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?[/
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.
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?