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

This study guide follows the IB Physics guide, first assessment 2025, and covers the additional higher level (AHL) part of D.1 Gravitational fields: potential energy and potential, the link between potential and field strength, equipotential surfaces, escape speed, orbital speed and the effect of atmospheric drag. Everything here is HL only. The standard level and higher level part of D.1 (Kepler’s laws, Newton’s law, field strength and field lines) is taught separately, so this page assumes you already know g = GM/r². Take the diagnostics first if you are unsure.

Related pages: revision notes, practice questions, the IB DP Physics hub, the checklist, forces and momentum and the syllabus guide.

Data used in the worked examples, as a question would supply it: G = 6.67 × 10⁻¹¹ N m² kg⁻². Masses and radii are given with each example.

What this unit covers

Guide ref	What you must be able to do	Level
D.1	Explain gravitational potential energy Ep as the work done to assemble a system from infinite separation	HL only
D.1	Use Ep = -G m1m2 / r for two bodies, with r measured between centres	HL only
D.1	Define and use gravitational potential Vg = -GM/r, zero at infinity	HL only
D.1	Use g = -ΔVg/Δr (field strength as potential gradient)	HL only
D.1	Use W = mΔVg for work done moving a mass	HL only
D.1	Describe equipotential surfaces and their relation to field lines; map a field using potential	HL only
D.1	Use vesc = √(2GM/r) and vorbital = √(GM/r)	HL only
D.1	Describe qualitatively the effect of a small drag force on an orbiting body	HL only
D.1	Apply these to circular orbits, orbit changes, launch from rest and escape	HL only

Potential energy and potential

The gravitational potential energy Ep of a system measures the work required to build it up, by bringing each part in from an infinite distance away. Gravity is attractive, so the bodies pull themselves together and you do negative work: Ep is negative at every finite separation and rises towards zero at infinity.

For two bodies, with r the distance between their centres:

Ep = -G m1 m2 / r

Gravitational potential Vg at a point equals the work done on each kilogram of a small test mass as it is carried from infinity to that point. For a point or spherical mass M:

Vg = -GM / r, with unit J kg⁻¹.

Vg is zero at infinity by definition. It is a scalar, so potentials from several bodies add as ordinary numbers. Ep = m Vg.

Worked example 1: Ep and Vg

Ganymede has mass 1.48 × 10²³ kg. A 900 kg probe is 1.00 × 10⁷ m from its centre.

GM = 6.67e-11 × 1.48e23 = 9.87 × 10^12 N m² kg⁻¹ (J m kg⁻¹)
Vg = -GM/r = -9.8716e12 / 1.00e7 = -9.87 × 10^5 J kg⁻¹
Ep = m Vg = 900 × (-9.8716e5) = -8.88 × 10^8 J

The probe needs 8.88 × 10⁸ J of external energy to reach infinity at rest. In the other direction, the field did that much work as the probe was brought in from infinity.

Work done and potential difference

Moving a mass slowly between two points, with no change in kinetic energy, needs an external work

W = m ΔVg

Positive ΔVg (moving to a less negative potential, that is away from the body) means you do positive work. Moving towards the body gives negative W: the field does work on the mass.

Worked example 2: lifting a mass through a large distance

Neptune has mass 1.02 × 10²⁶ kg and radius 2.46 × 10⁷ m. How much work lifts a 350 kg module from the surface to a distance of 2R from the centre?

GM = 6.803e15
Vg(surface) = -6.803e15 / 2.46e7 = -2.766 × 10^8 J kg⁻¹
Vg(2R)      = -6.803e15 / 4.92e7 = -1.383 × 10^8 J kg⁻¹
ΔVg = +1.383 × 10^8 J kg⁻¹
W = 350 × 1.383e8 = 4.84 × 10^10 J

Do not use mgh here: g falls with height, so it overestimates the work. For small height changes, mgh and mΔVg agree closely.

Field strength as potential gradient

g = -ΔVg / Δr

The magnitude of g is the steepness of the Vg against r graph. The minus sign records that g points towards decreasing potential, that is, towards the mass. On a graph of Vg against r (negative values, rising to zero), the gradient is positive and g points inward.

Worked example 3: gradient from data

Triton has mass 2.14 × 10²² kg. Vg is calculated at r = 2.00 × 10⁶ m and r = 2.20 × 10⁶ m, with GM = 1.4274 × 10¹².

Vg(2.00e6) = -7.137 × 10^5 J kg⁻¹
Vg(2.20e6) = -6.488 × 10^5 J kg⁻¹
ΔVg = +6.49 × 10^4 J kg⁻¹, Δr = 2.0 × 10^5 m
|g| = 6.49e4 / 2.0e5 = 0.324 N kg⁻¹
Check: GM/r² at 2.10e6 m = 1.4274e12 / (2.10e6)² = 0.324 N kg⁻¹

The gradient over an interval equals the field strength at about the midpoint. A smaller interval gives a better match.

Equipotential surfaces

An equipotential surface joins points of equal Vg. For a point or spherical mass these are concentric spheres. Moving a mass along one needs no work, because ΔVg = 0.

• Field lines cross equipotentials at right angles. If there were a component of g along the surface, moving along it would change Vg.
• Field lines point from higher to lower potential.
• Where equipotentials drawn at equal steps of Vg are close together, the field is strong. Where they are far apart, it is weak.

Near a planet’s surface the equipotentials are almost flat, parallel and evenly spaced, which matches a uniform field.

Worked example 4: mapping with potential

Ganymede has GM = 9.8716 × 10¹² N m² kg⁻¹. Find the radii of surfaces at Vg = -3, -2 and -1 × 10⁶ J kg⁻¹.

r = GM / |Vg|
-3e6: r = 3.29 × 10^6 m
-2e6: r = 4.94 × 10^6 m
-1e6: r = 9.87 × 10^6 m

The gaps are 1.65 × 10⁶ m, then 4.93 × 10⁶ m. Equal steps in Vg are spaced further apart as r increases, so the field gets weaker.

Potential from two bodies (along the line joining them)

Add the individual potentials. Two small asteroids of mass 2.0 × 10¹⁵ kg and 8.0 × 10¹⁵ kg are 9.0 × 10³ m apart. Where g = 0, the two fields cancel: x/(d - x) = √(M1/M2) = 1/2, so x = 3.0 × 10³ m from the lighter body. There Vg = -133 J kg⁻¹. Moving towards either body Vg falls (-148 J kg⁻¹ at the midpoint, -200 J kg⁻¹ at 1.0 × 10³ m from the lighter body), so along that line this point is a maximum of Vg. This is where the gradient is zero, which is why g = 0.

Escape speed

A body at distance r escapes if its total energy is at least zero, so it reaches infinity with no kinetic energy left:

½ m v² - GMm/r = 0, so vesc = √(2GM/r)

This is independent of the mass of the escaping object, assumes no drag, and applies at any distance r from the centre.

Worked example 5: escape and launch speed

Ceres has mass 9.4 × 10²⁰ kg and radius 4.7 × 10⁵ m.

vesc = √(2 × 6.2698e10 / 4.7e5) = √(2.668e5) = 5.17 × 10^2 m s⁻¹

A probe is launched from Ganymede’s surface (GM = 9.8716e12, R = 2.63 × 10⁶ m) at 6.0 × 10³ m s⁻¹. Its speed far away follows from energy conservation:

½v∞² = ½(6.0e3)² - GM/R = 1.8e7 - 3.7535e6 = 1.4247e7
v∞ = √(2.849e7) = 5.34 × 10^3 m s⁻¹

Escape speed is 2.74 × 10³ m s⁻¹ here, so the probe has plenty to spare.

Circular orbits

Set the gravitational force equal to the centripetal force needed: GMm/r² = mv²/r, so

vorbital = √(GM/r)

Larger orbits are slower: a larger r gives a smaller speed. Note vesc = √2 × vorbital at the same r.

Worked example 6: speed and period

A satellite orbits 300 km above Neptune (GM = 6.803 × 10¹⁵, R = 2.46 × 10⁷ m). Then r = 2.49 × 10⁷ m.

v = √(6.803e15 / 2.49e7) = 1.65 × 10^4 m s⁻¹
T = 2πr / v = 2π × 2.49e7 / 16530 = 9.46 × 10^3 s (2.63 h)

Energy of an orbiting satellite

From mv² = GMm/r:

• Ek = GMm / 2r
• Ep = -GMm / r
• Total E = -GMm / 2r

So Ek = -E and Ep = 2E. The total energy is negative because the satellite is bound. A larger orbit has a total energy closer to zero.

Worked example 7: changing orbit and launching

A 640 kg satellite moves from r = 2.49 × 10⁷ m to r = 4.00 × 10⁷ m around Neptune.

E1 = -6.803e15 × 640 / (2 × 2.49e7) = -8.74 × 10^10 J
E2 = -6.803e15 × 640 / (2 × 4.00e7) = -5.44 × 10^10 J
Energy to add = 3.30 × 10^10 J

Ep rises by 6.60 × 10¹⁰ J while Ek falls by 3.30 × 10¹⁰ J, so the energy supplied is only half the rise in Ep, and the satellite ends up slower (1.30 × 10⁴ m s⁻¹).

Launch from rest on a non-rotating planet. Energy needed = E(orbit) - E(surface) = -GMm/2r + GMm/R. For the same satellite from the surface of Neptune to r = 2.49 × 10⁷ m, this is 1.770e11 - 8.74e10 = 8.96 × 10¹⁰ J. Here almost all of it (8.74 × 10¹⁰ J) is the orbital Ek, because a low orbit raises Ep by only 2.2 × 10⁹ J; for higher orbits the Ep rise takes a larger share.

Escape from orbit. The extra energy to reach infinity from orbit is |E| = GMm/2r, here 8.74 × 10¹⁰ J. That is the same as the satellite’s Ek, which is why a speed increase by a factor √2 is enough.

Atmospheric drag

A thin atmosphere exerts a small drag force opposing the motion. Drag removes total energy, so E becomes more negative, and since E = -GMm/2r the orbit shrinks. The speed increases, because v = √(GM/r) rises as r falls. The kinetic energy rises while the potential energy falls by twice as much.

For example, a 640 kg satellite sinking from 2.49 × 10⁷ m to 2.45 × 10⁷ m around Neptune loses 1.43 × 10⁹ J of total energy and speeds up from 1.653 × 10⁴ to 1.666 × 10⁴ m s⁻¹. This is a slow spiral that can be treated as a series of near-circular orbits.

Common errors

• Using the surface height as r. Measure r from the centre of the body.
• Forgetting the minus sign, or giving a positive potential.
• Using mgh over large heights.
• Stating that drag slows the satellite. The speed increases as the orbit decays.
• Writing the escape speed as √(GM/r), which is the orbital speed.
• Applying W = mΔVg with the wrong sign for ΔVg (final minus initial).

Examiner insights

Show the substitution for Vg before the numbers. For “explain” questions on drag, a chain of linked statements (energy lost, smaller r, larger speed) tends to earn more credit than a single claim, and including units on every potential, energy and speed shows careful working.

Official syllabus

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