E = 1/2 mv ^ 2 ?

Question

Let's try this again. I typed in 'c' for 'v' by mistake in the eariler question. My bad.

So, about the equation ' E=1/2MV^2' ? Can the answer to practical space travel be found in this equation?

Eddie, what if I were thinking along the lines of harnessing gravity and turning it against itself?

What if this included ping-ponging a projectile into or away from separate gravitational fields?

James, you're close enough ... but are you not considering the possibility of manipulating or negating the gravitational forces, as I suggested, by using the gravitational forces against themselves?

Answer 1

This equation is for kinetic energy and is more for classical mechanics (Newtonian physics). In a simplified world there is no friction. An object in motion stays in motion forever

Space travel there is almost zero friction (you will go forever, in theory). However the distances are so great that you can't get V high enough to get anywhere fast enough.

So the energy it takes to make a 100 kg mass (about 220 lbs) the speed of light is =1/2*100 kg * 3x10^8 m/s = 1.5*10^10 J. To put it in perspective that's more energy than 1000 TONS of TNT (4.18x10^9 J). So in all practical (and controllable) purposes the speed of light takes infinite energy to move a mass of any size.

But let's say you can go that speed. You have a problem. Gravitational effects. These are related by the equation.

F=GMm/r^2 where M and m are the masses of the two objects and r^2 is the radius between the objects. F is the resulting attractive force between objects.

If you fly too close to a large mass you will either enter orbit or crater into the object.

E=mc^2 was Einstein's theory of realtivity (c=speed of light). Apparently in his hypothesis the 1/2 term used in Newtonian physics disappeared (I don't know why but it might be related to the loss of a constant gravity term in space).

Hope that made sense... :)
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You can slingshot off planets (that's currently the way JPL and NASA take significant amounts of time off of travel from here to Mars and other local bodies).

This works well within our solar system (where we (being the smart people doing this for a living) are familiar with the workings and have math models for solar system). However this is just a velocity boost.

You can only get so much bang for the buck out of doing this. Going into a decaying orbit would cause you to accelerate but then you need a monster about of energy to break away from the planet's gravitational pull - meaning you've just burned off some the energy you picked up.

Also the further out from our solar system the object you want to go fast, gets the more and more difficult the math gets since your error compounds over distance. It was only a very small unit conversion error that (supposedly) cratered a satellite into Mars.

It's great stuff for science fiction. But, for the moment, going lightspeed is just science fiction. That being said, many of the concepts in science fiction have become reality as technology advances.

So making a long answer short - No. The answer to space travel is not in that equation - there's a lot more to it. Check out nasa.gov or jpl.gov and see what they say about manned and unmanned space travel.

Answer 2

I couldn't find your previous question.

The correct form for the kinetic energy of a mass m moving with a speed V is

K.E = m C^2 {[1- y}^ [ -0.5] ─ 1 }

Where y = [V/C] ^2 in which V is the velocity of the object and C is the velocity of light and m is the mass when the body is at rest.

K.E =
m C^2 {1 + 0.5y + 0.375 y^2 - 0.3125 y^3 + 0.2734375 y^3 - 0.2460937y^4 etc. ─ 1}

= m C^2 {0.5y + 0.375 y^2 - 0.3125 y^3 + 0.2734375 y^3 - 0.2460937y^4 etc.}


In y = [V/C] ^2 if V/C is very small, [V/C] ^2 will be still small and hence except the first one, all terms in the bracket becomes negligible and the form reduces to
m C^2 * 0.5y = 0.5 m V^2.

Therefore, only when the speed V is very less compared to C, we can use the classical equation E = m V^2 /2.

Answer 3

This is simply the classical equation for the kinetic energy of a moving object. Naturally, this comes into serious account when considering an object moving through space, but one must also take potential energies into account as well, as gravitational potentials allow us to maneuver ourselves in space.

Answer 4

To broaden on taurianyoj's reply, mc^two is a time period that variety of dropped out of Einstein's paintings while he used to be mucking approximately with complete power after taking exact relativity under consideration. The complete power of some thing is its kinetic power plus its expertise power (to significantly simplify, of direction). However, as you method the velocity of sunshine, you get alteration of mass, distance and time with the intention to be certain that every body consistently measures the velocity of sunshine as being the identical. Einstein plugged those elements in after which did a few algebra to split the whole thing again out. Once he'd performed so, he determined that he had a time period left over, mc^two (strictly talking, m is m-sub-0, the "leisure mass"). After making definite he hadn't erred someplace, he learned that this used to be the power that represented simply current. Thus, complete power is established on kinetic (how rapid you are relocating), expertise (in which you're when it comes to different stuff) and leisure power (how a lot of you there's within the first situation). E=mc^two is not a declaration of the speculation of relativity, it is a very exciting result of that thought. As for why the a million/two in kinetic power, that is seeing that of calculus. :) The price of difference in kinetic power is vigour, that's drive instances speed (amongst different matters). dE/dt = d/dv(a million/two mv^two)*dv/dt = mv * dv/dt = mva = mav = Fv.

Answer 5

That equation is only valid for low velocities. For relativistic velocities the equation is
E^2 = (pc)^2 + (mc^2)^2
p = momentum, m = rest mass (usually written m0, where the 0 is a subscript).

Answer 6

no this only refers to general kinetic energy. objects here on earth.