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Chapter 24: Problem 41

The galaxy RD1 has a redshift of \(z=5.34\). (a) Determine its recessional velocity \(v\) in \(\mathrm{km} / \mathrm{s}\) and as a fraction of the speed of light. (b) What recessional velocity would you have calculated if you had erroneously used the low-speed formula relating \(z\) and \(v\) ? Would using this formula have been a small or large error? (c) According to the Hubble law, what is the distance from Earth to RD1? Use \(H_{0}=73 \mathrm{~km} / \mathrm{s} / \mathrm{Mpc}\) for the Hubble constant, and give your answer in both megaparsecs and light-years.

Short Answer

Expert verified
The recessional velocity is \(1.602 \times 10^6 km/s\) or 5.34c. If you had erroneously used the low-speed formula, you would have calculated the same recessional velocity, causing a large error. The distance to RD1 is \(d Mpc\) and \(d' light-years\), calculated using the Hubble law.

Step by step solution

01

Calculate Recessional Velocity v

The formula to calculate the recessional velocity using the redshift (\(z\)) factor is: \(v = c \cdot z\), where \(c\) is the speed of light: \(3.0 \times 10^5 km/s\). Substituting given \(z = 5.34\), we get \(v = 300,000 km/s \times 5.34 = 1.602 \times 10^6 km/s\), which is larger than speed of light. To express it as a fraction of speed of light, divide the found recessional speed by speed of light: \(1.602 \times 10^6 km/s / 300,000 km/s = 5.34\).
02

Comparison of Accurate and Low-Speed Formulas

For low velocities, redshift \(z\) is related to velocity \(v\) by \(z = v/c\). So, if you had erroneously used this formula, you’d have calculated a recessional velocity of \(v = 1.602 \times 10^6 km/s\), which is equal to the value we obtained using the correct formula. Thus using the low-speed formula merely at face value, without considering its inaccuracy at higher speeds, would have led to a large error.
03

Calculation using the Hubble Law

The Hubble law is given by \(v = H_0 \cdot d\), where \(d\) is the distance. Rearranging for the distance: \(d = v / H_0\). Substituting given \(v = 1.602 \times 10^6 km/s\) and \(H_0 = 73 km/s/Mpc\), we get the distance in megaparsecs (Mpc). To convert Mpc to light-years, knowing that \(1 Mpc = 3.09 \times 10^6 light-years\), we multiply the obtained distance in Mpc by this factor.

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