Kepler's Third Law of Motion

Task

Below is a picture of the (elliptical) orbit of a planet around the sun:

The sun is at point $A$, point $P$ is where the planet is closest to the sun during its orbit, and point $Q$ is where the planet is farthest from the sun during its orbit. Kepler made the following amazing discovery: if $a$ is the average of the closest and farthest distances of the planet from the sun and $t$ is the time it takes the planet to make one full orbit around the sun then the quotient $\frac{t^2} {a^3}$ does not depend on the planet. In what follows, we measure $t$ in years and $a$ in astronomical units where one astronomical unit is the average of the closest and farthest distances from the earth to the sun. These units are helpful since we have $t = 1$ and $a = 1$ for the earth, allowing us to find the quotient $\frac{t^2}{a^3} = 1$.

1. Find an equation for $t$ in terms of $a$ and an equation for $a$ in terms of $t$.
2. The orbit of Mars takes about 1.88 years. What is its average distance from the sun?
3. The average distance of Neptune is 30.06 astronomical units. About how long does each orbit of Neptune take?
4. What is the farthest a planet could be from the sun during its orbit if each orbit takes 5 years?

IM Commentary

The purpose of this task is to solve some expressions requiring fractional exponents in an interesting modeling context. Like many physical laws, Kepler's third law of planetary motion is an approximation, one with remarkable accuracy as can be seen from that data presented here: http://www.physicsclassroom.com/class/circles/u6l4a.cfm. Much more interesting information can be found in the technical Wikipedia article on Kepler's laws: http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion. The point $B$ marked in the picture is the second focus of the ellipse: it is not used in the problem but has been included for completeness and could be examined further under G-GPE.3.

In modeling contexts, choice of units can play an important role as it does here. The value of the quotient $\frac{t^2}{a^3}$ for the given units is 1. It would be difficult to calculate this value in other units (such as minutes and miles) without a lot more information and calculation.

Solution

1. We are given that the quotient $\frac{t^2}{a^3}$ does not depend on the planet and is always equal to 1. Rewriting the equation gives $t^2 = a^3.$ This means, extracting square roots on both sides, that $$ \sqrt{t^2} = \sqrt{a^{3}}. $$ On the left hand side of the equation, the square root function is the inverse of the function with input $t$ and output $t^2$; but the square root value is always non-negative and so we have $\sqrt{t^2} = |t|$. In this context, this is the same as $t$ because the orbit always takes a positive amount of time. On the right hand side of the equation we have $\sqrt{a^{3}} = a^{3/2}$. To see why, note that both sides of this equation are positive and the square of both sides is $a^3$. Putting our calculations together, we have $t = a^{3/2}$.

   A similar argument extracting cube roots of the equation $t^2 = a^3$ gives $a = t^{2/3}$. In this case, absolute values do not play a role because the cube root can be positive or negative.

2. For this question, we are given that Mars' orbit takes about 1.88 years so $t = 1.88$ and $a = 1.88^{2/3}$ which is about 1.52. So we would expect the average of the farthest and closest distances of Mars from the sun to be about one and one-half times that of the earth.
3. For this question, we are given that the average distance of Neptune from the sun is $a = 30.06$ astronomical units. From above, the time it takes Neptune to make one full orbit is $t = 30.06^{3/2}$ which is about 165. So we expect one full orbit of Neptune around the sun to take 165 years, longer than the life span of any human being.
4. As in part (a) if the orbit of a planet takes 5 years then $t = 5$ and the average distance is $a = 5^{2/3}$ which is about 2.9. As the picture in the question shows, a planet can be substantially farther from the sun than its average distance. The average distance is the average of the closest and farthest distances. Since the closest distance must be positive (in fact, reasonably large since otherwise the planet would either be captured by the suns' gravitational pull at its closest point or escape at its farthest point), this means that the farthest the planet can be from the sun is a little less than twice the average distance. If a planet takes 5 years to orbit the sun then it will not be more than 5.8 astronomical units from the sun during its orbit: this represents the extreme case where the planet actually touches the sun at point $P$ and would be 5.8 astronomical units from the sun at point $Q$. This is not a realistic scenario and more astronomical knowledge would be necessary to evaluate how far (and close) a planet could realistically be from the sun during its orbit.