The Luminosity of Alpha Centauri A. Alpha Centauri A lies at a
distance of 4.4 light-years and has an apparent brightness in
our night sky of 2.7 X 10-8 watt/m2. Recall that J light-year =
9.5 X 10 12 km = 9.5 X 1015 m.
a. Use the inverse square law for light to calculate the luminosity
of Alpha Centauri A.
b. Suppose you have a light bulb that
emits 100 watts of visible light. (Note: This is not the case for a
standard 100-watt light bulb, in which most of the 100 watts goes
to heat and only about l0-15 watts is emitted as visible light.)
How far away would you have to put the light bulb for it to have
the same apparent brightness as Alpha Centauri A in our sky?
(Hint: Use 100 watts as Lin the inverse square law for light, and
use the apparent brightness given above for Alpha Centauri A.
Then solve for the distance.)
The Correct Answer and Explanation is :
Part (a): Calculate the Luminosity of Alpha Centauri A
To calculate the luminosity ((L)) of Alpha Centauri A, we can use the inverse square law for light, which is given by the equation:
[
B = \frac{L}{4 \pi d^2}
]
Where:
- (B) is the apparent brightness (in watts per square meter),
- (L) is the luminosity (in watts),
- (d) is the distance to the star (in meters),
- (4 \pi d^2) is the surface area of a sphere with radius (d).
Step 1: Rearranging the equation to solve for luminosity:
[
L = B \cdot 4 \pi d^2
]
Step 2: Substitute the known values into the equation:
- The apparent brightness (B = 2.7 \times 10^{-8} \, \text{watt/m}^2),
- The distance (d = 4.4 \, \text{light-years}). To convert this into meters, use the conversion factor (1 \, \text{light-year} = 9.5 \times 10^{15} \, \text{meters}):
[
d = 4.4 \times 9.5 \times 10^{15} = 4.18 \times 10^{16} \, \text{meters}
]
Step 3: Plug these values into the equation for (L):
[
L = (2.7 \times 10^{-8}) \cdot 4 \pi (4.18 \times 10^{16})^2
]
Now, we calculate the luminosity:
[
L \approx 2.7 \times 10^{-8} \cdot 4 \pi \cdot (1.75 \times 10^{33}) \, \text{watts}
]
[
L \approx 2.7 \times 10^{-8} \cdot 2.2 \times 10^{34} \, \text{watts}
]
[
L \approx 5.94 \times 10^{26} \, \text{watts}
]
So, the luminosity of Alpha Centauri A is approximately (5.94 \times 10^{26} \, \text{watts}).
Part (b): Distance for a 100-watt light bulb to match Alpha Centauri A’s brightness
Next, let’s solve for the distance at which a 100-watt light bulb would need to be placed to have the same apparent brightness as Alpha Centauri A.
We will use the inverse square law again:
[
B = \frac{L}{4 \pi d^2}
]
We want the brightness (B) to be the same as that of Alpha Centauri A, and we know the luminosity of the light bulb (L = 100 \, \text{watts}).
Step 1: Rearranging to solve for (d):
[
d^2 = \frac{L}{4 \pi B}
]
[
d = \sqrt{\frac{L}{4 \pi B}}
]
Step 2: Substitute the values:
- (L = 100 \, \text{watts}),
- (B = 2.7 \times 10^{-8} \, \text{watt/m}^2).
[
d = \sqrt{\frac{100}{4 \pi (2.7 \times 10^{-8})}}
]
[
d = \sqrt{\frac{100}{3.4 \times 10^{-7}}}
]
[
d = \sqrt{2.94 \times 10^{8}} \, \text{meters}
]
[
d \approx 1.71 \times 10^{4} \, \text{meters}
]
So, the light bulb would need to be placed approximately (1.71 \times 10^4 \, \text{meters}) or about 17.1 kilometers away to match the apparent brightness of Alpha Centauri A in our sky.
Explanation
The inverse square law for light tells us that the brightness of a light source decreases as the square of the distance from the source. In part (a), we applied this law to calculate the luminosity of Alpha Centauri A by rearranging the formula to solve for luminosity. Given the apparent brightness and the distance to Alpha Centauri A, we were able to compute its luminosity, which turned out to be about (5.94 \times 10^{26} \, \text{watts}).
In part (b), we applied the inverse square law again to determine how far a 100-watt light bulb would need to be placed in order to match the apparent brightness of Alpha Centauri A. By plugging the luminosity of the light bulb and the apparent brightness of Alpha Centauri A into the equation, we calculated that the light bulb would need to be placed approximately 17.1 kilometers away to have the same brightness. This highlights the incredible luminosity of Alpha Centauri A, as the light bulb, despite emitting 100 watts, needs to be positioned much closer to match its brightness in the sky.