Chapter 23, Problem 7Q

To determine

(a)

The distance to M51 in $\text{Megaparsecs (Mpc)}$ when the brightness of the supernova 19941 in M51was measured and the distance modulus was determined to be $29.2$

Answer to Problem 7Q

Distance in $\text{Megaparsecs (Mpc)}$ when distance modulus is $29.2$ = $6.92\text{ Mpc}$

Explanation of Solution

Given data Distance modulus of the supernova 19941 in M51 = $29.2$

Distance modulus of the planetary nebulae in M51 = $29.6$

Distance to Andromeda Galaxy from Earth = $750\text{ kpc}$

Formula used

$\begin{array}{l}\text{Distance modulus = }m-M\\ m-M=5{\log }_{10}\left(d\right)-5\\ \text{Here, d is measured in parsecs}\end{array}$

Calculation

$\begin{array}{l}m-M=5{\log }_{10}\left(d\right)-5\\ 5{\log }_{10}\left(d\right)=\left(m-M\right)+5\\ {\log }_{10}\left(d\right)=\frac{\left(m-M\right)+5}{5}\\ d={10}^{\frac{\left(m-M\right)+5}{5}}\\ \text{When }\left(m-M\right)=29.2\\ d={10}^{\frac{\left(29.2\right)+5}{5}}\\ d={10}^{\frac{34.2}{5}}\\ d={10}^{6.84}\\ d=6918309.71\text{ pc}\\ d=6.91830971\text{ Mpc}\\ \text{Rounding off to the second decimal place, }\\ d=6.92\text{ Mpc}\end{array}$

Conclusion Distance in $\text{Megaparsecs (Mpc)}$ when distance modulus is $29.2$ = $6.92\text{ Mpc}\text{.}$

To determine

(b)

The distance in $\text{Megaparsecs (Mpc)}$ when a separate distance measurement involving the planetary nebulae in M51 gave a distance modulus of $29.6$.

Answer to Problem 7Q

Distance in $\text{Megaparsecs (Mpc)}$ when distance modulus is $29.6$ = $8.32\text{ Mpc}\text{.}$

Explanation of Solution

Given data Distance modulus of the supernova 19941 in M51 = $29.2$

Distance modulus of the planetary nebulae in M51 = $29.6$

Distance to Andromeda Galaxy from Earth = $750\text{ kpc}$

Formula used

$\begin{array}{l}\text{Distance modulus = }m-M\\ m-M=5{\log }_{10}\left(d\right)-5\\ \text{Here, d is measured in parsecs}\end{array}$

Calculation

$\begin{array}{l}m-M=5{\log }_{10}\left(d\right)-5\\ 5{\log }_{10}\left(d\right)=\left(m-M\right)+5\\ {\log }_{10}\left(d\right)=\frac{\left(m-M\right)+5}{5}\\ d={10}^{\frac{\left(m-M\right)+5}{5}}\\ \text{When }\left(m-M\right)=29.6\\ d={10}^{\frac{\left(29.6\right)+5}{5}}\\ d={10}^{\frac{34.6}{5}}\\ d={10}^{6.92}\\ d=8317637.711\text{ pc}\\ d=8.317637711\text{ Mpc}\\ \text{Rounding off to the second decimal place, }\\ d=8.32\text{ Mpc}\end{array}$

Conclusion

Distance in $\text{Megaparsecs (Mpc)}$ when distance modulus is $29.2$ = $8.32\text{ Mpc}\text{.}$

To determine

(c)

The difference between the calculated distances in part (a) and part (b).

Answer to Problem 7Q

Difference between the 2 calculated values = $1.4\text{ Mpc}$

Explanation of Solution

Given data Distance modulus of the supernova 19941 in M51 = $29.2$

Distance modulus of the planetary nebulae in M51 = $29.6$

Distance to Andromeda Galaxy from Earth = $750\text{ kpc}$

Calculation

$\begin{array}{l}\text{Difference between the 2 calculated values }=\left(8.32-6.92\right)\text{Mpc}\\ \text{Difference between the 2 calculated values }=1.4\text{ Mpc}\end{array}$

Conclusion

$\text{Difference between the 2 calculated values }=1.4\text{ Mpc}$

To determine

(d)

Compares differencecalculated value in part c, with the $750\text{ kpc}$ distance from Earth to Andromeda Galaxy.

Answer to Problem 7Q

$\text{Difference of calculated values to M51 galaxy}<\text{Distance to Andromeda Galaxy from Earth}$

Explanation of Solution

Given data Distance modulus of the supernova 19941 in M51 = $29.2$

Distance modulus of the planetary nebulae in M51 = $29.6$

Distance to Andromeda Galaxy from Earth = $750\text{ kpc}$

Calculation

$\begin{array}{l}\text{Distance to Andromeda Galaxy from Earth = }750\text{ kpc}\\ \text{Distance to Andromeda Galaxy from Earth = }750\times {\text{10}}^{-3}\text{ Mpc}\\ \text{Distance to Andromeda Galaxy from Earth = }0.75\text{ Mpc}\\ \text{Difference of calculated values to M51 galaxy = 1}\text{.4 Mpc}\\ \text{As 1}\text{.4 Mpc < }0.75\text{ Mpc}\\ \text{Difference of calculated values to M51 galaxy}<\text{Distance to Andromeda Galaxy from Earth}\end{array}$

Conclusion

$\text{Difference of calculated values to M51 galaxy}<\text{Distance to Andromeda Galaxy from Earth}$

It shows that when the distance increase, the error associated with the reading increases as well.