The expansion of the universe currently makes anything beyond ~16 billion light years away from us unreachable (because of the cosmic event horizon). This makes less than $3\%$ of the observable universe reachable, and even less of it available for 2-way trips.

If we send a futuristic spacecraft (approaching the speed of light) to some object that is near the edge of the reachable universe, by the time it gets there Earth will be far beyond it's reach due to the expansion of space, so it won't be able to get back to us.

I'm wondering what is the furthest object I can get a 2-way trip to, so by the time my spacecraft reaches that object it will still be inside the reachable universe so it can get back to Earth.

But approaching the longest possible trip means approaching a trip that would last an eternity. I want my spacecraft to get back to me at a reasonable time, so I need an equation that can tell me how long (from Earth's frame of reference) a 2-way trip will take given the current distance of the destination object from Earth. With it I can find out both the theoretical limit for the longest trip, and what is the furthest object that I can bring to Earth reasonably fast.

Since scientists are still not sure about how exactly the universe expands, I'd be happy with an answer that covers just the currently most favored theory.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Jan 30, 2019 at 3:57

1 Answer 1


The cosmic event horizon represents the furthest distance that light can travel due to the expansion of the universe. Or in other words, it represents the furthest distance which, when we send a light beam "now" the information can be received and send back to us in a finite amount of time. Further, then that distance, the light would never reach us back.

The only problem is that, the radius of the cosmic event horizon changes with respect to time. The reason is the change in the scale factor. The cosmological constant becomes dominant in the energy density while the matter density dilutes in time. In this sense scale factor also changes with respect to time and it affects the radius of the cosmic event horizon.

For this reason, the cosmic event horizon will increase a bit and will stop increasing likely in $\approx20$ billion years later at a distance of $\approx 17.5$ billion ly.

Proper distance to the cosmic event horizon is defined as,


So we can reach to any oject near to the cosmic horizon. The distance to the cosmic horizon is defined as above. But if you wait until the galaxy leaves the cosmic event horizon then the distance calculations becomes easier (which its what we assumed). Because we will be the near at the cosmic horizon with respect to the galaxy. So this proper distance would be,


So the total distance trip, in proper distance $$D=d_{hor}(t_0)+d_{hor}(t)$$

$$D=a(t)[\int_0^{\infty} \frac {cdt } {a(t)}+\int_{t}^{\infty} \frac {cdt } {a(t)}]$$

where $t$ is the time it took to reach us the galaxy at near horizon.

And the scale factor


for $t_{\Lambda}=\frac {2} {3H_0\sqrt {\Omega_{\Lambda}}}$

The first term represents the comoving distance to the cosmic event horizon and the second term represents the comoving distance between galaxy and earth. And in last to find the proper distance we are multiplying these values with $a(t)$.

To find the total time trip we just have to divide this distance by $c$,

$$T=\frac {D} {c}=a(t)[\int_0^{\infty} \frac {dt } {a(t)}+\int_{t}^{\infty} \frac {dt } {a(t)}]$$

So these are the general equations. The $D$ that I defined only represents the distance to the cosmic event horizon plus the returning distance.

There's no need to define a function $f(d,t)$ because you can send any object within the radius of the cosmic event horizon and you can wait until the galaxy is about to leave the cosmic horizon. And then you can send your space-ship back when the galaxy is about the cross the cosmic event horizon.

Since you are asking the furthest distance, the spacecraft can go furthest as the distance to the cosmic event horizon. I just added the returning distance to the equation.

But we can write an equation for that too. Lets suppose you wanted to go somewhere within the radius of the event horizon and then you wanted to wait until the galaxy crosses the event horizon.

So within the cosmic event horizon, pick a planet, measure the z and the proper distance to that planet is defined as $$R=a(t)[\frac {c} {H} \int_0^z \frac {dz} {E(z)}]$$

I write the equation in terms of redshift(z) but not in time (t) because the redshift is the observable value.

Then the total trip distance would be

$$D=a(t)[\frac {c} {H} \int_0^z \frac {dz} {E(z)}+\int_{t}^{\infty} \frac {dt } {a(t)}]$$

For $$E(z)=\sqrt{\Omega_{\Lambda}+\Omega_m(1+z)^3+\Omega_r(1+z)^4+\Omega_{\kappa}(1+z)^2}$$

or simplified version for current values,


  • $\begingroup$ Thanks for the reply, but it still doesn't answer my question. I wanted is something like f(d,t)=... where d is the proper distance of the destination object from Earth at the time of the departure, which is less than the distance to the event horizon (even if it will get to the event horizon by the time the light reaches it), t is the time the trip starts, and the result of the equation is how long from Earth's frame of reference this 2-way trip will take. (doesn't have to be squeezed all into one equation ofc, but just a way to get that result) $\endgroup$
    – potato
    Jan 31, 2019 at 3:20
  • $\begingroup$ The reason I want it that way is to be able to establish a limit for which objects we can theoretically bring to Earth, which is not all the objects that are currently in the reachable universe. And to get an idea of how many of them can be collected in a reasonable time frame. $\endgroup$
    – potato
    Jan 31, 2019 at 3:31
  • $\begingroup$ Also please take into account that I know nothing about the math here, so I don't know for example how to calculate a(t). $\endgroup$
    – potato
    Jan 31, 2019 at 4:01
  • $\begingroup$ @potato I edited my answer $\endgroup$
    – seVenVo1d
    Jan 31, 2019 at 6:17
  • $\begingroup$ Sorry for being so late with this, got carried away with something else and forgot about this, I marked your answer as accepted now. Thank you for your time! $\endgroup$
    – potato
    May 11, 2019 at 17:24

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