The statement of the law
Newton's reasoning, recounted in NCERT §7.3, began with the moon. He noted that the moon's centripetal acceleration in its orbit was much smaller than $g$ at the Earth's surface — about $1/3600$ of it. Since the moon is roughly 60 Earth-radii away, the factor $1/3600 = 1/60^2$ told him the gravitational force weakens as the inverse square of the distance from the centre of the Earth. From this single clue he generalised to every pair of bodies in the universe.
The verbatim NCERT statement is worth memorising:
Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
NIOS §5.1 phrases the same law for particles: "Every particle attracts every other particle in the universe with a force which varies as the product of their masses and inversely as the square of the distance between them." Two ideas are bundled here — the force grows with each mass, and it dies off as the square of separation. The constant that turns this proportionality into an equality is $G$, the universal gravitational constant, treated in detail on the gravitational constant page.
The formula and its vector form
For two point masses $m_1$ and $m_2$ separated by a distance $r$, the magnitude of the gravitational force is
$$F = G\,\frac{m_1 m_2}{r^2}.$$
NCERT writes the law as the force on $m_2$ due to $m_1$. To capture the direction as well as the magnitude, it is expressed in vector form. Let $\hat{r}$ be the unit vector pointing from $m_1$ to $m_2$, and $\vec{r} = \vec{r}_2 - \vec{r}_1$. Then the force on $m_2$ is
$$\vec{F} = -\,G\,\frac{m_1 m_2}{r^2}\,\hat{r}.$$
The minus sign is the whole story of attraction: although $\hat{r}$ points from $m_1$ toward $m_2$, the force on $m_2$ points the opposite way — back toward $m_1$. The force pulls the two masses together, never apart.
Two point masses attract along the line joining them. The force on each points toward the other, and the two forces are equal in magnitude and opposite in direction (Newton's third law). The separation $r$ is measured centre to centre.
The companion force on $m_1$ due to $m_2$ is, by Newton's third law, equal and opposite: $\vec{F}_{12} = -\vec{F}_{21}$. The two masses pull on each other with the same strength — the heavier one does not pull harder. What differs is the resulting acceleration, since $a = F/m$.
Attractive, along the line, and central
Three properties of the force follow directly from the vector form, and NEET tests all three.
It is always attractive. Both masses are positive quantities, so $G\,m_1 m_2/r^2$ is always positive and the $-\hat{r}$ direction always points each body toward the other. There is no repulsive gravity in classical mechanics. Two stationary astronauts adrift in space, with no other forces, will drift toward each other — never apart — under their mutual pull.
It acts along the line joining the masses. The force vector lies on the straight line connecting the two point masses; it has no sideways component. For point masses or spherically symmetric bodies, the force has no torque about either centre.
It is a central force. Because the force on a body points toward a fixed centre (the other mass), gravitation is a central force. NCERT §7.2 defines a central force as one for which "the force on the planet is along the vector joining the Sun and the planet." A central force conserves angular momentum, and that conservation is exactly what produces Kepler's law of areas. NCERT stresses that the law of areas "is valid for any central force" — it does not even require the inverse-square form. The inverse-square detail is what fixes the shape of the orbit and Kepler's third law; the centrality alone gives the second law.
The central-force property is the engine behind the planetary laws — see Kepler's laws for how angular-momentum conservation yields the law of areas.
The inverse-square dependence
The exponent on $r$ is the most heavily examined feature of the law. The force varies as $1/r^2$: double the separation and the force falls to a quarter; triple it and the force drops to a ninth. This is the same $1/60^2 = 1/3600$ ratio that first convinced Newton, comparing the moon's acceleration to surface gravity.
| Change in separation | New distance | Resulting force (relative to original $F$) |
|---|---|---|
| Distance doubled | $2r$ | $F/4$ |
| Distance tripled | $3r$ | $F/9$ |
| Distance halved | $r/2$ | $4F$ |
| One mass doubled, distance fixed | $r$ | $2F$ |
| Both masses doubled, distance doubled | $2r$ | $F$ (unchanged) |
The inverse-square law also makes gravitation extraordinarily weak at the everyday scale: $G$ is so small that two ordinary objects attract each other with a force far below anything we can feel — which is why Cavendish needed a delicate torsion balance to measure it at all. Yet because mass only ever adds (it never cancels, unlike charge), gravitation dominates at astronomical scales where enormous masses are involved.
Superposition for many masses
The law as stated applies to a single pair of point masses. When a body feels the pull of several masses at once, NCERT invokes the principle of superposition: each pairwise force is computed independently as if the others were not there, and the resultant is the vector sum.
$$\vec{F}_R = \vec{F}_1 + \vec{F}_2 + \cdots + \vec{F}_n = \sum_{i=1}^{n} \vec{F}_i.$$
The crucial physical content, in NCERT's words, is that "each force acts independently and uninfluenced by the other bodies." Placing a third mass nearby does not weaken or strengthen the force between the original pair; it merely adds another vector to the sum. Because forces are vectors, the addition is vectorial — magnitudes and directions both matter, and forces in opposite directions can cancel.
Superposition: the force on $m_1$ is the vector sum of the separate pulls of $m_2$, $m_3$ and $m_4$, each computed independently from $F = G\,m_1 m_i / r_i^2$ (NCERT Fig. 7.4).
A useful symmetry consequence: if equal masses sit at the vertices of a regular configuration and a test mass is at the centre, the individual pulls cancel by symmetry and the net force is zero. NCERT Example 7.2 shows this for three equal masses at the corners of an equilateral triangle with a mass at the centroid — the resultant is zero. Break the symmetry, say by doubling one vertex mass, and a net force survives, pointing toward the heavier vertex.
The two shell theorems
The law is written for point masses, but real bodies — planets, stars — have finite size. To bridge the gap, NCERT states two results, provable by calculus, for a uniform spherical shell. They are the reason a planet can be treated as a point at its centre.
Shell theorem 1 (left): a uniform shell pulls an external point mass as if all its mass sat at the centre. Shell theorem 2 (right): anywhere inside the shell, the pulls from all regions cancel and the net force is zero.
Theorem 1 (external point). In NCERT's words: "The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is concentrated at the centre of the shell." Qualitatively, the sideways components of the pulls from different patches of the shell cancel, leaving only a resultant along the line to the centre.
Theorem 2 (internal point). Again from NCERT: "The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it is zero." The pulls from the various regions of the shell cancel each other completely, no matter where inside the body sits. A solid uniform sphere is a stack of nested shells, so for an external point all of them act as if their mass were at the common centre — which is precisely why $r$ in the force law is measured centre to centre, and why $g = GM_E/R_E^2$ at the surface.
Worked examples
Two heavy spheres, each of mass $100\ \text{kg}$ and radius $0.10\ \text{m}$, are placed $1.0\ \text{m}$ apart (centre to centre) on a horizontal table. Find the magnitude of the gravitational force one sphere exerts on the other. (Based on NCERT Exercise 7.21.)
Set-up. The spheres are uniform, so by the shell theorem each behaves as a point mass at its centre, and $r = 1.0\ \text{m}$ is the centre-to-centre distance — the radii do not enter the force at all.
Apply the law. $F = G\,\dfrac{m_1 m_2}{r^2} = \dfrac{(6.67\times10^{-11})(100)(100)}{(1.0)^2}.$
Result. $F = 6.67\times10^{-7}\ \text{N}$, a minuscule attraction — illustrating just how weak gravitation is between ordinary objects. By the midpoint symmetry, an object placed exactly halfway between the two equal spheres feels equal and opposite pulls, so the net force on it is zero.
Three equal masses of $m$ kg each are fixed at the vertices of an equilateral triangle $ABC$. A mass $2m$ is placed at the centroid $G$, with $AG = BG = CG = 1\ \text{m}$. (a) What is the net force on the mass $2m$? (b) What is the force if the mass at vertex $A$ is doubled?
(a) Symmetry argument. Each vertex mass pulls $2m$ with the same magnitude $G\,(m)(2m)/(1)^2 = 2Gm^2$, directed toward that vertex. The three pulls point toward the three vertices of an equilateral triangle, $120^\circ$ apart and equal in size. By symmetry their vector sum is zero. Net force = 0. NCERT confirms: "on the basis of symmetry the resultant force ought to be zero."
(b) Break the symmetry. Doubling the mass at $A$ to $2m$ adds an extra pull toward $A$. The three original equal pulls still sum to zero, so the only surviving force is the extra contribution from the doubled vertex: an additional $2Gm^2$ directed toward $A$. Net force $= 2Gm^2$ directed from $G$ toward $A$.
The lesson: superposition lets you handle the balanced part and the excess separately, and symmetry does much of the arithmetic for you.
The gravitational intensity (force per unit mass) at the centre of a hemispherical shell of uniform mass density — which way does it point? (NCERT Exercise 7.10/7.11 setup.)
Use the shell result as scaffolding. For a complete uniform spherical shell, the field at the centre is exactly zero — the pull of every patch is cancelled by the diametrically opposite patch (shell theorem 2). Imagine the full sphere as two hemispheres: the upper and lower hemisphere fields at the centre must be equal in magnitude and opposite in direction so that together they give zero.
Remove one hemisphere. Take just the lower hemispherical shell. Its field at the centre is whatever cancelled the upper hemisphere's field — so it points away from the lower hemisphere's bulk, i.e. straight up along the axis of symmetry, toward the missing half.
Direction. The intensity at the centre of a hemispherical shell lies along the axis of symmetry, directed toward the open face (the side where mass is missing). This is the reasoning NEET rewards: invoke the zero-field-inside-a-full-shell theorem, then argue by the missing half. There is no sideways component, by symmetry about the axis.
The universal law in one breath
- Statement: every body attracts every other with $F = G\,m_1 m_2 / r^2$, along the line joining them, always attractive.
- Vector form: $\vec{F} = -\,G\,m_1 m_2\,\hat{r}/r^2$; the force on $m_1$ and on $m_2$ are equal and opposite ($\vec{F}_{12} = -\vec{F}_{21}$).
- $G = 6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$ is universal and constant; $g$ is the local acceleration due to gravity — different quantities.
- Inverse-square: double $r$ to quarter $F$; the distance $r$ is centre to centre for spheres.
- Superposition: net force is the vector sum of independent pairwise forces; symmetry can make it zero.
- Central force → angular momentum conserved → Kepler's law of areas (holds for any central force).
- Shell theorems: a uniform shell attracts an external point as a centre-point mass; the field anywhere inside it is zero. Gravitational shielding is impossible.