# Is there a derivation for the thrust formula?

Wikipedia defines thrust as $$T=v\frac{dm}{dt}$$

Is this something fundamental or can this be derived ?

I was not able to find it on the internet so I thought of asking if here .If proof questions are not allowed a link would be very helpful.

Thank you .

Thrust ($$T$$) is the force one receives by "expelling" some mass $$dm$$.

We start by conservation of momentum for a mass $$m$$ travelling in one dimension at speed $$v$$ which expels a smaller mass $$dm$$ and changes thus its velocity by $$dv$$. The expelled mass travels at speed $$-c$$ with respect to the moving mass, so that it has a speed $$(-c+v)$$ in the observer's frame of reference.

We have $$m v = (m - dm)(v+dv)+dm (-c+v)$$ which becomes

$$m v = mv +mdv -dmv -dmdv -dmc + dm v$$

and simplifies to

$$0 = mdv-dmdv -dm c$$ we neglect $$dmdv$$ as is the product of two small quantities and are left with

$$mdv = c dm$$

we divide everything by $$dt$$ and get

$$m {dv \over dt} = c {dm \over dt}$$

Finally, using Newton's second law, $$m {dv \over dt}=ma_T=T$$ is a force and indeed it is the force that "accelerated" the mass forward by a quantity $$dv$$ due to mass expulsion, which is indeed the thrust, so

$$T= c {dm \over dt}$$