Planetary motion

What you'll learn

Kepler’s three laws and what they say about real planetary orbits.
Why gravity provides the centripetal force for an orbit.
How to derive and use $T2=(4π2/GM)r3T^2 = (4\pi^2/GM)r^3$ .
What makes a satellite geostationary, and why that is useful.

The big picture

Planetary motion is a beautiful link between observation and theory. Kepler described patterns in planetary data. Newton then explained those patterns using gravitational force and circular motion.

For OCR A-Level Physics, you mainly use a circular orbit model. Real planetary orbits are often slightly elliptical, but the circular model is powerful and gives the key equation used for planets, moons, satellites, and exoplanets.

Kepler’s three laws

Definition

Orbit and orbital period

An orbit is the path followed by one body moving around another due to gravity. The orbital period, $T$ , is the time taken for one complete orbit, measured in seconds (s).

Kepler’s laws are empirical laws: they were found from astronomical observations before Newton’s theory of gravitation explained them.

Diagram showing Kepler’s three laws: elliptical orbit, equal areas in equal times, and circular-orbit approximation

First law: law of orbits

Planets move in elliptical orbits with the Sun at one focus of the ellipse.

Definition

Ellipse and focus

An ellipse is a stretched circle. A focus is one of two special points inside the ellipse; for a planet orbiting the Sun, the Sun is at one focus, not usually at the centre.

For many A-Level calculations, the orbit is treated as circular, so the orbital radius $r$ is constant.

Second law: law of areas

A line from the Sun to the planet sweeps out equal areas in equal times.

This means a planet moves faster when it is closer to the Sun and slower when it is further away.

Example

Explaining why a planet moves faster near the Sun

In equal time intervals, Kepler’s second law says the swept-out areas are equal.
Near the Sun, the planet is at a smaller distance from the Sun, so the same area must be swept out using a larger angle and a longer part of the orbit.
Since the planet covers a larger distance in the same time near the Sun, its speed is greater there.

Third law: law of periods

For planets orbiting the same central body, the square of the orbital period is proportional to the cube of the orbital size:

T2∝r3T^2 \propto r^3

For elliptical orbits, the more precise orbital size is the semi-major axis. In the circular orbit model used for the derivation, this becomes the orbital radius $r`.

Key Idea

Kepler’s third law

For objects orbiting the same central mass, larger orbits have longer periods, and the relationship is not linear: $T2∝r3T^2 \propto r^3$ .

Common Mistake

Circular model

The derivation you need assumes a circular orbit of radius $r$ . Do not try to derive the full elliptical version; that is beyond the required specification.

Gravity provides the centripetal force

An object in circular motion is constantly changing direction, so it must have an acceleration towards the centre of the circle.

Definition

Centripetal force

A centripetal force is the resultant force directed towards the centre of a circular path. It is not a new type of force; it is the role played by an actual force, such as gravity, tension, or friction.

For a planet orbiting the Sun, the gravitational force between the planet and the Sun acts towards the Sun. This inward gravitational force provides the required centripetal force.

Circular orbit diagram showing gravitational force towards the central mass acting as centripetal force

For a planet of mass $m$ orbiting a star of mass $M$ :

\frac{GMm}{r^2}

For circular motion:

\frac{mv^2}{r}

So, for a stable circular orbit:

GMmr2=mv2r\frac{GMm}{r^2} = \frac{mv^2}{r}

Common Mistake

Inventing an extra force

Do not draw both “gravity” and “centripetal force” as separate forces on the planet. Gravity is the centripetal force in this situation.

Deriving the orbital period equation

The specification expects you to derive:

T2=(4π2GM)r3T^2 = \left(\frac{4\pi^2}{GM}\right)r^3

where:

$T$ is the orbital period in seconds (s)
$G$ is the gravitational constant
$M$ is the mass of the central body in kilograms (kg)
$r$ is the orbital radius measured from centre to centre in metres (m)

Example

Deriving the orbital period equation

Equate gravitational force to centripetal force because gravity provides the inward resultant force:
$GMmr2=mv2r\frac{GMm}{r^2} = \frac{mv^2}{r}$
Cancel the orbiting mass $m ‘ an d o n e f a c t or o f$ r`:
$v2=GMrv^2 = \frac{GM}{r}$
For one circular orbit, the distance travelled is the circumference $2\pi r`, so the orbital speed is:
$\frac{2\pi r}{T}$
Substitute this into $v^2 = GM/r`:
$(2πrT)2=GMr\left(\frac{2\pi r}{T}\right)^2 = \frac{GM}{r}$
Rearrange:
$4π2r2T2=GMr\frac{4\pi^2r^2}{T^2} = \frac{GM}{r}$ $T2=4π2r3GMT^2 = \frac{4\pi^2r^3}{GM}$

Tip

Why the planet’s mass cancels

In this model, the period of a circular orbit does not depend on the mass of the orbiting object. A small satellite and a large satellite at the same orbital radius around Earth have the same orbital period.

Using $T2=(4π2/GM)r3T^2 = (4\pi^2/GM)r^3$

You can calculate an orbital period if you know the central mass and orbital radius.

Example

Calculating the period of a planet

Mars orbits the Sun at an approximate radius of $m2.28\times10^{11}\ \text{m}$ . The mass of the Sun is $kg1.99\times10^{30}\ \text{kg}$ . Calculate the orbital period of Mars.

Choose the orbital period equation and rearrange by square rooting:
$\sqrt{\frac{4\pi^2r^3}{GM}}$
Substitute the values, using $6.67\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}$ :
$\sqrt{\frac{4\pi^2(2.28\times10^{11}\ \text{m})^3}{(6.67\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2})(1.99\times10^{30}\ \text{kg})}}$
Calculate the period:
$5.94\times10^7\ \text{s}$
Convert to days for interpretation:
$\frac{5.94\times10^7\ \text{s}}{86400\ \text{s day}^{-1}} = 688\ \text{days}$

So the orbital period is about $5.94\times10^7\ \text{s}`, or 688 days.

Common Mistake

Using the wrong distance

In orbit equations, $r$ is measured from the centre of the central body to the centre of the orbiting body. For a satellite, this is not the height above the surface.

Applying Kepler’s third law beyond the Solar System

Kepler’s third law is not just for planets around the Sun. It applies to any system where a smaller object orbits a much larger central mass under gravity, such as:

moons orbiting a planet
artificial satellites orbiting Earth
exoplanets orbiting another star

For objects orbiting the same central mass:

T2r3=4π2GM\frac{T^2}{r^3} = \frac{4\pi^2}{GM}

The constant depends on the central mass $M`.

Example

Comparing two exoplanets around the same star

An exoplanet has orbital radius $m2.00\times10^{10}\ \text{m}$ and period 20.0 days. A second exoplanet orbits the same star at radius $m5.00\times10^{10}\ \text{m}$ . Estimate the second planet’s period.

Since both planets orbit the same star, use the ratio form of $T^2 \propto r^3`:
$(T2T1)2=(r2r1)3\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$
Square root both sides:
$T2T1=(r2r1)3/2\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}$
Substitute the radii:
$m)3/2T_2 = 20.0\ \text{days}\left(\frac{5.00\times10^{10}\ \text{m}}{2.00\times10^{10}\ \text{m}}\right)^{3/2}$
Calculate:
$daysT_2 = 20.0\ \text{days}(2.50)^{3/2} = 79.1\ \text{days}$

The second exoplanet has an orbital period of about 79.1 days.

Common Mistake

Comparing different stars

You can only use the simple ratio $T2∝r3T^2 \propto r^3$ directly if the objects orbit the same central mass. Different stars have different masses, so the constant changes.

Geostationary orbit

Definition

Geostationary orbit

A geostationary satellite is a satellite that remains above the same point on Earth’s equator. It must have a circular orbit in the equatorial plane, travel in the same direction as Earth’s rotation, and have the same angular speed as Earth.

Its orbital period is approximately 24 h. More precisely, it matches Earth’s rotation period relative to the stars, but 24 h is commonly used unless the question gives a more precise value.

Geostationary satellite above Earth’s equator, showing orbital radius and altitude

Uses of geostationary satellites

Geostationary satellites are useful because ground receivers can point at a fixed position in the sky. Uses include:

satellite television and radio broadcasting
telephone and internet communications
weather monitoring of the same region over time

Key Idea

Why geostationary satellites are predictable

Newtonian gravity and circular motion predict one particular orbital radius for a satellite with a 24 h period around Earth. The satellite’s mass does not affect this radius.

Example

Calculating the height of a geostationary satellite

Calculate the approximate altitude of a geostationary satellite. Use Earth’s mass $5.97\times10^{24}\ \text{kg}`, Earth’s radius$ R = 6.37\times10^6\ \text{m}, and $T = 24.0\ \text{h}.

Convert the period into seconds:
$\times 3600\ \text{s} = 86400\ \text{s}$
Rearrange the orbital period equation for $r`:
$\left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$
Substitute the values:
$\left(\frac{(6.67\times10^{-11})(5.97\times10^{24}\ \text{kg})(86400\ \text{s})^2}{4\pi^2}\right)^{1/3}$
Calculate the orbital radius from Earth’s centre:
$4.22\times10^7\ \text{m}$
Subtract Earth’s radius to find the altitude above the surface:
$4.22\times10^7\ \text{m} - 6.37\times10^6\ \text{m}$ $3.58\times10^7\ \text{m}$

The satellite’s altitude is about $3.58\times10^7\ \text{m}`, or 35 800 km.

Common Mistake

Period alone is not enough

A 24 h orbit is not automatically geostationary. The orbit must also be circular, above the equator, and in the same direction as Earth’s rotation.

Exam technique

In the exam

For orbit calculations, check whether $r$ is an orbital radius from the centre or an altitude above the surface.
When deriving $T2=(4π2/GM)r3T^2 = (4\pi^2/GM)r^3$ , start by equating $GMmr2\frac{GMm}{r^2}$ and $mv2r\frac{mv^2}{r}$ , then use $v = 2\pi r/T`.
For geostationary satellites, state all three conditions: 24 h period, equatorial circular orbit, and same direction as Earth’s rotation.

Self review

Check yourself

Why does the mass of the orbiting planet or satellite cancel in the derivation?
What is the difference between orbital radius and altitude?
Why can’t a satellite in a tilted 24 h orbit be geostationary?

What you'll learn

Kepler’s three laws and what they say about real planetary orbits.
Why gravity provides the centripetal force for an orbit.
How to derive and use $T2=(4π2/GM)r3T^2 = (4\pi^2/GM)r^3$ .
What makes a satellite geostationary, and why that is useful.

The big picture

Planetary motion is a beautiful link between observation and theory. Kepler described patterns in planetary data. Newton then explained those patterns using gravitational force and circular motion.

Kepler’s three laws

Definition

Orbit and orbital period

An orbit is the path followed by one body moving around another due to gravity. The orbital period, $T$ , is the time taken for one complete orbit, measured in seconds (s).

Kepler’s laws are empirical laws: they were found from astronomical observations before Newton’s theory of gravitation explained them.

Diagram showing Kepler’s three laws: elliptical orbit, equal areas in equal times, and circular-orbit approximation