Problem 35
Question
A galaxy in the constellation Pisces is 5210 Mly from the earth. (a) Use the Hubble law to calculate the speed at which this galaxy is receding from earth. (b) What redshifted ratio \(\lambda_{0} / \lambda_{s}\) is expected for light from this galaxy?
Step-by-Step Solution
Verified Answer
(a) The galaxy is receding at 111,816 km/s; (b) the redshifted ratio is 1.373.
1Step 1: Understand the Hubble Law
The Hubble Law describes the relationship between the distance of a galaxy from Earth and its recession speed. It is given by the equation \( v = H_0 \times d \), where \( v \) is the galaxy's recessional velocity, \( H_0 \) is the Hubble constant, and \( d \) is the distance to the galaxy in megaparsecs.
2Step 2: Convert Distance to Megaparsecs
The distance to the galaxy is given as 5210 million light-years. First, we need to convert this distance into megaparsecs. Using the conversion 1 Mly (million light-years) = 0.3066 Mpc, we get:\[ d = 5210 \times 0.3066 = 1597.386 \, \text{Mpc} \]
3Step 3: Calculate Recessional Speed Using Hubble Law
With the distance in megaparsecs and assuming a Hubble constant of \( H_0 = 70 \, \text{km/s/Mpc} \) (a common value used for rough calculations), we calculate the recessional speed:\[ v = H_0 \times d = 70 \, \text{km/s/Mpc} \times 1597.386 \, \text{Mpc} = 111816.02 \, \text{km/s} \]
4Step 4: Understand Redshift and the Redshifted Ratio
The redshifted ratio \( \lambda_{0} / \lambda_{s} \) represents the ratio of the observed wavelength \( \lambda_{0} \) to the source wavelength \( \lambda_{s} \). The redshift \( z \) is given by \( z = \frac{\lambda_{0} - \lambda_{s}}{\lambda_{s}} \), where \( z = \frac{v}{c} \) for non-relativistic speeds (\( c \) being the speed of light).
5Step 5: Calculate the Redshift
To find the redshift, using the velocity \( v = 111816.02 \, \text{km/s} \) and the speed of light \( c = 299792 \, \text{km/s} \), we find:\[ z = \frac{v}{c} = \frac{111816.02}{299792} \approx 0.373 \]
6Step 6: Determine the Redshifted Ratio
Using the definition of redshift, calculate the redshifted ratio:\[ \frac{\lambda_{0}}{\lambda_{s}} = 1 + z = 1 + 0.373 = 1.373 \]
Key Concepts
Recessional VelocityRedshiftMegaparsecs (Mpc)
Recessional Velocity
Recessional velocity refers to the speed at which a galaxy moves away from Earth. This concept is a cornerstone of Hubble's Law, which highlights the expanding nature of the universe. To calculate the recessional velocity (\( v \)) of a galaxy, we use the formula: \[v = H_0 \times d\]
- Hubble constant ( \( H_0 \) ): a value that represents the rate of expansion of the universe (commonly about \( 70 \text{ km/s/Mpc} \)).
- Distance ( \( d \) ): the galaxy's distance from the Earth measured in megaparsecs (Mpc).
Redshift
Redshift occurs when the light from a galaxy is stretched to longer wavelengths as the galaxy moves away from us. This shift is a visual clue of a galaxy's movement and speed.We calculate the redshift (\( z \)) using the formula:\[z = \frac{\lambda_{0} - \lambda_{s}}{\lambda_{s}}\]where:
- \( \lambda_{0} \): observed wavelength.
- \( \lambda_{s} \): source wavelength.
- \( v \): the recession velocity.
- \( c \): the speed of light (approximately \(299792\text{ km/s}\)).
Megaparsecs (Mpc)
Megaparsecs (Mpc) are a unit of distance commonly used in astronomy to measure large-scale distances in the universe. A parsec itself is defined by the distance at which one astronomical unit subtends an angle of one arcsecond, and is approximately \(3.26 \text{ light-years}\).1 Mpc is equivalent to 1 million parsecs, making it particularly useful for denoting intergalactic lengths. To convert from million light-years (Mly) to Mpc, one can use the conversion factor:\[\text{1 Mly} \approx 0.3066 \text{ Mpc}\]This means when we know the distance in million light-years, we multiply by \(0.3066\)to find the distance in megaparsecs.Megaparsecs help simplify calculations in astronomy, as they efficiently handle large numbers typical of cosmological measurements. This standardization plays a key role in utilizing Hubble’s Law effectively.
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