How this instrument works
Recession velocity here is not a speed anything travels through space; it is how fast the proper distance to an object grows because the metric of space itself is stretching. The general relation is dD/dt = H(t) × D, where H(t) is the Hubble parameter and D is proper distance at that instant. v = H₀ × d is that same relation evaluated right now, using H₀, today's value of H(t), for one object at distance d — which is why the multiplication carries no exponent and no offset: it is a rate times a distance, sampled at a single moment in cosmic time.
Because the relation is a straight multiplication, distance alone decides how large v gets, and push d far enough and v = H₀ × d exceeds the speed of light. At H₀ = 70 km/s/Mpc that crossover, called the Hubble radius, sits near 4,283 Mpc, roughly 14 billion light-years out. This is not a breach of relativity: nothing there is moving through its own local patch of space faster than light. It is the accumulated stretching of the space between here and that point doing the work, a distinction the cosmologists Tamara Davis and Charles Lineweaver spent a well-known paper untangling because so many textbooks get it backwards.
The straight line is only today's tangent to a curve that has been changing since the Big Bang. H(t) fell steeply through the radiation- and matter-dominated eras, when gravity's pull decelerated the expansion, and it keeps falling even now, only more slowly, because dark energy is pushing the scale factor's growth into acceleration even as the ratio H = ȧ/a continues to ease toward a constant far-future value. Because of that changing history, v = H₀ × d is trustworthy only for the nearby universe, roughly where redshift stays under about 0.1; farther out, a fixed rate is not enough and the full expansion history has to replace it.
- Type a value into Hubble constant, km/s/Mpc — the expansion rate for the epoch you are modelling; 70 is the standard round figure used in textbook problems.
- Type the object's proper distance into Distance, Mpc (megaparsecs); one Mpc equals roughly 3.2616 million light-years.
- Read the result in Recession velocity, km/s — a straight product of the two inputs, updated instantly as either one changes.
- Hold the Hubble constant fixed and try several distances in turn to see the linear relationship for yourself; velocity should scale exactly with distance.
- Treat the output as the smooth Hubble-flow speed only; it excludes any peculiar velocity the object has from nearby gravity, typically tens to a few hundred km/s.
Worked example — a galaxy at 10 megaparsecs, H₀ = 70
Set Hubble constant, km/s/Mpc to 70, the standard textbook value, and Distance, Mpc (megaparsecs) to 10 — a galaxy roughly 32.6 million light-years away, well outside the Local Group and squarely inside the smooth Hubble flow. The instrument multiplies directly: v = 70 × 10 = 700 km/s. That figure is entirely a consequence of the expanding metric; no rocket, explosion, or gravitational slingshot is involved, only space accumulating stretch between here and that galaxy.
Because the relation is strictly linear, moving that same galaxy out to 20 Mpc, twice as far, would raise the reading to exactly 1,400 km/s with no further correction needed. Ten megaparsecs is a small slice of the roughly 4,283 Mpc Hubble radius described above, so this example sits safely inside the regime where the simple product v = H₀ × d is essentially exact, far from the high-redshift territory where a changing expansion rate would start to matter.
Questions
Can recession velocity actually exceed the speed of light?
Yes, and it does not break relativity. Beyond the Hubble radius — about 4,283 Mpc for H₀ = 70 km/s/Mpc — the plain product v = H₀ × d exceeds c, but nothing there is moving through its own local space faster than light. It is the accumulated expansion of the space between here and that point doing the work, a distinction cosmologists Davis and Lineweaver devoted a well-known paper to clarifying.
Why is the relation a plain multiplication with no exponent?
Because it is one instant of a rate equation, dD/dt = H(t) × D, evaluated at today's value H₀ for a single distance d. A rate times a distance gives a speed with no curvature needed; curvature only enters once H(t) itself is allowed to change over cosmic time, which this snapshot deliberately ignores.
Does this formula work for every galaxy at every distance?
No, it is a local approximation, accurate only while redshift stays well under about 0.1. H(t) has changed considerably across cosmic history, falling steeply through the radiation- and matter-dominated eras and still easing downward now, so distant, high-redshift objects need the full expansion history rather than one fixed H₀.
What is peculiar velocity, and why isn't it in this formula?
It is an object's own motion through space, layered on top of the smooth Hubble flow by nearby gravity, typically a few hundred km/s for a galaxy pulled by its cluster. The Hubble constant, km/s/Mpc and Distance, Mpc (megaparsecs) fields produce only the pure expansion term; a real measured velocity is peculiar velocity plus this Hubble-flow value.
Where do astronomers actually get the distance to plug in?
Rarely from a ruler. Nearby distances come from standard candles like Cepheid variables or Type Ia supernovae; for remote galaxies the far more common approach runs this relation backwards. A measured redshift yields a velocity, and dividing that velocity by H₀ gives the distance, which is how most galaxy distances in catalogues such as NED are actually estimated.
Why use km/s/Mpc instead of a unit that cancels out cleanly?
Because it keeps both familiar scales visible at once: speeds in the kilometres astronomers already work in, distances in the megaparsecs that fit galaxy surveys. Converted to pure SI, H₀ = 70 km/s/Mpc becomes about 2.27 × 10⁻¹⁸ per second, a number carrying the same information but none of the working intuition.