Kepler's Laws of Planetary Motion

Introduction

Kepler's laws were based on emperical observations of planetary motions from the famously accurate astronomer Tycho Brae. Before 1605 it was thought that the orbits of planets around the sun were circular. Kepler was able to deduce from Brae's accurate observations that this was not the case. He then formulated the follwing mathematical laws to explain planetary motion.

Kepler's First Law

diagram of ellipse showing foci and axes

Kepler's first law states that objects in the solar system orbit the sun in an ellipse instead of a perfect circle with a certain amount of eccentricity. This eccentricity(e) determines the shape of the ellipse (the higher the value of e the longer the semi-major axis(a) and the shorter the semi-minor) . An ellipse can be defined as having 2 focal points with a distance f away from the centre of the ellipse.
We can use the equation:

$e=\frac{f}{a}$

to calculate the eccentricity, the parameters of an ellipse give us a good idea what the orbit of an object will look like.If we pick the focal point on the left(closest to A) to be the sun we can then name points A and B Perihelion(point of closest approach) and Aphelion(most distant point) respectively. We can find the orbital distance at these points using the formulae:

Kepler's Second Law

Kepler's second law states that the area of a sector defined by the arc of an objects orbit in a given period of time must be equal to the area within any sector for which the period of time is the same. This is due to the force upon the orbiting body due to gravity. As the object travels away from the centre of mass, gravity will be acting against its direction of travel hence slowing it down. When the object reaches the furthest point from the centre of mass(aphelion), the object will start to accelerate again as the force due to gravity is acting in the same direction as its travel. Thus the velocities at perihelion and aphelion can be related as below:

${\frac {v \left( { aph} \right) }{v \left( {per} \right) }}={ \frac {d \left( {per} \right) }{d \left( {aph} \right) }}$

Kepler's Third Law

If we equate the centripetal force required to keep an object in orbit and the gravitational force from the centre of mass we get:

$F(grav) = F(centripetal)$

${\frac {GMm}{d^2}}={\frac{mv^2}{d}}$

$v = \frac{2\pi d}{P}$

${\frac {GM}{d^2}}={\frac{4 \pi^2d}{P^2}}$

We can now rearrange to find P

$P^2={\frac{4 \pi^2}{GM}}d^3$

Kepler discovered through empirical observations that the period for an object orbiting in an eclipse is similar to that of a circular orbit but with the variable of a rather than d. Note: We can take $\frac{4 \pi^2}{GM}$ to be a constant when working in solar units(AU,years,Solar Masses)this simplifies to 1.
Thus we get the period of orbit for an ellipse to be:

$P^2={\frac{4 \pi^2}{GM}}a^3$

Example

Find the orbital period of a comet around the sun with a elliptical orbit with semi-major axis of 30AU? If when at its point of closest approach it is 0.5AU from the sun calculate its furthest distance from the sun and the force at this point that the suns gravitational field provides?

$P^2=(1)a^3$

$P = 30^{\frac{3}{2}}$

$P = 164years$

We can now calculate f as we know the distance at perihelion and the the semi-major axis a.

$f = a-d(per) = 30-0.5 = 29.5AU$

Now for the aphelion(farthest point)

$d(aph) = a+f = 30+29.5 = 59.5AU$

To give some comparison to the scale pluto's aphelion is about 50AU where the Earth is 1AU from the sun. Also the Earth's orbit is not very eccentric so aphelion and perihelion are rougly equal.
To be able to measure the force the sun acts on the comet at aphelion we need to know the Gravitational constant(G)($6.67x10^{-11}$),the mass of the Sun(M)($2x10^{30}kg$) and the value of 1AU(~$1.5x10^{11}m$) also we are assuming the mass of the comet is negliable compared to the Sun(this gets rid of (m)).

$F = {\frac{GM}{d^2}}$

$F = {\frac{6.67x10^{-11}x2x10^{30}}{(59.5x1.5x10^{11})^2}}$

$F = 1.67x10^{-6}N$