"Simple" Rocket Science: Where have I gone wrong?

  • 2 Replies

0 Members and 1 Guest are viewing this topic.


Offline harrogate22

  • First timers
  • *
  • 1
    • View Profile
This problem has nagged me for a long time, and yes; it IS rocket science!

Assume a rocket with mass m in free vacuum space switches the on its engine for time t at a set throttle level.  The fuel-burn is constant, so force developed is constant, and hence the rocket's velocity will rise in a linear fashion.

(For simplicity I have ignored the change in mass due to fuel useage; let's assume it's an ion drive)

So after time t the velocity is v, and the kinetic energy of the rocket is (1/2) * m * v * v

The rocket now coasts along for an indeterminate time using no fuel, and maintaining its kinetic energy.

The engine is now switched on again to repeat exactly the acceleration as above for the same time and with the same fuel useage.  It's velocity is now 2v (isn't it; it has, after all, carried out two bursts of identical acceleration)

The problem now is that the rocket's kinetic energy is (1/2 * m * (2v) *(2v)), and this is four times the kinetic energy it had after the first fuel-burn.

So the question is this: how can the rocket gain more (double, in fact) energy kinetically than has been used in the two fuel-burn sessions.

Clearly there is an error, and onecan easily demonstrate by other means how the books do actually balance.  The problem I find is in identifying the error in the above reasoning.


Offline Soul Surfer

  • Neilep Level Member
  • ******
  • 3345
  • keep banging the rocks together
    • View Profile
    • ian kimber's web workspace
"Simple" Rocket Science: Where have I gone wrong?
« Reply #1 on: 05/01/2008 23:41:58 »
The simplest answer is just "that's the way the equations work" but that does not solve your problem because you have shown it clearly with your mathematics.

When you fired the rocket motors what has increased linearly is the momentum.  The momentum is the mass times the velocity so firing the motors a second time you just doubled the momentum.

Now to understand energy properly we need to get involved with differential calculus.

Now burning the rocket motors produces a constant force and assuming a constant mass a constant acceleration because force equals mass times acceleration. Momentum is one of the important things in nature that are conserved like energy but most people latch on to the conservation of energy.

The definition of energy is the capacity to do work. Now work is not just a force (pushing against a brick wall as hard as I like does not do anything useful unless I manage to move that brick wall!)it is a force times a distance moved under the influence of that force.

Now think about your rocket when I accelerated it from 0 to v with the first burn it traveled say a distance S in time T to achieve a velocity v

The second time it was already moving with a velocity v so during the time t that it is accelerating it goes a lot further because during the period it starts at v and ends at 2 v in fact if you do the calculation it goes 3 times as far.

The reverse hapens in deceleration the first burn reduces it from 2v to v but the distance moved is 3 times greater than the second burn which reduces the speed from v to zero.

I hope this helps you to understand the difference between energy and momentum in moving bodies
« Last Edit: 05/01/2008 23:44:11 by Soul Surfer »
Learn, create, test and tell
evolution rules in all things
God says so!



  • Guest
"Simple" Rocket Science: Where have I gone wrong?
« Reply #2 on: 06/01/2008 16:33:26 »
There is no paradox if you realise that a huge amount of energy goes into the KE of the propellant. As you go faster, more of the total energy goes into your rocket's KE. And, as soul surfer says, the work done is force times distance - not force times time. Force times time is Impulse - change of momentum - and the impulse in each case IS the same. The total amount of momentum of the rocket plus fuel remains at Zero! The total KE increases as chemical energy is transferred in the rocket engine.