# Do I need to take weight of the rocket into account when calculating escape velocity?

Here there is the old problem. I know from the old problem that the work $W_v$ that I need to make a rocket fast enough to reach the escape velocity is $$W_v= G \frac{mM}{r}$$ therefore because $$W_v=F\cdot S = G \frac{mM}{r} \rightarrow F_v=\frac{W}{S}=G \frac{mM}{rS}$$ that is the force I need to make a rocket fast enough to reach the escape velocity BUT
Do I also have to count the weight of the rocket?

If yes then the equation will be like this: $$F_f=F_g - F_v= G \frac{mM}{r^2}-G \frac{mM}{rS}=G \frac{mM}{r}\biggl(\frac{1}{r} \cdot \frac{1}{S}\biggr) = G \frac{mM}{r}(rS)^{-1}$$

• escape velocity doesn't depend on mass (but work depends on mass) of a rocket, the equation for escape velocity looks like this: $v=\sqrt{\frac{2GM_{earth}}{r}}$, And Work is change in kinetic energy $\Delta KE$, So $W=\frac{mv_{last}^2}{2}-\frac{mv_{initial}^2}{2}$, And Remember you have to apply that your acceleration is changing, because mass of fuel inside a rocket decreases, and because of Newton's second law acceleration will increase $a=\frac{F}{m(t)}$, or $a=\frac{F}{m_{rocket} + m_{fuel}(t)}$. – Gigi Butbaia May 2 '14 at 19:59
• The last line of the Question has a fraction-subtraction error resulting in a value that isn't dimensionally correct – DavePhD May 2 '14 at 20:17
• @GigiButbaia: that looks like an answer to me. It answers both the question in the title and whether the energy requirement depends on mass. – Ross Millikan May 2 '14 at 20:18

Escape Velocity doesn't depend on mass(but work depends on mass) of a rocket, Escape Velocity equation looks like this: $$v=\sqrt{\frac{2GM_{earth}}{r}}$$ And Work is just change in kinetic energy $\Delta KE$ or $KE_{final}-KE_{initial}$, So Work equation looks like this: $$W=\frac{m_{rocket}v_{final}^2}{2}-\frac{m_{rocket}v_{initial}^2}{2}=\frac{m_{rocket}}{2}(v_{final}^2-v_{initial}^2)=\frac{m_{rocket}\Delta v^2}{2}$$ And Don't forget that Fuel mass is decreasing (When Engine is powered) So That means that acceleration is changing (It's just Newton's second law that says this: $a=\frac{F}{m(t)}$ or full equation looks like this: $a=\frac{F}{m_{rocket}+m_{fuel}(t)}$). In order to solve this you have to "construct" differential equation. (The aforementioned equation in differential form looks like this: $\ddot x= \frac{F}{m_{rocket}+m_{fuel}(t)}$ and $V=\int_{t_0}^{t_1} a(t)$ $dt$.
You already have! This is the $m$ in $G\frac{m M}{r}$.
I don't understand the line of reasoning for your second equation, and you made a subtraction/multiplication error as DavePHD pointed out. The value $F_g+F_v$ would be the total acceleration experienced by astronauts (or the payload) sitting in your rocket, but I don't know what $F_g-F_v$ would mean if you're accelerating away from your gravitational source, and I don't know what you want it to mean. So I can't really help you on that one.