In relativistic interstellar travel, would a speck of dust = total destruction?

[krash661]

I refer you to Einstein, my predecessor.

can you show the math on this please.

but you just made it obvious that you can not.

Edited May 4, 2013 by krash661

[Widdekind]

The average density of the ISM is approximately 0.3 particles per cubic cm. For a (hypothetical) starship traveling near C = 30 B cm / s, every square cm of frontal surface would encounter approximately 10B particles per square cm per second. Each of those particles would basically become a Cosmic Ray, with an energy approximately equal to its rest-mass energy (assuming a gamma ~ 2), i.e. ~GeV.

 

So, every front-facing square cm, would sweep up billions of GeV cosmic-ray-equivalents, per second. At earth's surface, about one GeV cosmic-ray collides per second (per square cm).

 

So, relativistic starships would encounter billions of times more cosmic-ray-equivalents per second vs. rock space-weathering on the surface of earth or its moon. Whatever material formed the frontal shield would be pulverized, into lunar regolith powder. After one year of space travel, the frontal shield would have suffered the equivalent of billions of world-years of space weathering, i.e. would become pulverized into powder, as seen on earth's moon (unless i'm mis-understanding why the moon is covered in powder).

 

Individual collisions would not be the problem -- billions of collisions per second (per square cm) would accumulate damage, to the craft.

 

 

http://en.wikipedia.org/wiki/Interstellar_medium

http://en.wikipedia.org/wiki/Cosmic_ray

 

 

 

Note, the ratio of speed-to-gamma-factor is maximum near 0.7C, where gamma ~ 1.4. So, you could reduce damage, by slowing speed, to (say) half to three-quarters C. At ~0.4C, gamma ~ 1.1, reducing (relative) particle energies to ~100 MeV. Perhaps half-light-speed is the maximum effective space speed limit.

Edited May 5, 2013 by Widdekind

[Dekan]

Thanks to everyone for your posts, I've read them, and have gained more understanding. Especially from Janus's #25, on the "meteor" analogy, and Widdekind's #27, on "space weathering".

 

Don't these show that friction can explain the phenomena well enough, without invoking relativistic ideas.

 

I mean, suppose a spaceship is travelling at extremely high speed, through interstellar space. As we all know, space isn't a true vacuum, it contains dust particles. These dust particles will keep hitting the ship, as it travels swiftly along. So the ship will eventually get worn down, and eroded, and its hull breached, and the crew unfortunately killed, by the constant impacts of the dust particles. Like sand particles, scouring and eroding a building.

 

That seems common-sense, do we need relativistic imports?

Edited May 7, 2013 by Dekan

[swansont]

 

That seems common-sense, do we need relativistic imports?

 

Yes, if you want the analysis done properly. If the ship is traveling at relativistic speeds, you have to use relativity in your calculations.

[Janus]

Thanks to everyone for your posts, I've read them, and have gained more understanding. Especially from Janus's #25, on the "meteor" analogy, and Widdekind's #27, on "space weathering".

 

Don't these show that friction can explain the phenomena well enough, without invoking relativistic ideas.

 

I mean, suppose a spaceship is travelling at extremely high speed, through interstellar space. As we all know, space isn't a true vacuum, it contains dust particles. These dust particles will keep hitting the ship, as it travels swiftly along. So the ship will eventually get worn down, and eroded, and its hull breached, and the crew unfortunately killed, by the constant impacts of the dust particles. Like sand particles, scouring and eroding a building.

 

That seems common-sense, do we need relativistic imports?

 

 

If you were traveling at 0.1c, a 1 microgram speck would have a non- relativistic kinetic energy impact energy of 4.5e8 joules, correcting for relativity, it would increase to 4.53e8 joules, not much.

 

At 0.5c it becomes 1.13e10 and 1.39e10 a 23% difference.

 

At 0.99c the difference is a factor of 13.

 

So if you are trying to determine what it would take to protect the ship moving at 0.99c and didn't factor relativity into it, you would have only 1/13 the protection you would need,

[Mr Monkeybat]

But isn't the energy of the impact still proportional to the fuel energy used to get up to that speed, along with the travel time according to the ships clock being the same as Newtonian calculations for the amount of reaction mass being expelled.

[Widdekind]

As crude estimate, the ram-pressure against a light-speed space-craft would be of order [math]P \approx \rho c^2 \approx 10^{-4} Pa \approx 10^{-9} atm[/math], which would perhaps require millions to billions of years, to slow a massive craft.

 

i guess people know the formula c = g y, i.e. accelerating at 1G for 1 yr. results in light-speed (semi-classically).

 

(Perhaps the more correct formula is [math]P = \gamma \rho c^2 \beta^2[/math] ? Relativistically, the swept-up momentum is not [math]\rho v[/math], but [math]\gamma \rho v[/math].)

 

Most of the ISM is ionized, so ring-shaped crafts, with an axial, solenoidal magnetic field, might funnel most particles safely through the center of the ring -- which also could be spun, for artificial gravity, essentially an annular space station habitat... accelerated to high speed.

 

Or, the ISM particles could be consumed, for fusion fuel, hypothetically, as an interstellar ramjet.

 

Still, cosmic rays, and "induced cosmic rays", would perhaps pose problems for people.

 

Inter-stellar space probes could only go to one star system. Whereas, spending the same amount of money, for super-sized space telescopes, near earth, would enable detailed surveys, of every star system in sight -- without risking burning up somewhere out in the ISM in interstellar space.

[swansont]

As crude estimate, the ram-pressure against a light-speed space-craft would be of order [math]P \approx \rho c^2 \approx 10^{-4} Pa \approx 10^{-9} atm[/math], which would perhaps require millions to billions of years, to slow a massive craft.

 

Slowing the craft via interaction with the interstellar background wasn't the issue. The pressure assumes you can ignore the individual interactions with more massive particles, and you can't.

[Delta1212]

 

Slowing the craft via interaction with the interstellar background wasn't the issue. The pressure assumes you can ignore the individual interactions with more massive particles, and you can't.

A bit like crushing up all the ice from a berg, evenly distributing it over the course of the Titanic's route and seeing whether it creates enough drag to stop it?

[swansont]

A bit like crushing up all the ice from a berg, evenly distributing it over the course of the Titanic's route and seeing whether it creates enough drag to stop it?

 

Sure, only more so. The interstellar background Widdekind is referencing is protons, and the rest of us are talking about a speck that's of order a microgram. That's only ~17 orders of magnitude different. That's the difference between discussing the impact of a 100 metric ton iceberg and nanogram specks of ice. Drag isn't what sunk the Titanic.

[Janus]

But isn't the energy of the impact still proportional to the fuel energy used to get up to that speed, along with the travel time according to the ships clock being the same as Newtonian calculations for the amount of reaction mass being expelled.

assume you are trying to reach 0.99c. Also assume an acceleration of 1g and a exhaust velocity of 0.5c for the Rocket,

 

Using purely Newtonian physics, assuming an acceleration of 1g and a exhaust velocity of 0.5c for the Rocket,

 

You would take 351 days to reach 0.99c and you would use something over six times your rocket's mass in fuel.

 

The energy of collision with a 1 kg mass would be 4.4e 16 J

 

Taking relativity into account and the same base assumptions,

 

it would take 2.6 years ship time to reach 0.99c and you would need to burn 198 times your rocket' mass in fuel.

 

The same collision would have an energy of 5.48e17 J.

[ccwebb]

Fanghur,

 

This dilemma has already been imaged by the creative mind of Gene Roddenberry. Thus his idea of needing a deflector dish, or something similar.

 

Also, this discussion has already been proven in the terrible and devastating destruction of the Columbia Space Shuttle. A piece of foam at low Mach numbers was able to create a hole in the protective shielding of the left wing. The foam was decelerating and the Shuttle was accelerating into it.

[Mr Monkeybat]

assume you are trying to reach 0.99c. Also assume an acceleration of 1g and a exhaust velocity of 0.5c for the Rocket,

 

Using purely Newtonian physics, assuming an acceleration of 1g and a exhaust velocity of 0.5c for the Rocket,

 

You would take 351 days to reach 0.99c and you would use something over six times your rocket's mass in fuel.

 

The energy of collision with a 1 kg mass would be 4.4e 16 J

 

Taking relativity into account and the same base assumptions,

 

it would take 2.6 years ship time to reach 0.99c and you would need to burn 198 times your rocket' mass in fuel.

 

The same collision would have an energy of 5.48e17 J.

 

As interesting as this is, is it actually supposed to be a response to the quoted post, it seem to completely miss the points the post you quoted. Notice that in your second scenario uses more fuel than the first a ridiculous amount in fact so of course there is more energy involved in the collision. If you had used the same amount of fuel as the first scenario in your relativistic scenario then of course to the Earthbound observer it would be slower. But manned interstellar travel only makes sense as a one way settler trip, so the earthbound observer time frame is irrelevant, and only the ships time frame matters which produces a similar trip time to Newtonian physics due to time dilation etc.

[Widdekind]

Dust grains are (essentially) "clumps of protons", one collision => many protons hit the craft all at once. But, the momentum imparted is still the same, (classically) M * v = N * m_p * v. The density of dust affects, and is already factored into, the overall density of the ISM.

 

Yes, dust grains would create larger (micro) craters in a craft, like a single solid slug vs. buckshot. But the ram-pressure still obeys P = p v². In a given interval of time, whether you sweep up N protons, all in one "solid slug" (dust grain), or as "buck shot" (ISM protons), the momentum transferred is the same. (Unless you're considering the minor loss in mass, from fusion, for multi-nucleon nuclei, <1%.)

 

correct ?

 

in analogy, cruising through the ISM, is like flying through dusty air. Most of what encounters the craft is "air" (ISM). And also, the dust matters too. (i understand, that "smoke" is a more correct analogy than "dust", typical ISM "dust" grains are more similar in size to "smoke" particles. Cruising through the ISM is like flying through smokey air. Cruising at even vaguely relativistic speeds, is like flying at nearly infinite mach-number, C => Mach # ~ 30K. So, you need an "air-craft" capable of flying through smokey air, at mach tens-of-thousands. In earth analogy, that'd be about 10K km per second, i.e. from one-side to the other of earth, in one second. Or, all the way around earth, up in the dusty stratosphere, in a few seconds.)

 

perhaps you could accelerate an asteroid, and cruise behind it, in its evacuated wake ? It would sweep up the dusty & gas, creating a conduit, through which to travel. And you would not need to decelerate its mass, at the other end.

[swansont]

Dust grains are (essentially) "clumps of protons", one collision => many protons hit the craft all at once. But, the momentum imparted is still the same, (classically) M * v = N * m_p * v. The density of dust affects, and is already factored into, the overall density of the ISM.

 

Yes, dust grains would create larger (micro) craters in a craft, like a single solid slug vs. buckshot. But the ram-pressure still obeys P = p v². In a given interval of time, whether you sweep up N protons, all in one "solid slug" (dust grain), or as "buck shot" (ISM protons), the momentum transferred is the same. (Unless you're considering the minor loss in mass, from fusion, for multi-nucleon nuclei, <1%.)

 

correct ?

 

No. You aren't comparing the same things. The density of the ISM being of order a proton per cm^3 will require interacting with 6 x 10^17 cm^3 (6 x 10^11 m^3) to reach a microgram. You can't use "same interval of time" because the impact with a single microgram particle doesn't take the same amount of time as hitting the equivalent amount of protons.

 

It's like saying that driving in the rain is like hitting a truck, because the accumulated mass of the water you collide with over some part of the life of your car is several tons.

 

[Widdekind]

huh ?

 

you made the analogy, of rain-drops to a truck, i.e. you said, that we pretend protons are raindrops, and smoke-sized dust grains are trucks.

 

fine. However, how big, in this analogy, is my space-craft ? If a proton is a rain-drop; and a dust-grain a truck; then, would not a macroscopic spacecraft be, in that analogy, the size of a world, or a star ??

 

And, how much would a world care, whether it ran into a scattered ten tons of rain water, or a single mac-truck ?

 

By implication, in that analogy, you & i are people, mid-way in size, between proton rain, and dust-grain trucks. So, evidently, we are large atoms, presumably in the forward facing "space umbrella". So, yes, the individual atoms, in the space umbrella, would react differently, to collisions w/ protons vs. dust. That would affect the nanoscopic details of the "space weathering" of the shield. In the meantime, macroscopically, momentum transfer, for large space-craft, would be the same, for the same amount of mass swept up. If our space-craft is the size of a world, then "rain-drops" would turn the surface into a fine powder; whereas "trucks" would scar & pit the surface, with visible AZ Barringer-sized craters. But from a macroscopic world-sized perspective, the momentum transfer would be the same.

 

correct ?

 

No. You aren't comparing the same things. The density of the ISM being of order a proton per cm^3 will require interacting with 6 x 10^17 cm^3 (6 x 10^11 m^3) to reach a microgram. You can't use "same interval of time" because the impact with a single microgram particle doesn't take the same amount of time as hitting the equivalent amount of protons.

 

It's like saying that driving in the rain is like hitting a truck, because the accumulated mass of the water you collide with over some part of the life of your car is several tons.

 

our sun's solar wind provides an "active defense" against incoming cosmic rays -- most are deflected away, all are decelerated to some degree, heading "up wind" towards the inner solar system. So, in analogy, perhaps an "active defense" would work well, for protecting craft, from the ISM ?

 

what if you had a "head-light", shining forward, tuned to the ionization frequencies, of H, He, dust ? then, you could ensure, that all material directly in front of the craft was ionized, and so affectable by the ship's EM field. For example, you could imagine some sort of "space ferry", whose bow & stern were identical, and which was hollow, although to accommodate the flow-thru of ISM, not cars. The cylindrically-shaped "space ferry" would have "head-lights" at the stern, for acceleration; and at the bow, for ionization. At destination, the bow head-light would increase in luminosity (and the stern would power down), to decelerate the ship. Ionized space material could be swept up, into a space ramjet.

 

Note, traveling to another star system is not like flying a plane to an airport -- the "airport" is itself moving. 'Tis more like a football quarterback throwing a pass (the ship) by leading the receiver (destination star system). The pass is thrown, to rendezvous with the receiver, "then there". But, before the ball connects with the receiver "then there", the receiver "here now" is running after the throw. So, if our football (the ship) had a headlight on it, none of the laser light would be seen by the receiver (until the ball was in their arms). So a spacecraft having a headlight would not be visible, to the destination star-system, until very near the time of arrival there. (you would have to be positioned, out in deep space, at the point of future rendezvous, at which position, you would see the inbound ship from one direction, and the oncoming star-system & planet, from some other direction, converging towards you, like standing on a football field, where the pass is thrown to, and where the receiver is running towards -- you'd see the football sailing straight at you from one side, and the receiver running towards you from some other direction.)

[Delta1212]

You are underestimating the comparative size of protons and dust particles. One microgram is approximately 1,000,000,000,000,000,000 protons, give or take a power of ten. That many raindrops is equivalent to a hell of a lot of trucks.

[swansont]

Widdekind, on 12 May 2013 - 15:01, said:

huh ?

 

you made the analogy, of rain-drops to a truck, i.e. you said, that we pretend protons are raindrops, and smoke-sized dust grains are trucks.

 

fine. However, how big, in this analogy, is my space-craft ? If a proton is a rain-drop; and a dust-grain a truck; then, would not a macroscopic spacecraft be, in that analogy, the size of a world, or a star ??

Go ahead and run the numbers for a spacecraft and interstellar protons. That was your example.

 

Widdekind, on 12 May 2013 - 15:01, said:

And, how much would a world care, whether it ran into a scattered ten tons of rain water, or a single mac-truck ?

But we're not talking about a world, are we?

Widdekind, on 12 May 2013 - 15:01, said:

By implication, in that analogy, you & i are people, mid-way in size, between proton rain, and dust-grain trucks. So, evidently, we are large atoms, presumably in the forward facing "space umbrella". So, yes, the individual atoms, in the space umbrella, would react differently, to collisions w/ protons vs. dust. That would affect the nanoscopic details of the "space weathering" of the shield. In the meantime, macroscopically, momentum transfer, for large space-craft, would be the same, for the same amount of mass swept up. If our space-craft is the size of a world, then "rain-drops" would turn the surface into a fine powder; whereas "trucks" would scar & pit the surface, with visible AZ Barringer-sized craters. But from a macroscopic world-sized perspective, the momentum transfer would be the same.

I (and others) have explained that this is not the case. How about addressing these arguments instead of simply repeating yours?

 

Widdekind, on 12 May 2013 - 15:01, said:

our sun's solar wind provides an "active defense" against incoming cosmic rays -- most are deflected away, all are decelerated to some degree, heading "up wind" towards the inner solar system. So, in analogy, perhaps an "active defense" would work well, for protecting craft, from the ISM ?

The OP did not ask about the ISM. Bringing it up (again) is moot.

[Widdekind]

a 1-ton spacecraft is ~e30 protons.

 

if each proton is a 1g raindrop, that's ~30g = e27kg

 

i.e. the mass of a world

 

correct ?

 

and so your analogy points to planets as the appropriate picture, for probes

 

Go ahead and run the numbers for a spacecraft and interstellar protons. That was your example.


But we're not talking about a world, are we?

I (and others) have explained that this is not the case. How about addressing these arguments instead of simply repeating yours?


The OP did not ask about the ISM. Bringing it up (again) is moot.

[swansont]

if each proton is a 1g raindrop, that's ~30g = e27kg

 

I have no clue what that is supposed to mean. A proton is not a raindrop.

[Widdekind]

using a crude single-burn-of-impulse approximation, you can calculate how much fusionable fuel is required, to accelerate the payload (plus fuel for deceleration) from departure, and the to decelerate the payload from destination.

 

With enough fuel mass (relative to payload mass), you can (theoretically) accelerate to any speed; and, the faster you go, the less fuel for deceleration (relative to total fuel) is required -- fuel usage is asymmetric, more is needed to accelerate (when you're pushing the fuel-for-deceleration too), than to decelerate (when you're simply stopping the payload part).

 

For fusion with e=0.007, you need total-fuel-to-payload ratios of hundreds to thousands, to reach semi-luminal speeds.

I have no clue what that is supposed to mean. A proton is not a raindrop.

 

you made the analogy, of rain-drops to a truck, i.e. you said, that we pretend protons are raindrops, and smoke-sized dust grains are trucks.

 

that is qualitatively accurate; quantitatively, as observed above, if we analogize protons to raindrops, as you yourself suggested, then microgram dustgrains would analogize to hundreds of billions of tons worth of water, i.e. hundreds of cubic km, i.e. "raindrops & asteroids" would be a more quantitatively correct comparison

[swansont]

using a crude single-burn-of-impulse approximation, you can calculate how much fusionable fuel is required, to accelerate the payload (plus fuel for deceleration) from departure, and the to decelerate the payload from destination.

 

With enough fuel mass (relative to payload mass), you can (theoretically) accelerate to any speed; and, the faster you go, the less fuel for deceleration (relative to total fuel) is required -- fuel usage is asymmetric, more is needed to accelerate (when you're pushing the fuel-for-deceleration too), than to decelerate (when you're simply stopping the payload part).

 

For fusion with e=0.007, you need total-fuel-to-payload ratios of hundreds to thousands, to reach semi-luminal speeds.

 

Which is still not what we're discussing.

 

you made the analogy, of rain-drops to a truck, i.e. you said, that we pretend protons are raindrops, and smoke-sized dust grains are trucks.

 

Yes. It was an analogy. Not a scale model. I never said pretend raindrops are protons. I implied there's a scale difference to be investigated, because hitting 10 tons worth of rain over the course of hours is not the same as hitting a 10 ton truck. Similarly, hitting protons is not the same as hitting a speck of dust. So we are ignoring the protons, just as we would ignore hitting raindrops if we investigated a collision with a truck.

 

that is qualitatively accurate; quantitatively, as observed above, if we analogize protons to raindrops, as you yourself suggested, then microgram dustgrains would analogize to hundreds of billions of tons worth of water, i.e. hundreds of cubic km, i.e. "raindrops & asteroids" would be a more quantitatively correct comparison

 

Again, it was an analogy, not meant as a quantitatively correct comparison.