500
What's the intuition behind the Tsiolkovsky rocket equation:
v−v0=vexlnmom.
What isThe equation is basically telling you that the change of velocity of the rocket, v−v0, is proportional to the expulsion velocity of the fuel vex.
But it is not quite equal, since as the rocket is burning fuel, it is getting lighter. This is why you also take into account the rate of change of the mass, ln(m0/m).
So, how do we derive this?
Consider an ideal rocket at two different points in time, say t and t+Δt.
We can start by giving some information about the rocket at t: it's flying with velocity v, and has mass m+Δm (where Δm is precisely the quantity of fuel that is going to be expelled between t and t+Δt). For an observer on the ground (rest frame), the total momentum is p(t)=(m+Δm)v.
Now, at t+Δt, the velocity of the rocket has increased by Δv, and is now v+Δv. Similarly, the mass of the rocket has decreased to m, while a mass Δm of fuel has been expelled with some velocity (negative, in the opposite direction) vex. Again, for an observer on the ground, the total momentum is p(t+Δt)=m(v+Δv)−Δmvex.
Now we make the assumption that there is no net force acting on the rocket (ideal). From Newton's second law, we have that ∑F⃗ =ma⃗ =dp⃗ dt=0⃗ .
Therefore we have: p(t)=p(t+Δt); or, after simplification: mΔv=−vexΔm.
We can move to calculus by taking the limit Δv→0, i.e. considering infinitesimal changes. That way, we can now write: dv=−vexdmm. After integration on both sides, this gives v−v0=−vexln(m0m) (where m0 is the total mass before combustion, and m1 is the mass left after combustion).