IB DP Physics: Gravitational potential, escape speed and orbital energy (DP Physics) – Revision Notes

These notes condense the additional higher level (AHL) part of D.1 Gravitational fields in the IB Physics guide, first assessment 2025. All of it is HL only. For full explanations and longer worked examples use the study guide. Test yourself with the practice questions, find your gaps with the diagnostics, and tick off outcomes on the checklist. The course page is the IB DP Physics hub. For the motion and force background see forces and momentum.

Definitions

• Gravitational potential energy Ep of a system: the work needed to build the system by bringing its parts in from infinite separation.
• Gravitational potential Vg at a point: work done per kilogram when a small test mass is carried in from infinity to that point. Zero at infinity.
• Equipotential surface: a surface on which Vg has the same value everywhere.
• Escape speed: the speed at a given point that lets a body reach infinity with no kinetic energy left (no drag, no further push).

Formulas

Quantity	Result	Notes
Ep, two bodies	-G m1 m2 / r	r between centres; negative
Vg	-GM / r	scalar; J kg⁻¹
Ep and Vg	Ep = m Vg
Field strength	g = -ΔVg / Δr	gradient; points to lower Vg
Work moving a mass	W = m ΔVg	ΔVg = final - initial
Escape speed	vesc = √(2GM/r)	independent of m
Orbital speed	vorbital = √(GM/r)	circular orbit
Orbital energies	Ek = GMm/2r; Ep = -GMm/r; E = -GMm/2r	Ek = -E, Ep = 2E

Method boxes

Work to move a mass between two points

1. Find Vg at both positions with r from the centre.
2. ΔVg = final - initial.
3. W = m ΔVg. Positive means you supply energy.

Escape or maximum height

1. Write total energy per kilogram at the start: ½v² - GM/r.
2. Escape: this is zero or more. Not escaping: set it equal to -GM/rmax and solve for rmax.
3. Subtract the radius if the question asks for height.

Orbit changes

1. Work out E = -GMm/2r at each radius.
2. Energy to supply = E(final) - E(initial).
3. Find speeds with √(GM/r); the larger orbit has the smaller speed.

Launch from rest on a non-rotating planet to a circular orbit

1. Energy needed = E(orbit) - E(surface) = GMm(1/R - 1/2r).
2. Split it: Ek at orbit = GMm/2r; the rest raises Ep.

Equipotentials and field lines

• Concentric spheres around a point or spherical mass, with centre-to-centre distance used for r.
• Field lines meet equipotentials at 90 degrees and point from high to low Vg.
• Equally spaced Vg steps: close spacing means a strong field; wide spacing means a weak field.
• Moving a mass along an equipotential costs no work.
• Near a surface the equipotentials are almost parallel and equally spaced, so the field is approximately uniform.
• Potentials of several bodies add as scalars. Along the line joining two masses, g = 0 where the gradient of the total Vg is zero.

Atmospheric drag on a satellite

Drag is small, so treat the orbit as a slow spiral through near-circular orbits.

Quantity	Change	Why
Total energy	decreases	drag does negative work
Orbit radius	decreases	E = -GMm/2r becomes more negative
Speed	increases	v = √(GM/r)
Ep	decreases (by twice the Ek gain)	smaller r
Ek	increases	v larger

Worked reminders

Ratio reasoning. Planet B has twice the mass and half the radius of planet A. vesc scales as √(M/R), so vesc(B) = √(2 ÷ 0.5) = 2 × vesc(A).

Equal-spaced check. For GM = 1.2 × 10¹³ N m² kg⁻¹, the surface Vg = -3.0 × 10⁶ J kg⁻¹ is at 4.0 × 10⁶ m. Halving r doubles |Vg|: the equal Vg steps are closer together near the planet.

Escape from orbit. At any circular orbit vesc = √2 × vorbital, so an increase in speed of 41% from orbital speed is enough to escape. The energy needed is |E| = GMm/2r, equal to the satellite’s Ek.

Launch energetics with numbers. For the same planet (GM = 1.2 × 10¹³, R = 4.0 × 10⁶ m), a 100 kg satellite is launched from rest at the surface into a circular orbit of radius 5.0 × 10⁶ m. The surface energy is -GMm/R = -3.0 × 10⁸ J. The orbit energy is -GMm/2r = -1.2 × 10⁸ J. The input is 1.8 × 10⁸ J. Of this, 1.2 × 10⁸ J is the orbital Ek and 6.0 × 10⁷ J is the rise in Ep (from -3.0 × 10⁸ J to -2.4 × 10⁸ J).

Reading graphs

• Vg against r: all values are negative and rise towards zero as r increases. The gradient at a point is the magnitude of g there, so the curve is steepest near the body.
• g against r: the area under the curve between two radii equals the change in Vg between them, in J kg⁻¹. Count squares or use trapezia and state the units.
• Ep against r: the same shape as Vg, scaled by m. The depth of the well at the starting radius is the energy needed to escape from rest.
• A tangent drawn on a Vg graph gives g at that point. Use a large triangle and show both sides of it.

Must-know distinctions

• Ep and Vg: energy of a system against potential per unit mass.
• Escape speed and orbital speed: factor √2 at the same r.
• Orbit energy and escape energy: E is negative; escape needs E raised to zero.
• Field strength and potential: g is a vector (gradient); Vg is a scalar (value).
• Speed and energy in a bigger orbit: the energy rises, the speed falls.

Command words in this unit

• Define: give the exact wording, including “per unit mass”, “from infinity” and the zero reference.
• Derive: start from a physical statement (centripetal force equal to gravity, or total energy equal to zero) and rearrange step by step.
• Explain: link cause to effect in order, for example drag, energy loss, smaller radius, larger speed.
• Sketch: label axes, show the sign of Vg and make the spacing of equipotentials match the field strength.
• Estimate: use a small interval for a gradient and state the assumption.

Quick self-test

Use GM = 1.2 × 10¹³ N m² kg⁻¹ for a planet, r measured from its centre, where a calculation is needed.

1. State what gravitational potential at a point means.
2. Calculate Vg at r = 4.0 × 10⁶ m.
3. Find Ep for a 50 kg mass at that distance.
4. Find the work to move the 50 kg mass from r = 4.0 × 10⁶ m to 8.0 × 10⁶ m.
5. Over a short distance Vg rises by 6.0 × 10⁴ J kg⁻¹ across 2.0 × 10⁴ m, moving away from the body. State |g| and its direction.
6. Find the escape speed at r = 4.0 × 10⁶ m.
7. Find the orbital speed at that radius.
8. A 200 kg satellite orbits at 4.0 × 10⁶ m. Find Ek, Ep and E.
9. How much energy is needed to move it to 8.0 × 10⁶ m?
10. State what happens to the orbit height and speed of a satellite in very thin atmosphere.
11. Why can no work be done by gravity along an equipotential?
12. By what factor must the speed of a satellite in circular orbit increase for it to escape?

Answers

1. The work done per kilogram of a small test mass carried from infinity to the point.
2. -GM/r = -1.2e13 / 4.0e6 = -3.0 × 10⁶ J kg⁻¹.
3. Ep = 50 × (-3.0e6) = -1.5 × 10⁸ J.
4. Vg at 8.0e6 is -1.5e6; ΔVg = +1.5e6; W = 50 × 1.5e6 = 7.5 × 10⁷ J.
5. |g| = 6.0e4 / 2.0e4 = 3.0 N kg⁻¹, directed towards the body (towards lower potential).
6. √(2 × 1.2e13 / 4.0e6) = √(6.0e6) = 2.4 × 10³ m s⁻¹ (2.45 × 10³).
7. √(1.2e13 / 4.0e6) = √(3.0e6) = 1.7 × 10³ m s⁻¹ (1.73 × 10³).
8. Ek = GMm/2r = 3.0 × 10⁸ J; Ep = -6.0 × 10⁸ J; E = -3.0 × 10⁸ J.
9. E at 8.0e6 is -1.5e8 J; energy to add = 1.5 × 10⁸ J.
10. The orbit shrinks and the speed increases.
11. ΔVg = 0, so W = mΔVg = 0; the field is perpendicular to the surface.
12. √2 (about 1.41).

Where marks are usually lost

• Substituting altitude for r, which must run to the planet centre.
• Dropping the minus sign on Vg or Ep.
• Using mgh for a large change in height.
• Saying drag makes the satellite slower.
• Confusing vesc with vorbital, or quoting vesc as √(GM/r).
• Using the wrong sign for ΔVg, so the work comes out negative when energy must be supplied.
• Leaving out units, or giving a potential in joules rather than J kg⁻¹.
• Forgetting that field lines cross equipotentials at right angles in a sketch.
• Reporting that a bigger orbit has a faster speed because it has more energy.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. Section D.1, additional higher level part.