In short

Every mass attracts every other mass with a force F = \frac{GMm}{r^2}, where G = 6.674 \times 10^{-11} N·m²/kg² is the universal gravitational constant, M and m are the two masses, and r is the centre-to-centre distance. The net force from multiple masses is the vector sum of the individual forces (superposition). A uniform spherical shell attracts an external mass as if all its mass were at its centre, and exerts zero net force on a mass inside it (shell theorem). Inside a uniform solid sphere, gravitational force is proportional to distance from the centre.

Drop a cricket ball from your balcony. It falls — accelerates toward the ground and smacks the pavement. Now look up at the Moon. It is 384,400 km away, tracing a circle around the Earth at about 1 km/s. Something is pulling it too — curving its path inward, preventing it from flying off in a straight line.

Is the force that yanks the cricket ball down the same force that holds the Moon in orbit?

Yes. The Moon is falling toward the Earth right now — just as your cricket ball fell. But the Moon is also moving sideways so fast that the curved surface of the Earth drops away beneath it at the same rate. It falls and falls and never hits the ground. An orbit is perpetual free fall where you keep missing the surface.

The equation that connects a ball falling 3 metres to a Moon orbiting 384,400 km away fits in a single line. This is the same equation ISRO's engineers use to plan every mission — Chandrayaan-3's trajectory to the Moon, Mangalyaan's path to Mars — applied millions of times along the route.

Every mass attracts every other mass

Before this law existed, "heaviness" was just a property of things near the ground. Heavy objects fall. That was about it. The radical insight was deeper: the Earth pulls the apple and the apple pulls the Earth. The Earth pulls the Moon and the Moon pulls the Earth. You pull your friend sitting next to you in class, and your friend pulls you. Every piece of mass in the universe attracts every other piece of mass. No exceptions.

The force is always attractive. Unlike electric charges, which can be positive or negative and therefore attract or repel, mass only comes in one sign. Two masses always pull toward each other, never push apart.

And the force is mutual. If the Earth pulls you down with 539 N (assuming you have a mass of 55 kg), you pull the Earth upward with exactly 539 N. Equal magnitude, opposite direction — Newton's third law. The reason you fall and the Earth does not noticeably move is the difference in mass: your acceleration is g = 9.8 m/s², while the Earth's acceleration toward you is F/M_{\text{Earth}} = 539/(6 \times 10^{24}) \approx 9 \times 10^{-23} m/s². That is so small that no instrument could ever detect it.

Newton's Law of Universal Gravitation

Every particle of matter attracts every other particle with a force directed along the line joining them, proportional to the product of their masses and inversely proportional to the square of the distance between them:

F = \frac{GMm}{r^2}

where G = 6.674 \times 10^{-11} N·m²/kg² is the universal gravitational constant, M and m are the two masses (in kg), and r is the distance between their centres (in m). The force F is in newtons.

Reading the law. Look at what each piece says:

Check the dimensions: G has units N·m²/kg², so G \times \text{kg} \times \text{kg}/\text{m}^2 gives newtons. The units work out exactly as they should.

Two masses attracting each other gravitationally Mass M on the left and mass m on the right, separated by distance r. Equal and opposite force arrows point toward each other. M m r F on m F on M
Two masses pull each other with equal and opposite forces. The force on $m$ (red arrow) points toward $M$; the force on $M$ (dark arrow) points toward $m$. Both have the same magnitude $F = GMm/r^2$.

The inverse-square law — how gravity weakens with distance

The r^2 in the denominator is the single most important feature of the law. To see why, work through some concrete numbers. Suppose two masses are separated by a distance r_0 and feel a gravitational force F_0. What happens as you change the distance?

Distance Force Factor
r_0 F_0 1
2r_0 F_0/4 ×0.25
3r_0 F_0/9 ×0.11
\tfrac{1}{2}r_0 4F_0 ×4
10r_0 F_0/100 ×0.01

The force drops steeply. At twice the distance, you do not get half the force — you get a quarter. At ten times the distance, you get a hundredth. This is why gravity dominates your life on Earth's surface (you are close to a very large mass) but is negligible between everyday objects.

This steep decline also explains why a spacecraft that escapes Earth's immediate gravitational grip can coast most of the way to Mars with engines off — once you are far enough away, the remaining pull is negligible compared to the Sun's influence at that distance.

Drag the red point in the figure below to explore the relationship between distance and force.

Interactive: gravitational force versus distance A curve showing F proportional to 1/r², with a draggable point to explore different distances. At r equals 1, the force is 1. At r equals 2, the force drops to 0.25. distance r / r₀ force F / F₀ 1 2 3 4 1 2 3 4 drag the red point along the axis
Drag the red point to change the distance. At $r = r_0$, the force is $F_0$. Double the distance to $2r_0$ and the force drops to $F_0/4$. The steep decline of the inverse-square curve is why gravity is a dominant force at planetary scales but negligible between everyday objects.

The gravitational constant G

The law says F = GMm/r^2, but to use it you need the value of G. And G is extraordinarily hard to measure, because gravitational forces between laboratory-sized objects are fantastically small.

Two 1 kg masses placed 1 m apart attract each other with a force of 6.674 \times 10^{-11} N. That is roughly the weight of a single human cell. You need extremely sensitive equipment to detect it.

The classic measurement uses a torsion balance: two small lead spheres sit at the ends of a light horizontal bar, suspended from a thin wire at its centre. Two large lead spheres are placed near the small ones, arranged diagonally so that both small-large pairs twist the bar in the same direction. The gravitational attraction twists the wire by a tiny angle. By measuring the twist and knowing the wire's torsion constant, you can calculate the force, and from it, G.

Torsion balance for measuring G Top-down view of a torsion balance: a thin wire suspends a horizontal bar with small masses at each end. Large masses placed diagonally attract the small masses, twisting the wire. thin wire (from above) s s L L twist small mass small mass large mass large mass
Top-down view of a torsion balance. The gravitational pull between each small–large pair (red arrows) twists the bar. Measuring the twist angle gives the force, and from that, $G$.

The accepted value is:

G = 6.674 \times 10^{-11} \text{ N·m}^2/\text{kg}^2

This value is universal — the same between any two masses anywhere in the universe, whether they are lead spheres in a laboratory or galaxies separated by millions of light-years. It has been measured to about five significant figures, making it one of the least precisely known fundamental constants. The difficulty is fundamental: gravity is the weakest of the four fundamental forces, and isolating a tiny gravitational signal from electrostatic, seismic, and thermal noise requires extraordinary care.

Superposition — forces from many masses

When more than two masses are present, the net gravitational force on any one mass is the vector sum of the individual forces from every other mass:

\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \cdots

Each \vec{F}_i = \frac{GM_i m}{r_i^2}\,\hat{r}_i is computed independently, as if the other masses did not exist. You calculate each pull separately and then add the vectors.

This is the principle of superposition, and it is what makes gravitational calculations tractable. Without it, the presence of a third body would change the interaction between the first two, and multi-body problems would be unsolvable even in principle.

Here is a concrete example. Imagine standing on a beach during a full Moon. The Earth pulls you downward with about 539 N (if you weigh 55 kg). The Moon, 384,400 km overhead, pulls you upward with:

F_{\text{Moon}} = \frac{6.674 \times 10^{-11} \times 7.35 \times 10^{22} \times 55}{(3.84 \times 10^8)^2} \approx 1.8 \times 10^{-3} \text{ N}

That is about 1.8 millinewtons — roughly 300,000 times weaker than the Earth's pull on you. Your net force is overwhelmingly downward. But this same tiny lunar pull, acting across the vast body of the oceans, creates the tides. The Moon pulls the near side of the ocean slightly more than the far side (because the near side is closer and F \propto 1/r^2), stretching the ocean into an oval bulge. That is superposition in action — each gravitational interaction calculated independently, then added.

The shell theorem — why spheres behave like point masses

The Earth is not a point. It is a sphere of radius 6,371 km with mass distributed throughout its volume. When you stand on the surface, some mass is directly below your feet (pulling you straight down), some is off to the side (pulling you at an angle), and some is on the far side of the planet (pulling you slightly upward). How do you add up all these contributions?

The answer is surprisingly clean. For a uniform spherical shell (a thin hollow ball):

  1. External point: the shell attracts a mass outside it exactly as if the shell's entire mass were concentrated at its centre.
  2. Internal point: the gravitational force on a mass anywhere inside the shell is zero.
Shell theorem: external and internal test masses Cross-section of a spherical shell. A test mass outside experiences a force toward the centre. A test mass inside experiences zero net force. centre m F = GMsm/d² m F = 0 d R uniform spherical shell
A uniform spherical shell. A test mass *outside* (red) feels a force as if the shell's entire mass were at the centre. A test mass *inside* (dark) feels zero net force — the pulls from all parts of the shell cancel exactly.

Since a solid sphere is just a stack of concentric shells, the shell theorem gives a powerful result:

A uniform solid sphere attracts an external mass as if the sphere's entire mass were concentrated at its centre.

This is why you can write F = GMm/r^2 for the Earth — treating all 5.97 \times 10^{24} kg as a point at the centre, with r measured from the centre. Without the shell theorem, you would need to integrate over the entire volume of the Earth for every calculation.

The zero-force result inside a shell is physically subtle. The nearby part of the shell is close and pulls strongly, but it subtends a small solid angle — a small patch of shell. The far part is distant and pulls weakly, but it subtends a much larger solid angle — a much bigger patch. These two effects cancel exactly. The mathematical proof (in the going-deeper section) shows this cancellation through integration.

The real Earth is not perfectly uniform — it has a dense iron core, a rocky mantle, and a thin crust. But as long as the mass distribution is approximately spherically symmetric (each layer has roughly uniform density), the shell theorem still applies. The real Earth bulges slightly at the equator, introducing small corrections — but the shell theorem gives you the right answer to within about 0.3%.

Gravity inside a uniform sphere

Now turn the shell theorem around. You are inside a solid uniform sphere of total mass M and radius R, at a distance r from the centre (where r < R).

Divide the sphere into two parts:

By the shell theorem, the outer shell contributes zero net force. Only the mass inside radius r pulls on you.

How much mass is that? For a uniform sphere with density \rho:

M_{\text{inside}} = \rho \cdot \frac{4}{3}\pi r^3 = M \cdot \frac{r^3}{R^3}

Why: the density is \rho = M / \left(\tfrac{4}{3}\pi R^3\right). The mass inside radius r is \rho times the volume \tfrac{4}{3}\pi r^3. The \tfrac{4}{3}\pi factors cancel, leaving M \cdot r^3/R^3 — the fraction of the total mass enclosed within radius r.

The gravitational force at distance r from the centre is:

F = \frac{G M_{\text{inside}} \cdot m}{r^2} = \frac{G M m}{R^3}\,r

Why: substitute M_{\text{inside}} = Mr^3/R^3 into F = GM_{\text{inside}}m/r^2. The r^3 in the numerator and r^2 in the denominator leave a single factor of r.

This is a striking result. Inside the sphere, gravity is proportional to r — linear, not inverse-square. At the centre (r = 0), the force is zero: every part of the sphere pulls equally in all directions, and the pulls cancel. At the surface (r = R), the force is GMm/R^2, matching the external formula. The transition at r = R is smooth.

Gravitational force versus distance from the centre of a uniform sphere Graph showing force versus distance. Inside the sphere (r less than R), force rises linearly. Outside (r greater than R), force drops as 1 over r squared. The peak is at the surface r equals R. distance from centre, r gravitational force, F 0 R 2R 3R GMm/R² F ∝ r F ∝ 1/r² inside outside surface
Gravitational force on a test mass versus its distance from the centre of a uniform sphere of radius $R$. Inside, the force grows linearly with $r$. At the surface, it peaks at $GMm/R^2$. Outside, it falls as $1/r^2$.

The linear proportionality F \propto r has a deep consequence. If you could drill a tunnel straight through the Earth from India to the other side and drop a ball into it (neglecting air resistance and assuming uniform density), the ball would oscillate back and forth through the centre. A restoring force proportional to displacement is exactly the condition for simple harmonic motion. The period of this oscillation would be:

T = 2\pi\sqrt{\frac{R}{g}} \approx 2\pi\sqrt{\frac{6.371 \times 10^6}{9.8}} \approx 5070 \text{ s} \approx 84.5 \text{ minutes}

Remarkably, this is the same as the orbital period of a hypothetical satellite skimming the Earth's surface. Both results come from the same physics — the gravitational force at radius R.

Worked examples

Example 1: Earth pulls the Moon

The Earth has mass M = 5.97 \times 10^{24} kg. The Moon has mass m = 7.35 \times 10^{22} kg. The centre-to-centre distance is r = 3.84 \times 10^8 m. Calculate the gravitational force between them.

Earth–Moon gravitational force diagram Earth (large circle, left) and Moon (small circle, right), separated by 3.84 times 10 to the 8 metres. Force arrows point toward each other. Earth 5.97 × 10²⁴ kg Moon 7.35 × 10²² kg 3.84 × 10⁸ m
The Earth and Moon attract each other with equal and opposite forces across 384,400 km of empty space.

Step 1. Write down the known quantities.

G = 6.674 \times 10^{-11} N·m²/kg², M = 5.97 \times 10^{24} kg, m = 7.35 \times 10^{22} kg, r = 3.84 \times 10^8 m.

Why: both Earth and Moon are approximately spherical, so the shell theorem lets you treat each as a point mass at its centre. The distance r is the centre-to-centre separation.

Step 2. Substitute into F = GMm/r^2.

F = \frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^8)^2}

Why: direct substitution — the law applies without modification because both bodies are spheres and you are computing the force between their centres.

Step 3. Compute the numerator.

G \times M \times m = 6.674 \times 5.97 \times 7.35 \times 10^{-11+24+22} = 292.7 \times 10^{35} = 2.927 \times 10^{37}

Why: multiply the decimal parts (6.674 \times 5.97 \times 7.35 \approx 292.7) and add the exponents (-11 + 24 + 22 = 35).

Step 4. Compute the denominator.

r^2 = (3.84)^2 \times 10^{16} = 14.75 \times 10^{16} = 1.475 \times 10^{17}

Why: square the decimal part (3.84^2 = 14.75) and double the exponent (8 \times 2 = 16).

Step 5. Divide.

F = \frac{2.927 \times 10^{37}}{1.475 \times 10^{17}} = 1.98 \times 10^{20} \text{ N}

Why: divide the decimal parts (2.927 / 1.475 \approx 1.98) and subtract the exponents (37 - 17 = 20).

Result: The gravitational force between the Earth and Moon is approximately \mathbf{1.98 \times 10^{20}} N — about 198 billion billion newtons.

This enormous force is what keeps the Moon in orbit. Every second, it curves the Moon's path inward by just enough to match the curvature of its circular orbit. Without gravity, the Moon would fly off in a straight line at 1 km/s and never return.

Example 2: Two students in a classroom

Two students, each of mass 55 kg, sit 1.5 m apart during a physics lecture. Calculate the gravitational force between them.

Gravitational attraction between two students Two students modelled as circles, each 55 kg, sitting 1.5 metres apart. Tiny force arrows point toward each other. 55 kg 55 kg 1.5 m F ≈ 9 × 10⁻⁸ N
The gravitational pull between two 55 kg students sitting 1.5 m apart. The arrows are exaggerated for visibility — the actual force is about 90 billionths of a newton.

Step 1. Identify the knowns.

m_1 = m_2 = 55 kg, r = 1.5 m, G = 6.674 \times 10^{-11} N·m²/kg².

Why: at a distance of 1.5 m, treating a person as a point mass introduces negligible error compared to modelling the actual mass distribution.

Step 2. Substitute into F = Gm_1 m_2/r^2.

F = \frac{6.674 \times 10^{-11} \times 55 \times 55}{1.5^2} = \frac{6.674 \times 10^{-11} \times 3025}{2.25}

Why: 55 \times 55 = 3025 and 1.5^2 = 2.25. Keep the powers of 10 separate for clarity.

Step 3. Compute.

F = \frac{6.674 \times 3025 \times 10^{-11}}{2.25} = \frac{20{,}189 \times 10^{-11}}{2.25} = 8.97 \times 10^{-8} \text{ N}

Why: 6.674 \times 3025 = 20{,}189. Dividing by 2.25 gives 8{,}973. Expressed properly: 8.97 \times 10^{-8} N.

Step 4. Put this in perspective.

Each student's weight is 55 \times 9.8 = 539 N. The gravitational pull between them is:

\frac{539}{8.97 \times 10^{-8}} \approx 6 \times 10^{9}

Their weight is about 6 billion times larger than the gravitational force between them.

Why: this ratio shows why you never feel the gravitational pull of the person next to you. The force exists, but it is so absurdly small compared to the Earth's pull that no classroom instrument could detect it.

Result: The gravitational force between two 55 kg students sitting 1.5 m apart is about \mathbf{9.0 \times 10^{-8}} N — roughly 6 billion times smaller than their weight, and completely undetectable without laboratory-grade torsion balances.

This is why gravity only matters when at least one mass is enormous. The constant G is so small that everyday masses produce forces lost in the noise. You need a planet, a star, or at the very least a small mountain before gravitational effects become measurable.

Common confusions

If you came here to learn the law of universal gravitation, apply it in calculations, and understand the shell theorem qualitatively, you have what you need. What follows is for readers who want the mathematical proof of the shell theorem and the geometric reason behind the inverse-square law.

Proof of the shell theorem — external point

Consider a thin uniform spherical shell of total mass M_s and radius R. A point mass m sits at distance d from the centre, with d > R.

Step 1. Slice the shell into thin rings. Take a ring at angle \theta from the axis connecting the centre to m. The ring has circumference 2\pi R\sin\theta, width R\,d\theta, and mass:

dM = \frac{M_s}{4\pi R^2} \cdot 2\pi R^2 \sin\theta\,d\theta = \frac{M_s}{2}\sin\theta\,d\theta

Why: the shell's total surface area is 4\pi R^2, and the ring's area is 2\pi R\sin\theta \times R\,d\theta. The mass of the ring is proportional to its fraction of the total area.

Step 2. Every point on the ring is at the same distance s from m. By the law of cosines:

s^2 = d^2 + R^2 - 2dR\cos\theta

The component of the ring's gravitational pull along the axis toward the centre is dF = (Gm\,dM/s^2) \cdot (d - R\cos\theta)/s, since the perpendicular components cancel by symmetry.

Step 3. Change variable from \theta to s. Differentiating the cosine-law expression gives s\,ds = dR\sin\theta\,d\theta. Rearranging the cosine law also gives d - R\cos\theta = (d^2 - R^2 + s^2)/(2d). Substituting both, the integral becomes:

F = \frac{GM_s m}{4d^2 R}\int_{d-R}^{d+R}\left(1 + \frac{d^2 - R^2}{s^2}\right)ds

Why: the change of variable absorbs the \sin\theta\,d\theta and the cosine factor into a clean algebraic integrand in s.

Step 4. Evaluate the integral.

F = \frac{GM_s m}{4d^2 R}\left[s - \frac{d^2 - R^2}{s}\right]_{d-R}^{d+R}

At s = d + R: the bracket gives (d + R) - (d - R) = 2R.

At s = d - R: the bracket gives (d - R) - (d + R) = -2R.

The difference is 2R - (-2R) = 4R. Therefore:

F = \frac{GM_s m}{4d^2 R} \times 4R = \frac{GM_s m}{d^2}

Why: all the geometric complexity — rings at varying angles, varying distances, varying projections — integrates out to the simplest possible result. The shell behaves exactly like a point mass M_s at the centre.

For the internal case (d < R), the limits of integration change to s \in [R - d,\; R + d]. The same bracket evaluates to 2R at both endpoints, so the difference is zero: F = 0.

Why inverse-square? The geometry of three dimensions

The 1/r^2 law is not arbitrary — it reflects the geometry of the space you live in. Imagine a point mass radiating gravitational influence equally in all directions. At distance r, that influence is spread over a sphere of surface area 4\pi r^2. The force per unit area — the intensity — drops as 1/r^2 because the same total influence is spread over a larger and larger sphere.

This geometric argument applies to every force or signal that spreads uniformly through three-dimensional space: the electrostatic force (Coulomb's law, also 1/r^2), light intensity from a point source (1/r^2), sound intensity from a point source (1/r^2). The inverse-square law is the signature of three spatial dimensions.

If space had four dimensions, the force would drop as 1/r^3 (the "sphere" in 4D has surface area proportional to r^3). In two dimensions, the force would drop as 1/r. The fact that gravity follows 1/r^2 is, at a deep level, telling you that you live in a universe with exactly three spatial dimensions.

Where this leads next