Relativistic Space Travel Calculator

How long a star journey takes, in two answers that are both right: the years that pass on Earth and the years the crew actually live through. Constant velocity is the headline; the constant-1g rocket, which is the one that makes the galaxy reachable, is worked out beside it.

Last updated 11 Nov 2026

Distance and cruise speed → Earth years and ship years

τ = T/γ for constant v, plus the constant-acceleration case

Distance to the destinationMeasured in the Earth’s frame, which is how every star catalogue quotes it. The default 4.25 light-years is the approximate distance to Proxima Centauri, rounded deliberately — nothing on this page depends on the third figure. Other useful numbers: Sirius 8.6 ly, the galactic centre about 26 000 ly, Andromeda about 2.5 million ly.

That distance is in1 parsec = 3.261 564 light-years; 1 AU = 1.495 978 707×10¹¹ m exactly by the 2012 IAU definition; 1 light-year = 9.4607304725808×10¹⁵ m exactly, being c times a Julian year.

Speed of the moving thingRelativity questions are almost always set in fractions of c, because that is the only number the formula cares about. If your question gives a speed in m/s, switch the unit below rather than dividing by 299 792 458 by hand — that division is where a tired reader loses a power of ten. Typing c itself, or anything faster, does not break the page: the velocity ratio is capped just below 1 and the answer comes back with a note saying so, because no object with mass can be given that speed.

The speed above is given asOne conversion worth carrying: 1% of c is 2 997 924.58 m/s, about 2998 km/s. A fighter jet is 0.000 002 c and a fast electron in a school tube is about 0.1 c, which is already a 0.5% gamma.

For the accelerating case: proper acceleration, in gUsed only by the four acceleration rows below the headline, not by the headline itself. 1 g is 9.806 65 m/s² exactly, and is the acceleration that would feel like standing on Earth for the whole voyage. The useful coincidence is that c/g is 0.9687 years, so at 1 g you reach a substantial fraction of c in about a year.

Three elapsed times for one journey, drawn to true relative length against the same end mark. The top bar is the years that pass on Earth. The middle bar is what the crew's own clocks record, and it really is 1/γ of the top one. The bottom bar is how long light would take, which is β of the Earth time — so above about 0.71c the crew's bar becomes SHORTER than light's, and a label appears to say so. Nothing has outrun light: the top bar is still the longest.

7.3612yearsExample

4.25 light-years to Proxima Centauri at a steady 0.5c

Constant speed in two lines, and the 1g rocket in four

T = d/v · τ = T/γ · at constant proper acceleration a: T = √((d/c)² + 2d/a) · τ = (c/a)·arcosh(ad/c² + 1) · arcosh x = ln(x + √(x² − 1))

T: Earth time, the coordinate time in the frame the distance was measured in. This is the one a star catalogue’s distance belongs to
τ: ship time, the proper time along the crew’s worldline. Always the smaller of the two, and the one that decides whether anybody arrives alive
a: PROPER acceleration — what an accelerometer on board reads, and what the crew feel as weight. It is NOT dv/dt in the Earth frame, which falls towards zero as the ship approaches c
arcosh: the inverse hyperbolic cosine, written here as ln(x + √(x²−1)) because that is the form a calculator without an arcosh key can evaluate. It is why ship time grows only LOGARITHMICALLY with distance

Worked example

4.25 light-years to Proxima Centauri at a steady 0.5c
EARTH TIME FIRST. T = d/v. Working in light-years and years makes this trivial: 4.25 / 0.5 = 8.5 years. The unit pair is the trick — a light-year divided by a fraction of c is a year, with no powers of ten anywhere.
GET γ. β = 0.5, so γ = 1/√(1 − 0.25) = 1/√0.75 = 1.154 701.
SHIP TIME. τ = T/γ = 8.5 / 1.154701 = 7.361 2 years. The crew ages 7.36 years while 8.5 years pass at home: a difference of 1.14 years, which is a real and permanent difference that survives their arrival.
CHECK IT FROM THE SHIP'S SIDE. In the ship's frame the ship is at rest and Proxima is coming towards it at 0.5c, with the distance contracted to 4.25/1.1547 = 3.6806 light-years. Time to cover it at 0.5c: 3.6806/0.5 = 7.3612 years. Same answer, different story — that agreement is the thing to show in an answer.
NOW TRY ACCELERATING INSTEAD. At a steady 1 g the same 4.25 light-years costs 5.128 years of Earth time and only 2.294 years of ship time, arriving at 0.9826c. Flipping the ship over halfway and decelerating so you can actually stop costs 5.876 Earth years and 3.543 ship years. All four numbers are in the rows under the answer.
THE LESSON IN THE LAST STEP. The accelerating trip beats the 0.5c cruise on BOTH clocks, and it does so without ever exceeding c. Ship time at constant acceleration grows like the logarithm of the distance, which is why the galactic centre (26 000 light-years) costs about 20 years aboard and 26 000 at home.

The same journey at different speeds, and the 1g alternative

The last column is the one that changes how the problem looks. At a constant 1 g — one Earth gravity, which the crew would simply experience as weight — ship time grows like the LOGARITHM of the distance, so Andromeda costs about 29 years aboard while costing two and a half million at home. Those are the honest numbers from the formula. They say nothing about whether such a ship could be built: the fuel required is the problem, and this page does not address it. The 1g figures here were computed with this calculator and cross-checked by hand for the first row.
Destination	Distance	At 0.5c: Earth / ship	At 0.99c: Earth / ship	At 1g, flip halfway: Earth / ship
Proxima Centauri	4.25 ly	8.5 / 7.36 yr	4.29 / 0.61 yr	5.88 / 3.54 yr
Sirius	8.6 ly	17.2 / 14.90 yr	8.69 / 1.23 yr	10.36 / 4.61 yr
Vega	25 ly	50 / 43.3 yr	25.3 / 3.56 yr	26.87 / 6.44 yr
Galactic centre	26 000 ly	52 000 / 45 000 yr	26 260 / 3700 yr	26 002 / 19.8 yr
Andromeda	2 500 000 ly	5 000 000 / 4 330 000 yr	2 525 000 / 356 000 yr	2 500 002 / 28.6 yr

Two clocks, two answers, and both of them are correct

A star journey has two durations and neither is the wrong one. Earth time is distance over speed, in the frame the distance was measured in — which is the frame every star catalogue uses. Ship time is that divided by γ. For Proxima Centauri at half light speed the figures are 8.5 years and 7.36 years, and when the crew radio home the people at the other end really have aged more.

The crew’s version of the story. Nobody aboard experiences time running slowly; their clocks, their heartbeats and their coffee all behave normally. What they see is a shorter journey: at γ = 1.1547 the distance to Proxima is contracted to 3.68 light-years, and at 0.5c that takes 7.36 years. Same number, different mechanism. Being able to produce both accounts and show they agree is the thing that distinguishes understanding relativity from substituting into it.

Why constant acceleration changes the problem. At a steady 1 g the ship’s γ grows without limit, so each extra light-year costs less and less ship time. The formula has an arcosh in it, which is to say the ship time rises like a logarithm: Proxima is about 3.5 ship years with a flip at halfway, the galactic centre is about 20, and Andromeda about 29. Earth time, meanwhile, can never be less than the light-travel time and so is 2.5 million years for Andromeda. The journey is survivable; the return is not, because there is nothing left at home to return to.

Where this goes wrong. Dividing the wrong way: ship time is Earth time divided by γ, so if your crew ages MORE than the people at home you have multiplied. Using the contracted distance with the Earth time, or the full distance with ship time — pick one frame and stay in it for the whole calculation. Treating a as dv/dt in the Earth frame, when the formula needs the PROPER acceleration the crew feel. And reading the accelerating rows as a prediction about engineering: they are pure kinematics, they assume the fuel arrives from nowhere, and the real obstacle to interstellar travel is entirely in that assumption.

Frequently asked questions

Which number do I quote as ‘the time taken’?

Say which clock. “Eight and a half years as measured on Earth, seven years and four months as measured by the crew” is a complete answer; “about eight years” is not. The two are different physical quantities, not two estimates of one quantity, and an examiner asking for one of them is usually checking that you know there are two.

Can ship time be less than the light-travel time?

Yes, above about 0.71c, and it is not a paradox. The crew’s bar on the diagram becomes shorter than light’s, and a label appears when it does. What may never happen is for EARTH time to be less than the light-travel time, and it never is: T = d/v is always more than d/c. The ship has not outrun anything. It has simply taken a shorter path through spacetime, and a shorter path is exactly what less proper time means.

Is this the twin paradox?

It is half of it. The outbound leg here is the travelling twin’s; a round trip doubles both numbers and the twin comes home younger. The reason the situation is not symmetric — why it is not equally true that the stay-at-home twin is younger — is that the traveller TURNED ROUND, and turning round means accelerating, which singles out one of the two worldlines. A clock in free fall always records more proper time than one that manoeuvres between the same two events.

Why does 1 g make such a difference?

Because c/g is about 0.97 years, so a ship accelerating at 1 g reaches a serious fraction of c inside the first year and keeps going. After that its γ grows steadily, so each further light-year costs less ship time than the one before. The arcosh in the formula is the mathematical form of that: ship time rises like the logarithm of distance. The catch is entirely about fuel, which this page does not model at all.

What distance should I type for a star?

The catalogue distance, which is measured in the Earth’s frame, and the page will do the contraction for you. Do not contract it yourself first and then type the contracted value — that applies γ twice. The unit selector takes parsecs directly if your source quotes those; 1 pc = 3.261 564 ly.

What happens if I set the cruise speed to zero?

There is no crossing time, so the page shows its empty message rather than infinity. If you set the speed to c it clamps to β = 0.999 999 999, which gives a ship time of the distance in light-years divided by 22 360.68 — effectively instantaneous aboard, and still the full light-travel time plus a sliver on Earth. That is as close as the arithmetic gets to the often-repeated line that a photon experiences no time at all, and the honest version is that a photon has no frame to experience anything in.

Related calculators

References

CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, NIST Standard Reference Database 121. Read on 11 November 2026 from the machine-readable table at physics.nist.gov/cuu/Constants/Table/allascii.txt, which prints “speed of light in vacuum 299 792 458 (exact)” — exact because the 1983 definition of the metre fixes it, so c² = 8.9875517873681764×10¹⁶ m²/s² carries no uncertainty either. The same table gives G = 6.674 30(15)×10⁻¹¹ m³kg⁻¹s⁻², which is the least precisely known of the constants on this page by a factor of ten thousand, standard acceleration of gravity 9.806 65 (exact), and the rest energies used here: electron 0.510 998 950 69(16) MeV, proton 938.272 089 43(29) MeV, muon 105.658 3755(23) MeV.
International Astronomical Union, Resolution B3 on recommended nominal conversion constants for selected solar and planetary properties, XXIX General Assembly, 2015. Read on 11 November 2026 from the IAU’s own PDF. The values this page uses are the nominal terrestrial mass parameter (GM)^N_E = 3.986 004×10¹⁴ m³s⁻², the nominal solar mass parameter (GM)^N_☉ = 1.327 124 4×10²⁰ m³s⁻² and the nominal terrestrial equatorial radius 6.3781×10⁶ m. Using GM rather than G×M matters here: GM is known to ten significant figures from spacecraft tracking while G itself is known to five, so a Schwarzschild factor built from GM is four orders of magnitude better determined than one built from a mass times G.
A. Einstein, “Zur Elektrodynamik bewegter Körper”, Annalen der Physik 17 (1905) 891–921 — the paper in which the two postulates, the Lorentz transformation, time dilation and length contraction all appear together, and the companion three-page note “Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig?”, Annalen der Physik 18 (1905) 639–641, which adds the mass-energy result. Cited for the result, not for a number: neither paper was read for this build, and nothing on this page rests on a figure taken from them.
AQA, AS and A-level Physics (7407, 7408) specification, section 3.12.3 Special relativity, inside the option 3.12 Turning points in physics (A-level only). Read from aqa.org.uk on 11 November 2026. It is worth knowing exactly what a board asks for: 3.12.3.3 requires proper time and time dilation in the form t = t₀ / √(1 − v²/c²), with evidence from muon decay; 3.12.3.4 requires length contraction as l = l₀√(1 − v²/c²); and 3.12.3.5 requires mass-energy equivalence, the form E = m₀c²/√(1 − v²/c²), graphs of mass and kinetic energy against speed, and Bertozzi’s experiment as direct evidence. Note what that implies: AQA still teaches relativistic mass. This page cites the specification; it does not claim the board endorses it.
CBSE / NCERT, Physics, Class XII, Curriculum 2025–26, read on 11 November 2026 from the government-hosted syllabus at edustud.nic.in (SYLLABUS_2025_26/12/12_Physics_EM.pdf). The nine units are Electrostatics, Current Electricity, Magnetic Effects of Current and Magnetism, Electromagnetic Induction and Alternating Currents, Electromagnetic Waves, Optics, Dual Nature of Radiation and Matter, Atoms and Nuclei, and Electronic Devices. Special relativity, time dilation, length contraction and the Lorentz factor appear nowhere in it. The only trace of the subject is “mass-energy relation, mass defect” in Chapter 13, Nuclei — and the document does not even write E = mc² out. So if you are on this syllabus, this topic is reading beyond it rather than revision for it, which is a useful thing to know the night before a paper.