- 05 Oct '09 03:29A trailer for a TV show had a woman saying kinetic weapons in space could hit something with the force of a small atomic bomb. My question is, what does that mean in terms of Joules? I know Ke=MV^2 and such but don't know how many Joules a small A bomb produces, say a 10 Kiloton job. I would assume a mass of a projectile of say 10 Kg, just a guess of course but using those figures, how fast would it have to be going to have kinetic energy = to a 10 Kiloton bomb?
- 05 Oct '09 03:37

Convert 20 kilotons of tnt energy into joules. Plug it into your kinetic energy formula making sure the units are correct and there is no way in heck fire we can fire something that weighs 10 kg that fast.*Originally posted by sonhouse***A trailer for a TV show had a woman saying kinetic weapons in space could hit something with the force of a small atomic bomb. My question is, what does that mean in terms of Joules? I know Ke=MV^2 and such but don't know how many Joules a small A bomb produces, say a 10 Kiloton job. I would assume a mass of a projectile of say 10 Kg, just a guess of course ...[text shortened]... hose figures, how fast would it have to be going to have kinetic energy = to a 10 Kiloton bomb?** - 05 Oct '09 12:00 / 7 edits

Well sure, but how many joules are there in 20,000 tons or 10,000 tons of dynamite? Or TNT? Are there more Joules in that much C4?*Originally posted by joe beyser***Convert 20 kilotons of tnt energy into joules. Plug it into your kinetic energy formula making sure the units are correct and there is no way in heck fire we can fire something that weighs 10 kg that fast.**

I got off my lazy asss and found a Wiki link:

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

It seems 10,000 tons of TNT=4.184 E 13 Joules, 41 trillion joules.

That's about a megawatt going for 1200 years. Wow.

Wasn't sure of the units in the MV^2 formula but it turns out to be Kg and Meters/sec.

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

So the example gives 80 Kg going 18 meters/second as having 12,960 joules. I am glad the units are Joules, that simplifies things!

So Ke=MV^2, solving for V, would be V=(Ke/M)^0.5 and we can also solve for M, Ke/V^2=M

So for 10 Kg, it would be (4.184E13/10) or 4.184E12^0.5 or about 2E6 meters/sec.

So 10Kg going 2,000 Km/sec. would have the energy of an atomic bomb of 10 Kilotons. While very high, you can see it is not a relativistic velocity, being less than 7% of C. Still a very high velocity. If the mass was smaller, say 1 Kg, it would be the sqr root of 4.184E13, it would have to go about 3 times faster or 6,500 Km/sec which is pushing on towards relativistic, at over 2% of C.

Lets look at the Kinetic energy of one gram going the speed of light:

C is close to 300 million Meters/sec, close enough anyway for our purposes so square that =9E16 times .001= 9E13 joules. That seems to be about the equivalent of a 20,000 ton bomb! One GRAM going at the speed of light or close.

So one gram going 0.75C would still have the equivalent of a 10,000 ton A bomb! Don't think I'd like to be on the receiving end of that weapon!

So if we had an electromagnetic rail gun, to get 10Kg to 2,000 Km/sec would require the same amount of energy but if it was available lets see what length of the rail you would need:

V=A*t so if you use 1,000 G's, or 9800M/sec^2, T=V/A, so 2,000,000/9800 would require 204 seconds, quite a length of time! so it seems to make it 200 milliseconds, you would need to accel at 1 million G's. So suppose you could do that, then the distance formula says:

S=(A(T^2))/2 which for one million G's for 200 Milliseconds=about 20 meters! But it would consume about 40 trillion watts doing it! 40 Terawatts/seconds! But not against the laws of physics! For the version going one megaG, at 200 milliseconds, that would require pumping in 200 Terawatts for that length of time, 5 times the actual joules, but it still works out to 40 terawatts/seconds of energy used.

If you used the 200 second at 1000 G version, it would require a rail gun 20 Km long but to have to wait over three minutes for a firing would not make it very valuable from a star wars perspective, eh.

So a realistic weapon like that aboard a spacecraft would appear to require a power source using anti-matter to get that much energy in a short period of time in a real life situation. I think that would put such a weapon out of the 21st century and into maybe the 23rd, for sure!

Would someone verify my calculations? - 06 Oct '09 01:17

I do not possess the mathamagical skills to check your work, but I do have some food for your calculating thought. What if the smallest mass of whatever fissionable material that would give the least amount of energy and still be critical mass was used as the basis for the small nuclear weapon that was used in the comparison of the kinetic energy weapon. Is it possible to fire any mass fast enough to equal its energy? In other words it may be somewhat less than 20 kilotons which would make it more likely. What is the smallest energy yield of a fission bomb? I am thinking it still is imposible.*Originally posted by sonhouse***Well sure, but how many joules are there in 20,000 tons or 10,000 tons of dynamite? Or TNT? Are there more Joules in that much C4?**

I got off my lazy asss and found a Wiki link:

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

It seems 10,000 tons of TNT=4.184 E 13 Joules, 41 trillion joules.

That's about a megawatt going for 1200 years. Wow.

Wasn't s ...[text shortened]... 21st century and into maybe the 23rd, for sure!

Would someone verify my calculations? - 06 Oct '09 04:57

I think the lowest yield is way under a kiloton, I think there are atomics that would fit inside a regular hand grenade. Cheery thought, eh. Still a hell of a lot of joules. I don't doubt for a minute that eventually a rail gun in space, say with superconductors to conduct the power will be built that can deliver that much accel, like I said, a million G's, and since it would be delivering a kinetic weapon, the actual mass would not need any electronics or optics and therefore could indeed stand the stress of such incredible forces.*Originally posted by joe beyser***I do not possess the mathamagical skills to check your work, but I do have some food for your calculating thought. What if the smallest mass of whatever fissionable material that would give the least amount of energy and still be critical mass was used as the basis for the small nuclear weapon that was used in the comparison of the kinetic energy weapon. ...[text shortened]... ikely. What is the smallest energy yield of a fission bomb? I am thinking it still is imposible.**

I am also just as sure it won't happen in the 21st century. Or maybe even the 22nd. They might even come up with a better technology like a field accelerator, that is to say a field surrounding the mass and accelerates all the atoms at exactly the same rate, something that does not happen to mass being accelerated by a regular rocket. Rockets slightly compress the bottom of the mass first and then proceeds through the mass being accelerated sort of like a sound wave, the point being it does not happen all at once everywhere in the accelerated mass but first in the portion of the mass closest to the rocket motor, then spreading upwards to the end of the mass. The field accelerator that I envision acts on every atom simultaneously so there is no differential field felt so a person could accel at one million g's with no effect, no squashing against the rear bulkhead, thank you very much

Right now it seems beyond any physics we know about but who knows in a hundred years, 2 hundred years.... - 07 Oct '09 03:02 / 1 edit

for starters*Originally posted by sonhouse***I think the lowest yield is way under a kiloton, I think there are atomics that would fit inside a regular hand grenade. Cheery thought, eh. Still a hell of a lot of joules. I don't doubt for a minute that eventually a rail gun in space, say with superconductors to conduct the power will be built that can deliver that much accel, like I said, a million G's, seems beyond any physics we know about but who knows in a hundred years, 2 hundred years....**

Ke = (1/2)(m)(v^2)

so

v = sqrt((2*Ke)/m) - 07 Oct '09 12:20

Oops, thanks! Lets see what the real # is: All my numbers are 2^0.5 off, or 1.414 times too high, so my first # of 2E6 Meters/second would be more like 1.414 E6 Meters/second, 0.707 times the original calc. So would the rest, so just multiplying by 0.7 would give the right #'s for velocity. Still the figures are accurate to within 30%. It looks like a million G launcher would then be 19 meters * 0.707 or a bit more than 13 meters long. Still in the ballpark, eh.*Originally posted by joe shmo***for starters**

Ke = (1/2)(m)(v^2)

so

v = sqrt((2*Ke)/m) - 07 Oct '09 15:21

actually your numbers are to low*Originally posted by sonhouse***Oops, thanks! Lets see what the real # is: All my numbers are 2^0.5 off, or 1.414 times too high, so my first # of 2E6 Meters/second would be more like 1.414 E6 Meters/second, 0.707 times the original calc. So would the rest, so just multiplying by 0.7 would give the right #'s for velocity. Still the figures are accurate to within 30%. It looks like a mill ...[text shortened]... r would then be 19 meters * 0.707 or a bit more than 13 meters long. Still in the ballpark, eh.**

the actual equation solved for velocity

v = sqrt(2)*sqrt(Ke/m)

now your equation was

V1 = sqrt(Ke/m)

so

V1*x = v

x*sqrt(Ke/m) = sqrt(2)*sqrt(Ke/m)

x = sqrt(2) - 08 Oct '09 01:34 / 2 editsWhen I see kinetic weapons being compared to nukes in this way it's usually with respect to Project Thor/kinetic bombardment.

*The most described system is 'an orbiting tungsten telephone pole with small fins and a computer in the back for guidance'...*

The idea is that the weapon would inflict damage because it moves at orbital velocities, at least 9 kilometers per second. The amount of energy released by the largest version when it hits the ground is roughly comparable to a small nuclear weapon

http://en.wikipedia.org/wiki/Kinetic_bombardment#Project_Thor - 09 Oct '09 03:17

density doesn't have to do with kinetic energy.*Originally posted by sonhouse***What dif does it make the density? I thought it was just the mass and velocity that made up kinetic energy.**

what ATHY did was this

density = mass/Volume

so

m = d*V

and

Ke = (1/2)(d*V)(v^2)

but he too forgot the factor of (1/2).... - 09 Oct '09 06:23

One thing is certain, and that is if an object is moving fast enough to have the kinetic energy of a small atom bomb, then at least that amount of energy is required to accelerate it. No matter if it is an ounce of tungsten or a couple of tons.*Originally posted by joe shmo***density doesn't have to do with kinetic energy.**

what ATHY did was this

density = mass/Volume

so

m = d*V

and

Ke = (1/2)(d*V)(v^2)

but he too forgot the factor of (1/2).... - 09 Oct '09 06:32

Not so sure about that...*Originally posted by joe beyser***One thing is certain, and that is if an object is moving fast enough to have the kinetic energy of a small atom bomb, then at least that amount of energy is required to accelerate it. No matter if it is an ounce of tungsten or a couple of tons.**

What space probe has highest velocity? How massive is it? What velocity has it? Use the formula and tell me its kinetic energy.

Is this energy really put into the space probe, or did it get its energy from somewhere else? - 09 Oct '09 19:38 / 1 edit

If you think about the kinetic energy of a falling object, accelerating in a gravity field, where does that energy come from? It's not free, of course, so what gets reduced when, say, an asteroid strikes the earth, it follows the geodesic of the earths mass and picks of 'free' energy.*Originally posted by FabianFnas*

[b]Not so sure about that...

What space probe has highest velocity? How massive is it? What velocity has it? Use the formula and tell me its kinetic energy.

Is this energy really put into the space probe, or did it get its energy from somewhere else?[/

For instance, there is a plan afoot, probably never to be used, but here is the theory: To get materials to and from space, say from earth orbit to the surface of the moon and back, you could of course use rockets just as we do today, but there is another way. That is a satellite that has a swinging cable that comes down to the top of the atmosphere, can't go lower or the tip would burn up.

Normally you have to burn enough energy to get into orbit in one go, but with the swinging cable affair, you have say, a big magnet on the tip or some kind of grabber, then a rocket comes up with its payload but without help would be a sub-orbital flight, like the first Burt Ratan spacecraft that got up to 70 miles or so. This time, instead of following the parabola back to earth, the spinning cable grabs the spacecraft and slings it up to full orbital speed. Naturally, this removes some kinetic energy from the orbiting space station at the center of the cable's swing.

So to make up for that energy, there is a scheme that launches stuff to the moon and there you make a solar powered electromagnetic launcher, say 1 Km long or so with the end pointed up, which can vary its angle and direction of launch. So, the energy required to launch from the moon is about 1/6th that of earth, an escape velocity of about 2 and1/2 odd Km/sec as opposed to the 11 or so Km/sec escape velocity of the earth.

The idea here is the space station in Earth orbit has a magnetic DEccelelerator that catches stuff launched from the surface of the moon.

Now since the escape velocity of the moon is so much less, when you aim the stuff at Earth, the Earths gravity well accelerates the payload to around earth's escape velocity, but here is where it gets interesting:

The space station has a magnetic track like an electromagnetic launcher in reverse(actually these launchers are reversable) so when a mass coming in from the moon VERY carefully aimed and timed to intercept the magnetic track on the space station, when it glides into the track, the magnetic fields interact like a linear electric motor in reverse and so generates an enormous amount of electricity in a very short time frame which theoretically can be stored, say in a superconducting power ring.

That means the space station has turned however much kinetic energy from the incoming moon launch into practical power which can be used to power the magnetic launcher in launch mode and send craft anywhere in the solar system or can use microwave beams to beam energy down to earth. The study was done with hypothetical 800 Kilogram launches from the moon and the conclusion was the energy recovered from the combination solar launch site and incoming acceleration of the Earths' gravity well imparts about 6 times as much energy in the same mass (800 Kg) of burned gasoline!

So the incoming mass can be used to expand the space station and the excess can be used to either ship energy to the ground VIA microwave radiation or to accelerate space craft to any part of the solar system, all for free, energy wise. So what does the system pay (that is to say, the Earths total energy system) to get that much energy?

But if you look at an EM rail gun, the rest energy (kinetic) could be said to be zero if the receiving end is say, in orbit around the sun going the same velocity as Earth, so any change in kinetic energy HAS to be inputted to the load by the energy put into the rail gun. Obviously that would not be 100% but if the rail gun was superconductive the efficiency of conversion of electrical energy to mechanical energy would be very close to 100% due to it being in space in the first place and on a frictionless magnetic linear bearing in the second place, making it a very good match of energy input and resultant kinetic energy. Remember, when it reaches the end of the rail gun and is away from the influence of the magnetic acceleration field, it is no longer being accelerated so it is stuck with the amount of relative kinetic energy that it had at that point till it strikes an object, enemy spacecraft, moon or whatever and will impact with the difference in velocities of the two masses.

It could for instance, ram into the moon like the LCROSS mission just did, we can see what the kinetic energy of that system is, I think they said it was about 700 Kg impacting the moon at about 1.5 Km/sec, if those #'s are close, the impact energy of that strike is about 700 million Joules or near a gigawatt second energy, Gigawatt for a second, or about 13 megawatts running for a minute or over 200,000 watts going for an hour. So around 200,000 Kilowatt/hours of energy in that one strike. Nothing like an atomic bomb but one hell of a lot of energy!

It got that energy from the rocket that accelerated it from earth but in that case the kinetic energy of Earths' escape velocity was SUBTRACTED from the total energy since it had to climb out of the Earths' gravity well so the total energy inputted was the escape velocity MV^2/2 minus the end velocity which was the positive acceleration provided by the moon, since the maximum velocity given by the original rockets was just the escape velocity plus a little bit to get into a minimum trans-lunar injection curve, I think it was en route for several months going in smaller and smaller lunar orbits ending in this impact.

So the energy gain from my theoretical space station/moon launcher would be this, lets us 11.4 Km for Earth's escape velocity and 2 Km/second for lunar escape velocity.

So 800 Kg going 2,000 meters/second is 4 E6* 800= 1.6 E 9 Joules, 1.6 Gigawatt/second or 444,444 Kilowatt hours of energy. That is the energy it takes to get it off the moon. So lets see how much it ends up with when it arrives near Earth's orbit ready to intercept with the theoretical space station EM rail gun running in reverse to slow it down to Earth's orbital velocity. That velocity is about 8 Km/sec but it would be coming in at about 11.4 Km/sec or 3.4 Km/sec relative, assuming they are going in the same direction. If in the opposite direction the difference would be 19.4 Km/sec, but we will deal with just the same direction version:

Remember, it started off with 1.6 gigajoules to get it on its way. It would have to go a bit faster to clear the moon's gravity and get to the point between the moon and earth where the gravity is equal and to enter the earths field but 1.6 gigajoules is good enough for us right now.

So it was given 1.6 Gigajoules which gets lost to the moon but gets back 3.4 Km/second when it reaches Earth orbit, so that is 3400 squared which is 11,560,000 times 800 divide by 2 which is now a rather outstanding 4.6 Gigajoules or over 1.2 MEGAWATT hours of energy, all seemingly for free. So where did THAT energy come from?