Chapter 8.2, Problem 55AYU

To determine

Find the possible distance between Earth and mercury.

Answer to Problem 55AYU

  $\boxed{187565036.69km}$ or $\boxed{101439973.74km}$

Explanation of Solution

Given information:

The distance from the Sun to Earth is approximately $149,600,000km$ . The distance from the Sun to Mercury is approximately $57,910,000km$ . The elongation angle $\alpha$ is the angle formed between the line of sight from Earth to the Sun and the line of sight from Earth to the mercury. The elongation angle for Mercury is $15°$ . Use this information to find the possible distance between Earth and mercury.

  

Calculation:

Consider the figure,

  

Let Earth is denoted by E, the Sun by S ang Mercury by M. So, that take $\Delta EMS$ . It is given,

The distance from the Sun to Earth is $149,600,000km$ .

The distance between Sun to Mercury is $57,910,000km$ .

  $\angle MES=\alpha =15°$

to find the distance between Earth to Mercury.

Using law of sine on $\Delta EMS$ ,

  $\begin{array}{l}\frac{\sin \angle MES}{MS}=\frac{\sin \angle EMS}{SE}=\frac{\sin \angle MSE}{EM}\\ \frac{\sin \left(15°\right)}{57,910,000}=\frac{\sin \angle EMS}{149,600,000}=\frac{\sin \angle MSE}{EM}\\ \frac{\sin \left(15°\right)}{57,910,000}=\frac{\sin \angle EMS}{149,600,000}\\ \sin \angle EMS=\frac{149,600,000}{57,910,000}\times \sin \left(15°\right)\\ \sin \angle EMS=\frac{149,600,000}{57,910,000}\times 0.2588190\\ \sin \angle EMS=0.66861\\ \angle EMS={\sin }^{-1}\left(0.66861\right)\\ or\\ 180°-{\sin }^{-1}\left(0.66861\right)\\ \angle EMS\approx 41.96004°or180°-41.96004°=138.03996°\\ if\angle EMS\approx 138.03996,then\angle MES+\angle EMS\approx 15°+41.96004°\approx 56.96004<180°\\ \angle E{M}_{1}S\approx 41.96004°and\angle E{M}_{2}S\approx 138.03996°,\end{array}$

Now apply property of triangle,

  $\begin{array}{l}\angle MES+\angle E{M}_{1}S+\angle MSE=180°\\ \angle MSE=180°-\left(\angle MES+\angle E{M}_{1}S\right)\\ \angle MSE=180°-\left(15°+41.96004\right)\\ \angle MSE\approx 123.03996°\end{array}$

For $\angle E{M}_{2}S=138.03996°$ apply property of triangle,

  $\begin{array}{l}\angle MES+\angle E{M}_{2}S+\angle MSE=180°\\ \angle MSE=180°-\left(\angle MES+\angle E{M}_{2}S\right)\\ \angle MSE=180°-\left(15°+138.03996°\right)\\ \angle MSE\approx 26.96004°\end{array}$

  $\begin{array}{l}\frac{\sin 15°}{57,910,000}=\frac{\sin 123.03996°}{EM}\\ EM=57,910,000\times \frac{\sin 123.03996°}{\sin 15°}\\ EM=187565036.6963km\end{array}$

Now,

  $\begin{array}{l}\frac{\sin 15°}{57,910,000}=\frac{\sin 26.96004°}{EM}\\ EM=57,910,000\times \frac{\sin 26.96004°}{\sin 15°}\\ EM=101439973.74km\end{array}$

Hence, the distance between Earth and mercury is, $\boxed{187565036.69km}$ or $\boxed{101439973.74km}$ .