<p>The necessary distance for the light bulb depends on the brightness of the light bulb.  The necessary brightness of the light bulb depends on the distance to the light bulb.</p>
<p>Part One:  Only Visible in One Location.</p>
<p>Do you imagine that after Betelgeuse becomes a supernova and then a white dwarf too dim to be seen, all humans will be living in a city ten miles wide on Earth?  And they will build a giant transparent dome with a diameter of a hundred miles around the city, and then build a giant track on the dome for the Betelgeuse impersonating light bulb to move on, duplicating the apparent motion of Betelgeuse as the Earth turns?</p>
<p>in that case the Betelgeuse impersonating light bulb will be about 50 miles, plus or minus five, from every place in the city.  And it seems like a simple matter to calculate how luminous it will have to be to have the apparent magnitude of Betelguese at a distance of 50 miles.  If the giant light bulb has the same surface temperature and thus color as Betelgeuse, it will be rather simple to calculate how wide it will have to be to have the same apparent magnitude as Betelgeuse.</p>
<p>And it is hoped that the giant light bulb will not have to be wide enough to be seen as a disc instead of a point at a distance of about 50 miles.</p>
<blockquote>
<p>The maximum angular resolution of the human eye is 28 arc seconds or 0.47 arc minutes,[22] this gives an angular resolution of 0.008 degrees, and at a distance of 1 km corresponds to 136 mm. This is equal to 0.94 arc minutes per line pair (one white and one black line), or 0.016 degrees. For a pixel pair (one white and one black pixel) this gives a pixel density of 128 pixels per degree (PPD)</p>
</blockquote>
<p><a href="https://en.wikipedia.org/wiki/Visual_acuity#Physiology" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Visual_acuity#Physiology</a></p>
<p>According to my rough calculations, the Betelgeuse impersonating light bulb should be less that about 0.816984 feet, or less than 9.80 inches, wide to appear as a dot instead of an extend object, at a distance of 45 miles.</p>
<p>45 miles would be 72.42 kilometers.  At a distance of 72.42 kilometers, the angular resolution of the human eye should be about 72.42 times 136 millimeters, or 9,849.12 millimeters, which should be 387.761 inches or 32.313 feet.  Which would be a much more reasonable diameter for the giant light bulb.</p>
<p>However, the Betelgeuse impersonating light bulb would have to be much farther than 45 to 55 miles away to appear at the same position in the sky as seen across the 10 mile wide city.</p>
<p>An object 206,265 Astronomical units (or AU) away would seem to shift its position in the sky by one arc second as view from points 1 AU apart.  So substituting 10 miles for 1 AU, an object would have to be 2,062,650 miles away to appear to shift position by 1 arc second as seen from points ten miles apart.  Since the human eye can only resolve angles to 28 arc seconds or more, the giant Betelgeuse impersonating light bulb could be only 73,666.07143 miles away to avoid appearing to be in a different direction from different parts of a ten mile wide city.</p>
<p>Betelgeuse is about 500 to 640 light years from Earth and has about 90,000 to 150,000 times the luminosity of the Sun.  Thus Betelgeuse is approximately 48.9 billion times as far away as the minimum distance for the giant Betelgeuse impersonating light bulb.  Thus the luminosity of the giant Betelgeuse impersonating light bulb should be approximately 90,000 to 150,000 times the luminosity of the Sun divided by approximately 48 billion squared.</p>
<p>And presumably there could be thousands of giant light bulbs in orbit around the Earth, each one lighting up when it came into approximately the right position to line up with Betelgeuse as seen from the city on Earth and turning off when the next one turned on. With enough giant light bulbs turning on and off fast enough, the fake Betelgeuse would not seem to twinkle much more than other stars in the sky.</p>
<p>Or maybe there would be one single giant light bulb hovering in position and constantly apply thrust to remain lined up between Betelgeuse and Earth.  And if the trust to keep it in the right position was supplied by rockets, maybe the exhaust gases would have the same temperature as Betelgeuse and so appear the same color.</p>
<p>Part Two: Visible from the Entire Planet Earth.</p>
<p>The planet Earth has a diameter of about 12,742 kilometers or 7,917.5 miles.</p>
<p>If an object will appear to move by one arc second if viewed from places separated by 1 Astronomical Unit (1 AU) at a distance of 206,265 AU, an object will appear to move by 1 arc second when viewed from places 12,742 kilometers apart at a distance of 206,265 times 12,742 kilometers, or 2,628,228,630 kilometers, or 1,633,105,557 miles.</p>
<p>But if the human eye can't resolve angles less than 28 arc seconds, the giant light bulb can be only 93,865,308.21  kilometers or 58,325,198.46 miles from Earth.</p>
<p>At that distance it can't have a natural orbit around either Earth or the Sun that will keep it lined up correctly, and so it will have to be kept in position between Earth and Betelgeuse by occasional orbital corrections.</p>
<p>Betelgeuse is about 500 to 640 light years from Earth and has about 90,000 to 150,000 times the luminosity of the Sun.  Thus Betelgeuse is approximately 61,722,893 times as far away as the minimum distance for the giant Betelgeuse impersonating light bulb.  Thus the luminosity of the giant Betelgeuse impersonating light bulb should be approximately 90,000 to 150,000 times the luminosity of the Sun divided by approximately 61,722,893 squared. That is approximately 90,000 to 150,000 times the luminosity of the Sun, divided by about 3,809,715,520,000,000.</p>
<p>Part Three: Visible from the Entire Solar System.</p>
<p>Or maybe when Betelgeuse becomes a supernova and then because a white dwarf too dim to be seen from Earth, Humans will be settled in scientific bases and space colonies  across the solar system.</p>
<p>The orbit of Neptune averages about 30 Astronomical Units (AU) in radius and thus has a diameter of about 60 AU.   So humans will be settled in an area with a radius of about 60 AU, and the Betelgeuse impersonating giant light bulb should be far enough away to appear at the same angle when seen from all parts of that area.</p>
<p>A parsec is defined as the distance at which one AU would appear about one arc second wide.  It also the distance at which an object's apparent position would change by one arc second when observed from places separated by 1 AU in a perpendicular direction.</p>
<p>Since the solar system is 60 AU in diameter (in this discussion), the Betelgeuse impersonating giant light bulb would have to be 60 parsecs from Earth to appear to move by no more than one arc second as seen from various places around the solar system.  But since human eyes can't resolve details finer than 28 arc seconds, the giant light bulb could be as close as about 2.14285 parsecs, or 6.989 light years, from the solar system.</p>
<p>That would put the giant light bulb in interstellar space, and probably closer to one or two other stars than it was to Earth.</p>
<p>Most people probably wouldn't notice if the fake &quot;Betelgeuse&quot; appeared to be a arc minute or two off position as viewed from different worlds, so you might be able to put the giant light bulb only about one light year from the solar system.</p>
<p>Betelgeuse is about 500 to 640 light years from Earth and has about 90,000 to 150,000 times the luminosity of the Sun.  Thus Betelgeuse is approximately 71.428 to 640 times as far away as the giant Betelgeuse impersonating light bulb if it is 1 to 7 light years from the Sun.  Thus the luminosity of the giant Betelgeuse impersonating light bulb should be approximately 0.3662 to 17.4352 times the luminosity of the Sun.</p>
<p>If the initial space velocity of the giant Betelgeuse impersonating light bulb is exactly that of the solar system gravitational interactions with the sun and nearby stars should pull it out of position very, very slowly, so corrections to keep it lined up accurately should be needed maybe only at intervals of tens or hundreds of thousands of years.</p>
<p>A giant Betelgeuse impersonating light bulb at a distance of light years would have to be so large and luminous to have the same apparent magnitude as Betelgeuse that it might almost make as much sense to tow a spectral class M1 or M2 main sequence red dwarf star into position between Betelgeuse and the solar system.</p>
<p>Conclusion.</p>
<p>So the size and the Luminosity of the giant Betelgeuse impersonating light bulb would vary by a vast amount - an &quot;astronomical&quot; amount one might say - depending on its distance from the viewing area it was designed to be seen from, which would depending on the size of the intended viewing area.</p>
<p>One way to reduce the energy requirements of the giant Betelgeuse impersonating light bulb would be to make it more like a searchlight focusing it beam of light in one direction instead of a star radiating equal amounts of light in one direction.</p>
<p>NO doubt different techniques would have different success in getting s much as possible of the radiation from the giant Betelgeuse impersonating light bulb to travel in a narrow beam.</p>

CC BY-SA 4.0
2023-10-30T08:37:38.170Z
Thanks, but I was looking for an answer that is more theoretically in reference to my previous solution method. But I appreciate it, though!
mathgirl752
CC BY-SA 4.0
2023-10-29T20:51:55.390Z
2023-10-29T21:35:54.140Z
2023-10-29T21:35:54.140Z
M. A. Golding
<p>The giant star Betelgeuse will develop into a supernova in the future. After this event, his remains will be far too dark to be observed from Earth. In order to maintain the view of the starry sky, it is considered to replace Betelgeuse with an appropriately positioned light bulb. In fact, a tungsten coil (coiled filament of wolfram) may be heated up to the current surface temperature of the Betelgeuse  ( surface temperature <span class="math-container">$T_B = 3600K$</span>; radius <span class="math-container">$R_B = 760*R_{⊙}$</span>; annual parallax <span class="math-container">$p=5.95$</span> mas). The coiled wire of the coil filament is 10 cm long and has a diameter of 50 μm. The entire heat output is delivered in accordance with the Stefan-Boltzmann law.</p>
<p>a) How far from the observer should the light bulb be placed so that its apparent brightness is similar to that of Betelgeuse?</p>
<p>b) At what wavelength would it shine brightest?</p>
<p>My proposition for b)</p>
<p>Wien's Displacement Law: <span class="math-container">$$λ_{max} = (b / T) = (2.898 \cdot 10^{-3} m K)/(3600 K) = 8.05 * 10^{-7} m$$</span></p>
<p>My suggestion to a)</p>
<p>The formula for apparent brightness (<span class="math-container">$B$</span>) is given by:</p>
<p><span class="math-container">$B = L/(4πd^2)$</span>, where <span class="math-container">$L$</span> is the luminosity of the light source and <span class="math-container">$d$</span> is the distance from the observer to the light source. On the other hand, the Stefan-Boltzmann law states:</p>
<p><span class="math-container">$$L = 4π(R^2)σ(T^4) \\ = 4π(760\cdot6.96\cdot10^8 m)^2(5.67 \cdot 10^{-8} W m^{-2} K^{-4})(3600K)^4 \\= 3.35\cdot10^{31} W$$</span></p>
<p>The light bulb's apparent brightness needs to be similar to that of Betelgeuse, therefore</p>
<p><span class="math-container">$$B_{light bulb} = B_{Betelgeuse}$$</span></p>
<p>So I need to solve for <span class="math-container">$d$</span>, but I do not know how to describe mathematically the left side of the equation (<span class="math-container">$B_{light bulb}$</span>). However, I know that:</p>
<p><span class="math-container">$$B_{Betelgeuse} = 3.35\cdot10^{31} W /(4πd^2)$$</span></p>
<p>(...)</p>

CC BY-SA 4.0
2023-10-28T17:56:58.160Z
Nice exercise. Please amend the question and elaborate on your own thoughts and approach of solving it. Please also see this sites guidelines on how to ask a good question
planetmaker
CC BY-SA 4.0
2023-10-29T10:52:44.230Z
Hello, @planetmaker Thanks for answering! I've just amended the question with my added thoughts and approach. I am even now not sure about b) and had to clue how to do a)...
mathgirl752
CC BY-SA 4.0
2023-10-29T11:35:13.380Z
For the light bulb you are given an effective temperature and via the dimensions of the filament the radiating surface. So you can calculate the radiated power and brightness similar to betelgeuze. Then you simply have to calculate the appropriate distance to get the same apparent brightness in order to make up for the difference in absolute brightness. Thus you can simply calculate the radiative power of betelgeuze on 1m^2 on earth (you basically did that, just did not enter 'd' yet into your equation). And calculate the distance for the filament to get the same power per m^2 on earth.
planetmaker
CC BY-SA 4.0
2023-10-29T17:51:59.903Z
@planetmaker But there is no given effective temperature of the light bulb... And how do I calculate its radiating surface??? Furthermore, may I use the same formulas for the light bulb from the star? (Because the light bulb is obviously not a star)?
mathgirl752
CC BY-SA 4.0
2023-10-29T20:54:26.253Z
@mathgirl  Thanks for the question.  Without it I would never have written the phrase "the giant Betelgeuse impersonating light bulb".  And if you asked in reference to a story you are thinking of writing, you should try the Worldbuilding Stack Exchange.
M. A. Golding
CC BY-SA 4.0
2023-10-29T21:34:14.820Z
@mathgirl752 from your question: "In fact, a tungsten coil may be heated up to the current surface temperature of the Betelgeuse". Both are black bodies for the sake of this exercise. And "coil filament is 10 cm long and has a diameter of 50 μm" gives you the size and thus ability to calculate surface
planetmaker
CC BY-SA 4.0
2023-10-30T08:42:15.387Z
@planetmaker But the diameter of 50 μm is already adequate for calculating the surface, so it is not necessary to take the length of the coil filament (10 cm) into account?
mathgirl752
CC BY-SA 4.0
2023-10-30T11:52:09.593Z
@mathgirl752 a light bulb filament, thus a wire, is basically a cylinder. So diameter and length suffice to get its surface. You might want to half that surface as you only see half of it.
planetmaker
CC BY-SA 4.0
2024-02-11T11:54:56.440Z
Your first steps are quite good. I would advise you to dig deeper and try to learn about black body radiation as well as the inverse square law. Also since this is a competition Problem from [IOAA Germany 2023](https://ioaa-germany.de), you are supposed to solve it on your own. Learning the Physics behind it will also help you for the next rounds. We also offer the IOAA Academy to teach you everything you need to know to solve the problems. For everyone who is interested in this kind of tricky astrophysics problems, feel free to check out our website.
IOAA-Police
<supernova><apparent-magnitude><parallax><betelgeuse><wavelength>
Betelgeuse, placing light bulb so that its apparent brightness is similar to that of Betelgeuse; wavelength where it shines brightest?
mathgirl752
# Betelgeuse, placing light bulb so that its apparent brightness is similar to that of Betelgeuse; wavelength where it shines brightest?

(Asked by: username_0 on 2023-10-28)

The giant star Betelgeuse will develop into a supernova in the future. After this event, his remains will be far too dark to be observed from Earth. In order to maintain the view of the starry sky, it is considered to replace Betelgeuse with an appropriately positioned light bulb. In fact, a tungsten coil (coiled filament of wolfram) may be heated up to the current surface temperature of the Betelgeuse ( surface temperature $T\_B = 3600K$; radius $R\_B = 760\*R\_{⊙}$; annual parallax $p=5.95$ mas). The coiled wire of the coil filament is 10 cm long and has a diameter of 50 μm. The entire heat output is delivered in accordance with the Stefan-Boltzmann law.

a) How far from the observer should the light bulb be placed so that its apparent brightness is similar to that of Betelgeuse?

b) At what wavelength would it shine brightest?

My proposition for b)

Wien's Displacement Law: $$λ\_{max} = (b / T) = (2.898 \cdot 10^{-3} m K)/(3600 K) = 8.05 \* 10^{-7} m$$

My suggestion to a)

The formula for apparent brightness ($B$) is given by:

$B = L/(4πd^2)$, where $L$ is the luminosity of the light source and $d$ is the distance from the observer to the light source. On the other hand, the Stefan-Boltzmann law states:

$$L = 4π(R^2)σ(T^4) \\ = 4π(760\cdot6.96\cdot10^8 m)^2(5.67 \cdot 10^{-8} W m^{-2} K^{-4})(3600K)^4 \\= 3.35\cdot10^{31} W$$

The light bulb's apparent brightness needs to be similar to that of Betelgeuse, therefore

$$B\_{light bulb} = B\_{Betelgeuse}$$

So I need to solve for $d$, but I do not know how to describe mathematically the left side of the equation ($B\_{light bulb}$). However, I know that:

$$B\_{Betelgeuse} = 3.35\cdot10^{31} W /(4πd^2)$$

(...)

**Tags:** supernova, apparent-magnitude, parallax, betelgeuse, wavelength

> **Comments:**
> **username_1**: Nice exercise. Please amend the question and elaborate on your own thoughts and approach of solving it. Please also see this sites guidelines on how to ask a good question  
> **username_0**: Hello, @username_1 Thanks for answering! I've just amended the question with my added thoughts and approach. I am even now not sure about b) and had to clue how to do a)...  
> **username_1**: For the light bulb you are given an effective temperature and via the dimensions of the filament the radiating surface. So you can calculate the radiated power and brightness similar to betelgeuze. Then you simply have to calculate the appropriate distance to get the same apparent brightness in order to make up for the difference in absolute brightness. Thus you can simply calculate the radiative power of betelgeuze on 1m^2 on earth (you basically did that, just did not enter 'd' yet into your equation). And calculate the distance for the filament to get the same power per m^2 on earth.  
> **username_0**: @username_1 But there is no given effective temperature of the light bulb... And how do I calculate its radiating surface??? Furthermore, may I use the same formulas for the light bulb from the star? (Because the light bulb is obviously not a star)?  
> **username_2**: @mathgirl  Thanks for the question.  Without it I would never have written the phrase "the giant Betelgeuse impersonating light bulb".  And if you asked in reference to a story you are thinking of writing, you should try the Worldbuilding Stack Exchange.  
> **username_1**: @username_0 from your question: "In fact, a tungsten coil may be heated up to the current surface temperature of the Betelgeuse". Both are black bodies for the sake of this exercise. And "coil filament is 10 cm long and has a diameter of 50 μm" gives you the size and thus ability to calculate surface  
> **username_0**: @username_1 But the diameter of 50 μm is already adequate for calculating the surface, so it is not necessary to take the length of the coil filament (10 cm) into account?  
> **username_1**: @username_0 a light bulb filament, thus a wire, is basically a cylinder. So diameter and length suffice to get its surface. You might want to half that surface as you only see half of it.  
> **username_3**: Your first steps are quite good. I would advise you to dig deeper and try to learn about black body radiation as well as the inverse square law. Also since this is a competition Problem from [IOAA Germany 2023](https://ioaa-germany.de), you are supposed to solve it on your own. Learning the Physics behind it will also help you for the next rounds. We also offer the IOAA Academy to teach you everything you need to know to solve the problems. For everyone who is interested in this kind of tricky astrophysics problems, feel free to check out our website.  

---

## 1 answers

### Answer by username_2 on 2023-10-29. Score: 1

The necessary distance for the light bulb depends on the brightness of the light bulb. The necessary brightness of the light bulb depends on the distance to the light bulb.

Part One: Only Visible in One Location.

Do you imagine that after Betelgeuse becomes a supernova and then a white dwarf too dim to be seen, all humans will be living in a city ten miles wide on Earth? And they will build a giant transparent dome with a diameter of a hundred miles around the city, and then build a giant track on the dome for the Betelgeuse impersonating light bulb to move on, duplicating the apparent motion of Betelgeuse as the Earth turns?

in that case the Betelgeuse impersonating light bulb will be about 50 miles, plus or minus five, from every place in the city. And it seems like a simple matter to calculate how luminous it will have to be to have the apparent magnitude of Betelguese at a distance of 50 miles. If the giant light bulb has the same surface temperature and thus color as Betelgeuse, it will be rather simple to calculate how wide it will have to be to have the same apparent magnitude as Betelgeuse.

And it is hoped that the giant light bulb will not have to be wide enough to be seen as a disc instead of a point at a distance of about 50 miles.

> The maximum angular resolution of the human eye is 28 arc seconds or 0.47 arc minutes,[22] this gives an angular resolution of 0.008 degrees, and at a distance of 1 km corresponds to 136 mm. This is equal to 0.94 arc minutes per line pair (one white and one black line), or 0.016 degrees. For a pixel pair (one white and one black pixel) this gives a pixel density of 128 pixels per degree (PPD)

<https://en.wikipedia.org/wiki/Visual_acuity#Physiology>

According to my rough calculations, the Betelgeuse impersonating light bulb should be less that about 0.816984 feet, or less than 9.80 inches, wide to appear as a dot instead of an extend object, at a distance of 45 miles.

45 miles would be 72.42 kilometers. At a distance of 72.42 kilometers, the angular resolution of the human eye should be about 72.42 times 136 millimeters, or 9,849.12 millimeters, which should be 387.761 inches or 32.313 feet. Which would be a much more reasonable diameter for the giant light bulb.

However, the Betelgeuse impersonating light bulb would have to be much farther than 45 to 55 miles away to appear at the same position in the sky as seen across the 10 mile wide city.

An object 206,265 Astronomical units (or AU) away would seem to shift its position in the sky by one arc second as view from points 1 AU apart. So substituting 10 miles for 1 AU, an object would have to be 2,062,650 miles away to appear to shift position by 1 arc second as seen from points ten miles apart. Since the human eye can only resolve angles to 28 arc seconds or more, the giant Betelgeuse impersonating light bulb could be only 73,666.07143 miles away to avoid appearing to be in a different direction from different parts of a ten mile wide city.

Betelgeuse is about 500 to 640 light years from Earth and has about 90,000 to 150,000 times the luminosity of the Sun. Thus Betelgeuse is approximately 48.9 billion times as far away as the minimum distance for the giant Betelgeuse impersonating light bulb. Thus the luminosity of the giant Betelgeuse impersonating light bulb should be approximately 90,000 to 150,000 times the luminosity of the Sun divided by approximately 48 billion squared.

And presumably there could be thousands of giant light bulbs in orbit around the Earth, each one lighting up when it came into approximately the right position to line up with Betelgeuse as seen from the city on Earth and turning off when the next one turned on. With enough giant light bulbs turning on and off fast enough, the fake Betelgeuse would not seem to twinkle much more than other stars in the sky.

Or maybe there would be one single giant light bulb hovering in position and constantly apply thrust to remain lined up between Betelgeuse and Earth. And if the trust to keep it in the right position was supplied by rockets, maybe the exhaust gases would have the same temperature as Betelgeuse and so appear the same color.

Part Two: Visible from the Entire Planet Earth.

The planet Earth has a diameter of about 12,742 kilometers or 7,917.5 miles.

If an object will appear to move by one arc second if viewed from places separated by 1 Astronomical Unit (1 AU) at a distance of 206,265 AU, an object will appear to move by 1 arc second when viewed from places 12,742 kilometers apart at a distance of 206,265 times 12,742 kilometers, or 2,628,228,630 kilometers, or 1,633,105,557 miles.

But if the human eye can't resolve angles less than 28 arc seconds, the giant light bulb can be only 93,865,308.21 kilometers or 58,325,198.46 miles from Earth.

At that distance it can't have a natural orbit around either Earth or the Sun that will keep it lined up correctly, and so it will have to be kept in position between Earth and Betelgeuse by occasional orbital corrections.

Betelgeuse is about 500 to 640 light years from Earth and has about 90,000 to 150,000 times the luminosity of the Sun. Thus Betelgeuse is approximately 61,722,893 times as far away as the minimum distance for the giant Betelgeuse impersonating light bulb. Thus the luminosity of the giant Betelgeuse impersonating light bulb should be approximately 90,000 to 150,000 times the luminosity of the Sun divided by approximately 61,722,893 squared. That is approximately 90,000 to 150,000 times the luminosity of the Sun, divided by about 3,809,715,520,000,000.

Part Three: Visible from the Entire Solar System.

Or maybe when Betelgeuse becomes a supernova and then because a white dwarf too dim to be seen from Earth, Humans will be settled in scientific bases and space colonies across the solar system.

The orbit of Neptune averages about 30 Astronomical Units (AU) in radius and thus has a diameter of about 60 AU. So humans will be settled in an area with a radius of about 60 AU, and the Betelgeuse impersonating giant light bulb should be far enough away to appear at the same angle when seen from all parts of that area.

A parsec is defined as the distance at which one AU would appear about one arc second wide. It also the distance at which an object's apparent position would change by one arc second when observed from places separated by 1 AU in a perpendicular direction.

Since the solar system is 60 AU in diameter (in this discussion), the Betelgeuse impersonating giant light bulb would have to be 60 parsecs from Earth to appear to move by no more than one arc second as seen from various places around the solar system. But since human eyes can't resolve details finer than 28 arc seconds, the giant light bulb could be as close as about 2.14285 parsecs, or 6.989 light years, from the solar system.

That would put the giant light bulb in interstellar space, and probably closer to one or two other stars than it was to Earth.

Most people probably wouldn't notice if the fake "Betelgeuse" appeared to be a arc minute or two off position as viewed from different worlds, so you might be able to put the giant light bulb only about one light year from the solar system.

Betelgeuse is about 500 to 640 light years from Earth and has about 90,000 to 150,000 times the luminosity of the Sun. Thus Betelgeuse is approximately 71.428 to 640 times as far away as the giant Betelgeuse impersonating light bulb if it is 1 to 7 light years from the Sun. Thus the luminosity of the giant Betelgeuse impersonating light bulb should be approximately 0.3662 to 17.4352 times the luminosity of the Sun.

If the initial space velocity of the giant Betelgeuse impersonating light bulb is exactly that of the solar system gravitational interactions with the sun and nearby stars should pull it out of position very, very slowly, so corrections to keep it lined up accurately should be needed maybe only at intervals of tens or hundreds of thousands of years.

A giant Betelgeuse impersonating light bulb at a distance of light years would have to be so large and luminous to have the same apparent magnitude as Betelgeuse that it might almost make as much sense to tow a spectral class M1 or M2 main sequence red dwarf star into position between Betelgeuse and the solar system.

Conclusion.

So the size and the Luminosity of the giant Betelgeuse impersonating light bulb would vary by a vast amount - an "astronomical" amount one might say - depending on its distance from the viewing area it was designed to be seen from, which would depending on the size of the intended viewing area.

One way to reduce the energy requirements of the giant Betelgeuse impersonating light bulb would be to make it more like a searchlight focusing it beam of light in one direction instead of a star radiating equal amounts of light in one direction.

NO doubt different techniques would have different success in getting s much as possible of the radiation from the giant Betelgeuse impersonating light bulb to travel in a narrow beam.

> **Comments:**
> **username_0**: Thanks, but I was looking for an answer that is more theoretically in reference to my previous solution method. But I appreciate it, though!  

---

