# Rocket propulsion

1. Dec 28, 2011

### aaaa202

Look at the attached picture, which is taken from my textbook. I don't understand the equation:

v = vex * ln(m0/m(t))
If the rocket looses mass at a constant rate, wouldn't that equation then say, that the acceleration decreases as time goes. Since the curve of ln(>1) flattens out. That doesn't match my intution behind conservation of momentum which says that the lighter you are pushing something away from you, the greather acceleration you will attain.

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2. Dec 28, 2011

### Tea Jay

Did you mean to say acceleration increases as time goes?

3. Dec 28, 2011

### D H

Staff Emeritus
That equation says nothing about acceleration, at least not directly. Differentiating it does say something about acceleration:
$$\frac{dv}{dt} = v_e\left(\frac 1{m_0/m(t)}\right)\left(\frac {-m_0}{m(t)^2}\frac{dm(t)}{dt}\right) = -v_e\frac{\dot m}{m(t)}$$
So assuming a constant burn rate, acceleration does increase as mass decreases. This is the reason that the Shuttle had to be throttled down near the end of the launch lest the astronauts and equipment be exposed to excessive acceleration.

What the ideal rocket equation (that's the name for this equation) says directly is that adding fuel has a small effect on final velocity. If you want to have a slightly faster final velocity when the fuel is depleted you need to add a considerable amount of fuel at launch. The relationship is not linear.