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. show that the space telescope can see about seven times farther than a ground - based telescope using the above equations.

Explanation:

Step1: Substitute the expressions for \(y\) and \(x\)

We know that \(y = \frac{k}{d_{1}^{2}}\) (space - telescope brightness) and \(x=\frac{k}{d_{2}^{2}}\) (ground - based telescope brightness), and \(y=\frac{1}{50}x\). Substitute \(y\) and \(x\) into \(y = \frac{1}{50}x\):
\(\frac{k}{d_{1}^{2}}=\frac{1}{50}\times\frac{k}{d_{2}^{2}}\)
Since \(k
eq0\) (because if \(k = 0\), there is no meaningful brightness), we can cancel out \(k\) from both sides of the equation.

Step2: Solve for the ratio \(\frac{d_{1}}{d_{2}}\)

After canceling \(k\), we get \(\frac{1}{d_{1}^{2}}=\frac{1}{50d_{2}^{2}}\). Cross - multiply to obtain \(50d_{2}^{2}=d_{1}^{2}\). Then, take the square root of both sides: \(d_{1}=\sqrt{50}d_{2}\).
Since \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\approx5\times1.414 = 7.07\approx7\)

Answer:

The space - telescope can see about seven times farther than a ground - based telescope.