# Thrust of rocket

1. Apr 20, 2005

### recon

A space-research rocket stands vertically on its launching pad. Prior to ignition, the mass of the rocket and its fuel is 1.9 X 103 kg. On ignition, gas is ejected from the rocket at a speed of 2.5 X 103 m/s relative to the rocket, and fuel is consumed at a constant rate of 7.4 kg/s. Find the thrust of the rocket.

Can someone please check my solution?

Let the mass of both the rocket and fuel be m, and let T be the thrust. Let 'a' be the acceleration of the rocket, and g be the acceleration due to gravity. Then,

$$T = \frac{d(mv)}{dt}$$

$$T = m\frac{dv}{dt} + v\frac{dm}{dt}$$

$$T = 7.4 kgs^{-1} \times 2.5 \times 10^3 ms^{-1}$$

$$T = 19.5 \times 10^3 N$$

2. Apr 20, 2005

### HallsofIvy

Staff Emeritus
Looks good to me.

3. Apr 20, 2005

### recon

I'm quite unfamiliar with problems involving a rate of change of momentum brought about by a rate of change of mass. In fact, this is the only problem of its kind that I've come across in my book.

Can someone please direct me to an online source of problems of this kind? I need more practise, thanks.

4. Apr 20, 2005

### arildno

You might want to look at the following thread:

Note that you are mixing together the rocket's acceleration $$a=dv/dt$$ while using "v" as the speed of the ejected fuel, relative to the rocket.
This is meaningless.

Although you get the right expression for the thrust (i.e, the force acting upon the rocket from the ejected fuel), your method of derivation is totally wrong.

now, the way by which i derive the rocket equation in post 4 is one way.
Later on in that thread, I'll derive it slightly differently, relying heavily upon Newton's 3.law (i.e, by relating the momentum change experienced by an ejected particle to the force acting upon it from the rocket system).

If you would be so kind to post your comments here, rather than in the thread in question, I would be grateful.