Answer :
To find the distance to a star using the Luminosity Distance Formula, let's follow these steps:
1. Understand the Given Data:
- Luminosity ([tex]\( L \)[/tex]) = [tex]\( 5.2 \times 10^{23} \)[/tex] watts
- Apparent Brightness ([tex]\( AB \)[/tex]) = [tex]\( 1.0 \times 10^{-10} \)[/tex] watts/m[tex]\(^2\)[/tex]
- Formula to use: [tex]\( AB = \frac{L}{4 \pi r^2} \)[/tex]
2. Rearrange the Formula to Solve for Distance ([tex]\( r \)[/tex]):
- The formula [tex]\( AB = \frac{L}{4 \pi r^2} \)[/tex] can be rearranged to solve for [tex]\( r \)[/tex]:
[tex]\[ AB \times 4 \pi r^2 = L \][/tex]
[tex]\[ 4 \pi r^2 = \frac{L}{AB} \][/tex]
[tex]\[ r^2 = \frac{L}{4 \pi AB} \][/tex]
[tex]\[ r = \sqrt{\frac{L}{4 \pi AB}} \][/tex]
3. Plug in the Given Values:
- [tex]\( L = 5.2 \times 10^{23} \)[/tex] watts
- [tex]\( AB = 1.0 \times 10^{-10} \)[/tex] watts/m[tex]\(^2\)[/tex]
4. Calculate the Intermediate Value [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = \frac{5.2 \times 10^{23}}{4 \pi \times 1.0 \times 10^{-10}} \][/tex]
[tex]\[ r^2 = \frac{5.2 \times 10^{23}}{4 \pi \times 10^{-10}} \][/tex]
Using the fact that [tex]\( 4 \pi \approx 12.5664 \)[/tex], we get:
[tex]\[ r^2 = \frac{5.2 \times 10^{23}}{12.5664 \times 10^{-10}} \][/tex]
[tex]\[ r^2 \approx 4.138 \times 10^{32} \][/tex]
5. Solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{4.138 \times 10^{32}} \][/tex]
[tex]\[ r \approx 2.034 \times 10^{16} \, \text{meters} \][/tex]
Thus, the distance to the star is approximately [tex]\( 2.034 \times 10^{16} \)[/tex] meters.
Given the options:
- [tex]\( 2.034 \times 10^{-16} \)[/tex] m
- [tex]\( 3.251 \times 10^{16} \)[/tex] m
- [tex]\( 2.034 \times 10^{16} \)[/tex] m
- [tex]\( 3.251 \times 10^{-16} \)[/tex] m
The correct answer is [tex]\( 2.034 \times 10^{16} \)[/tex] meters.
1. Understand the Given Data:
- Luminosity ([tex]\( L \)[/tex]) = [tex]\( 5.2 \times 10^{23} \)[/tex] watts
- Apparent Brightness ([tex]\( AB \)[/tex]) = [tex]\( 1.0 \times 10^{-10} \)[/tex] watts/m[tex]\(^2\)[/tex]
- Formula to use: [tex]\( AB = \frac{L}{4 \pi r^2} \)[/tex]
2. Rearrange the Formula to Solve for Distance ([tex]\( r \)[/tex]):
- The formula [tex]\( AB = \frac{L}{4 \pi r^2} \)[/tex] can be rearranged to solve for [tex]\( r \)[/tex]:
[tex]\[ AB \times 4 \pi r^2 = L \][/tex]
[tex]\[ 4 \pi r^2 = \frac{L}{AB} \][/tex]
[tex]\[ r^2 = \frac{L}{4 \pi AB} \][/tex]
[tex]\[ r = \sqrt{\frac{L}{4 \pi AB}} \][/tex]
3. Plug in the Given Values:
- [tex]\( L = 5.2 \times 10^{23} \)[/tex] watts
- [tex]\( AB = 1.0 \times 10^{-10} \)[/tex] watts/m[tex]\(^2\)[/tex]
4. Calculate the Intermediate Value [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = \frac{5.2 \times 10^{23}}{4 \pi \times 1.0 \times 10^{-10}} \][/tex]
[tex]\[ r^2 = \frac{5.2 \times 10^{23}}{4 \pi \times 10^{-10}} \][/tex]
Using the fact that [tex]\( 4 \pi \approx 12.5664 \)[/tex], we get:
[tex]\[ r^2 = \frac{5.2 \times 10^{23}}{12.5664 \times 10^{-10}} \][/tex]
[tex]\[ r^2 \approx 4.138 \times 10^{32} \][/tex]
5. Solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{4.138 \times 10^{32}} \][/tex]
[tex]\[ r \approx 2.034 \times 10^{16} \, \text{meters} \][/tex]
Thus, the distance to the star is approximately [tex]\( 2.034 \times 10^{16} \)[/tex] meters.
Given the options:
- [tex]\( 2.034 \times 10^{-16} \)[/tex] m
- [tex]\( 3.251 \times 10^{16} \)[/tex] m
- [tex]\( 2.034 \times 10^{16} \)[/tex] m
- [tex]\( 3.251 \times 10^{-16} \)[/tex] m
The correct answer is [tex]\( 2.034 \times 10^{16} \)[/tex] meters.