Universal Law Of Gravitation And Field Intensity

PhysicsGravitationFor JEE aspirants

Newton's law of gravitation states that every point mass attracts every other point mass with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them: $F = \frac{G m _{1} m _{2}}{r ^{2}}$ , where $G = 6.67 \times 1 0^{- 11} N m^{2} / kg^{2}$ . The gravitational field intensity at a point is the force per unit test mass placed at that point, $E = F / m_{0}$ . This concept covers the universal law, superposition and shell theorem, force due to extended bodies (rod, ring, sphere), and is foundational for both JEE Mains and NEET Gravitation.

Key Formulas - Quick Reference

Newton's law: $F = \frac{G m _{1} m _{2}}{r ^{2}}$ , with $G = 6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$
Vector form: $F_{12} = - \frac{G m _{1} m _{2}}{r ^{2}} \overset{r}{^}_{12}$ (attractive)
Dimensional formula of $G$ : $[M^{- 1} L^{3} T^{- 2}]$
Field intensity: $E = \frac{F}{m _{0}} = - \frac{GM}{r ^{2}} \overset{r}{^}$ ; dimensions $[M^{0} L T^{- 2}]$
Superposition: $F_{n e t} = \sum_{i} F_{i}$ (vector sum of pair-wise forces)
Shell theorem: Outside, sphere acts as a point mass at centre; inside a shell, field is zero
Force on point mass $m$ due to rod (mass $M$ , length $L$ , end at distance $r$ ): $F = \frac{GM m}{L} [\frac{1}{r} - \frac{1}{r + L}]$
Field on axis of ring (mass $M$ , radius $R$ , axial distance $r$ ): $E = \frac{GM r}{( R ^{2} + r ^{2} ) ^{3/2}}$
Field of solid sphere: outside $E = \frac{GM}{r ^{2}}$ ; inside $E = \frac{GM r}{R ^{3}}$

1. Newton's Law of Universal Gravitation

Between any two point particles of masses $m_{1}$ and $m_{2}$ separated by a distance $r$ , there exist mutually attractive forces $F_{12}$ and $F_{21}$ directed along the line joining their centres. The magnitude of these forces is directly proportional to the product of the masses and inversely proportional to the square of the distance between them:

F_{12} = F_{21} = \frac{G m _{1} m _{2}}{r ^{2}}

In vector form, $F_{12} = - F_{21}$ (Newton's third law pair).

Here $G$ is the universal gravitational constant, whose value is:

$G = 6.67 \times 1 0^{- 11} N m^{2} / kg^{2}$

Figure 1: Newton's law of gravitation. Force magnitude

F = \frac{G m _{1} m _{2}}{r ^{2}}

; the two forces are equal in magnitude and opposite in direction (Newton's third law pair).

Dimensional formula of G

Rearranging $F = G m_{1} m_{2} / r^{2}$ gives $G = F r^{2} / (m_{1} m_{2})$ . Substituting dimensions:

$[G] = \frac{[ M L T ^{- 2} ] [ L ^{2} ]}{[ M ^{2} ]} = [M^{- 1} L^{3} T^{- 2}]$

Key properties of the gravitational force:

It is a conservative and central force (always along the line joining the masses).
It is always attractive - no repulsive gravitation exists in classical theory.
It is the weakest of the four fundamental forces but dominates at large scales because it is always additive (no negative masses).
The law holds strictly for point masses; for extended bodies it is applied via integration or (for spherically symmetric shells) via the shell theorem.

2. Principle of Superposition

The gravitational force on a particle due to a number of other particles is the vector sum of the individual forces exerted by each other particle. The presence of a third mass does not alter the force between any two of them.

If a mass $m$ is surrounded by point masses $M_{1}, M_{2}, \dots, M_{n}$ located at distances $r_{1}, r_{2}, \dots, r_{n}$ , then the net force on $m$ is:

F_{n e t} = F_{1} + F_{2} + \dots + F_{n} = - i = 1 \sum n \frac{G m M _{i}}{r _{i}^{2}} \overset{r}{^}_{i}

Figure 2: Superposition principle. Net gravitational force

F_{n e t} = F_{1} + F_{2} + F_{3}

on mass

m

from sources

M_{1}, M_{2}, M_{3}

. Each individual force is unaffected by the presence of the others.

3. Shell Theorem

Extended spherically-symmetric bodies (like the Earth) obey two elegant results, together called the shell theorem. These were first proved by Newton using calculus.

(a) Field outside a uniform spherical shell

A uniform spherical shell attracts an external particle as if the entire mass of the shell were concentrated at its centre. For a point at distance $r$ from the centre ( $r > R$ ):

$E_{o u t s i d e} = \frac{GM}{r ^{2}}$

(b) Field inside a uniform spherical shell

A uniform spherical shell exerts zero net gravitational force on a particle located anywhere inside it. The pulls from all shell elements exactly cancel by symmetry.

$E_{in s i d e} = 0 (r < R)$

Figure 3: Shell theorem. Outside the shell (

r > R

): field is

\frac{GM}{r ^{2}}

as if all mass were at the centre. Inside the shell (

r < R

): net field is zero everywhere.

A uniform solid sphere can be treated as a nested set of thin shells. This makes it act like a point mass at the centre for all external points, and gives a linearly-rising field for internal points (see Section 6).

4. Gravitational Force due to a Uniform Rod

Consider a uniform rod of mass $M$ and length $L$ . A point particle of mass $m$ is placed at a distance $r$ from one end of the rod, along the axial line. We find the net gravitational force on $m$ by integrating over the rod.

Take a small element of width $d x$ on the rod at a distance $x$ from the point mass. Its mass is:

$d m = \frac{M}{L} d x$

Figure 4: Force on point mass

m

due to uniform rod of mass

M

and length

L

. Integrating the contribution from element

d x

at distance

x

gives

F = \frac{GM m}{L} [\frac{1}{r} - \frac{1}{r + L}]

The force on $m$ due to this element is:

$d F = \frac{G m d m}{x ^{2}} = \frac{GM m}{L x ^{2}} d x$

Integrating from $x = r$ to $x = r + L$ gives the total force:

$F = \int_{r}^{r + L} \frac{GM m}{L} \frac{d x}{x ^{2}} = \frac{GM m}{L} [\frac{1}{r} - \frac{1}{r + L}]$

In the limit

L \to 0

(rod shrinks to a point mass at distance

r

), the bracket becomes

L / r^{2}

and

F \to GM m / r^{2}

, recovering Newton's law - a useful sanity check.

Solved Example 1

A mass

M

is split into two parts

m

and

(M - m)

, which are then separated by a fixed distance

r

. What ratio

m / M

maximises the gravitational force between the parts?

Solution:

The force between them is:

$F = \frac{G m ( M - m )}{r ^{2}} = \frac{G}{r ^{2}} (m M - m^{2})$

For $F$ maximum, $\frac{d F}{d m} = 0$ (treating $M$ and $r$ as constants):

$\frac{d}{d m} [\frac{G}{r ^{2}} (m M - m^{2})] = 0 ⟹ M - 2 m = 0$

$\frac{m}{M} = \frac{1}{2}$

The force is maximum when the mass is split equally.

5. Gravitational Field Intensity

The gravitational field intensity (denoted $I$ or $E$ ) at a point is defined as the gravitational force experienced by a unit test mass placed at that point:

E = \frac{F}{m _{0}} = - \frac{GM}{r ^{2}} \overset{r}{^}

The negative sign indicates that the field points from the test point toward the source mass $M$ (attractive). The SI unit is $N/kg$ or equivalently $m/s^{2}$ .

Figure 5: Gravitational field intensity

E = - \frac{GM}{r ^{2}} \overset{r}{^}

at the location of test mass

m

. The negative sign indicates the field points from the test point toward the source mass

M

Dimensional formula of gravitational field intensity

$[E] = \frac{[ F ]}{[ M ]} = \frac{[ M L T ^{- 2} ]}{[ M ]} = [M^{0} L T^{- 2}]$

Same dimensions as acceleration - which is why $g$ is often called the "acceleration due to gravity" as well as the "field intensity of Earth at its surface".

Solved Example 2

The distance between the Earth and the Moon is about

3.8 \times 1 0^{5}

km. At what point(s) will the net gravitational field of the Earth-Moon system be zero? Take the mass of the Earth as 81 times that of the Moon.

Solution:

Treat Earth and Moon as point masses (since separation $≫$ radii). The field due to a point mass is $I = GM / r^{2}$ , which vanishes at infinity. In addition, there is a point between the two bodies where the pull of the Earth balances the pull of the Moon.

Let this point be at distance $x$ from the Earth (so distance $3.8 \times 1 0^{8} - x$ metres from the Moon). Equating field magnitudes:

$\frac{G M _{E}}{x ^{2}} = \frac{G M _{M}}{( 3.8 \times 1 0 ^{8} - x ) ^{2}}$

Using $M_{E} = 81 M_{M}$ :

$\frac{81}{x ^{2}} = \frac{1}{( 3.8 \times 1 0 ^{8} - x ) ^{2}}$

Taking square roots (positive branch, since $x < 3.8 \times 1 0^{8}$ ):

$9 (3.8 \times 1 0^{8} - x) = x ⟹ 10 x = 9 \times 3.8 \times 1 0^{8}$

$x = 3.42 \times 1 0^{8} m = 3.42 \times 1 0^{5} km$

This "null point" lies at about 9/10 of the way from Earth to Moon - much closer to the Moon because Earth is 81 times more massive.

6. Gravitational Field due to Common Shapes

The table below summarises the gravitational field intensity for standard mass distributions encountered in JEE / NEET problems. In every case the field points from the field point toward the source mass.

Source shape	Field intensity $E$	Notes
Point mass $M$ at distance $r$	$E = \frac{GM}{r ^{2}}$	Inverse-square, directed toward $M$
Ring (mass $M$ , radius $R$ ) on axis at distance $r$	$E = \frac{GM r}{( R ^{2} + r ^{2} ) ^{3/2}}$	$E = 0$ at centre; maximum at $r = R / 2$
Rod (mass $M$ , length $L$ ) on axial line, end at distance $r$	$E = \frac{GM}{r ( r + L )}$	Force per unit test mass, from rod formula above
Rod at equatorial point, distance $r$ from centre	$E = \frac{2 GM}{r L ^{2} + 4 r ^{2}}$	Perpendicular components cancel; net along the perpendicular bisector
Circular arc of length $L$ at its centre	$E = \frac{2 GM sin ( θ /2 )}{L \cdot R}$	For a full ring ( $θ = 2 π$ ), $E = 0$
Hollow sphere (mass $M$ , radius $R$ ) at distance $r$	$r \geq R$ : $E = \frac{GM}{r ^{2}}$ $r < R$ : $E = 0$	Shell theorem; discontinuous jump at surface
Solid sphere (mass $M$ , radius $R$ ) at distance $r$	$r \geq R$ : $E = \frac{GM}{r ^{2}}$ $r \leq R$ : $E = \frac{GM r}{R ^{3}}$	Continuous at surface; linear inside, inverse-square outside
Infinite thin rod (linear density $λ$ ) at perpendicular distance $r$	$E = \frac{2 G λ}{r}$	Note: $1/ r$ dependence, not $1/ r^{2}$

Figure 6: Field on the axis of a uniform ring of mass

M

and radius

R

E = \frac{GM r}{( R ^{2} + r ^{2} ) ^{3/2}}

, directed along the axis toward the ring centre.

E = 0

at the centre (

r = 0

) and maximum at

r = R / 2

Figure 7: Field of a uniform solid sphere. Inside (

r \leq R

E = \frac{GM r}{R ^{3}}

(linear). Outside (

r \geq R

E = \frac{GM}{r ^{2}}

(inverse square). Maximum at the surface

r = R

Notice how the geometry changes the distance dependence: point mass and sphere give

1/ r^{2}

, infinite rod gives

1/ r

, and the ring/finite rod give more complex expressions that reduce to

1/ r^{2}

far away.

7. Inertial Mass vs Gravitational Mass

Historically, mass appears in physics in two distinct ways:

Inertial mass ( $m_{i}$ ) - the property that resists acceleration in Newton's second law, $F = m_{i} a$ .
Gravitational mass ( $m_{g}$ ) - the property that produces and responds to a gravitational field, $F = G M_{g} m_{g} / r^{2}$ .

Experiment (Eötvös, Dicke, and modern lunar laser ranging) has confirmed to high precision (better than 1 part in $1 0^{13}$ ) that $m_{i} = m_{g}$ for every material tested. This equivalence is the empirical foundation of Einstein's general theory of relativity.

Practical consequence: for a body in free fall,

$m_{i} a = \frac{G M _{E} m _{g}}{R _{E}^{2}} ⟹ a = \frac{G M _{E}}{R _{E}^{2}} = g$

All bodies, regardless of mass, fall with the same acceleration $g$ at a given location - a fact famously demonstrated by Galileo and by the Apollo 15 hammer-and-feather drop on the Moon.

Solved Example 3

Three point masses each of

2 kg

are placed at the vertices of an equilateral triangle of side

1 m

. Find the net gravitational force on any one mass due to the other two.

Solution:

By symmetry, each of the two forces on the chosen mass has magnitude:

$F_{1} = F_{2} = \frac{G \cdot 2 \cdot 2}{1 ^{2}} = 4 G = 4 \times 6.67 \times 1 0^{- 11} N = 2.668 \times 1 0^{- 10} N$

The angle between the two force vectors (both attractive, toward the other two vertices) is $6 0^{\circ}$ . Using the parallelogram law:

$F_{n e t} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} cos 6 0^{\circ} = 3 F_{1}$

$F_{n e t} = 3 \times 2.668 \times 1 0^{- 10} N \approx 4.62 \times 1 0^{- 10} N$

The direction is along the line from the chosen mass toward the centroid of the triangle (bisector of the $6 0^{\circ}$ angle).

Common Mistakes to Avoid

Watch out

Confusing $G$ with $g$ : $G$ is the universal constant ( $6.67 \times 1 0^{- 11} N m^{2} / kg^{2}$ , same everywhere in the universe). $g$ is the local acceleration due to gravity (about $9.8 m/s^{2}$ on Earth, varies with location and altitude).
Applying $F = GM m / r^{2}$ to extended bodies without justification: The formula is exact only for point masses. It works for extended spherically-symmetric bodies (Earth, Sun) by the shell theorem, but not for rods, disks, or irregular shapes - those need integration.
Forgetting that the field inside a hollow shell is zero: If you are inside a hollow spherical shell (not on its surface), the net gravitational field from the shell alone is zero everywhere, not just at the centre.
Adding forces as scalars: Gravitational force is a vector. When two or more masses exert forces on a body, add them by the vector triangle/parallelogram rule, not by simple addition of magnitudes.
Using $F = GM m / r^{2}$ with $r$ as the surface-to-surface distance: The $r$ in Newton's law is the distance between the centres, not the surface separation.
Assuming gravitational force needs a medium: Gravity acts through vacuum. It requires no material medium and propagates at the speed of light.

Frequently Asked Questions

What is the universal law of gravitation?

Newton's universal law of gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. Mathematically, $F = G m_{1} m_{2} / r^{2}$ , where $G = 6.67 \times 1 0^{- 11} N m^{2} / kg^{2}$ .

What is the value and dimensional formula of the universal gravitational constant G?

$G = 6.67 \times 1 0^{- 11} N m^{2} / kg^{2}$ . Its dimensional formula is $[M^{- 1} L^{3} T^{- 2}]$ . The value was first measured experimentally by Henry Cavendish in 1798 using a torsion balance.

What is gravitational field intensity and how is it different from gravitational force?

Gravitational field intensity at a point is the force experienced per unit mass of a small test mass placed at that point: $E = F / m_{0}$ . Force depends on both the source mass and the test mass; field is a property of the source mass alone at that location. Units: $N/kg$ or $m/s^{2}$ .

What is the shell theorem in gravitation?

The shell theorem has two parts: (1) A uniform spherical shell attracts an external particle as if all of the shell's mass were concentrated at its centre. (2) A uniform spherical shell exerts zero net gravitational force on any particle located inside it. First proved by Newton, this theorem lets us treat spherical bodies like the Earth as point masses in most calculations.

Why is the gravitational force always attractive?

There is only one kind of mass (always positive) as far as gravity is concerned, and mass-mass interactions in Newton's theory are always attractive. This is different from electric charge, which comes in two signs and can therefore attract or repel. In general relativity, positive-energy sources always curve spacetime in the same attractive way for ordinary matter.

What is the gravitational field on the axis of a uniform ring?

For a ring of mass $M$ and radius $R$ , the field at a point on the axis distance $r$ from the centre is $E = GM r / (R^{2} + r^{2})^{3/2}$ , directed along the axis toward the centre of the ring. The field is zero at the centre ( $r = 0$ ), reaches a maximum at $r = R / 2$ , and falls off as $GM / r^{2}$ far from the ring.

Do the inertial mass and gravitational mass of a body have the same value?

To all experimental precision (better than 1 part in $1 0^{13}$ ), yes - inertial mass equals gravitational mass for every substance tested. This equivalence is the empirical foundation of Einstein's general relativity and is why all objects, regardless of composition or mass, fall with the same acceleration $g$ at a given location.

Is the universal law of gravitation valid for objects of any size?

The law $F = G m_{1} m_{2} / r^{2}$ is strictly valid only for point masses. For spherically symmetric extended bodies (like planets and stars), it also works using the distance between centres, thanks to the shell theorem. For irregular shapes (rods, discs), we must integrate the force from infinitesimal mass elements over the whole body.

Why is the gravitational force considered the weakest of the four fundamental forces?

Comparing the electric and gravitational forces between two protons, gravity is weaker by a factor of about $1 0^{36}$ . However, gravity dominates on astronomical scales because it is always attractive and cumulative - there are no negative masses to cancel it - while electric charges neutralise. So a large body like a planet has huge net gravity but essentially zero net electric force.

Previous year questions on Universal Law Of Gravitation And Field Intensity

3 questions from past papers, each with a step-by-step solution.

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