Chapter 09 · Gravitation · Full book

Gravitation: The Force That Governs Everything

Core intuition: Chapter Roadmap

Gravity is by far the weakest of the four fundamental forces—yet it alone reaches across the emptiness of space to bind apples to the Earth, the Moon to its orbit, and galaxies to one another. This chapter builds the whole of Newtonian gravitation from a single inverse-square law and two conservation principles.

The universal law of gravitation — \(F = Gm_1m_2/r^2\): why the force must weaken as the inverse square, what the constant \(G\) measures, and how superposition handles many masses at once.
Field and potential — recasting the force as a field \(\vec{g} = -GM\hat{r}/r^2\) and a scalar potential \(V = -GM/r\), the two languages that make multi-body problems tractable, linked by \(\vec{g} = -\nabla V\).
The shell theorem — the geometric result that lets a spherical body act as a point mass from outside while producing zero field in a hollow interior; with it follow the variations of \(g\) with altitude, depth, and latitude.
Energy and escape — gravitational potential energy \(U = -GMm/r\), the work to assemble a system, and the escape speed that separates capture from freedom.
Orbits and Kepler's laws — circular and elliptical motion, orbital energy \(E = -GMm/2a\), and the three laws Newton's force law explains: ellipses, equal areas, and \(T^2 \propto a^3\).
Satellites and manoeuvres — orbital period, the unique geostationary radius, and the counter-intuitive truth that a higher orbit is slower yet more energetic.
Tides and the equivalence principle — the differential pull that raises two bulges and locks moons, and the deep equality of inertial and gravitational mass that points the way to General Relativity.

Master one force law and two conservation laws, and the entire clockwork of the heavens—from a dropped stone to a transfer to Mars—follows.

Perplexing Questions

Perplexing questions

Before we begin, let these questions sit with you. Do not reach for formulas. You will return to each one—with full answers—before this chapter ends.

The Falling Moon: The Moon is in free fall toward Earth right now. Every second, gravity pulls it inward. Yet it has been “falling” for four billion years and has never arrived. How can something fall continuously and perpetually miss?
The Weightless Astronaut: An astronaut inside the International Space Station floats freely—no floor pressure, no sense of weight. Yet at 400 km altitude, Earth's gravitational field is still nearly ninety percent of its surface value. Gravity is very much present. So what, precisely, is “weightlessness”?
The Universal Equality: Drop a feather and a cannonball in a vacuum. They hit the ground at exactly the same instant. This works on Earth, on the Moon, on Mars—for every pair of objects ever tested. Why should the resistance of a body to being accelerated and its response to gravitational attraction be exactly equal, for every object in the universe, without a single known exception?
The Apple and the Planet: An apple falls from a tree under Earth's gravity. Jupiter orbits the Sun under the Sun's gravity. A globular cluster of a million stars orbits the centre of a galaxy under its gravity. Is the same force law—the same mathematical relationship, the same constant—responsible for all three?

Each question targets a trap that catches students every year.
By the end of this chapter, every one of them will be transparent.

Why Gravity Demands a Deeper Look

Every chapter in this book so far has quietly relied on one background fact: objects near Earth's surface fall with acceleration \(g \approx 9.8\,\mathrm{m/s^2}\). We used it in kinematics to trace parabolic trajectories. We used it in Newton's Laws to resolve contact forces on inclines. We used it in Chapter 4 to assign gravitational potential energy. We used it in circular motion when banking a curve or looping a loop. We even used it in the rocket equation to compute the gravity penalty on a vertical launch.

We never explained it.

That is about to change.

The number \(g = 9.8\,\mathrm{m/s^2}\) is not a fundamental constant of nature—it is a derived quantity, a consequence of Earth's mass, Earth's radius, and a force law that Newton discovered not by watching apples but by comparing the Moon's motion to a stone thrown horizontally. The same law that makes an apple fall also keeps the Moon in orbit, determines how stars collapse, and governs the expansion of the universe itself.

This is the most far-reaching force law in the history of physics. Before Newton, the heavens and the Earth operated under different rules. After Newton, they obeyed one.

Core intuition: The Central Insight

Every particle in the universe attracts every other particle. The force is proportional to each mass and inversely proportional to the square of the distance between them: \[ F = \frac{G m_1 m_2}{r^2} \] The word universal is not rhetorical. The same constant \(G\), the same form, the same \(1/r^2\)—whether the bodies are falling apples, orbiting moons, or colliding galaxies.

How Newton Connected the Apple to the Moon

The key step was recognising that an orbit is not a different kind of motion from falling—it is falling, with enough sideways speed that the ground curves away as fast as the object drops toward it.

Newton asked: if the force that pulls the apple downward also reaches out to the Moon, how must it weaken with distance? The Moon's centripetal acceleration can be measured from its known orbital period and radius. The result is roughly \(2.7 \times 10^{-3}\,\mathrm{m/s^2}\).

The Moon is approximately 60 Earth radii away. If the force weakens as the square of the distance, then the Moon's acceleration should be: \[ a_{\text{Moon}} = \frac{g}{60^2} = \frac{9.8}{3600} \approx 2.7\times10^{-3}\,\mathrm{m/s^2} \] The agreement was exact. An apple at one Earth radius and the Moon at sixty Earth radii both obeying the same inverse-square law: this single numerical check is one of the most consequential calculations in the history of science.

Why the Force Must Weaken as \(1/r^2\)

Fig 8.1 The inverse square is pure geometry. The same bundle of field lines that pierces area \(A\) at distance \(r\) must spread over \(4A\) at \(2r\) — so the intensity, lines per area, falls to a quarter: \(F \propto 1/r^{2}\).

The inverse-square dependence is not an arbitrary choice—it is forced by the geometry of three-dimensional space.

Imagine the gravitational influence of a mass spreading outward in all directions. At distance \(r\), that influence is distributed over the surface of a sphere of area \(4\pi r^2\). If no gravitational “charge” exists in the empty space between source and field point—if the total amount of influence is conserved—then the intensity at radius \(r\) must fall off exactly as \(1/r^2\).

This reasoning has a precise name: Gauss's law for gravity. It shows that the inverse-square law is not a discovery about gravity in particular but a consequence of living in three spatial dimensions with a field that has no sources in the vacuum. Change the number of dimensions and the exponent changes with it.

There is a spectacular consequence of the exact value of the exponent. For a force law \(F \propto 1/r^2\), bound orbits are closed ellipses that repeat identically forever. For any other power of \(r\), orbits precess—they drift open like a rosette and are not closed. The observed repeatability of planetary orbits is therefore direct experimental evidence that the exponent is exactly two.

Core intuition: The Principle That Runs This Chapter

The gravitational force between any two masses falls off as the square of the distance separating them. This single statement—once made precise with Newton's law and the shell theorem—will account for everything in this chapter: the surface value of \(g\), the motion of satellites, the shape of orbits, the energy needed to escape a planet, the origin of tides, and the reason astronauts feel weightless.

Three Ideas, One Framework

This chapter builds three tightly connected ideas.

The gravitational force law. A precise statement of Newton's law of universal gravitation, the meaning of the gravitational constant \(G\), and the tools for calculating forces between point masses, spherical shells, and extended spheres—culminating in the shell theorem.

Gravitational field and potential. A shift in language from forces between specific pairs of masses to the field that a source mass creates at every point in space, and the associated scalar potential. This framework makes superposition straightforward and connects cleanly to the energy methods developed in Chapter 4.

Orbital mechanics and energy. The conditions for circular and elliptical orbits, Kepler's three laws derived from Newton's law, escape speed, and the total mechanical energy of a bound gravitational system.

Together, these ideas allow you to analyse a satellite launch, predict the period of a newly discovered exoplanet, compute the minimum speed needed to leave Earth's surface, and explain why geostationary orbit sits at one unique altitude—all from one universal constant and one force law.

What Lies Ahead

We begin by stating and interpreting Newton's law of universal gravitation, connecting it to \(g\) and to Kepler's empirical patterns. We then prove the shell theorem, which is the bridge between point-mass gravity and spherical bodies. From there we introduce the gravitational field and potential, extend to gravitational potential energy at arbitrary distances, and use energy to derive escape speed and orbital energetics. Kepler's three laws are stated, derived, and connected to conservation laws. We then treat satellites and orbital mechanics in detail, discuss tides as differential gravity, and close with a treatment of the Equivalence Principle and a bridge toward General Relativity.

If at any point you feel lost, return to the single idea that anchors everything: every mass attracts every other mass with a force that falls off as the square of the distance—and from that one fact, all the rest follows. Every result in this chapter is a consequence of that one relationship.

✓ Checkpoint · Thinking About Gravity

The Moon is approximately 60 Earth radii from Earth's centre. Using only the inverse-square law and the surface value \(g \approx 9.8\,\mathrm{m/s^2}\), estimate the Moon's centripetal acceleration toward Earth. Do not look up the orbital period.
\(a = g/60^2 = 9.8/3600 \approx 2.7\times10^{-3}\,\mathrm{m\,s^{-2}}\). The factor of \(60^2 = 3600\) comes directly from the inverse-square scaling: multiplying the distance by 60 divides the field by \(3600\).
A spacecraft is in a stable circular orbit at altitude \(h\) above Earth's surface. The crew reports feeling completely weightless. A student argues: “They are weightless because gravity is weak at that altitude.” Identify the error and give the correct explanation in two sentences.
The error is equating weightlessness with absence of gravity. Gravity remains substantial in orbit—it is the force maintaining the orbit. Weightlessness arises because the spacecraft and crew are in free fall together: they share the same gravitational acceleration, so no contact force acts between crew and floor.
Jupiter's mass is about 318 times Earth's mass and its radius is about 11 times Earth's radius. Without computing numerically, determine whether surface gravity on Jupiter is greater or less than on Earth, and estimate the factor.
\(g \propto M/R^2\). Ratio: \(318/11^2 = 318/121 \approx 2.6\). Jupiter's surface gravity is about 2.6 times Earth's—large mass wins over large radius.

Newton's Universal Law of Gravitation

By the middle of the seventeenth century, the motion of the planets was already well described by Kepler's three empirical laws. Planets moved in ellipses; a line from the Sun swept equal areas in equal times; the square of the orbital period was proportional to the cube of the semi-major axis. These were patterns extracted from decades of observation by Tycho Brahe and codified by Kepler between 1609 and 1619. They were precise, they were universal across the known planets, and they were completely unexplained.

No one had asked the question that Newton asked: what force produces these patterns?

Newton's answer—reached over several years and published in the Principia Mathematica in 1687—was that Kepler's laws are not independent facts. They are all consequences of a single force law between any two masses anywhere in the universe.

Core intuition: The Law in Words

Every particle of matter in the universe attracts every other particle. The strength of the attraction is proportional to each mass and inversely proportional to the square of the distance separating them. The direction is always along the line joining the two particles, and it is always attractive—gravity never repels. This is not a law about planets and the Sun. It is a law about every pair of masses—apples and Earth, Earth and Moon, two people standing in a room, two stars at opposite ends of a galaxy.

Statement of the Law

Fig 8.2 Newton’s universal law: two masses attract along the line joining their centres with \(F = Gm_1m_2/r^{2}\). The two pulls form an action–reaction pair — equal in magnitude even when the masses are wildly unequal.

For two point masses \(m_1\) and \(m_2\) separated by a distance \(r\), the magnitude of the mutual gravitational force is:

\[ \boxed{F = \frac{G m_1 m_2}{r^2}} \]

where \(G\) is the universal gravitational constant: \[ G = 6.674 \times 10^{-11}\,\mathrm{N\,m^2\,kg^{-2}} \]

In vector form, the force exerted on \(m_2\) by \(m_1\) is: \[ \vec{F}_{12} = -\frac{G m_1 m_2}{r^2}\,\hat{r}_{12} \] where \(\hat{r}_{12}\) is the unit vector pointing from \(m_1\) toward \(m_2\). The negative sign encodes attraction: the force on \(m_2\) points toward \(m_1\), opposite to \(\hat{r}_{12}\).

Symmetry of the Gravitational Interaction

The force that \(m_1\) exerts on \(m_2\) is equal in magnitude and opposite in direction to the force that \(m_2\) exerts on \(m_1\). This is Newton's third law, built into the structure of the law itself.

Notice that both forces share the same formula \(Gm_1m_2/r^2\). Neither mass is privileged. Earth pulls the Moon and the Moon pulls Earth with forces of identical magnitude; it is only Earth's far greater mass that makes its acceleration imperceptibly small.

The symmetry also means there is no directionality to the source. A mass does not “emit” gravity in one direction more than another— the field it creates is spherically symmetric, pointing inward from all sides.

Why the Dependence Must Be Inverse-Square

The exponent \(2\) in \(1/r^2\) is not an empirical accident. It is forced by the geometry of three-dimensional space and a conservation principle about how fields spread through empty volume.

Birth of the formula: Inverse-Square from Spherical Spreading

Let a point mass \(M\) sit at the origin and produce a gravitational influence that spreads outward uniformly in all directions. Draw an imaginary sphere of radius \(r\) centred on \(M\). The total gravitational “flux” passing through this sphere must be the same for every radius—there is no source or sink of gravity in the empty space between \(M\) and the sphere. Call this conserved total flux \(\Phi\). The sphere has surface area \(4\pi r^2\). If the flux is distributed uniformly over the sphere's surface, the flux per unit area—which is what the gravitational field intensity represents—is: \[ g(r) = \frac{\Phi}{4\pi r^2} \propto \frac{1}{r^2} \] The inverse-square dependence follows directly from the conservation of total flux and the \(r^2\) growth of spherical surface area in three spatial dimensions. This argument is not an analogy. It is the content of Gauss's law for gravity: the gravitational flux through any closed surface is proportional to the enclosed mass, regardless of the shape of the surface. In \(d\) spatial dimensions, spherical surface area grows as \(r^{d-1}\), so the field would fall off as \(r^{-(d-1)}\). In three dimensions: \(r^{-2}\). The inverse-square law is therefore a signature of a point source in three-dimensional space with no intervening gravitational charge. Changing the exponent is equivalent to changing the number of dimensions.

There is a profound observational consequence. For a force law \(F \propto 1/r^2\) exactly, bound orbits are closed ellipses that repeat identically forever (Bertrand's theorem). For any other power of \(r\), orbits precess—they rotate slowly, tracing open rosette paths. The repeatability of planetary orbits year after year, millennium after millennium, is direct experimental evidence that the exponent in the force law is exactly two.

The Gravitational Constant \(G\)

The constant \(G = 6.674\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\) is extraordinarily small by everyday standards. Two people of mass \(70\,\mathrm{kg}\) standing \(1\,\mathrm{m}\) apart attract each other gravitationally with a force of roughly \(3.3\times10^{-7}\,\mathrm{N}\)—far too small to detect without a precision instrument.

Only when at least one body is astronomically massive does gravity dominate. The smallness of \(G\) is why gravity, despite being the most far-reaching force in nature, is the weakest of the four fundamental interactions at the scale of elementary particles.

Newton himself could not measure \(G\) directly—he could determine ratios of masses but not their absolute values. The first laboratory measurement came from Henry Cavendish in 1798, using a torsion balance: two small lead spheres attracted by two larger ones, the gravitational force detected through the twist of a thin fibre. Cavendish's experiment was described at the time as “weighing the Earth,” because once \(G\) is known and the surface value of \(g\) is measured, Earth's mass follows immediately from \(g = GM_E/R_E^2\).

Core intuition: Recovering \(g\) from First Principles

At Earth's surface, the gravitational force on a mass \(m\) is: \[ F = \frac{GM_E m}{R_E^2} \] Newton's second law gives the acceleration: \[ a = \frac{F}{m} = \frac{GM_E}{R_E^2} \] The mass \(m\) cancels completely. Every object gets the same acceleration at the same location, regardless of its mass, composition, or shape. Substituting \(G = 6.674\times10^{-11}\), \(M_E = 5.972\times10^{24}\,\mathrm{kg}\), \(R_E = 6.4\times10^{6}\,\mathrm{m}\) (rounded; precise value \(6.371\times10^6\,\mathrm{m}\)): \[ g = \frac{(6.674\times10^{-11})(5.972\times10^{24})}{(6.4\times10^{6})^2} \approx 9.8\,\mathrm{m\,s^{-2}} \] The number used since earlier chapters is now derived, not assumed. Note: \(R_E = 6.4\times10^6\,\mathrm{m}\) is the standard JEE round value. The precise value is \(6.4\times10^6\,\mathrm{m}\), used in problems that require numerical precision.

Superposition of Gravitational Forces

Fig 8.3 Superposition: each source mass pulls on \(m\) as if the others did not exist, and the net force is the vector sum \(\vec F_{net}=\vec F_1+\vec F_2+\cdots\) — completed here by the parallelogram.

When more than two masses are present, the total gravitational force on any one of them is the vector sum of the forces from all the others. This is the superposition principle for gravity, and it holds exactly in Newtonian mechanics.

For a mass \(m\) acted on by masses \(m_1, m_2, \ldots, m_n\): \[ \vec{F}_{\text{total}} = \sum_{i=1}^{n} \vec{F}_i = -Gm \sum_{i=1}^{n} \frac{m_i}{r_i^2}\,\hat{r}_i \] where \(\hat{r}_i\) points from \(m_i\) toward \(m\).

Superposition is what makes gravitation tractable for extended bodies. A continuous mass distribution is handled by replacing the sum with an integral over volume elements \(dM\): \[ \vec{F} = -Gm \int \frac{dM}{r^2}\,\hat{r} \] This integral underlies the shell theorem, which we prove later and which justifies treating spherical planets as point masses located at their centres.

Example 1

Gravitational Force Between Two Bodies Calculate the gravitational force between:

Two people, masses \(60\,\mathrm{kg}\) and \(80\,\mathrm{kg}\), standing \(1.5\,\mathrm{m}\) apart.
The Earth and the Moon. (\(M_E = 5.97\times10^{24}\,\mathrm{kg}\), \(M_M = 7.35\times10^{22}\,\mathrm{kg}\), centre-to-centre distance \(r = 3.84\times10^{8}\,\mathrm{m}\).)

Solution. (a) Applying the equation above directly: \[ F = \frac{G m_1 m_2}{r^2} = \frac{(6.674\times10^{-11})(60)(80)}{(1.5)^2} = \frac{3.204\times10^{-7}}{2.25} \approx 1.42\times10^{-7}\,\mathrm{N} \] This is roughly the weight of a particle of dust. The mutual gravitational pull between two people is utterly negligible in any mechanical context. It is detectable only with a Cavendish-type apparatus. (b) For the Earth–Moon system: \[\begin{aligned} F &= \frac{(6.674\times10^{-11})(5.97\times10^{24})(7.35\times10^{22})} {(3.84\times10^{8})^2} \\[4pt] &= \frac{6.674\times10^{-11} \times 4.388\times10^{47}} {1.475\times10^{17}} = \frac{2.928\times10^{37}}{1.475\times10^{17}} \approx 1.98\times10^{20}\,\mathrm{N} \end{aligned}\] This force—close to \(2\times10^{20}\,\mathrm{N}\)—is what keeps the Moon in its orbit and is the primary driver of ocean tides on Earth. By Newton's third law, the Moon pulls Earth with the same magnitude. The Moon's greater acceleration (compared to Earth's) reflects only the Moon's much smaller mass: \(a = F/m\).

✓ Checkpoint · Newton's Law of Gravitation

The distance between two masses is tripled while both masses are doubled. By what factor does the gravitational force between them change?
\(F \propto m_1 m_2 / r^2\). New force: \((2m_1)(2m_2)/(3r)^2 = 4m_1m_2/9r^2\). Factor: \(4/9\). The force decreases to four-ninths of its original value.
Two students argue about the gravitational force on a \(1\,\mathrm{kg}\) book resting on a table. Student A says Earth pulls the book down with force \(F\). Student B says the book also pulls Earth upward with force \(F\). Who is correct? What prevents Earth from accelerating noticeably upward?
Both are correct: Newton's third law guarantees the forces are equal in magnitude. Earth does accelerate upward, but by \(a = F/M_E \approx 10^{-24}\,\mathrm{m\,s^{-2}}\)— completely undetectable. Equal force does not imply equal acceleration when masses differ by 24 orders of magnitude.
The surface gravitational acceleration on a planet is \(g_p\). If the planet's radius is halved while its total mass is kept constant, what is the new surface acceleration?
\(g \propto M/R^2\). Halving \(R\) with fixed \(M\) gives \(g_{\text{new}} = M/(R/2)^2 \cdot G = 4GM/R^2 = 4g_p\). Surface gravity quadruples.

Gravitational Field

◈ Field & potential explorer

live bench

Switch between a solid sphere and a hollow shell and sweep a test point through every radius: the field arrows and the potential well redraw together, making \(g = -dV/dr\) something you can see.

◈ loading the interactive bench…

↗ open in the Simulations page

Every time we write \(F = GMm/r^2\), two masses appear in the formula. But consider what this means physically: Earth's gravitational influence exists at every point in space around it whether or not a satellite is orbiting, whether or not a person stands on its surface. The Moon did not create Earth's gravity by arriving. Something is already present at every point in space around a massive body—a capacity to exert force on anything placed there—even before any second mass appears.

This “something” is the gravitational field.

The field concept separates two things that the force law conflates: the source mass, which creates the field, and the test mass, which responds to it. Once the field of a source is known at every point in space, finding the force on any test mass placed at any point requires only a multiplication. The source does not need to be reconsidered.

This separation becomes especially powerful when the source is complicated—a planet with uneven mass distribution, a binary star, a galaxy cluster. Compute the field once; the force on any test mass follows immediately.

Core intuition: Force and Field Are Different Things

The gravitational force \(\vec{F}\) is a property of the interaction between two specific bodies. Remove either body and the force does not exist. It has units of newtons. The gravitational field \(\vec{g}\) is a property of a point in space, created by the source mass alone. It exists at every point whether or not a test mass is present. It has units of \(\mathrm{N\,kg^{-1}}\), equivalently \(\mathrm{m\,s^{-2}}\). Their relationship: \[ \vec{F} = m\,\vec{g} \] The field is the force per unit mass that the source would exert on a test mass placed at that point.

Gravitational Field of a Point Mass

Fig 8.4 The gravitational field of a point mass points radially inward everywhere. Its magnitude depends only on distance — the arrows on the outer ring are shorter by exactly the inverse-square factor.

Birth of the formula: Deriving the Field of a Point Mass

Let a point mass \(M\) be fixed at the origin. Place a test mass \(m\) at position \(\vec{r}\), so the separation is \(r = |\vec{r}|\). The gravitational force on \(m\) due to \(M\) is: \[ \vec{F} = -\frac{GMm}{r^2}\,\hat{r} \] where \(\hat{r}\) is the unit vector pointing from \(M\) toward \(m\), and the negative sign indicates attraction. The gravitational field at position \(\vec{r}\) is defined as the force per unit test mass: \[ \vec{g}(\vec{r}) \;\equiv\; \frac{\vec{F}}{m} = -\frac{GM}{r^2}\,\hat{r} \] The test mass \(m\) cancels completely from the definition. The field is a property of \(M\) and the position \(\vec{r}\) alone. The magnitude at distance \(r\) from a point mass \(M\): \[ \boxed{g(r) = \frac{GM}{r^2}} \] Direction: always radially inward, toward \(M\).

Three features of this result deserve emphasis before moving on.

The field diverges at \(r = 0\). As \(r \to 0\), \(g \to \infty\). This is an artefact of the point-mass idealisation. Real bodies have finite size; the field inside a body is finite and well-defined, computed using the shell theorem.

The field falls off as \(1/r^2\). This is the inverse-square law recast as a property of space rather than of a specific pair of masses. Moving twice as far from \(M\) reduces the field to one-quarter its previous value, regardless of what test mass is placed there.

The surface field equals the free-fall acceleration. At Earth's surface (\(r = R_E\)): \[ g_{\text{surface}} = \frac{GM_E}{R_E^2} \approx 9.8\,\mathrm{m\,s^{-2}} \] The gravitational field at Earth's surface is numerically equal to the free-fall acceleration used since earlier chapters. Note: \(9.8\,\mathrm{m\,s^{-2}}\) is the standard value used throughout this book. More precise calculations give \(9.81\,\mathrm{m\,s^{-2}}\); the difference is negligible for all problems at this level. This is not a coincidence: the field is the acceleration that any freely falling body experiences at that location, since \(a = F/m = g\).

How \(g\) Varies with Distance

Fig 8.5 Variation of \(g\) through a uniform Earth: inside, only the sphere beneath you counts, so \(g \propto r\); outside, the whole mass acts from the centre and \(g \propto 1/r^{2}\). The two branches meet at the surface value \(g_s\).

Two practically important cases follow from \(g = GM/r^2\).

Above the surface. At height \(h\) above Earth's surface the distance from the centre is \(r = R_E + h\), so: \[ g(h) = \frac{GM_E}{(R_E + h)^2} = \frac{g_0}{\bigl(1 + h/R_E\bigr)^2} \] For \(h \ll R_E\), expanding to first order gives the linear approximation: \[ g(h) \approx g_0\!\left(1 - \frac{2h}{R_E}\right) \] This is accurate to about 1% for altitudes below 30 km. Beyond that the full formula is required.

Below the surface. For a uniform-density sphere the shell theorem shows that only the mass enclosed within radius \(r\) contributes to the field. Since enclosed mass scales as \(r^3\): \[ g(r) = \frac{GM_{\text{enc}}}{r^2} = \frac{G(M_E/R_E^3)\,r^3}{r^2} = \frac{GM_E}{R_E^3}\,r = g_0\,\frac{r}{R_E} \] Inside a uniform sphere the field grows linearly from zero at the centre to \(g_0\) at the surface—the opposite behaviour from the exterior. The physical reason: as one descends, the enclosed mass decreases (weakening the field) but so does the distance to it (strengthening the field); for uniform density the distance effect wins and the field grows with depth.

Superposition of Fields

Because force obeys superposition, so does the field. The total gravitational field at a point due to several source masses is the vector sum of their individual fields.

For sources \(M_1, M_2, \ldots, M_n\) at distances \(r_1, r_2, \ldots, r_n\) from the field point, with unit vectors \(\hat{r}_i\) pointing from each source toward the field point: \[ \vec{g}_{\text{total}} = \sum_{i=1}^{n} \vec{g}_i = -G\sum_{i=1}^{n} \frac{M_i}{r_i^2}\,\hat{r}_i \] For a continuous distribution of mass with density \(\rho(\vec{r}')\): \[ \vec{g}(\vec{r}) = -G\int \frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|^2} \,\hat{r}'\,dV' \] This integral is the foundation for computing the field of shells and solid spheres later in this chapter.

Example 2

Gravitational Field at Orbital Altitude The International Space Station orbits at altitude \(h = 400\,\mathrm{km}\). Calculate the gravitational field there and express it as a fraction of the surface value. (\(R_E = 6.4\times10^6\,\mathrm{m}\), \(g_0 = 9.8\,\mathrm{m\,s^{-2}}\).)

Solution. The distance from Earth's centre to the station is: \[ r = R_E + h = 6.4\times10^6 + 0.4\times10^6 = 6.8\times10^6\,\mathrm{m} \] Since \(g \propto 1/r^2\), the ratio to the surface value is: \[ \frac{g(r)}{g_0} = \frac{R_E^2}{r^2} = \left(\frac{6.4}{6.8}\right)^{\!2} = (0.9412)^2 \approx 0.886 \] The gravitational field at the ISS is approximately \(0.886 \times 9.8 \approx 8.68\,\mathrm{m\,s^{-2}}\)—about 89% of its surface value. This resolves the weightlessness puzzle from the chapter opener. The ISS crew floats not because gravity is absent or even weak there, but because the station and everything inside it are in free fall together. In a freely falling reference frame every object shares the same gravitational acceleration, so there is no contact force between crew and floor. The sensation of weight requires a contact force; free fall eliminates it.

✓ Checkpoint · Gravitational Field

At what altitude above Earth's surface is the gravitational field exactly half its surface value? Express your answer in terms of \(R_E\).
\(g(r) = g_0/2\) requires \(R_E^2/r^2 = 1/2\), so \(r = R_E\sqrt{2}\). Altitude \(= r - R_E = R_E(\sqrt{2}-1) \approx 0.414\,R_E \approx 2640\,\mathrm{km}\).
A tunnel is drilled through the centre of a uniform-density Earth. An object is released from rest at the surface. Describe qualitatively how the gravitational field it experiences changes as it travels from the surface to the centre and back out to the other side.
From the surface to the centre, the field decreases linearly from \(g_0\) to zero—the field inside a uniform sphere is \(g = g_0 r/R_E\). At the centre the field is zero; past the centre it grows again, directed back toward the centre (always restoring). The object oscillates back and forth with simple harmonic motion.
Two identical stars each of mass \(M\) are separated by distance \(d\). At the midpoint between them, what is the magnitude of the gravitational field? Justify without algebra.
Zero. By symmetry, both stars produce fields of equal magnitude at the midpoint but directed in opposite directions—one pulling left, the other pulling right. They cancel exactly. The midpoint is a point of unstable equilibrium.

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Surface gravity of another planet

Mars has mass \(M = 6.39\times10^{23}\,\mathrm{kg}\) and radius \(R = 3.39\times10^{6}\,\mathrm{m}\). Find the acceleration due to gravity at its surface, and express it as a fraction of Earth's \(g\).

Solution. Given: \(M = 6.39\times10^{23}\,\mathrm{kg}\), \(R = 3.39\times10^{6}\,\mathrm{m}\).
Find: surface \(g_{\mathrm{Mars}}\), and \(g_{\mathrm{Mars}}/g_E\).
Setup: The surface field of a spherically symmetric body follows from treating its mass as concentrated at the centre (shell theorem): \(g = GM/R^2\).
Solve: \[ g_{\mathrm{Mars}} = \frac{GM}{R^2} = \frac{(6.674\times10^{-11})(6.39\times10^{23})}{(3.39\times10^{6})^2} = \frac{4.265\times10^{13}}{1.149\times10^{13}} = 3.71\,\mathrm{m/s^2}. \] Answer: \(\boxed{g_{\mathrm{Mars}} = 3.71\,\mathrm{m/s^2} \approx 0.38\,g_E}\)
Check: Order of magnitude: Mars has about \(0.11\) Earth masses and \(0.53\) Earth radii. Since \(g \propto M/R^2\), this predicts \(0.11/0.53^2 \approx 0.39\) of Earth's \(g\), close to the \(0.38\) found. ✓

Example 2 · Force between two masses

Two small spheres of mass \(5.0\,\mathrm{kg}\) and \(8.0\,\mathrm{kg}\) have their centres \(0.50\,\mathrm{m}\) apart. Find the gravitational force between them.

Solution. Given: \(m_1 = 5.0\,\mathrm{kg}\), \(m_2 = 8.0\,\mathrm{kg}\), \(r = 0.50\,\mathrm{m}\).
Find: the gravitational force \(F\).
Setup: Newton's law of gravitation, \(F = Gm_1m_2/r^2\), applies directly to point-like (or spherically symmetric) masses.
Solve: \[ F = \frac{Gm_1m_2}{r^2} = \frac{(6.674\times10^{-11})(5.0)(8.0)}{(0.50)^2} = \frac{2.670\times10^{-9}}{0.25} = 1.07\times10^{-8}\,\mathrm{N}. \] Answer: \(\boxed{F = 1.07\times10^{-8}\,\mathrm{N}}\)
Check: Magnitude sanity: gravity between everyday objects is famously tiny—about \(10^{-8}\,\mathrm{N}\) here, roughly the weight of a microgram. That smallness is exactly why \(G\) is hard to measure and why we never feel the pull of nearby furniture. ✓

Example 3 · Gravitational field above the Earth

Find the magnitude of Earth's gravitational field at a point \(4.0\times10^{7}\,\mathrm{m}\) from its centre.

Solution. Given: \(r = 4.0\times10^{7}\,\mathrm{m}\), \(M_E = 5.972\times10^{24}\,\mathrm{kg}\).
Find: field magnitude \(g(r)\).
Setup: Outside a spherical mass the field is that of a point mass: \(g = GM_E/r^2\). The point lies well above the surface (\(r \approx 6.3\,R_E\)), so we expect a value far below \(9.8\,\mathrm{m/s^2}\).
Solve: \[ g = \frac{GM_E}{r^2} = \frac{(6.674\times10^{-11})(5.972\times10^{24})}{(4.0\times10^{7})^2} = \frac{3.986\times10^{14}}{1.6\times10^{15}} = 0.249\,\mathrm{N/kg}. \] Answer: \(\boxed{g = 0.249\,\mathrm{N/kg}}\)
Check: Inverse-square scaling: \(r = 6.28\,R_E\), so \(g\) should be \(g_{\text{surf}}/6.28^2 = 9.82/39.4 = 0.249\,\mathrm{N/kg}\). ✓

Example 4 · Variation of \(g\) with altitude

A satellite orbits at an altitude of \(400\,\mathrm{km}\). Find the true gravitational acceleration there, and express it as a fraction of the surface value.

Solution. Given: \(h = 400\,\mathrm{km} = 4.00\times10^{5}\,\mathrm{m}\), \(R_E = 6.371\times10^{6}\,\mathrm{m}\).
Find: \(g(h)\) and \(g(h)/g_{\text{surf}}\).
Setup: The distance from the centre is \(r = R_E + h\); use \(g = GM_E/r^2\).
Solve: \[\begin{aligned} r &= 6.371\times10^{6} + 4.00\times10^{5} = 6.771\times10^{6}\,\mathrm{m},\\ g(h) &= \frac{(6.674\times10^{-11})(5.972\times10^{24})}{(6.771\times10^{6})^2} = \frac{3.986\times10^{14}}{4.585\times10^{13}} = 8.69\,\mathrm{m/s^2}. \end{aligned}\] Answer: \(\boxed{g(400\,\mathrm{km}) = 8.69\,\mathrm{m/s^2} \approx 0.885\,g_{\text{surf}}}\)
Check: Binomial estimate: for \(h \ll R_E\), \(g \approx g_{\text{surf}}(1 - 2h/R_E) = 9.82(1 - 2(0.0628)) = 9.82(0.874) = 8.59\,\mathrm{m/s^2}\), close to the exact \(8.69\) (the linear estimate slightly overshoots the correction, as expected). The ISS is not “beyond gravity”—\(g\) is still about \(89\%\) of its surface value. ✓

Example 5 · Escape speed from Earth

Find the speed with which a projectile must leave Earth's surface to escape its gravity, ignoring air resistance and Earth's rotation.

Solution. Given: \(M_E = 5.972\times10^{24}\,\mathrm{kg}\), \(R_E = 6.371\times10^{6}\,\mathrm{m}\).
Find: escape speed \(v_e\).
Setup: Escape means just reaching infinity with zero speed, so total mechanical energy is zero: \(\tfrac12 mv_e^2 - GM_Em/R_E = 0\).
Solve: \[ v_e = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{\frac{2(6.674\times10^{-11})(5.972\times10^{24})}{6.371\times10^{6}}} = \sqrt{1.251\times10^{8}} = 1.12\times10^{4}\,\mathrm{m/s}. \] Answer: \(\boxed{v_e = 11.2\,\mathrm{km/s}}\)
Check: Alternative form: \(v_e = \sqrt{2gR_E} = \sqrt{2(9.82)(6.371\times10^{6})} = \sqrt{1.251\times10^{8}} = 11.2\,\mathrm{km/s}\), since \(g = GM_E/R_E^2\). Both routes agree. ✓