Chapter 3.6, Problem 13Q

Solution

To calculate: The length of the orbit’s minor axis if the orbital period of Mars is about 1.88 years.

The length of the orbit’s minor axis is $228.2$ kilometers.

Given information:

The value of P is 1.88 years and the given function is $P=0.199a^{\frac{3}{2}}$.

According to Kepler’s third law of planetary motion, the function $P=0.199a^{\frac{3}{2}}$ relates a planet’s orbital period P (in days) to the length a (in millions of kilometers) of the orbit’s minor axis.

Formula used: Raise each side of the equation to the same power.

Calculation:

Write the original function:

  $P=0.199a^{\frac{3}{2}}$ .

It is given that $P=1.88\text{years}$ ,

  $\begin{array}{c}P\text{=1}\text{.88}\cdot \text{365}\\ P=686\end{array}$

Now substitute the value of P into the original function.

  $686=0.199a^{\frac{3}{2}}$ .

Divide each side of the equation by $0.199$ .

  $\begin{array}{l}\frac{686}{0.199}=\frac{0.199a^{\frac{3}{2}}}{0.199}\\ \frac{686}{0.199}=a^{\frac{3}{2}}\end{array}$

Raise each side to the power $\frac{2}{3}$ and simplify,

  $\begin{array}{c}{\left(\frac{686}{0.199}\right)}^{\frac{2}{3}}={\left(a^{\frac{3}{2}}\right)}^{\frac{2}{3}}\\ {\left(\frac{686}{0.199}\right)}^{\frac{2}{3}}=a\\ {\left(3447.236\right)}^{\frac{2}{3}}\approx a\\ 228.2\approx a\end{array}$

Therefore, the value of the length of the orbit’s minor axis is $228.2$ kilometers.