Classical Mechanics · Chapter 6

Gravitation

Newton's universal law of gravity, Kepler's laws, and orbital mechanics in three dimensions.

Prerequisites

Newton's laws \cdot Circular motion \cdot Energy conservation \cdot Basic calculus

Learning Goals

Apply Newton's law of universal gravitation to calculate the gravitational force between two masses.
Derive escape velocity from conservation of energy and calculate it for Earth.
State and apply Kepler's three laws to describe orbital shape, speed variation, and period.
Use Kepler's third law to find orbital periods and semi-major axes of planetary bodies.
Relate orbital speed, energy, and radius using the vis-viva equation.

6.1 Newton's Law of Universal Gravitation

In 1687, Newton proposed that every pair of massive objects attracts each other with a force proportional to their masses and inversely proportional to the square of the distance between them. This was a radical unification — the same force that pulls an apple to Earth governs planetary orbits.

F = G\frac{m_1m_2}{r^2} \qquad G = 6.674\times10^{-11}\,\mathrm{N\,m^2/kg^2}

(6.1)

Near Earth's surface (r ≈ R_E), this reduces to F = mg where g = GM_E/R_E² = 9.81 m/s². At the Moon's distance (r = 60 R_E), g drops by a factor of 3600.

Definition 6.1 — Gravitational Potential Energy

For objects separated by arbitrary distances, gravitational PE is:

U(r) = -Gm_{1}m_{2} / r

This is negative (gravity is attractive and doe

s positive work as objects approach). The escape velocity is the speed needed to reach r \to

Example 6.1 — Escape Velocity from Earth

$Find Earth's escape velocity. (M_{E} = 5.97 \times 10^{24} kg, R_{E} = 6.37 \times 10^{6} m$

Set KE = |PE|:½

mv^{2} = GMm/R \to v = \sqrt(2GM/R

Calculate:

v = \sqrt(2 \times 6.674\times10^{-11} \times 5.97\times10^{24} / 6.37\times10^{6}) = \sqrt(125.1\times10^{6}

Result:

v_{esc} =

11.2 km/s

\approx 40,000 km/h

6.2 Kepler's Three Laws

Johannes Kepler (1609–1619) derived three empirical laws from Tycho Brahe's astronomical observations. Newton later showed they follow directly from the inverse-square gravity law.

Theorem 6.1 — Kepler's First Law — Elliptical Orbits

Every planet orbits the Sun in an ellipse, with the Sun at one focus. An ellipse has two parameters:

Semi-major axis a — half the longest diameter
Eccentricity e — $shape parameter (0 = circle, 1 = parabola$

r(\theta) = a(1 - e^{2}) / (1 + e cos \theta) \qquad (polar equation of orbit

Theorem 6.2 — Kepler's Second Law — Equal Areas

A line segment joining a planet and the Sun sweeps o

ut equal areas in equal times. This is equivalent to conservation of angular momentum: L =

faster near perihelion (closest approach) andslower near aphelion (farthest point).

Theorem 6.3 — Kepler's Third Law — Orbital Period

The square of the orbital period T is proportional to the cube of the semi-major axis a:

T^{2} = (4\pi^{2}/GM) \cdot a^{3} \qquad or \qquad T^{2} \propto a^{3}

Earth: a = 1 AU, T = 1 year. Jupiter: a = 5.2 AU \to T = 5.2^{3/2} = 11.9 years

6.3 Orbital Simulation — 3D

The simulation below shows a planet orbiting a star using Kepler's equations of motion, solved via the eccentric anomaly. The velocity arrow (gold) grows near perihelion where orbital speed peaks, following the vis-viva equation:

v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right) \qquad \text{(vis-viva)}

(6.2)

Notice that inclination tilts the orbit out of the ecliptic plane. Real planetary orbits have inclinations of 0°–7° relative to Earth's orbital plane; comets can be inclined up to 90°+.

Loading 3D simulation…

Figure 6.1.

3D orbital simulation. Eccentricity controls shape (0 = circle, 0.95 = elongated ellipse

. Inclination tilts the orbit plane. The gold arrow is the velocity vector — watch it lengthen near perihelion.

Example 6.2 — Period of a Geostationary Satellite

$Find the altitude of a geostationary orbit (T = 24 h = 86,400 s). M_{E} = 5.97 \times 10^{24} kg$ .

Kepler's 3rd:

T^{2} = 4\pi^{2}a^{3}/GM_E \to a^{3} = GM_E T^{2}/(4\pi^{2}

Calculate:

a^{3} = (6.674\times10^{-11})(5.97\times10^{24})(86400)^{2} / (4\pi^{2}) = 7.54\times10^{22} m^{3}

Solve:

a = (7.54\times10^{22})^{1/3} = 4.22\times10^{7} m = 42,200 km

Altitude:

h = a - R_{E} = 42,200 - 6,370 =

35,830 km above Earth's surface

Definition 6.2 — Common Traps

Gravity is universal: both masses pull equally on each other, even when one acceleration is tiny.
Use center-to-center distance: orbital radius is measured from the central body's center, not its surface.
Weight is not mass: mass stays fixed; weight changes with local gravitational field strength.
Potential energy is negative for bound gravity: zero is chosen at infinite separation.
Kepler's third law uses semi-major axis: for ellipses, do not substitute perihelion or aphelion distance.

Exercises — 6.1–6.3 Gravitation

Calculate the escape velocity from Earth's surface.

km/s

Straightforward

Using Kepler's third law, find Mars's orbital period given its semi-major axis of 1.52 AU.

years

Straightforward

At what altitude above Earth (in terms of R_{E}) does gravitational acceleration equal g/4

? What is the actual g there?

Straightforward

4.Mars's orbital semi-major axis is 1.52 AU. Using Kepler's Third Law, find Mars's orbital period in Earth years.

Intermediate

A comet has orbital eccentricity e = 0.5. What is the ratio of its speed at perihelion to

its speed at aphelion?

Intermediate

Show that for a circular orbit, the total mechanical energy is E = -GMm/(2r). What is the

minimum additional kinetic energy needed to escape from that orbit?

Challenging

Key Takeaways

$Newton's gravity (F \propto 1/r^{2}) explains why planets follow ellipses — Kepler's laws are conse$ quences.
Kepler's 2nd law (equal areas) is conservation of angular momentum in disguise.
$Kepler's 3rd law (T^{2} \propto a^{3}) lets us calculate orbital periods from semi-major axis alone$ .
Orbital speed peaks at perihelion; the vis-viva equation gives speed at any point.
$Escape velocity from a body is v = \sqrt(2GM/R) — independent of escape direction$ .