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Jonathan's answer is essentially correct, but as Rob Jeffries comments, he doesn't take into account that the Universe is expanding during the journey.

The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be "the farthest you can go"). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).

So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is $$ \tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} \right), $$ where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.

If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly 45 years. To get to the edge of the Universe at 47 Gly actually only takes a few years more. The reason for this is simply that traveling at 1G gets you to (almost) the speed of light in only a couple of years, and hence you experience (almost) no time, no matter how far you go.

The time experienced for the Earthlings for the traveler at constant acceleration is given by $$ t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right), $$ which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.

Jonathan's answer is essentially correct, but as Rob Jeffries comments, he doesn't take into account that the Universe is expanding during the journey.

The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be "the farthest you can go"). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).

So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is $$ \tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} +1 \right), $$ where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.

If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly 45 years. To get to the edge of the Universe at 47 Gly actually only takes a few years more. The reason for this is simply that traveling at 1G gets you to (almost) the speed of light in only a couple of years, and hence you experience (almost) no time, no matter how far you go.

The time experienced for the Earthlings for the traveler at constant acceleration is given by $$ t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right), $$ which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.

Jonathan's answer is essentially correct, but as Rob Jeffrey comments, he doesn't take into account that the Universe is expanding during the journey.

The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be "the farthest you can go"). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).

So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is $$ \tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} \right), $$ where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.

If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly 45 years. To get to the edge of the Universe at 47 Gly actually only takes a few years more. The reason for this is simply that traveling at 1G gets you to (almost) the speed of light in only a couple of years, and hence you experience (almost) no time, no matter how far you go.

The time experienced for the Earthlings for the traveler at constant acceleration is given by $$ t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right), $$ which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.

Jonathan's answer is essentially correct, but as Rob Jeffries comments, he doesn't take into account that the Universe is expanding during the journey.

The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be "the farthest you can go"). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).

So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is $$ \tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} \right), $$ where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.

If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly 45 years. To get to the edge of the Universe at 47 Gly actually only takes a few years more. The reason for this is simply that traveling at 1G gets you to (almost) the speed of light in only a couple of years, and hence you experience (almost) no time, no matter how far you go.

The time experienced for the Earthlings for the traveler at constant acceleration is given by $$ t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right), $$ which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.

Jonathan's answer is essentially correct, but as Rob Jeffrey comments, he doesn't take into account that the Universe is expanding during the journey.

The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be "the farthest you can go"). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).

So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is $$ \tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} \right), $$ where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.

If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly 45 years. To get to the edge of the Universe at 47 Gly actually only takes a few years more.

The time experienced for the Earthlings for the traveler at constant acceleration is given by $$ t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right), $$ which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.

Jonathan's answer is essentially correct, but as Rob Jeffrey comments, he doesn't take into account that the Universe is expanding during the journey.

The edge of the observable Universe is 47 billion lightyears (Gly) away. Even if you are a lightbeam, you cannot reach that point. The farthest you can go if departing today is roughly 5 Gpc, or 17 Gly, but this journey would of course take infinitly long (or else it wouldn't be "the farthest you can go"). This distance is probably what the linked article is referring to (I didn't read the article; it's very, very long).

So, in order for the answer to be any fun, you have to freeze the Universe, using magic, which is what Jonathan's calculator is doing. Here I'll just provide the analytical solution: In that case, the proper time $\tau$ (i.e. the time experienced by the traveler) to reach a distance $x$ when traveling at a constant acceleration $a$ is $$ \tau = \frac{c}{a} \cosh^{-1} \left( \frac{ax}{c^2} \right), $$ where $c$ is the speed of light. If you wish to decelerate after having reached halfway, you just divide $x$ by $2$ and multiply the result by $2$.

If you plug in the $x=15\,\mathrm{Gly}$ you request, you get roughly 45 years. To get to the edge of the Universe at 47 Gly actually only takes a few years more. The reason for this is simply that traveling at 1G gets you to (almost) the speed of light in only a couple of years, and hence you experience (almost) no time, no matter how far you go.

The time experienced for the Earthlings for the traveler at constant acceleration is given by $$ t(\tau) = \frac{c}{a} \sinh \left( \frac{a \tau}{c} \right), $$ which works out to 15 Gyr for the 15 Gly, and 47 Gyr for the observable Universe. The reason is simply that the traveler, from the point of view of the Earthlings, extremely fast reaches a speed which is almost the speed of light.