Chapter 8: Gravitation

Gravitation is the oldest and most universal force we know — it controls falling apples and galactic clusters alike. Newton's law of universal gravitation unified terrestrial mechanics with celestial motion. In this chapter we extract from a single inverse-square law the orbits of planets, the rise and fall of projectiles, the orbital architecture of satellites, and the boundary between bound and free motion (escape velocity).

Concept Map

• 8.1 — Kepler's laws of planetary motion
• 8.2 — Newton's law of universal gravitation
• 8.3 — Gravitational constant: Cavendish experiment
• 8.4 — Acceleration due to gravity $g$ and its variation
• 8.5 — Gravitational potential energy
• 8.6 — Escape velocity
• 8.7 — Orbital velocity, period, and energy of a satellite
• 8.8 — Geostationary and polar satellites
• 8.9 — Weightlessness in satellites

8.1 Kepler's Laws of Planetary Motion

Statements

1. Law of Orbits: Every planet revolves around the Sun in an elliptical orbit with the Sun at one focus.
2. Law of Areas: The line joining a planet to the Sun sweeps out equal areas in equal intervals of time. (Areal velocity $dA/dt =$ constant.)
3. Law of Periods: The square of the orbital period is proportional to the cube of the semi-major axis: $T^2 \propto a^3$.

Geometrical Meaning of the Second Law

If $\vec{r}$ is the position of the planet relative to the Sun, the area swept in time $dt$ is

$dA = \tfrac{1}{2}|\vec{r}\times d\vec{r}| = \tfrac{1}{2}|\vec{r}\times\vec{v}|\,dt$

So $dA/dt = L/(2m)$. Constant areal velocity is equivalent to conservation of angular momentum — a direct consequence of the gravitational force being central ($\vec{F}\parallel\vec{r}$, so $\vec{\tau} = 0$).

Derivation of Third Law (Circular Orbit Approximation)

For a circular orbit of radius $r$, gravity provides the centripetal force:

$\frac{GMm}{r^2} = \frac{mv^2}{r} = m\omega^2 r$

$\omega^2 = \frac{GM}{r^3}, \quad T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{r^3}{GM}}$

$\boxed{T^2 = \frac{4\pi^2}{GM}r^3}$

So $T^2 \propto r^3$ (general result: $a^3$ for elliptic orbits, with $a$ the semi-major axis).

Worked Example

Earth's orbital radius $1\,\text{AU}$, period $1\,\text{yr}$. Mars at $1.524\,\text{AU}$. Find Mars's period.

$\left(\frac{T_M}{T_E}\right)^2 = \left(\frac{a_M}{a_E}\right)^3 = 1.524^3 = 3.54$

$T_M = \sqrt{3.54} \approx 1.88\,\text{yr}$

Common Mistakes

• Treating $r$ as the closest or farthest distance; for elliptic orbits use $a$ (semi-major axis = average of perihelion and aphelion).
• Assuming Kepler's second law implies constant speed (it doesn't — speed is higher near perihelion).

8.2 Newton's Law of Universal Gravitation

Statement

Every particle attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In vector form:

$\vec{F}_{12} = -\frac{Gm_1 m_2}{r^2}\hat{r}_{12}$

where $\hat{r}_{12}$ points from 1 to 2 and the minus sign means the force is attractive.

Magnitude

$F = \frac{Gm_1 m_2}{r^2}$

$G$ is the universal gravitational constant.

Superposition Principle

The force on a particle due to many others is the vector sum:

$\vec{F}_i = \sum_{j\ne i} \vec{F}_{ij}$

Shell Theorem (Statement)

For a spherically symmetric body of total mass $M$:

1. The gravitational force outside the shell is as if the entire mass were at the centre.
2. Inside a uniform thin shell, the force is zero.

This justifies treating the Earth as a point mass for satellites and external objects.

Worked Example

Two $1\,\text{kg}$ masses are $1\,\text{m}$ apart. Force between them?

$F = \frac{6.67\times 10^{-11}\cdot 1\cdot 1}{1^2} = 6.67\times 10^{-11}\,\text{N}$

Negligible — gravity between everyday objects is extraordinarily weak.

Common Mistakes

• Forgetting the inverse-square nature: doubling the distance reduces force by factor 4.
• Misapplying the shell theorem to non-spherical bodies.

8.3 Gravitational Constant — Cavendish Experiment

Summary

Cavendish (1798) used a torsion balance: two small lead spheres on a horizontal beam suspended by a thin fibre. Two large lead spheres were brought close to the small ones, causing the beam to rotate by a small angle due to gravitational attraction. The angle is measured by deflection of a light beam reflected off a mirror on the fibre.

From the angle $\theta$, the known torsion constant $\kappa$ of the fibre, the masses $m, M$ and distance $r$,

$\frac{GmM}{r^2}\cdot L = \kappa\theta$

where $L$ is the lever arm. Solving for $G$:

$G \approx 6.674\times 10^{-11}\,\text{N m}^2/\text{kg}^2$

This was the first lab-scale measurement of a "universal constant" and yielded the mass of the Earth.

Mass of the Earth (Derivation)

At the surface, $g = GM/R^2$, so

$M_E = \frac{gR^2}{G} = \frac{9.8\cdot (6.4\times 10^6)^2}{6.67\times 10^{-11}} \approx 6.0\times 10^{24}\,\text{kg}$

8.4 Acceleration Due to Gravity and its Variation

At the Surface

$g = \frac{GM}{R^2}$

with $M$ the mass of the Earth, $R$ its radius.

Variation with Altitude $h$ (Above Surface)

$g(h) = \frac{GM}{(R+h)^2} = g\left(1 + \frac{h}{R}\right)^{-2}$

For $h \ll R$, binomial expansion:

$g(h) \approx g\left(1 - \frac{2h}{R}\right)$

So at $h = R/100$, $g$ drops by about $2\%$.

Variation with Depth $d$ (Below Surface)

Assume uniform density $\rho$. By the shell theorem, only mass within radius $r = R - d$ matters.

Mass within $r$: $M' = M(r/R)^3$. So

$g(d) = \frac{GM'}{r^2} = \frac{GM r}{R^3} = g\left(1 - \frac{d}{R}\right)$

At $d = R$ (centre), $g = 0$.

Variation with Latitude (Due to Earth's Rotation)

At latitude $\lambda$, a body at the surface moves in a circle of radius $R\cos\lambda$. Pseudo (centrifugal) force in the rotating Earth frame reduces effective gravity:

$g_{\text{eff}} = g - \omega^2 R\cos^2\lambda$

(where $\omega$ is Earth's angular velocity)

• At equator ($\lambda = 0$): $g_{\text{eff}} = g - \omega^2 R$ (minimum).
• At poles ($\lambda = 90^\circ$): $g_{\text{eff}} = g$ (maximum).

Numerical: $\omega^2 R \approx 0.034\,\text{m/s}^2$, so $g$ at equator is about $0.3\%$ less than at the poles due to rotation alone.

Variation Due to Earth's Shape

The Earth is an oblate spheroid: $R_{\text{equator}} > R_{\text{pole}}$ by about $21\,\text{km}$. Since $g \propto 1/R^2$, this further increases $g$ at the poles.

Combined effect: $g_{\text{pole}} \approx 9.83$, $g_{\text{equator}} \approx 9.78\,\text{m/s}^2$.

Worked Example

A satellite at $h = 600\,\text{km}$. Find $g$. ($R = 6400\,\text{km}$, $g_0 = 9.8\,\text{m/s}^2$)

$g = 9.8\left(\frac{6400}{7000}\right)^2 = 9.8\cdot 0.836 \approx 8.19\,\text{m/s}^2$

Worked Example

A mine $1\,\text{km}$ deep. $g$ at the bottom?

$g = 9.8\left(1 - \frac{1}{6400}\right) \approx 9.798\,\text{m/s}^2$

Common Mistakes

• Mixing up altitude and depth variation formulae.
• Forgetting that the rotation contribution is a centrifugal effect (vanishes at poles).

8.5 Gravitational Potential Energy

Near the Surface

$U = mgh$ (taking $U = 0$ at ground level). This is the special case of the more general formula valid only when $h \ll R$.

General Formula (Two Point Masses)

For a particle of mass $m$ at distance $r$ from a particle of mass $M$, taking $U(\infty) = 0$:

$U(r) = -\frac{GMm}{r}$

The minus sign indicates that gravity is attractive — work must be done against gravity to separate the masses to infinity.

Derivation

Work done by gravity in bringing a particle from $\infty$ to $r$:

$W_{\text{grav}} = \int_\infty^r \vec{F}\cdot d\vec{r} = \int_\infty^r \left(-\frac{GMm}{r'^2}\right)(-dr') = -\frac{GMm}{r}$

(The signs: the force is radially inward; the displacement is also inward, hence work is positive in magnitude but we record it as $-GMm/r$ relative to infinity.) The potential energy is minus the work done by the force:

$U(r) = -W_{\text{grav}} = -\frac{GMm}{r}$

Wait — there's a sign subtlety. Standard convention: $U(r) - U(\infty) = -\int_\infty^r \vec{F}\cdot d\vec{r}$. With $\vec{F} = -GMm/r^2\,\hat{r}$ (attractive, pointing inward, i.e., toward decreasing $r$), and $d\vec{r} = dr\,\hat{r}$:

$\vec{F}\cdot d\vec{r} = -\frac{GMm}{r^2}\,dr$

$U(r) = -\int_\infty^r \left(-\frac{GMm}{r'^2}\right)dr' = \int_\infty^r \frac{GMm}{r'^2}dr' = \left[-\frac{GMm}{r'}\right]_\infty^r = -\frac{GMm}{r}$

Connecting to $mgh$

At the surface, $U_0 = -GMm/R$. At height $h$, $U_h = -GMm/(R+h)$.

$\Delta U = -\frac{GMm}{R+h} + \frac{GMm}{R} = \frac{GMmh}{R(R+h)} \approx \frac{GMm}{R^2}h = mgh$

for $h \ll R$. So $U = mgh$ is just the linear approximation of the inverse-square formula.

Worked Example

How much energy is required to lift a $1000\,\text{kg}$ satellite from Earth's surface to a height equal to $R$?

$\Delta U = -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{GMm}{2R} = \frac{mgR}{2}$

With $m = 1000$, $g = 9.8$, $R = 6.4\times 10^6$,

$\Delta U = 1000\cdot 9.8\cdot 6.4\times 10^6/2 = 3.14\times 10^{10}\,\text{J}$

Common Mistakes

• Using $U = mgh$ for satellite-altitude problems.
• Forgetting that $U < 0$ everywhere (until $r = \infty$).

8.6 Escape Velocity

Definition

The minimum speed needed at the Earth's surface for an object to escape to infinity (with zero residual KE).

Derivation

By energy conservation,

$\tfrac{1}{2}mv_e^2 - \frac{GMm}{R} = 0$

(KE at infinity = 0, PE at infinity = 0)

Solving,

$v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$

Numerical Value for Earth

$v_e = \sqrt{2\cdot 9.8\cdot 6.4\times 10^6} \approx 1.12\times 10^4\,\text{m/s} = 11.2\,\text{km/s}$

Properties

• Independent of mass of the escaping object.
• Independent of direction of launch (as long as it doesn't hit the Earth).
• Connected to orbital velocity: $v_e = \sqrt{2}\,v_{\text{orb}}$.

Worked Example

Find escape velocity from the Moon. ($g_M = 1.62\,\text{m/s}^2$, $R_M = 1.74\times 10^6\,\text{m}$)

$v_e = \sqrt{2\cdot 1.62\cdot 1.74\times 10^6} \approx 2.37\times 10^3\,\text{m/s} = 2.37\,\text{km/s}$

Common Mistakes

• Adding the surface rotation to escape velocity without specifying launch direction.
• Forgetting that "escape" means $E \ge 0$, not $v \ge v_e$ at every height.

8.7 Orbital Velocity, Time Period, and Energy

Orbital Velocity (Circular Orbit)

For a satellite at distance $r$ from Earth's centre,

$\frac{GMm}{r^2} = \frac{mv_o^2}{r} \Rightarrow v_o = \sqrt{\frac{GM}{r}}$

Just above the surface ($r \approx R$): $v_o = \sqrt{gR} \approx 7.9\,\text{km/s}$ — first cosmic velocity.

Time Period

$T = \frac{2\pi r}{v_o} = 2\pi\sqrt{\frac{r^3}{GM}}$

Kinetic Energy

$K = \tfrac{1}{2}mv_o^2 = \frac{GMm}{2r}$

Potential Energy

$U = -\frac{GMm}{r}$

Total Energy

$E = K + U = -\frac{GMm}{2r}$

Notice $E < 0$: the satellite is bound. Also $E = -K = U/2$ — a special property of inverse-square orbits (the virial theorem).

Binding Energy

The energy required to liberate the satellite (send it to infinity):

$E_b = |E| = \frac{GMm}{2r}$

Worked Example

A satellite at altitude $h = R$ (i.e., $r = 2R$). Find $v_o, T, E$.

$v_o = \sqrt{GM/(2R)} = \sqrt{gR/2} \approx 5.6\,\text{km/s}$.

$T = 2\pi\sqrt{(2R)^3/(GM)} = 2\pi\sqrt{8R/g}\approx 2.4\,\text{hr}$.

$E = -GMm/(4R) = -mgR/4$.

Common Mistakes

• Confusing orbital velocity with escape velocity.
• Forgetting that $r$ is measured from the Earth's centre, not its surface.

8.8 Geostationary and Polar Satellites

Geostationary Satellite

A satellite is geostationary if it appears motionless from a point on Earth — that is, it shares the Earth's rotation. Requirements:

1. Orbital period = $24\,\text{hr}$ = $86400\,\text{s}$.
2. Orbit in the equatorial plane.
3. Direction of revolution same as Earth's rotation.

From $T = 2\pi\sqrt{r^3/(GM)}$:

$r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$

Numerically, $r \approx 4.22\times 10^7\,\text{m}$, i.e., altitude $\approx 36000\,\text{km}$.

Polar Satellite

Orbits in a plane containing the Earth's poles. As the Earth rotates beneath, the satellite scans different longitudes — useful for weather and reconnaissance. Typically low-Earth-orbit ($500$–$800\,\text{km}$ altitude) with periods of ~$100$ min.

Worked Example

What is the orbital speed of a geostationary satellite?

$v_o = 2\pi r/T = 2\pi\cdot 4.22\times 10^7/86400 \approx 3.07\times 10^3\,\text{m/s}$ — much slower than low-orbit satellites.

Common Mistakes

• Calling any satellite at $36000\,\text{km}$ geostationary (it must also be equatorial and prograde).
• Confusing geosynchronous (same period) with geostationary (same period and equatorial and prograde).

8.9 Weightlessness in Satellites

Explanation

An astronaut in an orbiting satellite is in free fall — both the satellite and the astronaut accelerate toward the Earth at $g(r) = GM/r^2$. In the satellite's (non-inertial) frame, the pseudo (centrifugal) force exactly cancels gravity along the orbital direction, so the astronaut floats.

More carefully: the astronaut's net acceleration toward Earth equals the satellite's. There is no normal contact force between astronaut and floor — hence "weightlessness."

This is not zero gravity. At low orbit, $g \approx 8\,\text{m/s}^2$, very close to surface gravity. It is the absence of normal force (apparent weight = $N$) that produces the floating sensation.

Other Examples of Weightlessness

• Freely falling lift.
• Top of a projectile arc (instantaneously).
• Inside a parabolic-flight "vomit comet."

Worked Example

Astronaut on the ISS at $h = 400\,\text{km}$. What is $g$ there?

$g = 9.8\left(\frac{6400}{6800}\right)^2 \approx 9.8\cdot 0.886 \approx 8.68\,\text{m/s}^2$

Not "zero gravity" — the ISS is in free fall.

Common Mistakes

• Believing weightlessness means no gravity.
• Confusing apparent weight with actual gravitational force.

Solved Problems

Problem 1

The mass of the Earth is $6\times 10^{24}\,\text{kg}$, radius $6.4\times 10^6\,\text{m}$. Find $g$ at the surface.

$g = \frac{GM}{R^2} = \frac{6.67\times 10^{-11}\cdot 6\times 10^{24}}{(6.4\times 10^6)^2} \approx 9.77\,\text{m/s}^2$

Problem 2

At what altitude is $g$ reduced to $g/4$?

$g/4 = g\cdot \frac{R^2}{(R+h)^2} \Rightarrow R+h = 2R \Rightarrow h = R = 6400\,\text{km}$

Problem 3

Escape velocity from the surface of a planet of mass $4M_E$ and radius $2R_E$?

$v_e = \sqrt{\frac{2G\cdot 4M_E}{2R_E}} = \sqrt{2}\cdot v_e^{(E)} \approx 15.84\,\text{km/s}$

Problem 4

A satellite revolves at altitude $R$ (so $r = 2R$). Find its period in terms of $T_0$ (period of surface satellite).

$\frac{T}{T_0} = \left(\frac{r}{R}\right)^{3/2} = 2^{3/2} = 2.83$

If $T_0 \approx 84.4\,\text{min}$, then $T \approx 238\,\text{min}$.

Problem 5

A satellite of $1000\,\text{kg}$ moves in a circular orbit at altitude $R$ (so $r = 2R$). Find its total energy.

$E = -\frac{GMm}{2r} = -\frac{GMm}{4R} = -\frac{mgR}{4}$

With $m = 1000$, $g = 9.8$, $R = 6.4\times 10^6$:

$E = -1000\cdot 9.8\cdot 6.4\times 10^6/4 \approx -1.57\times 10^{10}\,\text{J}$

Problem 6

If the radius of the Earth shrinks by 1% while mass remains constant, how does $g$ change?

$g \propto 1/R^2$, so $\Delta g/g = -2\Delta R/R = +2\%$. $g$ increases by 2%.

Problem 7

Two satellites $A$ (radius $R$) and $B$ (radius $4R$) of same mass orbit the Earth. Ratio of their kinetic energies?

$K = GMm/(2r) \Rightarrow K_A/K_B = r_B/r_A = 4$.

JEE/NEET Edge Cases

1. Gravitational field inside a uniform sphere varies linearly with distance from the centre ($g(r) = gr/R$), while inside a hollow shell it is zero.
2. Two-body problem: the planet and the Sun orbit their common CM. For Earth–Sun, the Sun's wobble is tiny but in close binary stars it's appreciable.
3. Energy to launch a satellite: includes both $\Delta U$ to raise it and $K$ to give it orbital speed.
4. Lagrange points (advanced): five special points where small bodies orbit with the same period as Earth–Moon — relevant for telescopes (James Webb at L2).
5. Slingshot maneuvers: spacecraft gain energy in the heliocentric frame by passing close to a planet (a clever use of gravitational scattering — like elastic collision).
6. Geosynchronous vs geostationary: only the latter requires equatorial prograde orbit; geosynchronous orbits can be inclined.

Quick Recap

• Kepler's laws: ellipse, equal areas, $T^2 \propto a^3$.
• $F = GMm/r^2$; gravity is universal, inverse-square.
• Shell theorem: spherical body acts as point mass externally; field is zero inside hollow shell.
• $g$ varies with altitude ($g(1-2h/R)$), depth ($g(1-d/R)$), latitude (due to rotation), shape.
• $U(r) = -GMm/r$ (general); $U = mgh$ (near surface).
• Escape velocity: $v_e = \sqrt{2gR} \approx 11.2\,\text{km/s}$.
• Orbital velocity: $v_o = \sqrt{GM/r}$; period: $T = 2\pi\sqrt{r^3/GM}$.
• Total energy of orbit: $E = -GMm/(2r) < 0$ (bound).
• Weightlessness in satellite = free fall, not absence of gravity.

Formula Sheet