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Topic 14 of 19

Universal Gravitation

Newton's law of universal gravitation unifies terrestrial and celestial mechanics under one equation: every mass attracts every other with a force proportional to their masses and inversely proportional to the square of their separation. This single law explains free fall, tides, planetary orbits, and the structure of the Solar System.

Key Concepts

Newton's Law of Gravitation

Every pair of masses

m_1

and

m_2

attract each other with force

F_g = Gm_1m_2/r^2

, where

G = 6.674\times10^{-11}

N·m²/kg² and

r

is the center-to-center distance.

Gravitational Field

The gravitational field

\vec{g}

at a point is the force per unit mass felt by a test particle there:

\vec{g} = \vec{F}_g/m = -GM\hat{r}/r^2

. At Earth's surface,

g \approx 9.8

m/s².

Gravitational Potential Energy (general)

U = -Gm_1m_2/r

. Negative because gravity is attractive. Zero at infinity. Near Earth's surface:

U \approx mgh

(when

r \gg \Delta r

Escape Velocity

Minimum launch speed to escape a planet's gravity:

v_\text{esc} = \sqrt{2GM/R}

(from the surface). Derived from

K + U = 0

at launch.

Kepler's Laws

(1) Orbits are ellipses with the star at one focus. (2) A line from star to planet sweeps equal areas in equal times (consequence of angular momentum conservation). (3)

T^2 \propto a^3

where

a

is the semi-major axis.

Circular Orbits

For circular orbit of radius

r

: gravitational force provides centripetal force. Orbital speed

v = \sqrt{GM/r}

and period

T = 2\pi\sqrt{r^3/GM}

Key Equations

Gravitational force

F_g = \frac{Gm_1m_2}{r^2}

Attractive, along the line joining the centers. G = 6.674×10⁻¹¹ N·m²/kg².

Orbital speed (circular orbit)

v = \sqrt{\frac{GM}{r}}

Speed required to maintain a circular orbit of radius r around mass M.

Kepler's Third Law

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

Period squared proportional to orbital radius cubed. For elliptical orbits, replace r with semi-major axis a.

Escape velocity

v_\text{esc} = \sqrt{\frac{2GM}{R}}

Minimum speed to escape to infinity from the surface of a body of mass M and radius R.

Gravitational PE (general)

U = -\frac{Gm_1m_2}{r}

Zero at r → ∞. More negative = more tightly bound. Near-surface: U ≈ mgh.

Worked Example

Orbital Period of the International Space Station

Problem

The ISS orbits at altitude 408 km above Earth's surface. Find its orbital speed and period. ( $M_E = 5.97\times10^{24}$ kg, $R_E = 6.37\times10^6$ m.)

Solution

Orbital radius: $r = R_E + h = 6.37\times10^6 + 4.08\times10^5 = 6.778\times10^6$ m.

Orbital speed:

v = \sqrt{\frac{GM_E}{r}} = \sqrt{\frac{6.674\times10^{-11}\times5.97\times10^{24}}{6.778\times10^6}} \approx 7660 \text{ m/s}

Period:

T = \frac{2\pi r}{v} = \frac{2\pi\times6.778\times10^6}{7660} \approx 5560 \text{ s} \approx 92.7 \text{ min}

Answer ISS orbital speed ≈ 7.66 km/s; period ≈ 92.7 minutes.

Practice

Exercises

7 problems

Free preview

2 free · 5 Pro

Exercise 1 / 7 Free

+10 XP

Gravity obeys the inverse-square law: F ∝ 1/r². At r = 1 m, F = 16 N. Find F at r = 2 m. Use: F₂ = F₁(r₁/r₂)².

F at r=2m = N

Exercise 2 / 7 Free

+10 XP

A satellite orbits at r = 4×10⁶ m from a planet of mass M = 6×10²⁴ kg. Find the orbital speed using v = √(GM/r).

v = m/s

3 of 7

A satellite orbits Earth in a circular orbit at an altitude of $h = 800$ km above the surface. What is its orbital speed (in m/s)? Use $GM_E = 3.986\times10^{14}$ m³/s², $R_E = 6.37\times10^6$ m.

Unlock Exercise 3