Kepler's laws

In a nutshell

Masses in orbit can be assumed to undergo circular motion. Using this assumption, the orbital velocities and periods can be calculated from the mass of the object creating the gravitational field. Kepler's laws relate to planetary motion. The three laws are:

Law	Description
First	All planets move in an ellipse with the Sun at one foci.
Second	For a line drawn from the Sun to a planet, the planet will sweep out equal areas in an equal amount of time.
Third	For a planet orbiting a star, the square of the time period of the planet is proportional to the cube of the mean distance of the orbit from the star. In other words, $T^2 \propto r^3$ .

Equations

Description	Equation
Newton's law of gravitation	$F=(-)\frac {GMm}{r^2}$
Centripetal force	$F=\frac {mv^2}{r}$
Orbital velocity	$v=\sqrt {\frac{GM}{r}}$
Kepler's third law	$T^2=(\frac {4 \pi^2 }{GM}) r^3$

Constants

name	symbol	value
$Gravitational \ constant$	$G$	$6.67 \times 10^{-11} \, N\,m^2\,kg^{-2}$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI BASE Units
$gravitational \ force$	$F$	$N$	$kg\,m\,s^{-2}$
$larger \ point \ mass$	$M$	$kg$	$kg$
$smaller \ point \ mass$	$m$	$kg$	$kg$
$distance \ between \ centres \ of \ masses$	$r$	$m$	$m$
$orbital \ speed$	$v$	$ms^{-1}$	$ms^{-1}$
$orbital \ period$	$T$	$s$	$s$

Motion of masses

Object's in orbit can be assumed to be undergoing circular motion. Satellites are objects which orbit a larger mass.

Natural satellites are planets, asteroids and comets orbiting stars or moons orbiting planets. Artificial satellites are man made objects which are created to orbit another object.

Orbital speed

An object in orbit is undergoing circular motion. Therefore, the equations for circular motion can be used.

Physics; Gravitational Fields; KS5 Year 12; Kepler's laws

The moon is undergoing circular motion as it orbits the Earth:

$F=\frac {mv^2}{r}$

The centripetal force is provided by the Earth's gravitational force:

$F=\frac {GMm}{r^2}$

Equating the two equations:

$\frac {GMm}{r^2}=\frac {mv^2}{r} \newline \\[0.1in]\frac {GM}{r}= {v^2}$

Taking square roots to obtain $v$ :

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

Note: This equation gives the orbital speed of an object around a point mass $M$ at a distance of $r$ from the centre.

Orbital period

The orbital period of a satellite can also be calculated. Again, assuming circular motion, the speed is given by the orbital speed equation and the distance is equal to one circumference $2\pi r$ :

$speed=\frac {distance}{time} \rightarrow time =\frac {distance}{speed}$

Substituting the equation for orbital speed and distance into the equation to obtain the time period $T$ :

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

Squaring everything to remove the square root:

$T^2=\frac {4 \pi^2 r^2}{\frac {GM}{r}} \newline \\[0.1in]T^2=\frac {4 \pi^2 r^3}{GM} \newline \\[0.1in]T^2=(\frac {4 \pi^2 }{GM}) r^3$

Example

The moon orbits the Earth once every $27.3$ days. Calculate the distance of the orbit.

$M_{Earth}=5.97\times 10^{24} kg$

Firstly, write down the known values:

$T=27.3 \ days = 27.3 \times (60 \times 60 \times 24) =2\,358\,720 \, s \newline \\[0.1in]M_{Earth}=5.97\times 10^{24} \, kg$

Next, write down the equations needed. Rearrange the equation by dividing by $(\frac {4 \pi^2 }{GM})$ and cube rooting both sides:

$T^2=(\frac {4 \pi^2 }{GM}) r^3 \rightarrow r=\sqrt[3] {T^2( \frac {GM}{4 \pi^2 } )}$

Then, substitute the values into the equation:

$r=\sqrt[3] {(2\,358\,720)^2( \frac {6.67\times 10^{-11}\times 5.97\times 10^{24}}{4 \pi^2 } )} \newline \\[0.1in]r=382\,852\,131 \ m$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$orbital \ distance , \ r=3.83\times 10^8 \, m$

The orbital distance of the moon is $\underline{3.83\times 10^8 \, m}.$ 

Geostationary satellites

Geostationary satellites are a special category of artificial satellite which stay above the Earth in one fixed place.

The conditions for a geostationary satellite are:

The orbit has to be above the equator.
The time period (and orbital distance) are fixed with a time period of $24 \, h$ .

Curiosity: Geostationary satellites are very good at sending communication signals as the satellite never moves. You may have seen sky satellite dishes always pointing in the same direction and don't move. That is because the satellite it is communicating to doesn't move!

Kepler's laws

Kepler was an astronomer who played a key role in physics during the 17th century. Some of his works were his laws of planetary motion, now referred to as Kepler's laws.

There are three which you need to know.

Kepler's first law

Kepler's first law states that each planet moves around the Sun in an ellipse with the Sun at one foci. There are two foci when a planet orbits the Sun. The further apart the foci, the more elliptic the orbit.

Physics; Gravitational Fields; KS5 Year 12; Kepler's laws

Curiosity: Ellipses have the property of eccentricity. The greater the eccentricity the more oval shaped the orbit will be up to a value of $1$ . A circle has an eccentricity of $0$ .

Kepler's second law

Kepler's second law is sometimes referred to as the equal distance-equal time law. A line drawn from a star to a planet, will sweep out an equal area in an equal amount of time.

Physics; Gravitational Fields; KS5 Year 12; Kepler's laws

This is because when the planet is close to the star, it will be travelling at its quickest, so will move round its orbit quicker, but as it gets closer, not as much area is covered.

When the planet is furthest away, the planet is moving the slowest but as the distance is greater, the area of the triangle is still the same amount!

Curiosity: The technical word for the planet being closest to the star is called the perihelion and when it is furthest away is called the aphelion.

Kepler's third law

Kepler's third law relates the time period of an orbit to the mean distance of an orbit.

$T^2 \propto r^3$

So for a planet orbiting a star, the square of the time period of the planet is proportional to the cube of the mean distance of the orbit from the star.

Note: The first section of this summary shows the derivation of Kepler's third law and the constant of proportionality is given as $\frac {4 \pi^2 }{GM}$ .

Example

A star is found to have two exoplanets orbiting. The first planet $A$ has a time period of $89 \ days$ and an orbital distance of $8.46 \times 10^5 \, km$ . If planet $B$ has an orbital distance of $4.91 \times 10^5 \, km$ , calculate the time period of planet B to the nearest whole day.

Firstly, write down the known values:

$T_A=89 \, days \newline \\[0.1in]r_A=8.46 \times 10^5 \, km \newline \\[0.1in]r_B=4.91 \times 10^5 \, km$

Next, write down the equations needed and rearrange if necessary:

$T^2 \propto r^3 \newline \\[0.1in]\frac {T^2}{r^3}=constant \newline \\[0.1in]\frac {T_A^2}{r_A^3}=\frac {T_B^2}{r_B^3} \newline \\[0.1in]T_B=\sqrt{\frac {T_A^2 \times r_B^3}{r_A^3}}$

Then, substitute the values into the equation:

$T_B=\sqrt{\frac {89^2 \times (4.91 \times 10^5)^3}{(8.46 \times 10^5)^3}}\newline \\[0.1in]T_B=39.351.... \ days$

Note: As all the units are constant throughout in this equation, there is no need to convert into SI units as the question is asking for the time period in days.

Make sure to include units and round to the nearest day as asked for in the question:

$orbital \ period \ of \ planet \ B,\ T_B=39 \, days$

The orbital period of planet B is $\underline{39 \, days}.$