# Homework Help: Rocket Thrust Balancing Gravity

1. Sep 21, 2009

### Piamedes

1. The problem statement, all variables and given/known data
A rocket (initial mass m0) needs to use its engines ot hover stationary, just above the ground. If it can afford to burn no more than a mass (lamda)m0 of its fuel, for how long can it hover? If the exhaust velocity is 3000 m/s and lambda is 10% how long can the rocket hover?

2. Relevant equations
Thrust = $$v_{ex} \frac{dm}{dt}$$

$$v_{ex}$$ is the exhaust velocity

3. The attempt at a solution

So starting off with the basic sum of forces equals zero:

$$\dot{m}v_{ex} = mg$$

$$\frac{dm}{dt} = \frac{mg}{v_{ex}}$$

$$\frac{dm}{m} = \frac{g dt}{v_{ex}}$$

$$\frac{g}{v_{ex}} \int_{0}^{t} dt = \int_{m_{0}}^{m_{0} - \lambda m_{0}} \frac{dm}{m}$$

$$\frac{g}{v_{ex}} t = ln(m) ]_{m_{0}}^{m_{0} - \lambda m_{0}}$$

$$\frac{g}{v_{ex}} t = ln [m_{0} - \lambda m_{0}] - ln [m_{0}]$$

$$\frac{g}{v_{ex}} t = ln [\frac{m_{0} - \lambda m_{0}}{m_{0}}]$$

$$\frac{g}{v_{ex}} t = ln [ 1 - \lambda]$$

$$t = \frac{v_{ex}}{g} ln [ 1 - \lambda]$$

And here is where I encounter an issue. Since lambda will always be positive, regardless of how much fuel the rocket can expend this equation will give a negative time. Does anyone see where I went wrong with the calculations, or are my original assumptions wrong?

Thanks for any and all help

2. Sep 21, 2009

### jambaugh

The thrust equation just gives magnitude. Note that dm/dt should be a negative number since mass is decreasing.

3. Sep 21, 2009

### Piamedes

So if I carry along the negative associated with the decreasing mass, then the last few steps would be:

$$(-) t = \frac{v_{ex}}{g} ln [ 1 - \lambda]$$

$$t = - \frac{v_{ex}}{g} ln [ 1 - \lambda]$$

$$t = \frac{v_{ex}}{g} ln [ (1 - \lambda)^(-1)]$$

$$t = \frac{v_{ex}}{g} ln [ \frac{1}{1 - \lambda}]$$

And that would give a positive value for time. This equation makes sense, but is it a proper solution?

4. Sep 21, 2009

### D H

Staff Emeritus
That's correct.

5. Sep 21, 2009

Thanks!