Gravitational Potential and Potential Energy

Gravitational potential is one of the most important — and most commonly misunderstood — concepts in A-Level Physics. While gravitational field strength tells you about the force a mass would experience, gravitational potential tells you about the energy associated with a position in a gravitational field.

---

Gravitational Potential

Gravitational potential V at a point in a gravitational field is defined as the work done per unit mass in bringing a small test mass from infinity to that point.

$V = -\frac{GM}{r}$V=−rGM​

where:

• V is the gravitational potential (J kg⁻¹)
• G is the gravitational constant (6.674 × 10⁻¹¹ N m² kg⁻²)
• M is the mass creating the field (kg)
• r is the distance from the centre of mass M to the point (m)

The units of gravitational potential are joules per kilogram (J kg⁻¹).

---

Why Is Gravitational Potential Negative?

This is crucial and comes up repeatedly in exams. Gravitational potential is always negative (or zero at infinity). Here is why:

At infinity, a test mass is completely free of the gravitational influence of M. We define V = 0 at infinity. As the test mass moves from infinity towards M, gravity does positive work on it (the mass accelerates towards M). Since the field does work on the mass as it falls in, the mass must have had higher energy at infinity than at distance r.

If V = 0 at infinity and V decreases as you approach the mass, then V must be negative at every finite distance. The closer you are to the mass, the more negative the potential — you are deeper in the gravitational potential well.

Think of it as a debt: negative potential tells you how much energy per kilogram you would need to supply to move the mass from that point all the way to infinity (where it would be "free").

Worked Example: Gravitational Potential at Different Altitudes

Calculate the gravitational potential at the surface, at ISS altitude, and at geostationary orbit altitude for Earth (M = 5.97 × 10²⁴ kg):

Location	r (m)	V = −GM/r (J kg⁻¹)
Earth's surface	6.37 × 10⁶	−6.25 × 10⁷
ISS orbit (408 km)	6.78 × 10⁶	−5.88 × 10⁷
Geostationary orbit	4.22 × 10⁷	−9.44 × 10⁶
Moon's orbit	3.84 × 10⁸	−1.04 × 10⁶
Infinity	∞	0

Notice: the potential becomes less negative (closer to zero) as you move further from Earth. The surface is the deepest point in the potential well.

Common exam mistake: Writing that potential is zero at Earth's surface. It is zero at infinity. At the surface, V = −GM/R, which is a large negative number (about −62.5 MJ kg⁻¹ for Earth).

---

Gravitational Potential Energy

Gravitational potential energy E_p is the energy of a mass m at a point where the gravitational potential is V:

$E_p = mV = -\frac{GMm}{r}$Ep​=mV=−rGMm​

where:

• E_p is the gravitational potential energy (J)
• m is the mass of the object in the field (kg)
• M is the mass creating the field (kg)
• r is the distance between the centres of the two masses (m)

Gravitational potential energy is also always negative. This means the system is bound — the masses are gravitationally trapped together. To separate them to infinity, you must supply energy equal to |E_p|.

Worked Example: Binding Energy of the Moon

How much energy would be needed to move the Moon to infinity from its current orbit?

$|E_p| = \frac{GMm}{r} = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{3.84 \times 10^8}$∣Ep​∣=rGMm​=3.84×1086.674×10−11×5.97×1024×7.35×1022​

$|E_p| = \frac{2.93 \times 10^{37}}{3.84 \times 10^8} = 7.63 \times 10^{28} \text{ J}$∣Ep​∣=3.84×1082.93×1037​=7.63×1028 J

This is about 7.6 × 10²⁸ J — an enormous amount of energy, roughly equivalent to the Sun's total energy output over several hours.

---

The Potential Well

A useful visual model is the potential well. Imagine a funnel or bowl shape where the depth at any point represents the gravitational potential at that distance from the mass. The deeper the well, the more negative the potential, and the more energy is required to escape.

flowchart TD
    A["V = 0 at infinity"] --> B["V becomes more negative\nas you approach M"]
    B --> C["V = −GM/R at the surface\n(deepest point in the well)"]
    C --> D["To escape: must supply\nenergy = |mV| = GMm/r"]
    subgraph Well["Potential Well Properties"]
        E["Deeper well → more massive body"]
        F["Steeper sides → stronger field"]
        G["All values negative or zero"]
    end

Key features of the potential well:

• The well is deepest at the surface of the mass (closest approach)
• The well gets shallower (V becomes less negative) as r increases
• At infinity, the rim of the well is at V = 0
• A more massive object creates a deeper well
• A denser object (same mass, smaller radius) has a steeper well near its surface

---

Work Done Moving Between Points

The work done in moving a mass m between two points in a gravitational field is equal to the change in gravitational potential energy:

$W = \Delta E_p = m(V_2 - V_1) = m \Delta V$W=ΔEp​=m(V2​−V1​)=mΔV

where V₁ and V₂ are the gravitational potentials at the starting and finishing points.

If you move a mass away from the source (to less negative potential), ΔV is positive, and work must be done against the gravitational field (you supply energy). If a mass moves towards the source (to more negative potential), the field does the work (the mass gains kinetic energy).

Worked Example: Raising a Satellite to a Higher Orbit

How much energy is required to move a 500 kg satellite from an orbit at radius 7.0 × 10⁶ m to an orbit at radius 1.0 × 10⁷ m from Earth's centre (M = 5.97 × 10²⁴ kg)?

$W = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = 6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 500 \times \left(\frac{1}{7.0 \times 10^6} - \frac{1}{1.0 \times 10^7}\right)$W=GMm(r1​1​−r2​1​)=6.674×10−11×5.97×1024×500×(7.0×1061​−1.0×1071​)

$W = 1.99 \times 10^{17} \times (1.43 \times 10^{-7} - 1.00 \times 10^{-7}) = 1.99 \times 10^{17} \times 4.29 \times 10^{-8} = 8.5 \times 10^{9} \text{ J}$W=1.99×1017×(1.43×10−7−1.00×10−7)=1.99×1017×4.29×10−8=8.5×109 J

About 8.5 GJ of energy is needed. This is the energy the rocket engines must provide (ignoring kinetic energy changes for now — we will address total orbital energy in a later lesson).

Worked Example: Landing on Mars

A 2000 kg probe falls from a great distance to the surface of Mars (M = 6.42 × 10²³ kg, R = 3.39 × 10⁶ m). How much kinetic energy does it gain?

At a great distance, V ≈ 0 (effectively infinity). At the surface:

$V_{\text{surface}} = -\frac{GM}{R} = -\frac{6.674 \times 10^{-11} \times 6.42 \times 10^{23}}{3.39 \times 10^6} = -1.264 \times 10^7 \text{ J kg}^{-1}$Vsurface​=−RGM​=−3.39×1066.674×10−11×6.42×1023​=−1.264×107 J kg−1

$\Delta E_p = m \times (0 - V_{\text{surface}}) \text{ [lost as KE gained]}$ΔEp​=m×(0−Vsurface​) [lost as KE gained]

Wait — the probe moves from V ≈ 0 to V = −1.264 × 10⁷, so ΔV = −1.264 × 10⁷ J kg⁻¹.

The PE decreases (becomes more negative), so the KE gained equals:

$E_k = |m \Delta V| = 2000 \times 1.264 \times 10^7 = 2.53 \times 10^{10} \text{ J} = 25.3 \text{ GJ}$Ek​=∣mΔV∣=2000×1.264×107=2.53×1010 J=25.3 GJ

---

Escape Velocity

Escape velocity is the minimum speed an object must have at a given point to escape to infinity without any further propulsion. At the point of escape, all kinetic energy is converted to gravitational potential energy:

$\frac{1}{2}mv^2 = \frac{GMm}{r}$21​mv2=rGMm​

Solving for v:

$v_{\text{esc}} = \sqrt{\frac{2GM}{r}}$vesc​=r2GM​​

Key observations:

• Escape velocity depends on M and r, but not on the mass of the escaping object. A molecule of gas and a spacecraft need the same escape velocity.
• Escape velocity is √2 times the orbital speed at that radius (a useful relationship we will derive in the orbits lesson)
• Escape velocity decreases with altitude because r is larger

Body	Surface escape velocity (m s⁻¹)	In km s⁻¹
Moon	2,380	2.4
Mars	5,030	5.0
Earth	11,200	11.2
Jupiter	59,500	59.5
Sun	617,000	617

Exam tip: If an object is launched at exactly escape velocity, it reaches infinity with zero speed (all KE converted to PE). If launched faster, it still has KE at infinity. If slower, it will eventually fall back.

Worked Example: Escape Velocity from the ISS

What is the escape velocity from ISS altitude (r = 6.78 × 10⁶ m)?

$v_{\text{esc}} = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.78 \times 10^6}} = \sqrt{\frac{7.966 \times 10^{14}}{6.78 \times 10^6}} = \sqrt{1.175 \times 10^8} = 10,840 \text{ m s}^{-1}$vesc​=6.78×1062×6.674×10−11×5.97×1024​​=6.78×1067.966×1014​​=1.175×108​=10,840 m s−1

This is slightly less than the 11,200 m s⁻¹ needed from the surface, because the ISS is already 408 km above the surface (further from the centre, shallower in the potential well).

---

Gravitational Potential Gradient

There is a direct relationship between gravitational potential and gravitational field strength:

$g = -\frac{\Delta V}{\Delta r}$g=−ΔrΔV​

The gravitational field strength at a point equals the negative of the potential gradient at that point. On a graph of V against r, the field strength at any point is the magnitude of the gradient (slope) of the graph.

The minus sign indicates that g points in the direction of decreasing potential — masses naturally move from high potential to low potential (from less negative to more negative), which is towards the source mass.

Why the Minus Sign Matters

Consider the V–r graph for a planet. V is negative and increases (becomes less negative) as r increases. The gradient dV/dr is therefore positive (V is increasing). But the field points inward (towards smaller r). The minus sign in g = −dV/dr ensures g points in the correct direction.

---

Equipotential Surfaces

An equipotential surface is a surface on which the gravitational potential is the same at every point. No work is done moving a mass along an equipotential surface because ΔV = 0.

For a spherical mass, equipotential surfaces are concentric spheres. The field lines are always perpendicular to the equipotential surfaces. Near Earth's surface, where the field is approximately uniform, equipotentials are approximately flat horizontal planes.

---

Summary

• Gravitational potential V = −GM/r is always negative (zero at infinity)
• Gravitational potential energy E_p = mV = −GMm/r
• Work done between two points: W = mΔV
• Escape velocity: v_esc = √(2GM/r) — independent of the escaping mass
• The potential gradient links V and g: g = −dV/dr
• Equipotential surfaces are perpendicular to field lines; no work is done moving along them

---

A-Level Deep Dive: Gravitational Potential and Potential Energy

Spec mapping

Edexcel 9PH0 specification Topic 12 — Gravitational Fields covers gravitational potential $V_g = -GM/r$Vg​=−GM/r, gravitational potential energy $E_p = -GMm/r$Ep​=−GMm/r, the link $g = -\Delta V/\Delta r$g=−ΔV/Δr, equipotential surfaces, and escape velocity (refer to the official specification document for exact wording). Topic 12 sits inside the synoptic Year 2 "Fields" cluster alongside Topic 7 (Electric and Magnetic Fields) and Topic 11 (Circular Motion), and is examined principally on Paper 2 (which carries Topics 6–8 and 11–13) but can also surface as a synoptic strand on Paper 3 under the "extended writing and synoptic" assessment objectives. The Edexcel data and formulae booklet does list $V_g = -GM/r$Vg​=−GM/r, $g = -\Delta V/\Delta r$g=−ΔV/Δr and the inverse-square form $g = GM/r^2$g=GM/r2 — but it does not give escape velocity in closed form. Candidates must derive $v_{esc} = \sqrt{2GM/r}$vesc​=2GM/r​ from energy conservation in the exam.

---

Worked example with full mark scheme

Question (8 marks):

A satellite of mass $m = 850\,\text{kg}$m=850kg is to be lifted from the surface of the Earth (radius $R_E = 6.37 \times 10^6\,\text{m}$RE​=6.37×106m) to a circular orbit at altitude $h = 2.00 \times 10^7\,\text{m}$h=2.00×107m above the surface. Take $G = 6.67 \times 10^{-11}\,\text{N m}^2\,\text{kg}^{-2}$G=6.67×10−11N m2kg−2 and $M_E = 5.97 \times 10^{24}\,\text{kg}$ME​=5.97×1024kg.

(a) Calculate the change in gravitational potential energy of the satellite between the surface and the orbit. (5)

(b) Hence determine the minimum work that must be done against gravity to place the satellite in this orbit, and explain why the actual energy supplied by the launch vehicle must exceed this value. (3)

Solution with mark scheme:

(a) Step 1 — identify the two radii from Earth's centre.

$r_1 = R_E = 6.37 \times 10^6\,\text{m}$r1​=RE​=6.37×106m (surface) and $r_2 = R_E + h = 6.37 \times 10^6 + 2.00 \times 10^7 = 2.637 \times 10^7\,\text{m}$r2​=RE​+h=6.37×106+2.00×107=2.637×107m (orbit).

M1 — both radii correctly measured from the centre of the Earth, not from the surface. Common error: using $r_2 = h = 2.00 \times 10^7$r2​=h=2.00×107 alone, treating altitude as the orbital radius. That destroys the entire calculation.

Step 2 — write the potential energy formula at each radius.

$E_p(r) = -\dfrac{GM_E m}{r}$Ep​(r)=−rGME​m​

M1 — quoting the negative form with infinity as the zero. Writing $E_p = +GMm/r$Ep​=+GMm/r loses this mark immediately and propagates a sign error throughout.

Step 3 — evaluate $E_p$Ep​ at each radius.

$E_p(r_1) = -\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(850)}{6.37 \times 10^6} = -5.31 \times 10^{10}\,\text{J}$Ep​(r1​)=−6.37×106(6.67×10−11)(5.97×1024)(850)​=−5.31×1010J

$E_p(r_2) = -\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(850)}{2.637 \times 10^7} = -1.28 \times 10^{10}\,\text{J}$Ep​(r2​)=−2.637×107(6.67×10−11)(5.97×1024)(850)​=−1.28×1010J

A1 — both values to 3 s.f. with correct negative signs.

Step 4 — compute $\Delta E_p$ΔEp​.

$\Delta E_p = E_p(r_2) - E_p(r_1) = (-1.28 \times 10^{10}) - (-5.31 \times 10^{10}) = +4.03 \times 10^{10}\,\text{J}$ΔEp​=Ep​(r2​)−Ep​(r1​)=(−1.28×1010)−(−5.31×1010)=+4.03×1010J

M1 — correct order of subtraction (final minus initial).

A1 — final answer $+4.03 \times 10^{10}\,\text{J}$+4.03×1010J with positive sign and unit.

(b) Step 1 — identify minimum work.

Minimum work against gravity equals $\Delta E_p = +4.03 \times 10^{10}\,\text{J}$ΔEp​=+4.03×1010J.

B1 — stated.

Step 2 — explain why launch energy is greater.

The launch vehicle must additionally supply: (i) the orbital kinetic energy $\tfrac{1}{2}mv^2$21​mv2 at the target radius (the satellite must be moving at orbital speed when it arrives, not stationary); (ii) energy lost to atmospheric drag, especially in the lower atmosphere; (iii) the kinetic energy of the spent rocket stages and unburnt propellant carried during ascent.

B1 — at least one of: orbital KE / atmospheric drag.

B1 — at least one further valid energy sink (rocket stages, propellant mass, gravity losses during slow ascent).

Total: 8 marks (5 + 3 as shown).

---

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): Show that the escape velocity from the surface of a spherical body of mass $M$M and radius $R$R is $v_{esc} = \sqrt{2GM/R}$vesc​=2GM/R​, stating clearly the energy assumptions you make. Hence calculate the escape velocity from the surface of a planet with $M = 6.42 \times 10^{23}\,\text{kg}$M=6.42×1023kg and $R = 3.39 \times 10^6\,\text{m}$R=3.39×106m.

Mark scheme decomposition by AO:

• M1 (AO1.1) — equating total mechanical energy at the surface to total mechanical energy at infinity: $\tfrac{1}{2}mv_{esc}^2 - GMm/R = 0$21​mvesc2​−GMm/R=0.
• B1 (AO1.2) — explicit statement of the assumptions: kinetic energy at infinity is zero (just escapes), gravitational PE at infinity is zero (chosen reference), no atmospheric or other resistive losses.
• M1 (AO2.1) — algebraic rearrangement: $\tfrac{1}{2}v_{esc}^2 = GM/R$21​vesc2​=GM/R, so $v_{esc}^2 = 2GM/R$vesc2​=2GM/R.
• A1 (AO2.1) — taking the positive square root and presenting $v_{esc} = \sqrt{2GM/R}$vesc​=2GM/R​.
• M1 (AO1.2) — substitution of values: $v_{esc} = \sqrt{2(6.67 \times 10^{-11})(6.42 \times 10^{23})/(3.39 \times 10^6)}$vesc​=2(6.67×10−11)(6.42×1023)/(3.39×106)​.
• A1 (AO1.2) — $v_{esc} \approx 5.03 \times 10^3\,\text{m s}^{-1}$vesc​≈5.03×103m s−1 (or $5.0\,\text{km s}^{-1}$5.0km s−1) to 2–3 s.f. with units.

Total: 6 marks split AO1 = 4, AO2 = 2. This is a classic 9PH0 "show that" derivation — the AO2 marks attach to the algebraic manipulation, the AO1 marks reward correct recall of the energy framework and arithmetic.

---

Synoptic links

Connects to:

1. Topic 11 — Circular motion (orbital mechanics): the orbital speed of a satellite at radius $r$r comes from $GMm/r^2 = mv^2/r$GMm/r2=mv2/r, giving $v_{orbit} = \sqrt{GM/r}$vorbit​=GM/r​. Note the elegant relationship $v_{esc} = \sqrt{2}\,v_{orbit}$vesc​=2​vorbit​ at the same radius — escape velocity is exactly $\sqrt{2}$2​ times orbital velocity. This links potential and kinetic descriptions of bound versus unbound trajectories.

2. Escape velocity and total energy: an object with total energy $E < 0$E<0 is bound (closed orbit, ellipse or circle); $E = 0$E=0 corresponds to a parabolic trajectory just escaping; $E > 0$E>0 gives a hyperbolic, unbound trajectory. The sign of the total mechanical energy $E = \tfrac{1}{2}mv^2 - GMm/r$E=21​mv2−GMm/r classifies orbits — a synoptic insight that connects Topic 12 to Topic 11 and to Kepler's laws.