# Thrust question

1. Mar 25, 2005

### tony873004

At launch, the space shuttle’s engines develop 3.0 * 10^7 N of thrust. If the speed of the exhaust gasses is 15 km/s, what mass of gas is exhausted per second?

$$F=ma\Rightarrow m=\frac{F}{a}$$

$$kg=\frac{kg\cdot m/s^2}{m/s^2}$$

$$kg=\frac{kg\cdot m/s^{\rlap{--} {2}}}{m/s^{\rlap{--} {2}}}\Rightarrow m=\frac{3.0\cdot 10^7kg\cdot \rlap{--} {m}/\rlap{--} {s}}{15(\rlap{--} {k}\rlap{--} {m}/\rlap{--} {s})\ast 1000\rlap{--} {m}/\rlap{--} {k}\rlap{--} {m}}$$

$$m=2000kg$$

I'm not sure if I'm allowed to cancel out the squared part of s squared, and still associate the 3.0e7 N with units that no longer represent it. Is my answer even correct?

2. Mar 25, 2005

### dextercioby

$$F=\frac{\Delta p}{\Delta t} =v\frac{\Delta m}{\Delta t}$$

That "Delta t" is obviously 1sec,since we're asked the flow in one second...

Therefore the mass exhausted in one second is:

$$\Delta m=\frac{F}{\frac{v}{\Delta t}}=...$$

Daniel.

P.S.That force corresponds to a loss in momentum in unit time.

Last edited: Mar 25, 2005
3. Mar 25, 2005

### tony873004

$$m=\frac{F}{\frac{\Delta v}{\Delta t}}=...$$
so...

$$m=F\frac{\Delta t}{\Delta v}$$

$$m=3*10^7kg(m/s^2)\frac{1s}{15000m/s}$$

$$m=2000 kg$$

Amazing I came up with the right answer anyway, because I think I could have been called for illegal use of units. Your way is more elegant. It make sense since this question is in the momentum chapter. Delta P in your example is change in momentum, right?

Thanks, Dex. You've bailed me out again

4. Mar 25, 2005

### dextercioby

Sorry,it's the other way around with the variation of momentum.It's due to the mass flow,and not the velocity variation.I'll edit the post...

Daniel.

Last edited: Mar 25, 2005