## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Section 2.8 - Modeling Using Variation - Exercise Set - Page 426: 56

#### Answer

The space telescope is able to see stars 7 times farther than the ground-based telescope.

#### Work Step by Step

Variation formulas are the formulas that define the relation between two quantities such that if one quantity varies we will know how it affects another quantity The illumination generated from a light source varies inversely as the square of the distance from the light source. The space telescope is able to see the faintest objects like stars / galaxies whose brightness is $\frac{1}{50}$ times that of the faintest object observed using ground-based telescopes. Inverse variation of illumination l and distance d can be defined as: $l=\frac{k}{{{d}^{2}}}$ , where k is the constant of variation. Say the distance of the faintest star seen from the space telescope is ${{d}_{1}}$ and the distance of the faintest star seen from the ground-based telescope is ${{d}_{2}}$. ${{l}_{1}}$ is the illumination of the faintest star observed by the space telescope and ${{l}_{2}}$ is the illumination of the faintest star observed by the ground based telescope. The illumination of the space telescope is 50 times better than the illumination of ground-based telescopes. Thus, ${{l}_{1}}=\frac{1}{50}{{l}_{2}}$ Now, ${{l}_{1}}=\frac{k}{{{d}_{1}}^{2}}$ and ${{l}_{2}}=\frac{k}{{{d}_{2}}^{2}}$ gives: $\frac{k}{{{d}_{1}}^{2}}=\frac{1}{50}\frac{k}{{{d}_{2}}^{2}}$ Cancel k from both sides: $\frac{1}{{{d}_{1}}^{2}}=\frac{1}{50{{d}_{2}}^{2}}$ Take square root on both sides: \begin{align} & \sqrt{\frac{1}{{{d}_{1}}^{2}}}=\sqrt{\frac{1}{50{{d}_{2}}^{2}}} \\ & {{d}_{1}}=\sqrt{50}{{d}_{2}} \end{align} Thus, ${{d}_{1}}\approx 7{{d}_{2}}$ This shows that the space telescope is able to see stars 7 times farther than ground-based telescopes.

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