Gravitational Potential, Potential Energy And Work Done

PhysicsGravitationFor NEET aspirants

The gravitational potential energy of a two-mass system is $U = - \frac{GM m}{r}$ , taking $U = 0$ at infinite separation. The gravitational potential at a point is the potential energy per unit test mass: $V = U / m = - \frac{GM}{r}$ for a point source. Field and potential are related by $E = - \nabla V$ , and the work done in moving a mass through a gravitational field equals the mass times the change in potential: $W = m (V_{B} - V_{A})$ . This concept covers PE, potential, work done, potential of extended bodies (including sphere and shell), self-energy, and $N$ -particle interaction energy for JEE Mains and NEET.

Key Formulas - Quick Reference

PE of two point masses: $U = - \frac{GM m}{r}$ (with $U = 0$ at $r = \infty$ )
Potential at distance $r$ from point mass $M$ : $V = - \frac{GM}{r}$
Relation: $V = U / m$ ; unit J/kg; dimensions $[M^{0} L^{2} T^{- 2}]$
Field-potential relation: $E = - \frac{d V}{d r} \overset{r}{^}$ ; equivalently $V = - \int E \cdot d r$
Work done shifting mass $m$ from A to B: $W = m (V_{B} - V_{A}) = U_{B} - U_{A}$
Escape energy: $∣ U ∣ = \frac{GM m}{R}$ ; escape speed $v_{e} = 2 GM / R$
Interaction energy of $N$ particles: $U_{sy s} = - \frac{1}{2} \sum_{i \neq = j} \frac{G m _{i} m _{j}}{r _{ij}}$
Self-energy of uniform solid sphere: $U_{se l f} = - \frac{3 G M ^{2}}{5 R}$
Self-energy of uniform spherical shell: $U_{se l f} = - \frac{G M ^{2}}{2 R}$
Potential of solid sphere at centre: $V_{ce n t r e} = - \frac{3 GM}{2 R}$
Potential of solid sphere inside: $V (r) = - \frac{GM}{2 R ^{3}} (3 R^{2} - r^{2})$ , valid for $r \leq R$

1. Gravitational Potential Energy

The gravitational potential energy of a body at a point in a gravitational field is defined as the work done by an external agent in bringing the body from infinity to that point, moving quasi-statically (no change in kinetic energy).

Consider a source mass $M$ fixed at the origin and a test mass $m$ moved from infinity to a distance $r$ . The gravitational force on $m$ at distance $r^{'}$ is $F = - GM m / r^{'2}$ (attractive, directed toward $M$ ). The work done against this force from infinity to $r$ is:

$U = - W_{\infty \to r} = - \int_{\infty}^{r} F \cdot d r$

Since $F$ and $d r$ point in opposite directions along the line joining the masses:

$U = \int_{\infty}^{r} \frac{GM m}{r ^{'2}} d r^{'} \cdot (- 1) \cdot (- 1) = \int_{\infty}^{r} - \frac{GM m}{r ^{'2}} d r^{'}$

Evaluating:

$U = - GM m [- \frac{1}{r ^{'}}]_{\infty}^{r} = - GM m (\frac{1}{r} - 0)$

U = - \frac{GM m}{r}

Figure 1: Potential energy

U = - \frac{GM m}{r}

for two point masses. Negative everywhere; approaches zero from below as

r \to \infty

; deep well near

r = 0

. Convention:

U = 0

at infinite separation.

Two key points about this sign convention:

$U < 0$ everywhere: the two masses are more strongly bound (lower energy) when close together than when far apart. To separate them requires positive external work.
$U = 0$ at $r = \infty$ by convention: this is the natural zero because gravity vanishes there.

On or near Earth's surface for small height changes, we often use the simpler

U_{g r a v} = m g h

. This is a linear approximation of the exact

- GM m / r

, valid when

h ≪ R

. The gap

m g h

is really the change in

U

Δ U = U (R + h) - U (R) \approx m g h

for small

h

2. Gravitational Potential

The gravitational potential at a point is the potential energy per unit mass placed at that point:

V = \frac{U}{m}

For a point source $M$ at distance $r$ :

$V = - \frac{GM}{r}$

Potential is a scalar quantity - much easier to work with than the vector field $E$ , because contributions from multiple source masses simply add algebraically:

$V_{t o t a l} = i \sum V_{i} = - i \sum \frac{G M _{i}}{r _{i}}$

Dimensional formula

$[V] = \frac{[ Work ]}{[ Mass ]} = \frac{[ M L ^{2} T ^{- 2} ]}{[ M ]} = [M^{0} L^{2} T^{- 2}]$

SI unit: joules per kilogram (J/kg), equivalently m²/s². (Same dimensions as the square of speed - a hint at how energy/mass appears in relativity.)

Relation between field and potential

For a spherically symmetric field:

E = - \frac{d V}{d r} \overset{r}{^} or E = - grad (V)

Verification: $V = - GM / r$ , so $d V / d r = GM / r^{2}$ , and $E = - d V / d r = - GM / r^{2}$ (negative sign indicating $E$ points inward). Magnitudes match: $∣ E ∣ = GM / r^{2}$ . Consistent.

Solved Example 1

At a point above the Earth's surface, the gravitational potential is

- 5.12 \times 1 0^{7} J/kg

and the acceleration due to gravity is

6.4 m/s^{2}

. Assuming the mean radius of the Earth is

6400 km

, calculate the height of this point above the surface.

Solution:

Let $r$ be the distance from the centre of Earth to the point. Then:

$V = - \frac{GM}{r} = - 5.12 \times 1 0^{7} J/kg \dots (1)$

$g = \frac{GM}{r ^{2}} = 6.4 m/s^{2} \dots (2)$

Dividing (1) by (2) (ignoring the negative sign in $V$ ):

$\frac{∣ V ∣}{g} = \frac{GM / r}{GM / r ^{2}} = r$

$r = \frac{5.12 \times 1 0 ^{7}}{6.4} = 8 \times 1 0^{6} m = 8000 km$

Height above the surface:

$h = r - R = 8000 - 6400 = 1600 km$

3. Potential and Field due to Common Shapes

The gravitational potential due to standard mass distributions is tabulated below. Field values are the negative derivative of potential; both are listed for reference.

Source shape	Potential $V$	Field $E$
Point mass $M$ at distance $r$	$V = - \frac{GM}{r}$	$E = \frac{GM}{r ^{2}}$
Ring (mass $M$ , radius $R$ ) on axis at distance $r$	$V = - \frac{GM}{R ^{2} + r ^{2}}$	$E = \frac{GM r}{( R ^{2} + r ^{2} ) ^{3/2}}$
Rod (mass $M$ , length $L$ ), axial: end at distance $r$	$V = - \frac{GM}{L} ln (\frac{r + L}{r})$	$E = \frac{GM}{r ( r + L )}$
Rod, equatorial point at distance $r$ from centre	$V = - \frac{GM}{L} ln (\frac{L + L ^{2} + 4 r ^{2}}{- L + L ^{2} + 4 r ^{2}})$	$E = \frac{2 GM}{r L ^{2} + 4 r ^{2}}$
Hollow sphere (mass $M$ , radius $R$ )	$r \geq R$ : $V = - \frac{GM}{r}$ $r \leq R$ : $V = - \frac{GM}{R}$ (constant)	$r \geq R$ : $E = \frac{GM}{r ^{2}}$ $r < R$ : $E = 0$
Solid sphere (mass $M$ , radius $R$ )	$r \geq R$ : $V = - \frac{GM}{r}$ $r \leq R$ : $V = - \frac{GM}{2 R ^{3}} (3 R^{2} - r^{2})$	$r \geq R$ : $E = \frac{GM}{r ^{2}}$ $r \leq R$ : $E = \frac{GM r}{R ^{3}}$

Special result: Potential of a hollow spherical shell

By the shell theorem, the potential due to a hollow shell of mass $M$ and radius $R$ is:

Outside ( $r \geq R$ ): $V = - \frac{GM}{r}$ (same as a point mass at the centre)

Inside ( $r \leq R$ ): $V = - \frac{GM}{R}$ (constant throughout the interior, equal to the surface value)

Inside the shell, potential is constant because the field is zero (no work is needed to move a test mass around inside). But the potential is not zero - it retains the surface value, since bringing a test mass in from infinity to any interior point requires work only up to reaching the surface.

Figure 2: Potential of a uniform hollow shell. Inside (

r \leq R

V = - \frac{GM}{R}

(constant). Outside (

r \geq R

V = - \frac{GM}{r}

. Continuous at the surface but slope has a kink (field jumps from 0 to

GM / R^{2}

Notice the kink at the surface: potential is continuous (

V \to - GM / R

from both sides) but its slope jumps abruptly. Inside, the slope is zero (

E = 0

); just outside, the slope is

GM / R^{2}

(

E = GM / R^{2}

). This mirrors the field discontinuity at the surface of a hollow shell.

Special result: Potential of a uniform solid sphere

Applying the shell theorem shell-by-shell to a uniform solid sphere of mass $M$ and radius $R$ :

Outside ( $r \geq R$ ): $V = - \frac{GM}{r}$ (same as a point mass at the centre)

Inside ( $r \leq R$ ): $V = - \frac{GM}{2 R ^{3}} (3 R^{2} - r^{2})$

At the surface ( $r = R$ ): $V_{s u r f} = - \frac{GM}{R}$

At the centre ( $r = 0$ ): $V_{ce n t r e} = - \frac{3 GM}{2 R}$ (deepest)

The centre value is $1.5 \times$ the surface value - the potential well is deepest at the centre. For Earth, $V_{ce n t r e} \approx - 9.4 \times 1 0^{7} J/kg$ .

Figure 3: Potential of a uniform solid sphere. Outside:

V = - \frac{GM}{r}

. Inside:

V = - \frac{GM}{2 R ^{3}} (3 R^{2} - r^{2})

. At centre:

V_{ce n t r e} = - \frac{3 GM}{2 R}

(deepest); at surface:

V_{s u r f} = - \frac{GM}{R}

Even though the field

E

is zero at the centre of a solid sphere (or anywhere inside a hollow shell), the potential

V

is not zero. Field measures the slope of

V

, so a flat (zero slope) potential still has a nonzero value. This is the same as a valley floor: the ground is flat (zero gradient) but the elevation is not zero.

4. Work Done in a Gravitational Field

If a body of mass $m$ is moved by an external agent from point A to point B in a gravitational field (with negligible change in kinetic energy), the work done equals the change in gravitational potential energy:

W_{e x t} = U_{B} - U_{A} = m (V_{B} - V_{A})

In words: work done = mass of the body × potential difference between the terminal points. Note the strong analogy with the electric case ( $W = q Δ V$ ).

Because gravity is conservative, this work depends only on the endpoints A and B - not on the path taken between them.

Solved Example 2

Find the work done in shifting a body of mass

m

from a height

h

above the Earth's surface to a height

2 h

above the surface.

Solution:

Gravitational potentials at the two points:

$V_{h} = - \frac{G M _{E}}{R _{E} + h}, V_{2 h} = - \frac{G M _{E}}{R _{E} + 2 h}$

Work done:

$W = m (V_{2 h} - V_{h}) = m [- \frac{G M _{E}}{R _{E} + 2 h} + \frac{G M _{E}}{R _{E} + h}]$

$W = G M_{E} m [\frac{1}{R _{E} + h} - \frac{1}{R _{E} + 2 h}]$

Simplifying with a common denominator: $W = \frac{G M _{E} m \cdot h}{( R _{E} + h ) ( R _{E} + 2 h )}$ . Positive, as expected - external work must be done against gravity to lift the body higher.

5. Interaction Energy of a System of Particles

For a system of $N$ point masses $m_{1}, m_{2}, \dots, m_{N}$ with pair-wise separations $r_{ij}$ , the total gravitational potential energy is the sum of the pair energies over all distinct pairs:

U_{sy s} = - \frac{1}{2} i = 1 \sum N j = 1 j \neq = i \sum N \frac{G m _{i} m _{j}}{r _{ij}} = - pairs \sum \frac{G m _{i} m _{j}}{r _{ij}}

The factor of $\frac{1}{2}$ in the double sum avoids double-counting each pair (once as $ij$ and once as $j i$ ). Equivalently, sum over each pair exactly once.

For $N$ particles, the number of distinct pairs is $(2 N) = N (N - 1) /2$ .

Solved Example 3

Three equal point masses each of mass

M

are placed at the vertices of an equilateral triangle of side

a

. Find the total gravitational potential energy of the system.

Solution:

Figure 4: Three equal masses

M

at corners of an equilateral triangle of side

a

. Three pairs, each with

U_{p ai r} = - \frac{G M ^{2}}{a}

. Total:

U_{sy s} = - \frac{3 G M ^{2}}{a}

There are three pairs: $(1, 2), (2, 3), (1, 3)$ . Each pair has separation $a$ , and each contributes:

$U_{p ai r} = - \frac{GM \cdot M}{a} = - \frac{G M ^{2}}{a}$

Total potential energy:

$U_{sy s} = 3 \times (- \frac{G M ^{2}}{a}) = - \frac{3 G M ^{2}}{a}$

This is the energy released (or, equivalently, the negative of the work required to disassemble the system to infinity).

Solved Example 4

Four equal masses

m

are placed at the corners of a square of side

a

. Find the total gravitational potential energy of the system.

Solution:

Pairs are of two kinds:

4 pairs along the sides, each at separation $a$ : contribute $4 \times (- \frac{G m ^{2}}{a}) = - \frac{4 G m ^{2}}{a}$
2 pairs along the diagonals, each at separation $a 2$ : contribute $2 \times (- \frac{G m ^{2}}{a 2}) = - \frac{2 G m ^{2}}{a}$

$U_{sy s} = - \frac{4 G m ^{2}}{a} - \frac{2 G m ^{2}}{a} = - \frac{G m ^{2}}{a} (4 + 2)$

$U_{sy s} = - \frac{( 4 + 2 ) G m ^{2}}{a} \approx - \frac{5.41 G m ^{2}}{a}$

6. Gravitational Self-Energy

The self-energy of a body is the total work done in assembling all its particles from infinity to their final configuration - equivalently, the energy possessed by the body due to internal gravitational interactions between its own parts.

Self-energy of a uniform solid sphere

Consider building a uniform sphere of density $ρ$ shell by shell from the centre outward. At an intermediate stage the assembled inner sphere has mass:

$m (r) = \frac{4}{3} π r^{3} ρ$

Figure 5: Self-energy calculation. Build the sphere shell by shell. When the inner sphere has radius

r

, adding a shell of thickness

d r

requires work

d U = - \frac{G m ( r ) d m}{r}

where

m (r) = \frac{4}{3} π r^{3} ρ

. Integrating:

U_{se l f} = - \frac{3 G M ^{2}}{5 R}

The next shell to be added has mass:

$d m = 4 π r^{2} ρ d r$

The work done in bringing this shell from infinity to the surface of the inner sphere:

$d U = - \frac{G m ( r ) d m}{r} = - \frac{G ( \frac{4}{3} π r ^{3} ρ ) ( 4 π r ^{2} ρ d r )}{r} = - \frac{16 π ^{2} G ρ ^{2}}{3} r^{4} d r$

Integrating from $r = 0$ to $r = R$ :

$U_{se l f} = - \frac{16 π ^{2} G ρ ^{2}}{3} \int_{0}^{R} r^{4} d r = - \frac{16 π ^{2} G ρ ^{2}}{3} \cdot \frac{R ^{5}}{5} = - \frac{16 π ^{2} G ρ ^{2} R ^{5}}{15}$

Now substitute $ρ = \frac{M}{( 4/3 ) π R ^{3}} = \frac{3 M}{4 π R ^{3}}$ :

$U_{se l f} = - \frac{16 π ^{2} G R ^{5}}{15} (\frac{3 M}{4 π R ^{3}})^{2} = - \frac{16 π ^{2} G R ^{5}}{15} \cdot \frac{9 M ^{2}}{16 π ^{2} R ^{6}}$

U_{se l f}^{solid sphere} = - \frac{3 G M ^{2}}{5 R}

Self-energy of a uniform spherical shell

A similar shell-by-shell argument (or direct integration) for a thin uniform shell of mass $M$ and radius $R$ gives:

$U_{se l f}^{shell} = - \frac{G M ^{2}}{2 R}$

Comparing the two: for the same total mass and radius, a solid sphere is more tightly bound than a shell (by a factor of $6/5$ ), because in the shell all mass is at the maximum possible distance from every other bit of mass.

Astrophysical application: The self-energy

- \frac{3}{5} G M^{2} / R

is the amount of gravitational binding energy released when a mass

M

collapses from a diffuse cloud to a compact body of radius

R

. For the Sun this is about

2.3 \times 1 0^{41}

J - enough to power the Sun's current luminosity for roughly 20 million years. This "Kelvin-Helmholtz timescale" was one of the great puzzles that pointed toward nuclear fusion as the true stellar energy source.

Common Mistakes to Avoid

Watch out

Dropping the negative sign in $U$ and $V$ : For gravity (always attractive), potential energy and potential are always negative with the standard convention ( $U = 0$ at infinity). Missing the sign changes bound (negative $E$ ) to unbound (positive $E$ ) and gives nonsense answers.
Using $U = m g h$ far from Earth's surface: $m g h$ is only the local approximation valid for $h ≪ R$ . For satellites or interplanetary distances, use $U = - GM m / r$ .
Confusing potential $V$ with potential energy $U$ : $V$ is per unit test mass (units J/kg); $U$ is total energy of a specific system (units J). Their relation: $U = mV$ .
Forgetting the factor $\frac{1}{2}$ in $N$ -particle interaction energy: The double sum $\sum_{i} \sum_{j \neq = i}$ counts each pair twice, so divide by 2. Equivalently, sum only over pairs where $i < j$ (no factor 2 needed).
Thinking $E = 0$ implies $V = 0$ : The field is the slope of the potential, not the potential itself. Inside a hollow shell (or at the centre of a solid sphere) $E = 0$ but $V$ is nonzero and constant/deepest respectively.
Assuming self-energy of a shell equals self-energy of a solid sphere: For the same mass and outer radius, a shell has self-energy $- G M^{2} /2 R$ while a solid sphere has $- 3 G M^{2} /5 R$ . Solid sphere is more tightly bound (mass is closer together).
Sign errors in the work formula: Work done by an external agent moving mass $m$ from A to B is $W_{e x t} = m (V_{B} - V_{A})$ . Work done by gravity is the negative: $W_{g r a v} = - m (V_{B} - V_{A}) = m (V_{A} - V_{B})$ .

Frequently Asked Questions

What is gravitational potential energy?

Gravitational potential energy of a system of two masses is the work done by an external agent in bringing them from infinite separation to their current positions (with no change in kinetic energy). For two point masses $M$ and $m$ separated by distance $r$ , $U = - \frac{GM m}{r}$ . It is always negative for gravity because the force is always attractive.

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

Gravitational potential $V$ is a property of the field at a point - the potential energy per unit test mass placed there: $V = U / m$ . It has units J/kg. Gravitational potential energy $U$ is the energy of a specific mass-and-source system, measured in joules. In a given field, $U = mV$ .

Why is gravitational potential always negative?

Because gravity is always attractive, the potential energy of two masses is lower when they are close together than when they are far apart. Choosing $U = 0$ at infinity (natural because gravity vanishes there) makes $U$ negative at every finite separation. Since $V = U / m$ , the potential $V$ also comes out negative.

What is the relation between gravitational field and gravitational potential?

They are related by $E = - grad (V)$ , i.e. the field is the negative gradient of the potential. For a spherically symmetric field, $E = - \frac{d V}{d r}$ . Physically: the field points in the direction of steepest decrease of the potential, and its magnitude equals the rate of that decrease.

What is the gravitational potential at the centre of a uniform solid sphere?

$V_{ce n t r e} = - \frac{3 GM}{2 R}$ for a uniform solid sphere of mass $M$ and radius $R$ . This is 1.5 times as deep as the surface potential $- GM / R$ . Note: even though the field is zero at the centre, the potential is nonzero and reaches its deepest (most negative) value there.

What is the work done in moving a body in a gravitational field?

The work done by an external agent in moving a mass $m$ (quasi-statically) from point A to point B is $W_{e x t} = m (V_{B} - V_{A}) = U_{B} - U_{A}$ , where $V$ is the gravitational potential. Because gravity is conservative, this work depends only on the endpoints, not on the path taken.

What is gravitational self-energy?

Self-energy is the total gravitational energy of a body due to interactions between all its own parts - equivalently, the work an external agent would do to assemble the body from parts brought in from infinity. For a uniform solid sphere: $U_{se l f} = - \frac{3 G M ^{2}}{5 R}$ . For a thin spherical shell: $U_{se l f} = - \frac{G M ^{2}}{2 R}$ .

How do you calculate the potential energy of a system of N particles?

Sum the potential energies of every distinct pair: $U_{sy s} = - \sum_{pairs} \frac{G m _{i} m _{j}}{r _{ij}}$ , where the sum is over all $(2 N) = N (N - 1) /2$ pairs. Equivalently, $U_{sy s} = - \frac{1}{2} \sum_{i} \sum_{j \neq = i} \frac{G m _{i} m _{j}}{r _{ij}}$ , with the factor $\frac{1}{2}$ correcting for double-counting.

Can gravitational potential be zero at a point where the field is nonzero, or vice versa?

Yes to both. (1) At the midpoint between two equal masses, both fields cancel (net $E = 0$ ) but potentials add (giving $V = - 2 GM / (d /2) = - 4 GM / d$ , nonzero). (2) Along the perpendicular bisector far from a finite dipole-like source, $V \to 0$ but the field can still be nonzero at intermediate points. Field is the slope of $V$ , not $V$ itself.

Previous year questions on Gravitational Potential, Potential Energy And Work Done

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