# Accelerating mass: how to compute

1. Aug 25, 2011

### BitWiz

Hi,

Say I have a machine with a mass of 10 kg at rest (with respect to an observer) in frictionless, gravity-free space. This machine now takes 1 kg of itself, and using the energy of 1 joule, flings this 1kg chunk toward the observer. How do I compute the final velocity of the remaining 9kg mass, given that I used a unit of energy as the impetus?

Ignoring Lorentz effects, say a 10 kg machine is in motion at an arbitrary velocity, moving directly away from an independent observer. I fling the 1kg chunk toward the observer using the energy of 1 J. Given that the machine had initial velocity > 0, is the difference in the initial and final velocities the same for the observer as it was when the machine was at rest? How about from the machine's frame looking at the observer in those two situations?

Ultimately, this is what I'm trying to determine: If I have a machine capable of harvesting the energy from a nuclear reactor in space, and I use that energy to accelerate the reactor by accelerating mass in the opposite direction (a mass driver), how do I compute the velocity difference -- from the reactor's frame -- from a quantity of energy? If that is possible, and, if the reactor's power is measured in joules/sec, can I then directly determine the acceleration 1:1? Again, ignoring Lorentz.

Thanks and warm regards,
Chris

2. Aug 25, 2011

### K^2

Kinetic energy of the mass you fired, computed in frame of the rocket, is equal to the energy expended (assuming 100% efficiency). The rest is conservation of momentum. In the continuous case, you should get rocket formula.

$$V_f = V_i + V_p ln\left(\frac{m_i}{m_f}\right)$$

Vp is the velocity with which you are expelling mass, mi is the mass of the rocket with the stuff you are going to expel, and mf is the mass of the rocket after you expelled all that stuff.

Total energy you are going to use up for all of this is just the kinetic energy formula.

$$E = \frac{1}{2}V_p(m_i-m_f)$$

So you can figure out how high Vp can be with whatever energy is available.

3. Aug 25, 2011

### BitWiz

Thanks, K^2. This seems so pleasingly simple. Just to make sure I understand:

$$E = \frac{1}{2}V_p(m_i-m_f)$$
then
$$V_p = \frac{2E}{m_i-m_f}$$
If the units agree, then if I expel a 1 kg mass from a 10 kg body with 1 joule of energy expended
$$V_p = \frac{2}{1}= 2 m/s ?$$
and then the rocket (mf) final velocity is$$V_r = V_p\frac{m_i-m_f}{m_f} = \frac{2}{9} m/s?$$

Thanks,
Chris