Question

Distance Using Hubble's Law II.Find the percent difference (% = |A-O|/A × 100) between the actual (A) distance values and calculated (O) distance values using the recessional velocities for the Virgo and Corona Borealis clusters (vVirgo = 1,200 km/s, dVirgo = 17 Mpc; vCorona Borealis = 22,000 km/s, dCorona Borealis = 310 Mpc). Use H = 70 km/s/Mpc.the percent difference for the Virgo = _________%the percent difference for the Corona Borealis =_________ %At which distance, the closer or further one, is Hubble's law more accurate for the objects? Closer or further

Answer 1

Answer:

Explanation:

Let us first calculate for Virgo

$A= d_{virgo} = 17;M_{pc}, V_{virgo} = 1200 km/s,$

Using Hubble's law

$v = H_{0}D$ For Virgo

$V_{virgo} = H_{0}D_{virgo}$

$O = D_{virgo} = \frac{v_{virgo}}{H_{0}} = \frac{1200 km/s}{70 km/s/Mpc}= 17.143 Mpc$

Percentage difference for the Virgo

$\% = \frac{|A-O|}{A}\times 100 = \frac{|17 Mpc-17.143 Mpc|}{17 Mpc}\times 100 = 0.84 \%$

Now for calculate for Corona Borealis

$A= d_{Corona Borealis} = 310 Mpc, v_{Corona Borealis} = 22000 km/s,$

Using Hubble's law

$v = H_{0}D$ For Corona Borealis

$v_{Corona Borealis } = H_{0}D_{Corona Borealis } \\O = D_{Corona Borealis } = \frac{v_{Corona Borealis }}{H_{0}} = \frac{22000 km/s}{70 km/s/Mpc}= 314.286 Mpc$

Percentage difference for the Virgo

$\% = \frac{|A-O|}{A}\times 100 = \frac{|310 Mpc-314.286 Mpc|}{310 Mpc}\times 100 = 1.3825 \%$

So clearly  Hubble's law is more accurate for the closer objects