Chapter 9.5, Problem 49PS

Solution

To find: The distance between the Earth and Mercury planet.

On that certain day the distance between the Earth and the Mercury is nearly 161 miles.

Given data:

The distance between the Sun and the Mercury is $37\text{}\text{million}\text{miles}$ and between the Sun and the Earth $93\text{}\text{million}\text{miles}$ . At a certain day angle between the Sun and the Mercury measured from the Earth is $22°$ .

Method/Formula used:

  

The ‘Law of Cosines’ states for a triangle ABC , Fig. 1:

  $\cos A=\frac{b^2+c^2-a^2}{2bc}$

The solution of quadratic equation, $x^2-bx+c=0$ is

  $x=\frac{b\pm \sqrt{b^2-4c}}{2}$ … (1)

Calculations:

The positions of the Sun, the Earth, and the Mercury with the triangle EMS is shown in Fig.2.

  

Here, the distance between the Sun and the Mercury ( SM ) is $37\text{}\text{million}\text{miles}$ and between the Sun and the Earth ( ES ) $93\text{}\text{million}\text{miles}$ . At a certain day angle between the Sun and the Mercury measured from the Earth is $22°$ .

Using ‘Law of Cosine’ in triangle EMS,

  $\cos \angle E=\frac{E{M}^2+E{S}^2-M{S}^2}{2\left(EM\right)\cdot \left(ES\right)}$

Substitute, $\text{22}°$ for $\angle E$ , 37 for SM , and 93 for ES ,

  $\begin{array}{c}\cos 22°=\frac{E{M}^2+{93}^2-{37}^2}{2\left(EM\right)\cdot \left(93\right)}\leftarrow \cos 22°=0.9272\\ 0.9272=\frac{E{M}^2+7280}{186\left(EM\right)}\\ E{M}^2+7280=172.46\left(EM\right)\\ 0=E{M}^2-172.46\left(EM\right)+7280\end{array}$

Substitute x for $EM$ , get quadratic equation

  $x^2-172.46x+7280=0$ ... (2)

  $\begin{array}{c}x=\frac{172.46\pm \sqrt{{\left(172.46\right)}^2-4\cdot \left(7280\right)}}{2}\\ =\frac{172.46\pm 149.86}{2}\\ =\frac{172.46+149.86}{2}\frac{172.46-149.86}{2}\\ \approx 161.36,-11.3\end{array}$

The distance cannot be negative, therefore, on that certain day the distance between the Earth and the Mercury is nearly 161 miles.