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Kepler and Gravitation

Kepler's Third Law

Problems for Kepler's Second Law

Problems for Kepler's Third Law

Statement of Kepler's Third Law

From observations collected over many centuries, and especially data compiled by the Danish astronomer Tycho Brahe, Kepler deduced a relationship between the orbital period and the radius of the orbit. Precisely:

the square of the period of an orbit is proportional to the cube of the semimajor axis length $a$.
Although Kepler never expressed the equation in this way, we can write down the constant of proportionality explicitly. In this form, Kepler's Third Law becomes the equation: \begin{equation} T^2 = \frac{4\pi^2 a^3}{GM} \end{equation} where $G$ is the Gravitational Constant that we shall encounter in Newton's Law, and $M$ is the mass about which the planet is rotating (usually the sun for our purposes). This relationship is extremely general and can be used to calculate rotational periods of binary star systems or the orbital periods of space shuttles around the earth.

A problem involving Kepler's Third Law

The orbit of Venus around the sun is roughly circular, with a period of 0.615 years. Suppose a large asteroid crashed into Venus, instantaneously decelerating its motion, such that it was thrown into an elliptical orbit with aphelion length equal to the radius of the old orbit, and with a smaller perihelion length equal to $98 \times 10^6$ kilometers. What is the period of this new orbit?

First we must calculate the radius of the original orbit: \begin{eqnarray*} r &=& \left(\frac{GM_sT^2}{4\pi^2}\right)^{1/3} \\ &=& \left(\frac{6.67\times 10^{-11}\times 1.989 \times 10^{30} \times (1.94 \times 10^7)^2}{4\pi^2}\right)^{1/3} \\ &=& 108 \times 10^9 \rm{meters} \end{eqnarray*} where $1.94 \times 10^7$ is the period expressed in seconds. The period of the new orbit is once again given by Kepler's Third Law but now with the semimajor axis length $a$ replacing $r$. This length is given by half the sum of the aphelion and perihelion lengths: \begin{equation} a = \frac{(98 + 108) \times 10^9}{2} = 103 \times 10^{9} \rm{meters} \end{equation} The new period is then given by: \begin{eqnarray*} T_{new} &=& \sqrt{\frac{4\pi^2a^3}{GM_s}} \\ &=& \sqrt{\frac{4\pi^2 \times (103 \times 10^9)^3} {6.67 \times 10^{-11} \times 1.989 \times 10^{30}}} \\ &=& 1.80 \times 10^7 \rm{secs} \end{eqnarray*} Although the asteroid slowed the planet down, we see that it now circles the sun in a shorter time. This is because the collision caused the planet to move faster at the perihelion, shortening the overall orbital distance.

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