Question

a telescope brought about revolutionary changes in astronomy. a comparable leap in our ability to observe the universe took place as a result of a space telescope. the space telescope can see stars and galaxies whose brightness is \\( \frac { 1 } { 50 } \\) of the faintest objects now observable using ground-based telescopes. use the fact that the brightness of a point source, such as a star, varies inversely as the square of its distance from an observer to show that the space telescope can see about seven times farther than a ground-based telescope. now write the equation using the same fact when the observer is a ground-based telescope. let \\( d _ { 2 } \\) be the distance between the object and ground-based telescope and \\( k \\) be the constant of variation. \\( x = \frac { k } { ( d _ { 2 } ) ^ { 2 } } \\) show that the space telescope can see about seven times farther than a ground-based telescope using the above equations. \\( y = \frac { 1 } { 50 } x \\) \\( \frac { k } { ( d _ { 1 } ) ^ { 2 } } = \frac { 1 } { 50 } \left( \frac { k } { ( d _ { 2 } ) ^ { 2 } } \
ight) \\) substitute \\( \frac { k } { ( d _ { 1 } ) ^ { 2 } } \\) for \\( y \\) and \\( \frac { k } { ( d _ { 2 } ) ^ { 2 } } \\) for \\( x \\). \\( ( d _ { 1 } ) ^ { 2 } = 50 ( d _ { 2 } ) ^ { 2 } \\) multiply both sides by \\( 50 ( d _ { 1 } ) ^ { 2 } ( d _ { 2 } ) ^ { 2 } \\) and simplify. \\( d _ { 1 } = \square \\) (type an exact answer, using radicals as needed.)

Explanation:

Step1: Solve for \(d_1\)

We have the equation \((d_1)^2 = 50(d_2)^2\).
Take the square root of both sides: \(d_1=\sqrt{50(d_2)^2}\).

Step2: Simplify the square root

Using the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (\(a = 50\), \(b=(d_2)^2\)), we get \(d_1=\sqrt{50}\cdot\sqrt{(d_2)^2}\).
Since \(\sqrt{(d_2)^2}=d_2\) (distance is non - negative) and \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\approx5\times1.414 = 7.07\approx7\).

Answer:

\(d_1 = 5\sqrt{2}d_2\approx7d_2\)