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As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using this table from Wikipedia. I'm going to go no further on the list than Sirius, and assume in each instance we are heading strait towards the Star. The formula for calculating the apperant magnitude given absolute magnitude, which is provided, is:

$m = M - 5 (1-\log_{10}{d})$

Setting up for our situation, the problem becomes:

$ 4.85 - 5(1-\log_{10}({d_{sun})})= M - 5 (1-\log_{10}({d-d_{sun}}))$

Or:

$\frac{M-4.85}{5}=\log_{10}{\frac{d_{sun}}{d-d_{sun}}}$

Continuing to solve for $d_{sun}$

$d_{sun}=\frac{d\cdot10^{\frac{M-4.85}{5}}}{10^{\frac{M-4.85}{5}}+1}$

Plugging that into a spreadsheet gives the following distances where the two stars are equally bright (Only including the strongest contenders)

• α Centauri A- 1.94 Light Years
• α Centauri B- 2.61 Light Years
• Sirius A- 1.46 light years

Bottom line, heading 1.46 light years towards Sirius, you'll see both Sirius and the Sun as equally bright. This is approximately the edge of the Oort Cloud, and still within the gravitational influence of the Sun, but is well on our way to another star system.

As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using this table from Wikipedia. I'm going to go no further on the list than Sirius, and assume in each instance we are heading strait towards the Star. The formula for calculating the apperant magnitude given absolute magnitude, which is provided, is:

$$m = M - 5 (1-\log_{10}{d})$$

Setting up for our situation, the problem becomes:

$$4.85 - 5(1-\log_{10}({d_{☉})})= M - 5 (1-\log_{10}({d-d_{☉}}))$$

Or:

$$\frac{M-4.85}{5}=\log_{10}{\frac{d_{☉}}{d-d_{☉}}}$$

Continuing to solve for $d_{☉}$

$$d_{☉}=\frac{d\cdot10^{\frac{M-4.85}{5}}}{10^{\frac{M-4.85}{5}}+1}$$

Plugging that into a spreadsheet gives the following distances where the two stars are equally bright (Only including the strongest contenders)

• α Centauri A- 1.94 Light Years
• α Centauri B- 2.61 Light Years
• Sirius A- 1.46 light years

Bottom line, heading 1.46 light years towards Sirius, you'll see both Sirius and the Sun as equally bright. This is approximately the edge of the Oort Cloud, and still within the gravitational influence of the Sun, but is well on our way to another star system.

As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using this table from Wikipedia. I'm going to go no further on the list than Sirius, and assume in each instance we are heading strait towards the Star. The formula for calculating the apperant magnitude given absolute magnitude, which is provided, is:

$m = M - 5 (1-\log_{10}{d})$

Setting up for our situation, the problem becomes:

$ 4.85 - 5(1-\log_{10}({d_{sun})})= M - 5 (1-\log_{10}({d-d_{sun}}))$

Or:

$\frac{M-4.85}{5}=\log_{10}{\frac{d_{sun}}{d-d_{sun}}}$

Continuing to solve for $d_{sun}$

$d_{sun}=\frac{d\cdot10^{\frac{M-4.85}{5}}}{10^{\frac{M-4.85}{5}}+1}$

Plugging that into a spreadsheet gives the following distances where the two stars are equally bright (Only including the strongest contenders)

• α Centauri A- 1.94 Light Years
• α Centauri B- 2.61 Light Years
• Sirius A- 1.46 light years

Bottom line, heading 1.46 light years towards Sirius, you'll see both Sirius and the Sun as equally bright. This is still well out of the solar system.

As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using this table from Wikipedia. I'm going to go no further on the list than Sirius, and assume in each instance we are heading strait towards the Star. The formula for calculating the apperant magnitude given absolute magnitude, which is provided, is:

$m = M - 5 (1-\log_{10}{d})$

Setting up for our situation, the problem becomes:

$ 4.85 - 5(1-\log_{10}({d_{sun})})= M - 5 (1-\log_{10}({d-d_{sun}}))$

Or:

$\frac{M-4.85}{5}=\log_{10}{\frac{d_{sun}}{d-d_{sun}}}$

Continuing to solve for $d_{sun}$

$d_{sun}=\frac{d\cdot10^{\frac{M-4.85}{5}}}{10^{\frac{M-4.85}{5}}+1}$

Plugging that into a spreadsheet gives the following distances where the two stars are equally bright (Only including the strongest contenders)

• α Centauri A- 1.94 Light Years
• α Centauri B- 2.61 Light Years
• Sirius A- 1.46 light years

Bottom line, heading 1.46 light years towards Sirius, you'll see both Sirius and the Sun as equally bright. This is approximately the edge of the Oort Cloud, and still within the gravitational influence of the Sun, but is well on our way to another star system.

As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using this table from Wikipedia. I'm going to go no further on the list than Sirius, and assume in each instance we are heading strait towards the Star. The formula for calculating the apperant magnitude given absolute magnitude, which is provided, is:

$m = M - 5 (1-\log_{10}{d})$

Setting up for our situation, the problem becomes:

$ 4.85 - 5(1-\log_{10}({d_{sun})})= M - 5 (1-\log_{10}({d-d_{sun}}))$

Or:

$\frac{M-4.85}{5}=\log_{10}{\frac{d_{sun}}{d-d_{sun}}}$

Continuing to solve for $d_{sun}$

$d_{sun}=\frac{d\cdot10^{\frac{M-4.85}{5}}}{10^{\frac{M-4.85}{5}}+1}$

Plugging that into a spreadsheet gives the following distances where the two stars are equally bright (Only including the strongest contenders)

• α Centauri A- 1.94 Light Years
• α Centauri B- 2.61 Light Years
• Sirius A- 1.46 light years

Bottom line, heading 1.46 light years towards Sirius, you'll see both Sirius and the Sun as equally bright.

As Mark Adler mentioned, the best way is to compare the brightness to other nearby stars. I'm going to assume that you have instantaneous travel time, and also take into account that you are actually getting closer to stars depending on the direction you go. I'm using this table from Wikipedia. I'm going to go no further on the list than Sirius, and assume in each instance we are heading strait towards the Star. The formula for calculating the apperant magnitude given absolute magnitude, which is provided, is:

$m = M - 5 (1-\log_{10}{d})$

Setting up for our situation, the problem becomes:

$ 4.85 - 5(1-\log_{10}({d_{sun})})= M - 5 (1-\log_{10}({d-d_{sun}}))$

Or:

$\frac{M-4.85}{5}=\log_{10}{\frac{d_{sun}}{d-d_{sun}}}$

Continuing to solve for $d_{sun}$

$d_{sun}=\frac{d\cdot10^{\frac{M-4.85}{5}}}{10^{\frac{M-4.85}{5}}+1}$

Plugging that into a spreadsheet gives the following distances where the two stars are equally bright (Only including the strongest contenders)

• α Centauri A- 1.94 Light Years
• α Centauri B- 2.61 Light Years
• Sirius A- 1.46 light years

Bottom line, heading 1.46 light years towards Sirius, you'll see both Sirius and the Sun as equally bright. This is still well out of the solar system.