Gravitational Potential and Energy

Spec mapping: OCR H556 Module 5.4 — Gravitational fields (gravitational potential $V$V defined as work done per unit mass to bring a test mass from infinity; the convention $V(\infty) = 0$V(∞)=0 and the consequence that $V$V is negative everywhere outside a source mass; gravitational potential energy $E_{p} = mV = -GMm/r$Ep​=mV=−GMm/r; gravitational potential difference and work done; escape velocity $v_{\text{esc}} = \sqrt{2GM/R}$vesc​=2GM/R​; equipotentials and their geometric relationship to field lines). Refer to the official OCR H556 specification document for exact wording.

Lesson 8 gave us Newton's law of gravitation and the spherical-mass field $g = GM/r^{2}$g=GM/r2. To handle the energetic side of gravitational problems — dropping probes onto Mars, launching satellites, computing escape velocities, deciding whether a meteor will impact the Earth or fly past — we need the partner concept: gravitational potential. This lesson develops the idea, derives $V = -GM/r$V=−GM/r and $E_{p} = -GMm/r$Ep​=−GMm/r, distinguishes potential from potential energy, sketches $V(r)$V(r), and applies the formulae to several specimen problems including escape velocity.

This lesson is the single most A discriminator-rich* lesson of Module 5.4. The sign of $V$V is the discriminator: by convention $V(\infty) = 0$V(∞)=0 and $V$V becomes more negative as you approach a source mass (you fall into a deeper potential well). Every escape-velocity, energy-conservation and orbital-energy calculation hinges on getting these signs right. If you take only one principle from this lesson, take this: gravitational potential is negative; zero is at infinity; getting closer to a mass makes V more negative. Reading "more negative" as "smaller" rather than "larger" is the single most common A-Level error in Module 5.4.

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From Force to Energy — Why a Potential Exists

An inverse-square force field has a special property: the work done in moving a test mass from one point to another depends only on the start and end points, not on the path taken. This defines a conservative force field. (Friction is the canonical non-conservative force, where path matters.)

For a conservative field we can attach a scalar number $V(\vec{r})$V(r) — the potential — to every point, with the property that the work done by the field on a test mass moving from $A$A to $B$B equals $m(V_{A} - V_{B})$m(VA​−VB​), and the work done against the field equals $m \Delta V$mΔV. The vector field is recovered as $\vec{g} = -\nabla V$g​=−∇V. OCR uses the symbol $V$V (sometimes $V_{g}$Vg​ for clarity against electric potential).

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Gravitational Potential — Definition

The gravitational potential at a point is the work done per unit mass by an external agent in bringing a small test mass from infinity to that point against the gravitational field.

In symbols:

$V = \frac{W_{\text{ext}}}{m}$V=mWext​​

Units: joules per kilogram, $\text{J kg}^{-1}$J kg−1.

Two pieces of bookkeeping fix the sign:

1. Infinity is the reference point. $V(\infty) \equiv 0$V(∞)≡0 — chosen because it is the only point where a test mass experiences zero gravitational attraction.
2. The work is done by an external agent, slowly. To escape from a source, the agent must do positive work against the inward pull; equivalently, falling in releases energy, so $V$V becomes negative everywhere outside an isolated source.

Potential $V$V (J kg⁻¹) and potential energy $E_{p}$Ep​ (J) of a mass $m$m are related by $E_{p} = mV$Ep​=mV. Confusing the two is the biggest source of unit and sign errors in this lesson.

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Why $V$V is Negative — the Single Most Important Conceptual Point

This is the subtlety that decides A* from A grades at OCR. Take it slowly.

Step 1. Start with a test mass at infinity, where $V = 0$V=0 by convention and the gravitational pull from any source is zero.

Step 2. Let the test mass drift inward under the gravitational pull of a source mass $M$M at the origin. Gravity is attractive, so the field $\vec{g}$g​ points inward — along the direction of motion of the falling mass. The field therefore does positive work on the test mass; its kinetic energy increases.

Step 3. By energy conservation, if the field does positive work on the mass, the mass's gravitational potential energy must decrease. It started at zero (at infinity) so it ends up negative.

Step 4. Per unit mass, $V$V inherits the same negative sign: $V$V starts at 0 at infinity and becomes more and more negative as we approach $M$M. As $r \to 0$r→0, $V \to -\infty$V→−∞ (the potential well becomes bottomless for a true point mass).

Putting it all together with $\vec{g} = -\nabla V$g​=−∇V and the spherical-mass field $g = GM/r^{2}$g=GM/r2 (Lesson 8), the integral $V(r) - V(\infty) = -\int_{\infty}^{r} \vec{g} \cdot d\vec{r}$V(r)−V(∞)=−∫∞r​g​⋅dr gives

$V(r) = -\frac{GM}{r}$V(r)=−rGM​

The minus sign is not a typo; it is the signature of an attractive field with the reference at infinity. Any point at finite $r$r has a lower (more negative) potential than infinity, which is at zero.

Picturing the Potential Well

Imagine $V$V as the depth of a well below the flat plain $V = 0$V=0 at infinity. The Earth's well has its rim at infinity and the surface at $V_{\text{surface}} = -6.25 \times 10^{7}$Vsurface​=−6.25×107 J kg⁻¹ — 62.5 MJ/kg below the rim. To escape, you need to gain that much KE per kg. "Falling in" makes $V$V more negative; "climbing out" makes $V$V less negative.

A discriminator.* OCR mark schemes award the minus sign explicitly. A response that writes "$V = GM/r$V=GM/r" (no minus) earns the structure mark but loses the sign mark — a guaranteed grade-band slip.

Derivation Outline (for completeness)

The field is the negative gradient of the potential, $g_{r} = -dV/dr$gr​=−dV/dr. Integrating $\vec{g} = -(GM/r^{2})\hat{r}$g​=−(GM/r2)r^ from infinity:

$V(r) - V(\infty) = -\int_{\infty}^{r} \vec{g} \cdot d\vec{r} = \int_{\infty}^{r} \frac{GM}{r'^{2}} dr' = -\frac{GM}{r}$V(r)−V(∞)=−∫∞r​g​⋅dr=∫∞r​r′2GM​dr′=−rGM​

so $V(r) = -GM/r$V(r)=−GM/r with $V(\infty) = 0$V(∞)=0. A-Level does not require the integration, but the structural idea — "potential is the negative integral of the field from infinity" — must be in your conceptual toolkit.

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Gravitational Potential Energy

A test mass $m$m at a point where the potential is $V$V has gravitational potential energy

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

This replaces the GCSE formula $E_{p} = mgh$Ep​=mgh — not because $mgh$mgh is wrong, but because $mgh$mgh assumes $g$g is constant over the height range $h$h. That assumption is excellent near the Earth's surface (where $g \approx 9.81$g≈9.81 N kg⁻¹ everywhere) but fails badly at altitude. The exact formula $E_{p} = -GMm/r$Ep​=−GMm/r is universal.

Consistency with $mgh$mgh — the Small-$h$h Limit

For an altitude $h$h that is small compared with $R_{E}$RE​, the change in the exact formula reduces to $mgh$mgh. Let us check.

$\Delta E_{p} = E_{p}(R_{E} + h) - E_{p}(R_{E}) = -\frac{GMm}{R_{E} + h} + \frac{GMm}{R_{E}} = GMm \left(\frac{1}{R_{E}} - \frac{1}{R_{E} + h}\right)$ΔEp​=Ep​(RE​+h)−Ep​(RE​)=−RE​+hGMm​+RE​GMm​=GMm(RE​1​−RE​+h1​)

Common-denominator the bracket:

$\Delta E_{p} = GMm \times \frac{(R_{E} + h) - R_{E}}{R_{E}(R_{E} + h)} = \frac{GMm \, h}{R_{E}(R_{E} + h)}$ΔEp​=GMm×RE​(RE​+h)(RE​+h)−RE​​=RE​(RE​+h)GMmh​

For $h$h small compared with $R_{E}$RE​, $R_{E} + h \approx R_{E}$RE​+h≈RE​ in the denominator, so

$\Delta E_{p} \approx \frac{GMm \, h}{R_{E}^{2}} = m \left(\frac{GM}{R_{E}^{2}}\right) h = m g_{s} h$ΔEp​≈RE2​GMmh​=m(RE2​GM​)h=mgs​h

— recovering $mgh$mgh with $g_{s} = GM/R_{E}^{2}$gs​=GM/RE2​, the surface field strength. So $mgh$mgh is just the first-order Taylor expansion of $-GMm/r$−GMm/r about $r = R_{E}$r=RE​; the two formulae do not disagree, they are consistent in the regime where $mgh$mgh applies.

The takeaway: use $mgh$mgh near the surface where $g$g is approximately constant; use $\Delta E_{p} = m \Delta V = -GMm(1/r_{2} - 1/r_{1})$ΔEp​=mΔV=−GMm(1/r2​−1/r1​) at any other range. OCR will set questions that deliberately require the full formula (a satellite raised to 36 000 km altitude, a probe sent to Mars) and questions where $mgh$mgh is fine (a ball dropped from a tower). Recognise which is which.

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Graph of $V$V versus $r$r

The graph of $V = -GM/r$V=−GM/r for $r \ge R$r≥R is a negative hyperbola: it sits below the $r$r-axis, approaches zero from below as $r \to \infty$r→∞, and plunges to $-GM/R$−GM/R at the surface (with a continuation to $-\infty$−∞ as $r \to 0$r→0 for an idealised point mass; real planets have a smoothed-out interior).

Three features to read off:

1. Asymptote at $V = 0$V=0 as $r \to \infty$r→∞. The reference convention.
2. Most negative at the surface ($r = R$r=R), where $V_{\text{surface}} = -GM/R$Vsurface​=−GM/R.
3. The graph never crosses zero for finite $r > 0$r>0 around an isolated source mass. $V$V is everywhere negative.

The gradient of the $V$V-vs-$r$r graph at any point $r$r equals $-g(r)$−g(r) (the field strength with a sign flip). Near $r = R$r=R the graph is steep (large $g$g); far away it is shallow (small $g$g). This is the geometric statement of $\vec{g} = -\nabla V$g​=−∇V.

OCR questions often draw $V(r)$V(r) and $g(r)$g(r) on adjacent axes and ask candidates to relate the two. The key fact: $g = -dV/dr$g=−dV/dr, so steep parts of $V(r)$V(r) correspond to large $g$g, flat parts to small $g$g. As $r \to \infty$r→∞, $V \to 0^{-}$V→0− from below and $g \to 0$g→0 from above (a $1/r^{2}$1/r2 tail in $g$g, a $1/r$1/r tail in $V$V).

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Equipotentials

An equipotential surface is a surface on which $V$V is constant. The defining property: no work is done in moving a test mass along an equipotential, because $\Delta V = 0$ΔV=0 along the surface.

For a spherically symmetric source, $V = -GM/r$V=−GM/r depends only on $r$r, so the equipotentials are concentric spheres centred on the source. Two consequences:

• Field lines (radial) cross equipotentials (spherical shells) perpendicularly at every point. This is general — for any conservative field, the gradient $\vec{g} = -\nabla V$g​=−∇V is everywhere perpendicular to surfaces of constant $V$V.
• Each successive equipotential sphere outward represents a less negative $V$V (closer to zero). Climbing from one equipotential to the next outward requires positive work; sliding inward releases positive work to the test mass.

Near the Earth's surface (uniform-field approximation), the equipotentials are horizontal planes and the field lines are vertical; the perpendicularity is preserved. Moving horizontally along a constant-altitude path is moving along an equipotential — no work is done against gravity. Moving vertically up or down crosses equipotentials at right angles.

A discriminator.* When OCR asks you to sketch both field lines and equipotentials around a planet, draw radial field lines (arrows inward toward the planet) and concentric circles as equipotentials, and make sure the two are perpendicular where they cross. The combined sketch is worth a whole exam mark on its own.

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Work Done and Potential Difference

The gravitational potential difference between two points $A$A and $B$B is

$\Delta V = V_{B} - V_{A}$ΔV=VB​−VA​

and the work done by an external agent in moving a mass $m$m from $A$A to $B$B (slowly, no change in kinetic energy) is

$W_{\text{ext}} = m \, \Delta V = m(V_{B} - V_{A})$Wext​=mΔV=m(VB​−VA​)

The work done by gravity on the same motion is the negative of this:

$W_{\text{grav}} = -W_{\text{ext}} = -m \, \Delta V = m(V_{A} - V_{B})$Wgrav​=−Wext​=−mΔV=m(VA​−VB​)

Sign-check using a concrete case. Lifting from $A$A (closer, more negative $V_{A}$VA​) to $B$B (further, less negative $V_{B}$VB​): $V_{B} > V_{A}$VB​>VA​ so $\Delta V > 0$ΔV>0 and $W_{\text{ext}} = m\Delta V > 0$Wext​=mΔV>0. You must do positive work to lift, in agreement with everyday intuition. The work done by gravity is correspondingly $-W_{\text{ext}} < 0$−Wext​<0 (gravity opposes the motion). Dropping from $B$B to $A$A: $\Delta V < 0$ΔV<0, $W_{\text{ext}} < 0$Wext​<0 (the external agent does negative work; equivalently, gravity does positive work on the falling mass).

A discriminator.* Always state who is doing the work — external agent or gravity. The two signs are always opposite. OCR mark schemes often phrase the question as "work done in moving the mass against gravity", which means $W_{\text{ext}}$Wext​ (positive when lifting). If the question says "work done by gravity" it means $W_{\text{grav}}$Wgrav​ (negative when lifting).

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Worked Example 1 — Potential at the Earth's Surface

Calculate $V$V at the Earth's surface. $M_{E} = 5.97 \times 10^{24}$ME​=5.97×1024 kg, $R_{E} = 6.37 \times 10^{6}$RE​=6.37×106 m.

$V_{\text{surface}} = -\frac{GM_{E}}{R_{E}} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^{6}} = -\frac{3.98 \times 10^{14}}{6.37 \times 10^{6}} = -6.25 \times 10^{7} \text{ J kg}^{-1}$Vsurface​=−RE​GME​​=−6.37×1066.67×10−11×5.97×1024​=−6.37×1063.98×1014​=−6.25×107 J kg−1

Interpretation: it would take $+6.25 \times 10^{7}$+6.25×107 J of work per kilogram to lift a test mass from the Earth's surface all the way out to infinity against gravity. Equivalently, the Earth's gravitational well is 62.5 MJ/kg deep at the surface. This number is the escape-energy budget — multiplied by the mass of a payload it gives the minimum kinetic energy needed at the surface to launch the payload to infinity (the escape-velocity computation below).

Note the minus sign and the units. A response without the minus loses the sign-convention mark; a response in J (not J kg⁻¹) confuses potential with potential energy and earns a unit-error penalty.

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Worked Example 2 — Work Done Lifting a Satellite to LEO

A 500 kg satellite is to be raised from the Earth's surface to a circular orbit at 1000 km altitude. Calculate the work done (against gravity) to lift it — without yet giving it orbital velocity.

The change in potential energy is $\Delta E_{p} = m \, \Delta V$ΔEp​=mΔV. We need $V$V at both radii.

At the surface ($r_{1} = 6.37 \times 10^{6}$r1​=6.37×106 m): $V_{1} = -GM_{E}/r_{1} = -6.25 \times 10^{7}$V1​=−GME​/r1​=−6.25×107 J kg⁻¹.

At the orbit ($r_{2} = 6.37 \times 10^{6} + 1.00 \times 10^{6} = 7.37 \times 10^{6}$r2​=6.37×106+1.00×106=7.37×106 m):

$V_{2} = -\frac{GM_{E}}{r_{2}} = -\frac{3.98 \times 10^{14}}{7.37 \times 10^{6}} = -5.40 \times 10^{7} \text{ J kg}^{-1}$V2​=−r2​GME​​=−7.37×1063.98×1014​=−5.40×107 J kg−1

Potential difference — note carefully: $V_{2}$V2​ is less negative than $V_{1}$V1​, so $\Delta V > 0$ΔV>0:

$\Delta V = V_{2} - V_{1} = (-5.40 \times 10^{7}) - (-6.25 \times 10^{7}) = +0.85 \times 10^{7} = 8.5 \times 10^{6} \text{ J kg}^{-1}$ΔV=V2​−V1​=(−5.40×107)−(−6.25×107)=+0.85×107=8.5×106 J kg−1

Work done by the external agent (the rocket):

$W_{\text{ext}} = m \, \Delta V = 500 \times 8.5 \times 10^{6} = 4.25 \times 10^{9} \text{ J} = 4.25 \text{ GJ}$Wext​=mΔV=500×8.5×106=4.25×109 J=4.25 GJ

About 4.25 GJ — comparable to the chemical energy in 100 kg of TNT — just to raise the satellite. Circularising the orbit then needs an additional $\tfrac{1}{2}mv_{\text{orbit}}^{2}$21​mvorbit2​ at $v_{\text{orbit}} \approx 7.35$vorbit​≈7.35 km s⁻¹, roughly $1.35 \times 10^{10}$1.35×1010 J — more than three times the lifting work. Most of a rocket's energy budget goes into KE, not altitude.

The sign of $\Delta V$ΔV is the key A* discriminator here. Mid-band candidates often subtract in the wrong order and conclude the work is negative — physically wrong. Always check: lifting outward means $V$V less negative, so $\Delta V = V_{2} - V_{1} > 0$ΔV=V2​−V1​>0.

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Escape Velocity

The escape velocity is the minimum launch speed from the surface of a body that allows a projectile to just reach infinity (with zero residual kinetic energy). It is the speed at which the launch kinetic energy exactly equals the depth of the surface potential well.

Derivation

Conservation of total energy from surface (speed $v_{\text{esc}}$vesc​, $E_{p} = -GMm/R$Ep​=−GMm/R) to infinity (speed $0$0, $E_{p} = 0$Ep​=0):

$\tfrac{1}{2} m v_{\text{esc}}^{2} + \left(-\frac{GMm}{R}\right) = \tfrac{1}{2} m (0)^{2} + 0$21​mvesc2​+(−RGMm​)=21​m(0)2+0

Cancelling $m$m and rearranging:

$\tfrac{1}{2} v_{\text{esc}}^{2} = \frac{GM}{R} \quad \Rightarrow \quad v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$21​vesc2​=RGM​⇒vesc​=R2GM​​

Equivalent form using $g_{s} = GM/R^{2}$gs​=GM/R2 (surface gravity): $v_{\text{esc}} = \sqrt{2 g_{s} R}$vesc​=2gs​R​.

Two A* observations from the derivation:

1. The minus sign in $E_{p} = -GMm/R$Ep​=−GMm/R is essential. It is the depth of the well; positive $\tfrac{1}{2}v_{\text{esc}}^{2}$21​vesc2​ exactly cancels it for the projectile to just reach infinity at zero KE.
2. The mass $m$m of the escapee cancels. Escape velocity depends only on the planet's $M$M and $R$R, not on what you are launching. A ball bearing and a Saturn V need the same speed at launch to escape; they require very different forces and energies to reach that speed, of course.

Escape Velocity of the Earth

$v_{\text{esc, Earth}} = \sqrt{\frac{2GM_{E}}{R_{E}}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.37 \times 10^{6}}}$vesc, Earth​=RE​2GME​​​=6.37×1062×6.67×10−11×5.97×1024​​

The numerator under the root is $2GM_{E} = 7.96 \times 10^{14}$2GME​=7.96×1014 m³ s⁻². Dividing by $R_{E} = 6.37 \times 10^{6}$RE​=6.37×106 gives $1.25 \times 10^{8}$1.25×108 m² s⁻². Taking the square root:

$v_{\text{esc, Earth}} = \sqrt{1.25 \times 10^{8}} = 1.12 \times 10^{4} \text{ m s}^{-1} \approx 11.2 \text{ km s}^{-1}$vesc, Earth​=1.25×108​=1.12×104 m s−1≈11.2 km s−1