KE= 1/2 mv^2

m = mass of a hydrogen atom (not a molecule, interstellar hydrogen is about one atom per cubic meter, the last time I saw a figure quoted)

v = 2.85 x 10^8 m/s

KE units are joules

Calculate m by taking the mass of a mole of

*molecular*hydrogen, which is 2 grams and contains 6.022x10^23 (Avogadro's number) molecules. That means each molecule of hydrogen weighs 3.32 x 10-24 grams, so each hydrogen atom weighs 1.66 x 10-24 grams.

KE of each atom hitting is therefore 1/2 * 1.66 x10^-24 * (2.85 x 10^8)^2 or 1 x 10-7 joules.

The area of the face of a 500 meter diameter cylinder is pi*r^2 or 1.96 x 10^5 meters

Each meter forward in interstellar space (means therefore 1.96 x 10^5 hydrogen atom impacts per second.

Total energy per second (in spaceship time, which is relativistically altered of course by the speed) is 1.97 x 10-2 joules. Over an hour, this is 1.17 joules.

A 9mm Parabellum has about 600 joules of energy. Interstellar hydrogen atom collisions even at .95c are essentially trivial.

Anyone see any errors? Really important if you plan to build a light-speed spaceship.

In case it isn't obvious, I am a fan of hard sci-fi like Asimov and Larry Niven.

Excellent and useful comments.

And yes, I tried to type 95%, not 05%.

1/2mv^2 is just an approximation that's pretty good near zero velocity.

ReplyDeleteat relativistic speeds, you've got to use a slightly different formula.

https://en.wikipedia.org/wiki/Relativistic_kill_vehicle discusses it.

at 0.95c, i get that the right answer is ~4.88 times larger than what the approximation produces.

looking at your question, your doing an interstellar drag problem more then an impact question. In short how much energy you will have to constantly apply from the engines to maintain forward momentum at 0.5% of the speed of light.

ReplyDeleteOf course your first error is that your velocity value is actually 95% of the speed of light: unless that is what you wanted. The second error is that I think most engineers in the Space program are more concerned as to what a micro-meteor will do to your surface plate. At say 1 X10-9 grams, collision speed of 5% of light the impact computes to some 112,344 joules or almost 190 times the impact of a 9 mm. At 95% velocity of light, the impact energy is 40,612,500 joules.

5% or 0.5%? The former probably requires the relativistic formula (which I can't remember) be used, although the difference isn't much.

ReplyDeleteAt those speeds, you have to deal with special relativity. Kinetic energy approaches infinity as velocity approaches the speed of light.

ReplyDeleteSorry, my math is too rusty, so don't have the formula, but you can get it from the following discussion on Wikipedia:

https://en.wikipedia.org/wiki/Kinetic_energy

Way back when I ran a dimensional analysis of interstellar flight taking the phenomenon of skin heating on the SR-71 as a base and extrapolating to the interstellar density of 1 atom/cu meter. That suggested that comparable skin heating would show up on your space ship at about 1/3c. It also implies that streamlining would be desirable and external flight control surfaces would serve at that speed.

ReplyDeleteThe more we learn about interstellar space the more it looks like flying at relativistic speeds, >.25c, would be a lot like flying through a sand storm. Armor begins to look pretty attractive at those speeds.

I believe (it's been 40 years since my Aerospace Eng degree) that the issue isn't energy imparted, it's abrasion by whatever interstellar particles you hit. This requires the generic 'particle shielding' most serious space SF writers come up with.

ReplyDeleteOf course, I could be wrong.

At 95% of the speed of light, you need to factor in relativistic mass increases. There's a complicated formula for calculating relativistic acceleration and changes in momentum, which I can never remember.

ReplyDeleteSo I cheat.

I calculate the change in relativistic mass at velocities of interest and recognize that every bit of mass above the rest mass is the result of energy added to the system. This change is simply the total relativistic mass minus the rest mass, or (γ - 1)m.

The gamma factor, γ = 1/(1 - v²/c²)

At v/c = 0.95, γ = 10.26

Of this, 9.26 times the mass of the hydrogen is relativistic mass increase due to the change in velocity. So each gram of hydrogen is requiring 9.26 mc² ergs of energy to bring up to the speed of the ship.

Given a cross-sectional area of roughly 2 * 10^5 square meters, and a speed of 2.85 * 10^8 m/s, your ship is passing through a volume of 5.7 * 10^13 cubic meters every second. At one atom per cubic meter, that's 5.7 * 10^13 atoms per second.

A mole of atomic hydrogen, 6.02 * 10^23 atoms, has a mass of one gram, give or take. The ship is colliding with 0.95 * 10^-10 moles or 9.5 * 10^-11g of hydrogen per second.

The total relativistic kinetic energy transferred per second is

E = (γ-1)mc² = 9.26 * 9.5 * 10^-11 * (3 * 10^8)²

= 9.26* 9.5 * 9 * 10^(16-11)

= 792 * 10^5

= 7.95 J/s

I have not adjusted for relativistic length contraction from the ship's point of view. Because of time dilation, the ship will experience interstellar gas having a density higher by a factor of γ, increasing the energy transfer by another factor of 10.26.

Call it 81 J/s in energy transfer. This is the amount of energy you have to apply as thrust to keep the ship at speed.

If you're calculating the amount of slowing effected on a space ship by this energy transfer, remember that the ship also has a lot of relativistic kinetic energy. It will have to shed an amount of energy equal to 9.26c² times the mass of the ship in order to come to rest.

I don't vouch for it, but the first Google hit was here.

ReplyDeleteThe risk, he seems to indicate, is not so much kinetic energy transfer but plain old radiation effects.

"The second error is that I think most engineers in the Space program are more concerned as to what a micro-meteor will do to your surface plate." Yes micrometeors are the bigger problem.

ReplyDelete