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
The surface area of skin on a human body can be approximated by the equation $A=0.007184h^{0.725}w^{0.425}$A=0.007184h0.725w0.425, where $A$A is the skin area in m2 of a person who is $h$h cm tall, and weighs $w$w kg.
In total, $\frac{2}{5}$25 of Britney’s skin is covered in sun spots and sun damage.
Given that Britney weighs $60$60 kg and is $152$152 cm tall, determine the total surface area of her skin that is covered in sun spots and sun damage.
Give your answer correct to three decimal places.
QUESTION 2
A scientist named Kepler found that the number of days $T$T a planet takes to complete one full revolution around the sun (period of orbit) is related to its distance $D$D (measured in millions of kilometres) from the sun. He found for a given planet, the square of its period of orbit is proportional to the cube of its average distance from the sun.
Using $k$k as the constant of variation, form an equation for $T$T in terms of $D$D.
Mercury is an average $57.9$57.9 million km away from the sun, and takes $88$88 days to orbit the Sun.
Solve for the value of $k$k to four decimal places.
Saturn is an average of $1427$1427 million km from the Sun. Using the rounded value of $k$k found in the previous part, determine the period of orbit, $T$T, of Saturn.
Give your answer to the nearest number of days.