Applications of Power Functions

Recall that a power function is a function expressible as $f\left(x\right)=ax^r$f(x)=axr where $r$r is a real number. 

We encounter many laws of science expressible as power functions, and one famous one was discovered by Johannes Kepler (1571-1630).

Kepler's third law

Kepler's third law of planetary motion states that the time taken for a planet to orbit the Sun (known as the planets period) is related to its distance away from the Sun.

Specifically, if $p$p is the period of the planet and $d$d is the average distance away from the Sun, then Kepler's 3rd law states that $d^3\propto p^2$d3∝p2.

By taking cube roots, and introducing a constant of variation $k$k,  the law can be mathematically written as $d=k\times p^{\frac{2}{3}}$d=k×p23​ .

Finding k

We can find the constant of variation by considering the data for the planet we live on.

Assuming that the period of Earth is given by $p=365.25$p=365.25 Earth days, and that its distance from the Sun is given by  $d=1.496\times10^8$d=1.496×108 km, then we can write:

$d$d	$=$=	$k\times p^{\frac{2}{3}}$k×p23​
$1.496\times10^8$1.496×108	$=$=	$k\times365.25^{\frac{2}{3}}$k×365.2523​
$\therefore k$∴k	$=$=	$\frac{1.496\times10^8}{365.25^{\frac{2}{3}}}$1.496×108365.2523​​
	$=$=	$2.928\times10^7$2.928×107

So, based on the Earth's period and radial distance, we have developed a power model for other planets.

Kepler's 3rd law becomes:

 $d=\left(2.928\times10^7\right)\times p^{\frac{2}{3}}$d=(2.928×107)×p23​

where $d$d is the mean distance from the Sun in km and $p$p is the period of the orbit.

 

The Mars-Sun distance

We now can predict the distance Mars' is away from our Sun.  

Mars has a period of $687$687 Earth days and so using this number, our function predicts  $d=\left(2.928\times10^7\right)\times687^{\frac{2}{3}}=2.2797\times10^8$d=(2.928×107)×68723​=2.2797×108 km. 

From the internet, Mars has a mean distance from the Sun of $2.279\times10^8$2.279×108 km, so the prediction is quite close.

 

Alternative approach to the 3rd law

Since $d^3\propto p^2$d3∝p2, we could just as easily made the period $p$p the subject, so that for a different constant of variation, say $k_2$k2​,  we have $p=k_2\times d^{\frac{3}{2}}=k_2\times d\sqrt{d}$p=k2​×d32​=k2​×d√d. 

Using the same strategy above, this new constant would have the approximate value of $k_2=1.996\times10^{-10}$k2​=1.996×10−10, so that the rearranged law would given as:

$p=\left(1.996\times10^{-10}\right)\times d\sqrt{d}$p=(1.996×10−10)×d√d.

 

Period of Venus

Considering Venus this time, with a average distance from the Sun of $1.082\times10^8$1.082×108 km, we would estimate the orbital period as:

$p=\left(1.996\times10^{-10}\right)\times\left(1.082\times10^8\right)\sqrt{\left(1.082\times10^8\right)}$p=(1.996×10−10)×(1.082×108)√(1.082×108)

When simplified, we find that $p=224.6$p=224.6 days, which is close to the accepted value of $225$225 days. 

 

 

Worked Examples

QUESTION 1

QUESTION 2