries the sign of the source charge. Potential energy follows immediately: Ep = mVg or qVe. Because potentials are scalars, multi-body problems simplify enormously — compute V from each source, add, then convert to energy or force information as needed.

Field strength is the (negative) gradient of potential: g = −ΔVg/Δr and E = −ΔVe/Δx. On a V–r graph the gradient at each radius gives the local field; equipotential surfaces are spaced closely where the field is strong. Between parallel plates the linear V–x graph gives the uniform E = V/d you met in Lesson 9 — a special case of the same rule.

Orbital mechanics with energy

For a circular orbit of radius r around mass M: kinetic energy Ek = +GMm/(2r), potential energy Ep = −GMm/r, total energy E_total = −GMm/(2r). The total is negative (bound system) and exactly half the potential energy. Escape speed from a point: set total energy to zero at infinity: ½mv² = GMm/r → v_esc = √(2GM/r) = √2 × circular orbit speed at that radius. Energy reasoning replaces the force analysis of Lesson 5 for anything beyond circular orbits — transfers, captures, and escape all become bookkeeping in Vg.

Worked problem 1 — potentials at Earth's surface

Calculate the gravitational potential at Earth's surface and the potential energy of a 450 kg satellite there and in orbit at r = 7.0 × 10⁶ m.

Vg(surface) = −GM/R = −(6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/6.37 × 10⁶ = −6.25 × 10⁷ J kg⁻¹
Ep(satellite at surface) = 450 × (−6.25 × 10⁷) = −2.81 × 10¹⁰ J
At r = 7.0 × 10⁶ m: Vg = −3.98 × 10¹⁴/7.0 × 10⁶ = −5.69 × 10⁷ J kg⁻¹
Total orbital energy = ½ m Vg(r) = ½ × 450 × (−5.69 × 10⁷) = −1.28 × 10¹⁰ J

Energy needed to launch from rest on the surface into that orbit: ΔE = (−1.28 × 10¹⁰) − (−2.81 × 10¹⁰) = 1.53 × 10¹⁰ J — the real cost driver of spaceflight, dwarfing the satellite's orbital kinetic energy alone.

Worked problem 2 — electric potential and ion energy

Two charges +4.0 μC and −4.0 μC are 0.20 m apart (an electric dipole arrangement). (a) Find the potential at the midpoint. (b) Find the potential at a point 0.20 m from both charges on the perpendicular bisector's far side — simplify by noting symmetry along the bisector: every point on the perpendicular bisector is equidistant from both charges. (c) A Ca²⁺ ion (charge +2e) moves from the midpoint to infinity: how much energy does the field take from it?

(a) V = k(4.0 × 10⁻⁶ − 4.0 × 10⁻⁶)/0.10 = 0 V
(b) Any point on the perpendicular bisector is equidistant from +Q and −Q, so V = 0 everywhere on it.
(c) ΔEp = qΔV = 2e × (0 − 0) = 0: the field does no net work along this path's endpoints —
    though the ion's path would curve under the (non-zero) field forces at the midpoint plane.

Point (a)/(b) together teach the mantra: zero potential does not mean zero field (recall Lesson 9: at the dipole midpoint E is large). Potential is energy bookkeeping, not force.

Worked problem 3 — escape speed from the Moon

M_moon = 7.35 × 10²² kg, R_moon = 1.74 × 10⁶ m.

v_esc = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 7.35 × 10²²/1.74 × 10⁶) = √(5.64 × 10⁶) = 2.37 × 10³ m s⁻¹

The Moon's escape speed (2.4 km s⁻¹) is below the typical thermal speed of light gas molecules at lunar daytime temperatures — which is why the Moon has essentially no atmosphere: its gravity cannot hold one. Fields, energy and kinetic theory in one argument.

Practical worked example — verifying V ∝ 1/r for a charged sphere

Using a flame probe/electrometer (or a simulation bench dataset supplied by the teacher), a student measures the potential at distances from a charged conducting sphere: r = 0.10, 0.15, 0.20, 0.30 m → V = 270, 181, 134, 90 V. Plotting V against 1/r gives a straight line through the origin (gradient ≈ 27 V m), confirming V ∝ 1/r and giving kQ = 27 → Q = 27/8.99 × 10⁹ = 3.0 × 10⁻⁹ C. The intercept is within uncertainty of zero — if it were not, a contact potential or stray charge would be suspected. Uncertainty in r (±0.005 m, i.e. 5% at the closest point) dominates; error bars on 1/r grow at small r, which the max/min line method handles honestly.

Nature of Science

Potential was adopted because it made calculation easier, yet it turned out to be the deeper quantity: quantum mechanics is formulated in terms of potential energy (the Schrödinger equation contains V, not F), and in relativity gravity itself is recast as geometry. A bookkeeping device became the foundation — a reminder that formalisms are chosen, then vindicated.

Real-world application

Ion thrusters and mass spectrometers are potential machines: ions fall through a known ΔV and gain qΔV of energy regardless of path (Lesson 9's electron gun scaled up). In radiotherapy, proton beams are prescribed an energy by the accelerating potential, and their range in tissue follows from it — dose planning is applied potential theory.

Common mistakes (examiner style)

• Dropping the negative sign of gravitational potential, or placing the zero at the planet's surface instead of infinity.
• Calculating orbital energy as Ek + Ep with Ek from ½mv² using a linear (non-orbital) speed, instead of Ek = +GMm/2r for circular orbits.
• Concluding E = 0 from V = 0 (dipole midpoint) or the reverse.
• Escape problems using gΔh for the potential change over planetary distances — mgh is a local approximation only.
• Mixing up the gradient relation's sign: field points from high to low potential (for positive charges), so E = −ΔV/Δx.

IB Exam Experience

Paper 1-style questions. 1. The gravitational potential at a point is −4.0 × 10⁷ J kg⁻¹. The potential energy of a 2.0 kg mass there is: A. −2.0 × 10⁷ J B. −8.0 × 10⁷ J C. −4.0 × 10⁷ J D. +8.0 × 10⁷ J. Key: B — Ep = mV. 2. For a satellite in a circular orbit, Ep : Ek : E_total is in the ratio: A. −2 : 1 : −1 B. 2 : −1 : 1 C. −1 : 2 : 1 D. 1 : 1 : −2. Key: A — Ep = −GMm/r, Ek = +GMm/2r, total = −GMm/2r. 3. Escape speed from a planet is 11 km s⁻¹. The circular orbit speed just above its surface is: A. 5.5 B. 7.8 C. 11 D. 16 km s⁻¹. Key: B — v_orbit = v_esc/√2 = 7.8.

Paper 2-style structured question (8 marks). A probe of mass 300 kg is in a circular orbit of radius 8.0 × 10⁶ m around a planet of mass 6.4 × 10²³ kg. (a) Show that the probe's orbital speed is about 2.3 × 10³ m s⁻¹. [3] (b) Calculate the probe's total energy in this orbit. [2] (c) The probe fires its engine briefly to escape the planet from this orbit. Calculate the minimum additional kinetic energy required. [2] (d) Explain why the required energy in (c) does not depend on the direction in which the probe is travelling after the burn. [1] *Marking notes:* (a) v = √(GM/r) = √(6.67 × 10⁻¹¹ × 6.4 × 10²³/8.0 × 10⁶) = √(5.34 × 10⁶) = 2.31 × 10³ m s⁻¹ ✓. (b) E_total = −GMm/(2r) = −(6.67 × 10⁻¹¹ × 6.4 × 10²³ × 300)/(1.6 × 10⁷) = −8.0 × 10⁸ J. (c) Escape means total energy ≥ 0: ΔE = +8.0 × 10⁸ J. (d) Energy is a scalar — only the final speed's magnitude matters, not its direction. Candidates who answer (c) with the kinetic energy itself earn 1 of 2 (the probe already has Ek = 8.0 × 10⁸ J; it needs that much *more*).

IA idea: Use a simulation dataset of a comet's elliptical orbit to test the constancy of total energy (Ek + Ep) around the orbit and evaluate the numerical errors of the simulation.

Elliptical orbits and transfer energy

For an ellipse, the total energy depends only on the semi-major axis a: E = −GMm/(2a). A satellite moving from a circular orbit of radius r₁ to one of radius r₂ (via an elliptical transfer) needs ΔE = (GMm/2)(1/r₁ − 1/r₂) — positive when moving outward, and it must be supplied by the engine in two burns (into the transfer ellipse, then circularising). For the Lesson 22 satellite (450 kg, Earth) moving from r = 7.0 × 10⁶ m to r = 1.4 × 10⁷ m: ΔE = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 450/2) × (1/7.0 × 10⁶ − 1/1.4 × 10⁷) = 8.96 × 10¹⁶ × (1.43 × 10⁻⁷ − 7.14 × 10⁻⁸) = 8.96 × 10¹⁶ × 7.14 × 10⁻⁸ = 6.4 × 10⁹ J. Note it finishes *slower* in the higher orbit (v ∝ 1/√r) yet has *more* total energy — the extra potential energy exceeds the lost kinetic energy by a factor of two. This counterintuitive bookkeeping is a favourite HL discussion point: firing a rocket "forward" to speed it up raises the orbit and, once circularised, the satellite moves more slowly.

Two-mass potential problem

At what point between the Earth (5.97 × 10²⁴ kg) and the Moon (7.35 × 10²² kg), separated by 3.84 × 10⁸ m, is the gravitational field zero? Set GM_E/x² = GM_M/(d − x)², so (d − x)/x = √(M_M/M_E) = √(0.01231) = 0.111, giving x = d/1.111 = 3.46 × 10⁸ m from Earth's centre. The point is much closer to the Moon — as expected, since the fields must match and the Moon is weaker. (This is near, but not exactly, the L1 Lagrange point; rotation of the frame shifts the true L1.) Field-zero and potential-zero are different places: the potential is negative everywhere between them, a distinction the examiners probe relentlessly.