Gravitational potential energy

This topic sits inside the CBSE Class 11 Physics chapter on Gravitation, immediately after acceleration due to gravity and its variation with height and depth, and immediately before escape speed. The chapter's prescribed scope explicitly lists $U = -GMm/r$ and the gravitational potential $V = -GM/r$ as examinable formulas, alongside the universal law of gravitation and Kepler's laws. This section stays strictly within that scope: no field-theory treatment of potential as a vector flux, no Gauss's law for gravity, and no discussion of tidal or perturbative effects — those are explicitly out of scope for CBSE 2025-26 and are university-level topics. What is expected of you is to state the general formula confidently, justify the negative sign, connect it back to $mgh$ as a special case, and use it correctly in numerical problems involving satellites and planetary distances.

Think of gravity as carving a bowl-shaped "well" in space around any mass $M$ — deep near the surface, shallower and shallower as you move away, and perfectly flat (zero) only infinitely far away. Any second mass $m$ placed in this well has a potential energy that depends only on how deep in the well it sits, which is to say, on its distance $r$ from the centre of $M$. The general expression for that potential energy is

where: $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ is the universal gravitational constant, $M$ is the mass creating the field (for example, the Earth), $m$ is the mass placed in the field, and $r$ is the centre-to-centre separation between them. Read physically, this says: at the surface of a planet $r$ is small, so $GMm/r$ is a large number, and $U$ is a large negative number — deep in the well, strongly bound. As $r$ grows, $GMm/r$ shrinks, so $U$ climbs steadily toward zero — climbing out of the well. It never becomes positive for an attractive force; it only asymptotes to zero as $r \to \infty$, which is exactly the point where the two masses no longer influence each other at all. Closely related is the idea of gravitational potential, the potential energy per unit mass that a location in space possesses, independent of what (if anything) you place there:

where: $V$ depends only on the source mass $M$ and the distance $r$ — it is a property of space itself around $M$, measured in joules per kilogram. Multiplying $V$ by any test mass $m$ you drop into that location instantly gives you that mass's potential energy there, $U = mV$. This is the same relationship that force and field ($F$ and $g$) have with each other: $g$ is force per unit mass, $V$ is energy per unit mass, and both are properties of the source alone.

Near a planet's surface, moving a mass $m$ from radius $R$ up to $R+h$ changes its potential energy by

where: $R$ is the planet's radius, $h$ is the height gained, and the two terms are simply $U$ evaluated at the final and initial radii. When $h \ll R$, this expression simplifies (using $GM = gR^2$, the relation between $G$, $M$, and surface gravity) to $\Delta U \approx mgh$. Physically, the smooth curved well looks almost perfectly flat over a short stretch near the surface — just as the Earth's curved surface looks flat over the length of a football field — and that local flat approximation is exactly $mgh$.

Choosing $U = 0$ at $r = \infty$ is a deliberate bookkeeping decision, not a fact forced on us by nature. Potential energy of any kind is only ever defined up to an arbitrary additive constant; only differences in $U$, which equal the negative of the work done by gravity, are physically meaningful. The Gravitation chapter fixes the zero at infinity because it makes $U$ naturally negative for every bound system and naturally zero for two masses that no longer interact — a convention that turns out to be extraordinarily convenient just one section later, when escape speed is defined as exactly the speed needed to make total mechanical energy reach zero. This differs from the $mgh$ convention, which fixes zero at the ground; the two agree only in the local, near-surface limit already derived above.

Board level - potential energy of an orbiting satellite

Find the gravitational potential energy of a $500\ \text{kg}$ satellite when it is at a distance of $7.0\times10^{6}\ \text{m}$ from the centre of the Earth. Take $M_E = 6.0\times10^{24}\ \text{kg}$ and $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$.

Step 1. Write down the formula and identify the values: $U = -\dfrac{GM_Em}{r}$, with $M_E = 6.0\times10^{24}\ \text{kg}$, $m = 500\ \text{kg}$, and $r = 7.0\times10^{6}\ \text{m}$.

Step 2. Compute the numerator. $GM_E = 6.67\times10^{-11} \times 6.0\times10^{24} = 4.0\times10^{14}$. Then $GM_Em = 4.0\times10^{14} \times 500 = 2.0\times10^{17}$.

Step 3. Divide by $r$ and attach the sign. $U = -\dfrac{2.0\times10^{17}}{7.0\times10^{6}} \approx -2.86\times10^{10}\ \text{J}$.

The final answer is $U \approx -2.86\times10^{10}\ \text{J}$. The negative sign confirms the satellite is gravitationally bound to Earth; about $2.86\times10^{10}\ \text{J}$ of energy would need to be supplied (ignoring any kinetic energy it carries) to move it all the way out to infinite separation, where $U$ would finally reach zero.

Competitive level - comparing exact PE change against the mgh approximation, then finding the crossover height

Part A: A $2\ \text{kg}$ object is lifted from Earth's surface ($R_E = 6.4\times10^{6}\ \text{m}$) to a height of $10\ \text{km}$. Compare the exact change in gravitational PE with the $mgh$ approximation, taking $g = 9.8\ \text{m/s}^2$ and $M_E = 6.0\times10^{24}\ \text{kg}$. Part B: estimate roughly how large $h$ would need to be, as a fraction of $R_E$, before the two values differ by about $5\%$, to build intuition for when $mgh$ stops being trustworthy.

Step 1. Compute the approximate value first. $\Delta U \approx mgh = 2 \times 9.8 \times 10{,}000 = 1.96\times10^{5}\ \text{J}$.

Step 2. Compute the exact value using $\Delta U = GMm\left(\dfrac{1}{R} - \dfrac{1}{R+h}\right)$, with $R = 6.4\times10^{6}\ \text{m}$ and $R+h = 6.41\times10^{6}\ \text{m}$. The bracket is $\dfrac{1}{6.4\times10^{6}} - \dfrac{1}{6.41\times10^{6}} \approx 1.5625\times10^{-7} - 1.5601\times10^{-7} = 2.4\times10^{-10}\ \text{m}^{-1}$.

Step 3. Multiply through. $GMm = 6.67\times10^{-11} \times 6.0\times10^{24} \times 2 = 8.0\times10^{14}$, so $\Delta U = 8.0\times10^{14} \times 2.4\times10^{-10} \approx 1.92\times10^{5}\ \text{J}$. The exact answer for Part A is $\Delta U \approx 1.92\times10^{5}\ \text{J}$, extremely close to the $1.96\times10^{5}\ \text{J}$ from $mgh$ — a difference of only about $2\%$, because $h \ll R$ here ($h/R \approx 0.0016$).

Step 4. For Part B, notice that the fractional error between the exact formula and $mgh$ scales roughly with $h/R$: doubling $h/R$ roughly doubles the percentage gap. Since $h/R \approx 0.0016$ gives about a $2\%$ gap, reaching a $5\%$ gap needs $h/R$ roughly $2$–$3$ times larger, so $h/R \sim 0.004$ to $0.005$, meaning $h$ around $25$–$32\ \text{km}$. This is the answer for Part B: the $mgh$ approximation stays trustworthy to a few per cent only up to roughly $h \sim 0.005\,R_E$, tens of kilometres — well below typical satellite altitudes of hundreds of kilometres, where the exact formula becomes essential. Timing note: in an exam, do not recompute the bracket difference by long division under time pressure — recognise the pattern $\dfrac{1}{R} - \dfrac{1}{R+h} \approx \dfrac{h}{R^2}$ for $h \ll R$, which gives $\Delta U \approx \dfrac{GMmh}{R^2} = mgh$ directly (since $GM/R^2 = g$), letting you jump straight to the approximate answer and only fall back to full precision if the question explicitly demands the exact value.

Students routinely lose marks by treating "negative" as meaning something has gone wrong, when it is simply the sign that follows automatically from the chosen zero at infinity — a negative $U$ is completely normal and expected for any bound pair of masses. Another frequent slip is measuring $r$ from the planet's surface rather than from its centre; the formula $U = -GMm/r$ always uses centre-to-centre distance, so a satellite at "height $h$ above the surface" sits at $r = R + h$, not $r = h$. Students also often blur gravitational potential $V$ (a property of a location, energy per unit kilogram) with gravitational potential energy $U$ (the energy of one specific two-mass system) — remember that $U = mV$ links them, and only $U$ depends on the particular mass $m$ you put there. Finally, many assume $mgh$ works at any altitude; it is only the small-height limit of the true formula, and using it for satellites, rockets far from Earth, or any situation where $h$ is a meaningful fraction of $R$ gives a wrong answer that a well-set exam question is specifically designed to catch.

• State the general formula $U = -GMm/r$ and explain clearly why the negative sign appears from the zero-at-infinity convention.
• Derive $U = -GMm/r$ from the inverse-square gravitational force using the same area-under-the-force-graph reasoning used for $mgh$.
• Define gravitational potential $V = -GM/r$ and correctly distinguish it from gravitational potential energy $U$.
• Show algebraically how $U = -GMm/r$ reduces to $\Delta U \approx mgh$ for small heights near a planet's surface, and judge when that approximation stops being reliable.
• Calculate gravitational potential energy for a mass at a specified distance from a planet or star, always measuring $r$ from the centre.
• Explain, using the idea of work done by an external agent, why PE is conventionally taken to be zero only at infinite separation.

Ask the tutor: if the negative sign or the jump from $mgh$ to $-GMm/r$ still feels shaky, ask the AI tutor to walk through the area-under-the-curve derivation step by step with a picture.