Starship Asterisk*

Thought I'd start a new topic about astronomical distances. Feel free to ask anything you wish about distance in space, such as ...

(1) At what distance from the Sun would it no longer be the brightest star in the sky?

BDanielMayfield wrote: ↑Thu Dec 27, 2018 7:00 pm
Thought I'd start a new topic about astronomical distances.
Feel free to ask anything you wish about distance in space, such as ...

(1) At what distance from the Sun would it no longer be the brightest star in the sky?

That would probably have to be some point between either:

• 1) The Sun and Sirius or
  2) The Sun and Alpha/Beta Centauri

The point P at which the Sun becomes as dim as a distant star of luminosity L & distance D
is determined by the formula: P = D/[sqrt(L) +1]

For Sirius: (L= 23.48 , D= 8.6 ly) that point P = 1.48 ly
For Alpha/Beta Centauri: (L= 1.964 , D= 4.37 ly) that point P = 1.82 ly

The Sun is much brighter from 1.48 ly than from 1.82 ly so one can easily ignore Alpha/Beta Centauri
and the Sun & Sirius would tie for the brightest star in the sky at magnitude: -1.88 [= 4.83 - 5 log(32.6 ly/1.48 ly)]

BDanielMayfield wrote: ↑Thu Dec 27, 2018 7:00 pm Thought I'd start a new topic about astronomical distances. Feel free to ask anything you wish about distance in space, such as ...

(1) At what distance from the Sun would it no longer be the brightest star in the sky?

Well, the Sun is 25.27 (apparent) magnitudes brighter than Sirius, or a factor of 1.28x10¹⁰. Take the square root of that and multiply it by 1 AU, and you get 1.8 light years.

What an interesting question, Bruce! Thanks for your answers, Art and Chris. I note that Art and Chris arrive at different answers when it comes to the distance at which the Sun would look fainter than Sirius.

Ann

Ann wrote: ↑Fri Dec 28, 2018 5:45 am
What an interesting question, Bruce! Thanks for your answers, Art and Chris.

I note that Art and Chris arrive at different answers when it comes to the distance at which the Sun would look fainter than Sirius.

We get essentially the same answer if one travels perpendicularly vs-a-vs Sirius in which case:

The point P at which the Sun becomes as dim as a star of luminosity L & distance D
is determined by the formula: P = D/[sqrt(L)]

For Sirius: (L= 23.48 , D= 8.6 ly) that point P = 1.78 ly
------------------------------------------------------------------------
Now if one travels directly away from Sirius:

The point P at which the Sun becomes as dim as a star of luminosity L & distance D
is determined by the formula: P = D/[sqrt(L) -1]

For Sirius: (L= 23.48 , D= 8.6 ly) that point P = 2.24 ly
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Since gravity falls off inversely with the square of the distance like light by substituting
mass M for luminosity L one obtains the point at which gravitational forces match:

The point P at which the Sun's gravitational pull becomes as weak as a that of a star of mass M & distance D
is determined by the formula: P = D/[sqrt(M) +1]

For the Alpha Centauri system: (M= 2.129 , D= 4.37 ly) that point P = 1.78 ly

A space ship traveling to the Alpha Centauri system has truly left the solar system after 1.78 ly.
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Ann wrote: ↑Fri Dec 28, 2018 5:45 am What an interesting question, Bruce! Thanks for your answers, Art and Chris. I note that Art and Chris arrive at different answers when it comes to the distance at which the Sun would look fainter than Sirius.

That's because there are two ways of looking at the question. I took the simple interpretation: how far away would the Sun have to be in order to have the same brightness as the otherwise brightest star, Sirius. But if you take the question as moving away from the Sun, it becomes more complex, since as you move the brightness of all the other stars in the sky shift, as well. Then the direction you move matters, which is how Art approached it.

Thanks for the replies. Art, you also anticipated the next logical question too.

If one were to make a 3d representation of the distance from which the Sun is no longer the brightest object in the sky, it'd probably be a pretty cool looking potato.

Hmm. I'd grown rather fond of the Sun's current position in the Solar System.

Happy New Year.

geckzilla wrote: ↑Mon Dec 31, 2018 2:40 am
If one were to make a 3d representation of the distance from which the Sun is no longer
the brightest object in the sky, it'd probably be a pretty cool looking potato.

The Brightest Stars, as Seen from the Earth all have a luminosity L>1.00.
.............................................................................
A 3d representation of the distance from which the Sun is no longer
the nth brightest object in the sky would consist of roughly n+1
overlapping(/intersecting) spheres of different radii R & different center P_c locations:
.............................................................................
The 2 points P_t & P_a (for towards & away along a star's line of sight)
at which the Sun becomes as dim as that star of luminosity L & distance D
are determined by the formulae:

• P_t = D/[sqrt(L) +1]
  P_a = -D/[sqrt(L) -1]

These two points define the ends of an off-center sphere centered at

• P_c = -D/(L -1) along the line of sight
  with radius R = -[sqrt(L)]P_c

--------------------------------------------------------------------------------
Such overlapping(/intersecting) spheres would probably include:
.............................................................................
For Sirius: (L= 23.48 , D= 8.6 ly) that off-center point

• P_c = -0.383 ly along the Sirius line of sight
  with radius R = 1.854 ly

......................................................
For Castor: (L= 363 , D= 49 ly) that off-center point

• P_c = -0.135 ly along the Castor line of sight
  with radius R = 2.579 ly

......................................................
For Rigel: (L= 145,000 , D= 1400 ly) that off-center point

• P_c = -0.0097 ly along the Rigel line of sight
  with radius R = 3.677 ly

......................................................
For Alpha Centauri: (L= 1.964 , D= 4.37 ly) that off-center point

• P_c = -4.533 ly along the Alpha Centauri line of sight
  with radius R = 6.353 ly

.............................................................................

It occurred to me that an enterprising young person could use Art's math for a nice science fair or steam (Science, Technology, Engineering, Art and Math) project to produce an accurate 3D map of the space dominated by our star's light and/or gravity.

It would have been just the kind of thing I'd have loved to have done back in my high school days. But on the other hand, we didn't have PCs back then. Hand calculators were just replacing the slide rule back in them thar olden days.

Bruce