- #1

Pianodan

- 6

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## Homework Statement

A rocket starts from rest in free space by emitting mass at a constant rate [itex]\alpha[/itex] and with ejection velocity u relative to the rocket. At what fraction of the original rocket mass is the kinetic energy of the rocket a maximum. (Assume the mass of the empty rocket to be negligible)

## Homework Equations

Just kinetic energy and momentum, as far as I can tell. Forgive me, I'm not great with LaTex yet, so I'm trying to minimize typing it here.

## The Attempt at a Solution

Using conservation of momentum, reach the basic rocket equation:

M dv = u dm

where M is the inital mass of the rocket. Separate variables and integrate:

v(t) = u ln (M / m(t))

where m(t) is the mass of the rocket at time (t)

The Kinetic Energy, T(t), of the rocket is:

1/2 (m(t) ) v(t)^2

Substitute m(t) = M - [itex]\alpha[/itex]t and the value of v(t) from above to get:

T = M u^2 ln (M/m(t)) - [itex]\alpha[/itex] t u^2 ln (M/m(t))

The kinetic energy should be at a maximum when the time derivative of the KE is zero, so differentiate (and this is the most likely place I screwed up) the above equation with respect to time to get:

[itex]\frac{dT}{dt}=ln(\frac{M}{m(t)})[/itex]

Setting that equal to zero, you find only one extremum in T at m(t) = M, or at the point of total fuel consumption. Given that the KE of a massless rocket is zero, I'm at a loss.

Just thinking about this in the abstract, it seems that the velocity over the entire period of thrust, the mass should decrease, and the KE, as the product of the mass and the square of the velocity (and 1/2), should increase to a maximum and then decrease to zero as all the mass moves to the ejecta.

What am I missing here?