Question

Distances to the nearest stars (up to\({\rm{500 ly}}\)away) can be measured by a technique called parallax, as shown in Figure\({\rm{34}}{\rm{.26}}\). What are the angles\({{\rm{\theta }}_{\rm{1}}}\)and\({{\rm{\theta }}_{\rm{2}}}\)relative to the plane of the Earth’s orbit for a star\({\rm{4}}{\rm{.0 ly}}\)directly above the Sun?

Short answer

The angle \({{\rm{\theta }}_{\rm{1}}}{\rm{ = }}{{\rm{\theta }}_{\rm{2}}}\) is equal to \({\rm{89}}{\rm{.99977}}^\circ \).

Step-by-step solution

Parallax method

Parallax method help us to measure the distance of the nearby starts with the help of mathematical method of trigonometry.

Evaluating the angles

The diagram we have is:

Now, taking into consideration that the right triangle Earth - Sun - Star, we have:

\(\begin{array}{c}{\rm{tan}}{{\rm{\theta }}_{\rm{2}}}{\rm{ }} = {\rm{ }}\frac{{{\rm{4 ly}}}}{{{\rm{1}}{\rm{.0 au}}}}\\ = {\rm{ }}\frac{{{\rm{4}}{\rm{.0 \times 3}}{\rm{.0 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{ \times 360 \times 24 \times 3600 m}}}}{{{\rm{1}}{\rm{.50 \times 1}}{{\rm{0}}^{{\rm{11}}}}{\rm{ m}}}}\\ = {\rm{ 2}}{\rm{.488 \times 1}}{{\rm{0}}^{\rm{5}}}\\{{\rm{\theta }}_{\rm{2}}}{\rm{ }} = {\rm{ 89}}{\rm{.99977}}^\circ \end{array}\)

Then, by symmetry:\({{\rm{\theta }}_{\rm{1}}}{\rm{ = }}{{\rm{\theta }}_{\rm{2}}}\).

Therefore, the angle is \({\rm{89}}{\rm{.99977}}^\circ \).