# Second order approximation in rocket equation

The simple rocket equation with no external forces can be derived in multiple ways.

Assume a rocket with mass $$m$$ and velocity $$v$$ at time $$t$$ that looses the mass $$-dm > 0$$ with relative velocity $$-u< 0$$ until time $$t+dt$$. First, from conservation of momentum, in physics it is often written as:

$$m v = (m+dm)(v+dv) + (-dm)(v-u)$$

which simplifies to

$$0 = m \cdot dv + u \cdot dm + dm \cdot dv$$

Now, the second order term is said to be smaller and therefore can be neglected. Following one can use separation of variables to solve the equation.

However, when deriving the equation based on the force

$$F = \frac{dp}{dt} = -u \frac{dm}{dt} = m \frac{dv}{dt}$$

one arrives at

$$-u \frac{dm}{dt} = m\frac{dv}{dt}$$.

This yields the same result after integrating but feels not as sloppy.

How can it be that using the first method one has to approximate, whereas the second method is exact?

I have a general confusion about the different usage of differentials in physics and mathematics, which are in the first solution treated to be infinitesimal but are said to be finite in mathematics.