Word Problem sun, mars, moon

Question

The sun has a diameter of $8.65 \times 10^5 \text{mi}$. Mars is $1.42 \times 10^8 \text{mi}$ from the sun. Its moon, Phobos, has a diameter of $17.4 \text{mi}$. What is the maximum distance that Phobos can be from Mars and still cause a total eclipse of the sun on Mars?

Answer 1

As suggested by TonyK, we consider the following situation

                                |
                 |              |
Mars  X----------|--------------|
                 |              |
                                |

               Phobos          Sun

We denote as $D_{P}$ the diameter size of Phobos, $D_{S}$ the diameter of the Sun and $L_{S}$ the distance between the Sun and Mars. We want to determine $L_{P}$ the distance between Mars and Phobos so that the eclipse is total.

The easiest way to interpret this condition of full eclipse is to introduce the angles $\theta_{P}$ and $\theta_{S}$ under which Phobos and the Sun are seen from Mars.

Trigonometry in the upper right-angled triangles tells us that we have (I can add more details if this step is unclear !)

$$ \begin{cases} \tan \left( \frac{\theta_{S}}{2} \right) = \frac{D_{S}}{2 L_{S}} \\ \tan \left( \frac{\theta_{P}}{2} \right) = \frac{D_{P}}{2 L_{P}} \end{cases}$$

and similarly for the variables corresponding to Phobos.

We want to have $\theta_{P} = \theta_{S}$. As $\tan$ is an increasing function, this condition is identical to $\tan\left(\frac{\theta_{S}}{2}\right) = \tan\left(\frac{\theta_{P}}{2}\right)$, so that finally we obtain

$$ \frac{D_{S}}{2 L_{S}} = \frac{D_{P}}{2 L_{P}}$$

We therefore immediately obtain the maximal distance between Mars and Phobos as

$$ L_{P} = \frac{D_{P}}{D_{S}} L_{S}$$

Some remark :

• No need to use the approximation of the small angles for which $\tan (\theta) \simeq \theta$.

• You wanted to use trigonometry, but the same result can also be immediately obtained using Thales theorem !