# Formula to calculate Rocket Thrust

1. Dec 4, 2005

### Serena_Greene

I'm tyring to find the formula that is used to calculate rocket thrust.

Thanks

Serena

2. Dec 5, 2005

### Tide

I don't know how detailed you want to be but a simple approximation would be

$$\vec F = \rho v_e^2 A$$

where $\rho$ is the density of the exhaust, $v_e$ is the speed of the exhaust and A is the cross-sectional area of the exhaust at the aperture. This doesn't take into account the lateral expansion of the gas nor details of nozzle design and so on.

3. Dec 5, 2005

### andrevdh

For a rocket fired vertically upwards and neglecting variation in g
The resultant force $F_R$ that a rocket experiences is given by
$$F_R=F_T-W$$
where $F_T$ is the thrust reaction force as a result of ejecting the rocket fuel at a speed of $v_r$ relative to the rocket.
The thrust can be shown to be given by
$$F_R=v_r\frac{dm}{dt}$$
and is therefore proportional to the rate at which the mass of the rocket is changing as a result of the burnt fuel. The acceleration of the rocket is therefore given by
$$a=v_r\frac{\dot m}{m}-g$$
For example if the rocket loses one sixty-ith of it's mass per second as a result of burning the rocket fuel which is ejected at a speed of 2400 m/s from it the acceleration of the rocket will be $30m/s^2$.

Last edited: Nov 29, 2006
4. Dec 8, 2005

### Duder

I'm currently working on a rocket/variable mass problem that includes linear air resistance, and I'm having difficulty solving the differential equation because it's non-separable (or so it seems). Any hints as to how to proceed?

5. Dec 8, 2005

### Tide

Write out your equation and we'll have a look at it.

6. Dec 8, 2005

### Duder

The initial equation I had was:
m*dv/dt - V*dm/dt = mg + kv

where big V is the relative exhaust speed and kv is the linear air resistance term. Without air resistance it's relatively easy to separate and integrate, but that kv is really mucking things up. Here's where I'm at now:

dv/V = -dm/m -(g/V)*dt - (kv/mV)*dt

..which takes up as the positive direction, btw. That last term is all sorts of trouble.

7. Dec 8, 2005

### Tide

Try setting dm/dt to a constant value to handle the special case where the exhaust leaves at a constant rate.

8. Dec 8, 2005

### FredGarvin

Tide, Just as a quick aside, I would think that your assertion is not a special case at all, but the norm. Most rockets do not have throttling capabilities or variable exit geometries so the assumption of constant exhaust mass flow, I think, is a good one.

9. Dec 8, 2005

### Duder

Ah okay, that worked. Good conceptual point. I was trying to solve the equation in the most generalized sense because less thrust is required as gravity decreases with altitude. It seems that effect is negligible here and dm/dt can be taken as constant.

Thanks Tide and Fred, much appreciated.