# When and How far out will the Voyager twins come to a full-stop?

So space is mostly, but not entirely, empty. Every few (dunno the order) cubic metres of space there is some cosmic dust.

Assuming that

• Cosmic dust exerts some friction on passing bodies,
• The Voyager twins are merely coasting along and
• Not subject to capture by any known bodies in their path

How far out before the friction exerted by cosmic dust reduces the velocity of either spaceraft to 0m/s in whatever direction it is headed - dead-in-space so to speak? Will either one even reach the nearest star it is headed towards?

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A full-stop relative to what?. –  Alfred Centauri Oct 26 '12 at 20:24
@AlfredCentauri: relative to the dust and cosmic background photons, whichever has more friction. –  Ron Maimon Oct 26 '12 at 21:24
@RonMaimon, I think it is the case that the Voyagers are bound to the Milky Way and the Milky Way has a relative motion w.r.t. the CMB. –  Alfred Centauri Oct 26 '12 at 21:40
@AlfredCentauri: ok, relative to whatever. It's still clear. –  Ron Maimon Oct 26 '12 at 22:03

The answer to this question depends on what your assumptions about how the vector velocities of the interstellar particles might be, of what kind of particles you might think you would encounter. Voyager does track the protons encountered by the space craft, and is currently colliding with 1-2 protons per second. My initial thought is that the particles would have velocity that is Maxwell-Boltzmann distributed which would lead to a type of Brownian Motion which would make its presence known by a type of wobble in the trajectory of the spacecraft.

Putting that thought aside, let's consider the case where the hydrogen is in-elastically colliding with Voyager. Mass of the Voyager space craft is ~733 kg and velocity is 17,000 m/s. The rest mass of the proton is $1.67 \times 10^{-27} kg$. Average velocity of protons enter the solar system from galactic sources is 26,000 m/s. We'll assume all the particles are impacting Voyager head on, which should approximate the worst case scenario.

The inelastic equation is:

$$\dfrac{m_{voy}v_{voy} + m_{p}v_{p}}{m_{voy}+ m_{p}} = v_{voy}'$$

Inputting our values:

$$\dfrac{(733kg)(17,000\dfrac{m}{s}) + (1.67\times10^{-27}kg)(-26,000\dfrac{m}{s})}{(733kg)+ (1.67\times10^{-27}kg)} = (17,000 - 9.79\times10^{-26})\dfrac{m}{s}$$

Or, in other words, under the inelastic assumption the decrease in velocity per collision is $$~9.79\times10^{-26}\dfrac{m}{s}$$

At 1-2 protons per second, the craft would require ~$10^{26}$ seconds to slow down by 1 meter per second. This is several orders of magnitude greater than the life of the universe.

This simple calculation of course does not take into account collisions with larger objects, or dust particles, but a simple adjustment in mass parameter plus knowledge of the number of collisions would allow for a reasonable estimate. A slightly more sophisticated calculation would consider elastic collisions, which I might add later.

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+1: Only I was hoping for dust particles to figure too ... given that Gaia collects around 30,000 tons annually. Then again, Gaia's mass is greater than that of Earth - hence gravitational pull proportionally greater too –  Everyone Oct 27 '12 at 16:18