Apply Newton's law of gravitation and the concept of gravitational field strength, derive orbital relationships, and account for geostationary orbits and Kepler's third law

What this dot point is asking

SEAB wants you to apply Newton's law of gravitation, work with gravitational field strength and potential, combine gravitation with circular motion to derive orbital speed and period, justify Kepler's third law, and explain the special case of geostationary orbits. This thread also previews the parallel structure of electric fields later in the course.

The answer

Newton's law of gravitation

Two point masses $M$ and $m$ separated by distance $r$ attract each other with a force:

where $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ is the gravitational constant. The force is attractive, acts along the line joining the masses, and follows an inverse-square law.

Gravitational field strength

The gravitational field strength at a point is the force per unit mass:

with units $\text{N kg}^{-1}$. Near a planet's surface this is the familiar acceleration of free fall. It is a vector directed toward the mass producing the field.

Gravitational potential

The gravitational potential at a point is the work done per unit mass to bring a small mass from infinity to that point:

It is negative because gravity is attractive and the reference (zero) is taken at infinity. The gravitational potential energy of a mass $m$ is $E_p = m\phi = -\dfrac{GMm}{r}$.

Orbital motion

For a circular orbit, gravity provides the centripetal force:

The orbital period follows from $v = \dfrac{2\pi r}{T}$:

Kepler's third law

The relation $T^2 \propto r^3$ is Kepler's third law, here derived from Newtonian gravity. It applies to all bodies orbiting the same central mass and lets you compare orbits without knowing $G$ or $M$ directly.

Geostationary orbits

A geostationary satellite stays fixed above a point on the equator. This requires three conditions: a period of one sidereal day (about $24$ hours), an orbit in the equatorial plane, and motion from west to east. Because the period is fixed, Kepler's third law fixes the radius at a single value of about $4.2 \times 10^7\ \text{m}$ from the Earth's centre.

Examples in context

Example 1. Weighing the Earth. Knowing $g = 9.81\ \text{N kg}^{-1}$ at the surface and the radius $R_E$, the relation $g = \dfrac{GM_E}{R_E^2}$ rearranges to $M_E = \dfrac{g R_E^2}{G} = 5.97 \times 10^{24}\ \text{kg}$. This is how the Earth's mass is determined without ever placing it on a scale.

Example 2. Comparing planetary orbits. Kepler's third law lets you compare two planets orbiting the Sun without knowing $G$ or the Sun's mass: $\dfrac{T_1^2}{r_1^3} = \dfrac{T_2^2}{r_2^3}$. Given Earth's year and orbital radius, the period of any other planet follows from its orbital radius alone.

Try this

Q1. State Newton's law of gravitation and define each symbol. [2 marks]

• Cue. $F = \dfrac{GMm}{r^2}$: $F$ the attractive force, $G$ the gravitational constant, $M$ and $m$ the masses, $r$ the separation of their centres.

Q2. Show that the orbital speed of a satellite is independent of its own mass. [2 marks]

• Cue. $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$; the satellite mass $m$ cancels, leaving $v = \sqrt{\dfrac{GM}{r}}$.

Q3. Explain why a geostationary satellite has only one possible orbital radius. [3 marks]

• Cue. Its period is fixed at one sidereal day; by $T^2 \propto r^3$, a fixed period gives a single $r$ (about $4.2 \times 10^7\ \text{m}$), with the orbit also required to be equatorial and eastward.