Why is gravitational potential energy negative?

Because we choose the zero of potential energy to be at infinite separation. Gravity is attractive, so as two masses move from infinity to a finite separation r the gravitational force does positive work and the stored potential energy falls below zero. The expression U = −GMm/r therefore comes out negative for every finite r. The negative sign means the system is bound: external work must be supplied to pull the masses back apart to infinity.

What is the difference between U = −GMm/r and U = mgh?

U = −GMm/r is the exact gravitational potential energy of a two-body system, valid at any separation r. U = mgh is only an approximation to the DIFFERENCE in potential energy between two points, and it holds only near the Earth's surface, where the height h is much smaller than the Earth's radius so that g can be treated as constant. NCERT states verbatim that the expression mgh is an approximation to the difference in the gravitational potential energy.

What is the difference between gravitational potential and gravitational potential energy?

Gravitational potential V = −GM/r is a property of the field at a point, defined as the potential energy of a particle of unit mass placed there; its SI unit is J/kg. Gravitational potential energy U = −GMm/r belongs to the two-body system and depends on the test mass m; its SI unit is the joule. They are linked by U = mV. Potential is field per unit mass; potential energy is the actual energy of the chosen mass in that field.

How do you find the gravitational potential due to several masses?

Gravitational potential is a scalar, so you add the contributions of each mass arithmetically. The potential at a point due to masses M1, M2, M3 at distances r1, r2, r3 is V = −G(M1/r1 + M2/r2 + M3/r3). There are no vector components and no angles to resolve — this superposition is far simpler than adding gravitational field vectors.

What does the work done in moving a mass in a gravitational field equal?

Because gravity is conservative, the work done by the gravitational force in moving a mass from point 1 to point 2 depends only on the endpoints, not the path. The work done BY gravity equals the drop in potential energy, U1 − U2. Equivalently, the work an external agent must do, moving the mass slowly without changing its speed, equals the rise in potential energy U2 − U1 = m(V2 − V1).

Is the U-versus-r curve a straight line or a curve?

It is a curve. U = −GMm/r is a rectangular hyperbola lying entirely below the r-axis. It is most negative (a deep well) at small r near the surface, rises steeply at first, then flattens and approaches zero from below as r tends to infinity. The mgh straight-line approximation is only the gentle, almost-linear stretch of this curve very close to the surface.

Does the choice of where U is zero change the physics?

No. Only differences in potential energy have physical meaning, so the additive constant is arbitrary. Choosing U = 0 at infinity simply fixes that constant at zero, which gives the clean form U = −GMm/r. The gravitational force, and every measurable quantity such as work done or escape speed, is unchanged by this choice.

Gravitational Potential Energy — NEET Physics

Why gravity stores energy

Potential energy is the energy stored in a body on account of its position, measured by the work done in bringing the body from a chosen standard position to its present one. For this idea to be well-defined the force must be conservative — the work it does between two points must be independent of the path taken. NCERT §7.7 states plainly that the force of gravity is a conservative force, so a gravitational potential energy function exists.

The consequence is the central rule of this topic: when a body moves in a gravitational field, the change in its potential energy equals the work done on it by the gravitational force. Two positions, one number — and that number does not care which route the body took between them.

The near-surface result: U = mgh

Start close to the Earth's surface, at heights much smaller than the Earth's radius. Here the gravitational force is practically constant, equal to $mg$ and directed towards the centre. Lifting a mass $m$ from a height $h_1$ to a height $h_2$ costs work

$$W_{12} = mg\,(h_2 - h_1).$$

Associate with each height a potential energy $W(h) = mgh + W_0$, where $W_0$ is a constant. Then $W_{12} = W(h_2) - W(h_1)$, and the constant $W_0$ cancels — confirming that only the difference matters. Setting $h=0$ shows $W_0$ is simply the potential energy at the surface. Dropping that constant gives the schoolbook form $U = mgh$.

This is clean, but it rests on one assumption: that $g$ does not change over the height involved. The moment $h$ becomes comparable to the Earth's radius, that assumption collapses and $mgh$ is no longer valid. For that regime we need the exact expression.

The exact two-body form: U = −GMm/r

For points at arbitrary distance from the Earth, the force is no longer constant; it follows the inverse-square law $F = \dfrac{GM_E m}{r^2}$ directed towards the centre. Integrating this force to find the work done in moving the particle from $r_1$ to $r_2$ gives

$$W_{12} = \int_{r_1}^{r_2}\frac{GM_E m}{r^2}\,dr = -\,GM_E m\left(\frac{1}{r_2}-\frac{1}{r_1}\right).$$

So we may associate a potential energy $W(r) = -\dfrac{GM_E m}{r} + W_1$, valid for $r > R$, such that once again $W_{12} = W(r_2) - W(r_1)$. Setting $r \to \infty$ makes the first term vanish, so $W_1$ is the potential energy at infinity. Conventionally one sets $W_1 = 0$: the potential energy at a point becomes exactly the work done in bringing the particle from infinity to that point. With that choice, for a two-body system of masses $M$ and $m$ separated by $r$,

$$\boxed{\,U = -\dfrac{GMm}{r}\,}\qquad (U \to 0 \text{ as } r \to \infty).$$

The $U$-versus-$r$ curve is a rectangular hyperbola lying entirely below the axis. It is deepest (most negative) near the surface and climbs steeply, then flattens and approaches $0$ from below as $r \to \infty$. The zero of energy sits at infinite separation; every bound configuration lies in the well beneath it.

Why U is negative

The negative sign is not a quirk of algebra; it carries physical meaning. Because the zero of potential energy sits at infinity and gravity is attractive, bringing two masses together from infinity lets gravity do positive work, draining the system's energy below zero. Every finite separation therefore gives a negative $U$.

The deeper (more negative) the value, the more tightly bound the system: more external work is needed to separate the masses back to infinity. This is exactly why the total mechanical energy of an orbiting satellite is negative — the system is gravitationally bound.

Gravitational potential V and the relation U = mV

The gravitational potential at a point is defined as the potential energy of a particle of unit mass placed there. Dividing the two-body energy by the test mass gives

$$V = \frac{U}{m} = -\frac{GM}{r}.$$

Potential $V$ describes the field itself, independent of any test mass; potential energy $U$ describes a particular mass $m$ sitting in that field. They are linked by $U = mV$. Crucially, $V$ is a scalar — its SI unit is J/kg, while $U$ is measured in joules.

Quantity	Symbol & formula	Depends on	SI unit
Gravitational potential	$V = -\dfrac{GM}{r}$	Source mass and position only	J kg$^{-1}$
Gravitational potential energy	$U = -\dfrac{GMm}{r}$	Both masses and separation	J
Link between them	$U = mV$	Test mass $m$	—

Because $V$ is a scalar, the potential due to several masses is the simple arithmetic sum of their individual contributions — the superposition principle. For masses $M_1, M_2, M_3,\dots$ at distances $r_1, r_2, r_3,\dots$ from a point,

$$V = -G\left(\frac{M_1}{r_1}+\frac{M_2}{r_2}+\frac{M_3}{r_3}+\cdots\right).$$

No angles, no vector components — that is the chief advantage of working with potential rather than field. The total potential energy of an isolated system of particles is likewise the sum over all distinct pairs.

Work done in a gravitational field

Since gravity is conservative, the work done depends only on the endpoints. Two equivalent statements are worth keeping straight. The work done by gravity as a mass moves from point 1 to point 2 equals the drop in potential energy, $W_{\text{grav}} = U_1 - U_2$. The work an external agent must do, moving the mass slowly so its speed does not change, equals the rise in potential energy:

$$W_{\text{ext}} = U_2 - U_1 = m\,(V_2 - V_1).$$

If the mass is carried between two points at the same potential, the net work is zero regardless of path — the defining mark of a conservative field. These relations turn every "work to move a mass" problem into a difference of two potential energies, no integration required once $U=-GMm/r$ is in hand.

Related drill

The conservative-force machinery here is the same as in mechanics — revisit potential energy and conservation for the general work–energy framework.

Worked example 1 — energy to lift a mass to height h

Worked Example

A body of mass $m$ is lifted from the Earth's surface (radius $R$, mass $M$) to a height $h$ above the surface. Find the exact increase in its gravitational potential energy, and the work an external agent must supply.

Set up the two endpoints. Initial separation from the centre is $r_1 = R$; final separation is $r_2 = R+h$. Use the exact form $U=-GMm/r$ with the zero at infinity.

Take the difference. $$\Delta U = U_2 - U_1 = -\frac{GMm}{R+h} - \left(-\frac{GMm}{R}\right) = GMm\left(\frac{1}{R}-\frac{1}{R+h}\right).$$

Simplify. $$\Delta U = GMm\cdot\frac{(R+h)-R}{R(R+h)} = \frac{GMm\,h}{R(R+h)}.$$ This is positive, as it must be: lifting against gravity raises the potential energy. The external agent, moving the mass slowly, supplies exactly $W_{\text{ext}} = \Delta U = \dfrac{GMm\,h}{R(R+h)}$.

Sanity check. For $h \ll R$ the denominator $R(R+h)\to R^2$, so $\Delta U \to \dfrac{GMm}{R^2}\,h = mgh$, since $g = GM/R^2$ at the surface. The exact result collapses to $mgh$ precisely in the near-surface limit.

Worked example 2 — mgh versus the exact difference

Worked Example

Compare the potential-energy increase given by $mgh$ with the exact value for a mass raised to (a) $h = R/100$ and (b) $h = R$ above the surface. Express the exact value as a fraction of $mgh$.

Form the ratio. From Example 1, the exact increase is $\Delta U = \dfrac{GMm\,h}{R(R+h)}$. Using $mgh = \dfrac{GMm}{R^2}h$, $$\frac{\Delta U_{\text{exact}}}{mgh} = \frac{R^2}{R(R+h)} = \frac{R}{R+h} = \frac{1}{1 + h/R}.$$

(a) $h = R/100$. Ratio $= \dfrac{1}{1+0.01} \approx 0.9901$. The $mgh$ estimate overstates the true value by only about $1\%$ — perfectly acceptable for near-surface problems.

(b) $h = R$. Ratio $= \dfrac{1}{1+1} = 0.5$. Now $mgh$ gives twice the true potential-energy increase. At heights comparable to $R$, the $mgh$ formula is grossly wrong, and only $U=-GMm/r$ is reliable.

Reading. The error in $mgh$ is governed entirely by the ratio $h/R$. This is the quantitative face of NCERT's statement that $mgh$ is merely an approximation to the difference in gravitational potential energy.

Worked example 3 — potential due to several masses

NCERT Example 7.3 (adapted)

Four particles, each of mass $m$, sit at the corners of a square of side $l$. Find the gravitational potential at the centre of the square.

Find the distance from centre to each corner. The half-diagonal of a square of side $l$ is $\dfrac{l\sqrt{2}}{2} = \dfrac{l}{\sqrt{2}}$. All four corners are equidistant from the centre at $r = \dfrac{l}{\sqrt{2}}$.

Add the scalar potentials. Potential is a scalar, so the four contributions add directly: $$V_{\text{centre}} = 4\times\left(-\frac{Gm}{r}\right) = -\frac{4Gm}{\,l/\sqrt{2}\,} = -\frac{4\sqrt{2}\,Gm}{l}.$$

Reading. No vector resolution was needed — that is the whole point of working with potential. By symmetry the gravitational field at the centre is zero, yet the potential there is decidedly non-zero. Field and potential are different objects, and a zero field never implies a zero potential.

Quick recap

Gravitational PE in one breath

Gravity is conservative, so a position-dependent potential energy exists; only differences in $U$ are physical.
Exact two-body PE: $U = -GMm/r$, with the zero set at infinite separation.
$U$ is negative because gravity is attractive and the zero is at infinity — the system is bound; more negative means more tightly bound.
$mgh$ is only the near-surface approximation, valid when $h \ll R$; the exact difference is $-GMm\left(\tfrac{1}{r_2}-\tfrac{1}{r_1}\right)$.
Gravitational potential $V = -GM/r$ is PE per unit mass (J/kg, a scalar); $U = mV$.
Potentials of several masses add arithmetically; work to move a mass $= m(V_2 - V_1)$, path-independent.

Gravitational Potential Energy