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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: $$F_f=F_g - F_v= G \frac{mM}{r^2}-G \frac{mM}{rS}=G \frac{mM}{r}\biggl(\frac{1}{r} \cdot \frac{1}{S}\biggr) = G \frac{mM}{r}(rS)^{-1} $$

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    $\begingroup$ 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 $a=\frac{F}{m_{rocket} + m_{fuel}(t)}$. $\endgroup$ – Gigi Butbaia May 2 '14 at 19:59
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    $\begingroup$ The last line of the Question has a fraction-subtraction error resulting in a value that isn't dimensionally correct $\endgroup$ – DavePhD May 2 '14 at 20:17
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    $\begingroup$ @GigiButbaia: that looks like an answer to me. It answers both the question in the title and whether the energy requirement depends on mass. $\endgroup$ – Ross Millikan May 2 '14 at 20:18
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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 $V=\int_{t_0}^{t_1} a(t)$ $dt$.

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  • $\begingroup$ Yeah you're right... I like physics but I'm not really good at it :( $\endgroup$ – Peterix May 3 '14 at 9:03
  • $\begingroup$ @Peterix if you want to know more about this you can learn it by playing, there is very good game called "Kerbal Space Program". This game is good for learning Newton Mechanics. $\endgroup$ – Gigi Butbaia May 3 '14 at 16:21
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Do I also have to count the weight of the rocket?

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.

The force needed to reach escape velocity over a certain distance does depend on the object's mass. The velocity needed to escape a planet is independent of the mass of the rocket.

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