Gravitational potential and energy

Lift a book onto a shelf and the work you do does not vanish — it is stored, ready to be released the instant the book falls. Near the ground we call that store $mgh$ , but $mgh$ is only a local approximation. A satellite raised thousands of kilometres, or a rocket escaping Earth altogether, needs a formula that works at any distance, not just "near the floor." That general formula is built from two related ideas: gravitational potential $V$ , the work done per unit mass in bringing a test mass in from infinity, and gravitational potential energy $U = mV$ , the actual energy stored for a mass $m$ . Both come out negative — a deliberate, physically meaningful choice — and both reduce exactly to the familiar $mgh$ when the height involved is small compared with the Earth's radius. This article derives $V = -GM/r$ and $U = -GMm/r$ from first principles, shows precisely how $\Delta U \to mgh$ falls out of the general result, and works through the numbers for a satellite being raised into orbit.

The core insight

Gravitational potential $V = -GM/r$ and gravitational potential energy $U = mV = -GMm/r$ are both negative because they are measured relative to zero at infinity, and the general $\Delta U$ collapses to the familiar $mgh$ only when the height is much smaller than the Earth's radius.

The core idea

Gravitational potential at a point $P$ , distance $r$ from a source mass $M$ , is defined as the work done in bringing a unit test mass from infinity to $P$ :

V_P = \dfrac{W_{\infty P}}{m}

where: $W_{\infty P}$ is the work done by an external agent moving the unit mass from infinity to $P$ , and $m$ is the (small) test mass used to probe the field. Gravity is attractive, so as the test mass falls in toward $M$ , the field does positive work on it; the external agent holding it back does negative work. By convention $V=0$ at infinity, so $V$ is negative everywhere closer in:

V(r) = -\dfrac{GM}{r}

where: $G = 6.67\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ is the universal gravitational constant, $M$ is the source mass, and $r$ is the distance from its centre. Potential is measured in $\text{J/kg}$ .

Potential difference between two points $A$ and $B$ follows by integrating the field intensity $E_g = GM/r^2$ along the radial path (using $dV = -E_g\,dr$ ):

\Delta V = V_B - V_A = GM\left(\dfrac{1}{r_A}-\dfrac{1}{r_B}\right)

Gravitational potential energy of an actual mass $m$ sitting at distance $r$ is just $m$ times the potential there — the total work needed to assemble the pair, starting from infinite separation where $U=0$ :

U(r) = mV(r) = -\dfrac{GMm}{r}

$U(r)$ is always negative for finite $r$ : to pull the masses apart to infinity you must add energy, which correctly makes $U$ increase (become less negative) as $r$ grows. A satellite deep in a planet's gravity well is more tightly bound (more negative $U$ ) than one in a distant orbit — the same negative, zero-at-infinity convention reappears later for electrons bound to a nucleus, so it is worth internalising now rather than memorising as a stray minus sign.

Figure

A graph of gravitational potential energy against distance from the planet's centre, a curve starting very negative near the surface and rising smoothly toward zero as distance increases without limit

Reducing to $mgh$ near the surface. For two points at radius $R$ (surface) and $R+h$ with $h \ll R$ :

\Delta U = U(R+h)-U(R) = GMm\left(\dfrac{1}{R}-\dfrac{1}{R+h}\right) = \dfrac{GMm\,h}{R(R+h)} \approx \dfrac{GMm\,h}{R^2} = mgh

where: the last step uses $(R+h)\approx R$ for $h \ll R$ , and $g = GM/R^2$ is the familiar surface gravitational acceleration. This is why $mgh$ "just works" for everyday heights, and why it silently fails for satellites and large altitudes.

Derivation APotential energy of a mass at distance r from a source mass

The goal is to derive $U(r) = -GMm/r$ starting only from the definition of work and the inverse-square law of gravitational force, and to show explicitly why the sign comes out negative rather than assuming it. We build the potential $V(r)$ first, by integrating the field intensity, then multiply by $m$ to get $U$ .

Step 1. Start from the gravitational field intensity at distance $r$ from mass $M$ : $E_g(r) = GM/r^2$ , directed toward $M$ . This is the force per unit mass on a test mass placed at $r$ , and it is what an external agent must exactly balance (in magnitude, opposite direction) to move the test mass quasi-statically.

Step 2. Write the potential as the work per unit mass done bringing the test mass from infinity to $r$ , using $dV = -E_g\,dr$ (the minus sign because moving inward, $dr<0$ , corresponds to the field doing positive work, so external work and hence $dV$ is negative). Integrating from $\infty$ to $r$ : $V(r) = -\displaystyle\int_{\infty}^{r} \dfrac{GM}{r'^2}\,dr' \cdot(-1)$ , which after evaluating the standard integral $\int GM/r'^2\,dr' = -GM/r'$ gives $V(r) = -GM/r$ , with $V(\infty)=0$ as required.

Step 3. Convert potential to potential energy for an actual mass $m$ placed at $r$ : $U(r) = mV(r) = -GMm/r$ . This is the work needed to assemble the two-mass system starting from infinite separation, and it is negative because gravity does positive work as the masses come together — an external agent restraining that fall does negative work.

Step 4. Check the sign physically. As $r\to\infty$ , $U\to0^-$ (approaches zero from below); as $r$ decreases, $U$ becomes more negative. So $U(r)$ is a monotonically increasing function of $r$ — moving a mass farther away always raises its potential energy, exactly as it should for an attractive force, confirming the derivation is self-consistent. Result: $U(r) = -GMm/r$ , negative for all finite $r$ , zero only at $r\to\infty$ , and increasing (less negative) as $r$ grows.

Worked example

A satellite of mass $500\ \text{kg}$ is raised from Earth's surface ( $R = 6.4\times10^6\ \text{m}$ ) to an orbit at $r = 7.2\times10^6\ \text{m}$ from Earth's centre. Take $GM = 4.0\times10^{14}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}$ . Find (a) the gravitational potential at the surface, and (b) the increase in gravitational potential energy.

Step 1. Find the potential at the surface using $V = -GM/R$ . With $GM = 4.0\times10^{14}$ and $R = 6.4\times10^6\ \text{m}$ :

V = -\dfrac{4.0\times10^{14}}{6.4\times10^6} = -6.25\times10^{7}\ \text{J/kg}

Step 2. Set up the potential energy change using $\Delta U = GMm\left(\dfrac{1}{R}-\dfrac{1}{r}\right)$ , since $\Delta U = U(r) - U(R) = -\dfrac{GMm}{r} + \dfrac{GMm}{R}$ .

Step 3. Compute the reciprocals. $\dfrac{1}{R} = \dfrac{1}{6.4\times10^6} = 1.5625\times10^{-7}$ , and $\dfrac{1}{r} = \dfrac{1}{7.2\times10^6} = 1.3889\times10^{-7}$ , so $\dfrac{1}{R}-\dfrac{1}{r} = 1.736\times10^{-8}$ .

Step 4. Substitute and finish: $\Delta U = (4.0\times10^{14})(500)(1.736\times10^{-8}) = (2.0\times10^{17})(1.736\times10^{-8}) = 3.47\times10^{9}\ \text{J}$ . So the surface potential is $-6.25\times10^7\ \text{J/kg}$ , and raising the satellite requires $\Delta U = 3.47\times10^9\ \text{J}$ — the work a rocket must supply against gravity.

Competitive stretch

The satellite's fractional change in $|U|$ is $\dfrac{|U(R)|-|U(r)|}{|U(R)|}$ . Using $U(R) = -GMm/R = -(4.0\times10^{14})(500)/(6.4\times10^6) = -3.125\times10^{10}\ \text{J}$ and $U(r) = -(4.0\times10^{14})(500)/(7.2\times10^6) = -2.778\times10^{10}\ \text{J}$ , the fraction is $(3.125-2.778)/3.125 \approx 11\%$ — a useful cross-check: $\Delta U = U(r)-U(R) = 3.47\times10^9\ \text{J}$ , matching Step 4.

Common traps

Dropping the negative sign in $V=-GM/r$ and $U=-GMm/r$ : the sign is essential, encoding the zero-reference at infinity, not an error to tidy away.
Confusing potential $V$ (energy per unit mass, $\text{J/kg}$ ) with potential energy $U=mV$ (total energy, $\text{J}$ ) — always check which one a question asks for.
Using $\Delta U = mgh$ at heights comparable to $R$ : this is only an approximation valid for $h\ll R$ ; for satellites and large altitudes use the full $U=-GMm/r$ .
Treating $U=0$ as "no energy" in an absolute sense: it is simply the chosen reference at infinite separation, not a special physical zero.

Ask the tutor: Want to walk through why the minus sign in $U=-GMm/r$ makes physical sense, or see the $\Delta U \to mgh$ reduction worked out one more time with numbers? Ask the AI tutor about Gravitational potential and energy.

Why is potential energy negative instead of just using mgh with a starting height of zero at the surface?

Because $mgh$ only works for small heights near one reference surface; it silently breaks down for satellites and large altitudes, where the field intensity itself changes noticeably with distance. The general formula $U=-GMm/r$ is valid at any distance, from planetary surfaces to interplanetary space, and it reduces exactly to $mgh$ when $h\ll R$ . Choosing infinity (rather than the surface) as the zero reference is what makes the single formula work everywhere, at the cost of $U$ being negative near the source.

Is gravitational potential the same thing as gravitational potential energy?

No, though they are closely related. Potential $V=-GM/r$ is a property of the field alone, in $\text{J/kg}$ — it tells you the potential energy per unit mass that any test mass would have at that point. Potential energy $U=mV=-GMm/r$ is specific to an actual mass $m$ placed there, in joules. Multiplying $V$ by the mass you are interested in converts one into the other.

Why does raising a satellite need positive energy input if U becomes less negative, which sounds like a decrease?

$U$ becoming less negative is itself an increase in $U$ — think of it on a number line, where $-2.78\times10^{10}$ is greater than $-3.13\times10^{10}$ . Energy conservation then says that increase in stored potential energy must come from somewhere, and that somewhere is the work the rocket engine does against gravity. The "less negative" phrasing describes the direction of change correctly; it is easy to misread it as a decrease if you forget negative numbers grow toward zero as they rise.

Gravitational potential and energy

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads