- 30 Jan '10 12:24

My thoughts:*Originally posted by wolfgang59***Consider a very fast (non-relativisic) space-ship travelling at 1,000,000 m/s.**

An astronaut inside of mass 100 kg walks towards the front at 1m/s.

a) What was his KE standing still?

b) What is his KE walking?

c) What is the increse in his Kinetic Energy and where does it come from?

1) K = 0.5*m*v*v = 0.5 * 100 * 1,000,000*1,000,000 = 50,000,000,000,000 J

2) K = 0,5*m*v*v = 0,5 * 100 * 1,000,001*1,000,001 = 50,000,100,000,050 J

3) delta K = 50,000,100,000,050 - 50,000,000,000,000 = 100,000,050 J. The energy comes from the person walking, but also from the spacecraft engine. The person walkking forward exerts a force on the craft slowing it down. THe engine has to do more work to maintain its velocity of 1,000,00 m/s. - 30 Jan '10 12:47

That is what I thought. But when I did the calculations it didnt stack up!*Originally posted by TheMaster37***My thoughts:**

1) K = 0.5*m*v*v = 0.5 * 100 * 1,000,000*1,000,000 = 50,000,000,000,000 J

2) K = 0,5*m*v*v = 0,5 * 100 * 1,000,001*1,000,001 = 50,000,100,000,050 J

3) delta K = 50,000,100,000,050 - 50,000,000,000,000 = 100,000,050 J. The energy comes from the person walking, but also from the spacecraft engine. The person walkking forward exerts a ...[text shortened]... craft slowing it down. THe engine has to do more work to maintain its velocity of 1,000,00 m/s.

Assume engines are OFF. assume Ship mass of 1,000,000 kg

Determine new velocity and calculate ship's loss in KE . - 30 Jan '10 16:12

At these velocities the graph of Ke vs. V is practically a vertical line, increasing the velocity by slight amount yeilds a massive increase in kinetic energy. So im afraid im not seeing the conundrum?*Originally posted by wolfgang59***Consider a very fast (non-relativisic) space-ship travelling at 1,000,000 m/s.**

An astronaut inside of mass 100 kg walks towards the front at 1m/s.

a) What was his KE standing still?

b) What is his KE walking?

c) What is the increse in his Kinetic Energy and where does it come from? - 30 Jan '10 16:40

Well I'm obviously doing something wrong!*Originally posted by joe shmo***At these velocities the graph of Ke vs. V is practically a vertical line, increasing the velocity by slight amount yeilds a massive increase in kinetic energy. So im afraid im not seeing the conundrum?**

My problem is at huge speeds the Energy equation involves squaring the new velocity

(V+v)^2 = V^2 + 2vV + v^2

The 2vV term means that for large V a small increase in velocity equates to a huge amount of Energy ... wher's it coming from? - 30 Jan '10 17:14 / 1 edit

Sorry - I prefer my calculations without so many numbers in them ðŸ™‚*Originally posted by wolfgang59***That is what I thought. But when I did the calculations it didnt stack up!**

Assume engines are OFF. assume Ship mass of 1,000,000 kg

Determine new velocity and calculate ship's loss in KE .

OK, the one thing we know is that momentum is conserved.

v = Initial spaceship velocity = 1,000,000 m/s

V = Final spaceship velocity

u = Final astronaut velocity wrt ship = 1m/s

m = astronaut mass = 100 kg

M = spaceship mass (without astronaut) = 1,000,000 kg

OK, we know is that momentum is conserved.

(M + m)v = MV + m(V + u)

=> V = v - mu/(M + m)

=> V = 1000000 - 1/10001 (~ 999999.9999)

Increase in KE of astronaut

= 0.5m(V + u)^2 - 0.5mv^2

= 0.5m[v + Mu/(M + m)]^2 - 0.5mv^2

= 0.5mMu[2v + Mu/(M + m)]/(M + m)

"Increase" in KE of ship (going to be -ve)

= 0.5MV^2 - 0.5Mv^2

= 0.5M[v - mu/(M + m)^2] - 0.5Mv^2

= 0.5Mmu[mu/(M + m) - 2v]/(M + m)

And the increase in total KE:

= 0.5Mmu^2/(M + m)

This is independent of v - which it has to be, because the increase in energy must be frame-independent. The energy comes from the work done by the astronaut in walking. - 30 Jan '10 21:06 / 1 edit

It seems to me there should be no net increase or decrease in the KE of the total system, the spacecraft. Think about a solenoid. Say there is a solenoid with a 100 Kg mass being driven, up a rod the full length of the craft, whatever that is, say 10 meters.*Originally posted by mtthw***Oh - and you'll notice that since M >> m this is approximately 0.5mu^2.**

So an electromagnet cuts in at the back of the ship, accelerates the 100 Kg to its max velocity, whatever that is, and at the other end there is another magnet and some kind of damper that slows down the mass and the reverse happens, the solenoid magnet pushes the mass back to the starting point at the back of the craft. You do that, you can see that you subtracted KE enough to get the 100 Kg moving to whatever it maxed out at, then decelerated at the other end, conservation of momentum would say while there would be a minor oscillation in the group velocity of the spacecraft, if it took say, 10 trillion microseconds exactly to do the trip it was sent to do with the 100 Kg not moving, the trip time would still be exactly 10 trillion microseconds with the mass moving from back to front and to the back again, no matter how many times that took. Even only one transfer of mass from front to back would not change the KE of the craft because the energy it took to get the mass moving would be given back to the system when it reaches the other end, unless the mass went through a door and was ejected into space. In that case there would be a slight change in the total time it took the system to get to its destination as compared to the time it took without the ejection, depending on whether the ejection was in front or back, just like a rocket motor shooting out the front to slow it down or shooting out the back to speed it up, all at the expense of mass being ejected from the system. - 30 Jan '10 22:22
*Originally posted by mtthw***Sorry - I prefer my calculations without so many numbers in them ðŸ™‚**

OK, the one thing we know is that momentum is conserved.

v = Initial spaceship velocity = 1,000,000 m/s

V = Final spaceship velocity

u = Final astronaut velocity wrt ship = 1m/s

m = astronaut mass = 100 kg

M = spaceship mass (without astronaut) = 1,000,000 kg

OK, we know is th ...[text shortened]... y must be frame-independent. The energy comes from the work done by the astronaut in walking.**OK, the one thing we know is that momentum is conserved.**

No momentum is not conserved ... the astronaut exerts a force on the spaceship in order to walk forward. If there was no force there could not be any Energy involved.

Anyway - I am interested in the KE before and after the astronaut increases his speed. It seems huge. - 30 Jan '10 22:24

As Master37 pointed out energy is being put into the system - you cannot asssume conservation of KE.*Originally posted by sonhouse***It seems to me there should be no net increase or decrease in the KE of the total system, the spacecraft. Think about a solenoid. Say there is a solenoid with a 100 Kg mass being driven, up a rod the full length of the craft, whatever that is, say 10 meters.**

So an electromagnet cuts in at the back of the ship, accelerates the 100 Kg to its max velocity, wh ...[text shortened]... shooting out the back to speed it up, all at the expense of mass being ejected from the system. - 31 Jan '10 09:12

THIS POST IS RUBBISH*Originally posted by wolfgang59***[b]OK, the one thing we know is that momentum is conserved.**

No momentum is not conserved ... the astronaut exerts a force on the spaceship in order to walk forward. If there was no force there could not be any Energy involved.

Anyway - I am interested in the KE before and after the astronaut increases his speed. It seems huge.[/b]

Dont know what I was thinking (or drinking!!) - 31 Jan '10 14:09

If all the matter involved was inside the craft, the change in KE of the inner mass would HAVE to come from the KE of the craft, law of action and reaction shows that so it would have to gain energy at the expense of changing the KE however slight, of the whole craft, both plus and minus. When the mass gets shot forward, it gets that at the expense of slightly slowing down the main craft and when stopped by the front of the craft, gives that KE back to the ship. It cannot be any other way other than maybe converting some of the acceleration energy into heat to be dissipated into space eventually.*Originally posted by wolfgang59***As Master37 pointed out energy is being put into the system - you cannot asssume conservation of KE.** - 31 Jan '10 14:10 / 2 edits

sorry for the double post, hit the wrong button. The point of this post is to show you could slow down a craft by continually moving a mass up and down the length of the craft and converting that energy into heat, it would not be particularly fast but seems like it could eventually shed all the extra KE.*Originally posted by sonhouse***If all the matter involved was inside the craft, the change in KE of the inner mass would HAVE to come from the KE of the craft, law of action and reaction shows that so it would have to gain energy at the expense of changing the KE however slight, of the whole craft, both plus and minus. When the mass gets shot forward, it gets that at the expense of sligh ...[text shortened]... e eventually. Sounds like a way to slow down the craft though, turn internal movement into heat.** - 31 Jan '10 15:54 / 1 edit

No, I think that's wrong.*Originally posted by sonhouse***sorry for the double post, hit the wrong button. The point of this post is to show you could slow down a craft by continually moving a mass up and down the length of the craft and converting that energy into heat, it would not be particularly fast but seems like it could eventually shed all the extra KE.**

Unless there's an external influence (gravity, friction, etc) then momentum must be conserved. When the person in the craft starts moving they put energy into the system (as per the calculation above). When they stop moving the energy gets returned as heat. These exactly balance each other.

It's important to remember that the living human can be a source of energy. - 31 Jan '10 16:08

Yes that is totally true but like an open door in front or back of the craft where mass can be ejected rocket-like is the only case where actual matter leaves the craft. In the case of heat, it is eventually like an open door to the outside, heat loss cannot be avoided, no matter what kind of technology you put in place to conserve heat, some of it will eventually escape and whatever thrust the evolution of heat provides would most likely be isotropic, going out in all directions equally, so there would be no net thrust from radiation and even if there was, the actual thrust would be way more than the loss of energy inherent in escaping radiation. In the case I described, KE would be turned into heat which would eventually be radiated, thus slowing down the craft if that continued. There is no way around that fact.*Originally posted by mtthw***No, I think that's wrong.**

Unless there's an external influence (gravity, friction, etc) then momentum must be conserved. When the person in the craft starts moving they put energy into the system (as per the calculation above). When they stop moving the energy gets returned as heat. These exactly balance each other.

It's important to remember that the living human can be a source of energy.