91CP - University Physics with Modern Physics

Question

Navigating in the Big Dipper. All of the stars of the Big Dipper (part of the constellation Ursa Major) may appear to be the same distance from the earth, but in fact, they are very far from each other. Figure P1.91 shows the distances from the earth to each of these stars. The distances are given in light-years (ly), the distance that light travels in one year. One light-year equals $\mathbf{9}\mathbf{.}\mathbf{461}\mathbf{\times }{\mathbf{10}}^{\mathbf{15}}\mathbf{ }\mathbf{m}$  (a) Alkaid and Merak are data-custom-editor="chemistry" $\mathbf{25}\mathbf{.}{\mathbf{6}}^{\mathbf{o}}$ apart in the earth’s sky. In a diagram, show the relative positions of Alkaid, Merak, and our sun. Find the distance in light-years from Alkaid to Merak. (b) To an inhabitant of a planet orbiting Merak, how many degrees apart in the sky would Alkaid and our sun be?

Step-by-Step Solution

Answer

1. The distance between Merak and Alkaid is $76.2 \mathrm{ly}$.
2. The required angle is $25.6^{\mathrm{o}}$.

Step 1: Given data

The angle is $\mathrm{\theta }=25.6^{\mathrm{o}}$ 

One light-year is equal to $9.461\times {10}^{15 }\mathrm{years}$. 

Step 2: Introduction

A light-year, alternatively spelled light year, is a large unit of length used to express astronomical distances and is equivalent to about $\mathbf{9}\mathbf{.}\mathbf{46}\mathbf{ }\mathbf{trillion}\mathbf{ }\mathbf{kilometers}\mathbf{ }$, or 5.88 trillion miles. 

Step 3: (a) Find the distance in light-years from Alkaid to Merak

Calculate the distance in this manner.

Here there are two sides of triangles and angle,  determine the third side as follow

 

$\begin{array}{rcl}{\mathrm{c}}^2& =& {\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\times \mathrm{cos\theta }\\ & =& {\left(138\right)}^2+{\left(77\right)}^2-2\times 138\times 77\times \cos \left(25.6^{\mathrm{o}}\right)\\ & =& 19044+5929-\left(21252\times 0.902\right)19169.304\\ & =& 5803.696\\ \mathrm{c}& =& \sqrt{5803.969}\\ & =& 76.2 \mathrm{ly}\end{array}$ 

  

Hence, the distance between Merak and Alkaid is 76.2 ly.

Step 4: (b) Find how far is it from sun

Here we have to calculate the value of angle $\mathrm{\alpha }$. 

 

Hence using the same formula here,

  

 $\begin{array}{rcl}{\mathrm{c}}^2& =& {\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\times \cos \left(\mathrm{\alpha }\right)\\ {\left(76.2\right)}^2& =& {\left(138\right)}^2+{\left(77\right)}^2-2\times 138\times 77\times \cos \mathrm{\alpha }\\ 21252\times \cos \mathrm{\alpha }& =& 19044+5929-5806.44\\ \cos \mathrm{\alpha }& =& \frac{19166.56}{21252}\\ \mathrm{\alpha }& =& {\cos }^{-1}\left(0.9018\right)\\ & =& 25.6^{\mathrm{o}}\end{array}$ 

 

Hence, the required angle is 25.6^o.